The Hungry Mouse











up vote
66
down vote

favorite
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Sixteen piles of cheese are put on a 4x4 square. They're labeled from $1$ to $16$. The smallest pile is $1$ and the biggest one is $16$.



The Hungry Mouse is so hungry that it always goes straight to the biggest pile (i.e. $16$) and eats it right away.



After that, it goes to the biggest neighboring pile and quickly eats that one as well. (Yeah ... It's really hungry.) And so on until there's no neighboring pile anymore.



A pile may have up to 8 neighbors (horizontally, vertically and diagonally). There's no wrap-around.



Example



We start with the following piles of cheese:



$$begin{matrix}
3&7&10&5\
6&8&12&13\
15&9&11&4\
14&1&16&2
end{matrix}$$



The Hungry Mouse first eats $16$, and then its biggest neighbor pile, which is $11$.



$$begin{matrix}
3&7&10&5\
6&8&12&13\
15&9&🐭&4\
14&1&color{grey}uparrow&2
end{matrix}$$



Its next moves are $13$, $12$, $10$, $8$, $15$, $14$, $9$, $6$, $7$ and $3$ in this exact order.



$$begin{matrix}
🐭&color{grey}leftarrow&smallcolor{grey}swarrow&5\
smallcolor{grey}nearrow&smallcolor{grey}swarrow&color{grey}uparrow&color{grey}leftarrow\
color{grey}downarrow&smallcolor{grey}nwarrow&smallcolor{grey}nearrow&4\
smallcolor{grey}nearrow&1&color{grey}uparrow&2
end{matrix}$$



There's no cheese anymore around the Hungry Mouse, so it stops there.



The challenge



Given the initial cheese configuration, your code must print or return the sum of the remaining piles once the Hungry Mouse has stopped eating them.



For the above example, the expected answer is $12$.



Rules




  • Because the size of the input matrix is fixed, you may take it as either a 2D array or a one-dimensional array.

  • Each value from $1$ to $16$ is guaranteed to appear exactly once.

  • This is code-golf.


Test cases



[ [ 4,  3,  2,  1], [ 5,  6,  7,  8], [12, 11, 10,  9], [13, 14, 15, 16] ] --> 0
[ [ 8, 1, 9, 14], [11, 6, 5, 16], [13, 15, 2, 7], [10, 3, 12, 4] ] --> 0
[ [ 1, 2, 3, 4], [ 5, 6, 7, 8], [ 9, 10, 11, 12], [13, 14, 15, 16] ] --> 1
[ [10, 15, 14, 11], [ 9, 3, 1, 7], [13, 5, 12, 6], [ 2, 8, 4, 16] ] --> 3
[ [ 3, 7, 10, 5], [ 6, 8, 12, 13], [15, 9, 11, 4], [14, 1, 16, 2] ] --> 12
[ [ 8, 9, 3, 6], [13, 11, 7, 15], [12, 10, 16, 2], [ 4, 14, 1, 5] ] --> 34
[ [ 8, 11, 12, 9], [14, 5, 10, 16], [ 7, 3, 1, 6], [13, 4, 2, 15] ] --> 51
[ [13, 14, 1, 2], [16, 15, 3, 4], [ 5, 6, 7, 8], [ 9, 10, 11, 12] ] --> 78
[ [ 9, 10, 11, 12], [ 1, 2, 4, 13], [ 7, 8, 5, 14], [ 3, 16, 6, 15] ] --> 102
[ [ 9, 10, 11, 12], [ 1, 2, 7, 13], [ 6, 16, 4, 14], [ 3, 8, 5, 15] ] --> 103









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  • 27




    +1 for that mouse character
    – Luis Mendo
    2 days ago








  • 2




    ...make that 103: [[9, 10, 11, 12], [1, 2, 7, 13], [6, 16, 4, 14], [3, 8, 5, 15]]
    – Jonathan Allan
    2 days ago








  • 7




    What a nicely written challenge! I'll keep it in mind for the best-of nominations.
    – xnor
    2 days ago






  • 6




    After misreading I was a little sad that this was not a hungry moose.
    – akozi
    2 days ago






  • 1




    @akozi Sandboxed challenge "The Hungry Moose": codegolf.meta.stackexchange.com/a/17123/39328
    – lirtosiast
    yesterday

















up vote
66
down vote

favorite
7












Sixteen piles of cheese are put on a 4x4 square. They're labeled from $1$ to $16$. The smallest pile is $1$ and the biggest one is $16$.



The Hungry Mouse is so hungry that it always goes straight to the biggest pile (i.e. $16$) and eats it right away.



After that, it goes to the biggest neighboring pile and quickly eats that one as well. (Yeah ... It's really hungry.) And so on until there's no neighboring pile anymore.



A pile may have up to 8 neighbors (horizontally, vertically and diagonally). There's no wrap-around.



Example



We start with the following piles of cheese:



$$begin{matrix}
3&7&10&5\
6&8&12&13\
15&9&11&4\
14&1&16&2
end{matrix}$$



The Hungry Mouse first eats $16$, and then its biggest neighbor pile, which is $11$.



$$begin{matrix}
3&7&10&5\
6&8&12&13\
15&9&🐭&4\
14&1&color{grey}uparrow&2
end{matrix}$$



Its next moves are $13$, $12$, $10$, $8$, $15$, $14$, $9$, $6$, $7$ and $3$ in this exact order.



$$begin{matrix}
🐭&color{grey}leftarrow&smallcolor{grey}swarrow&5\
smallcolor{grey}nearrow&smallcolor{grey}swarrow&color{grey}uparrow&color{grey}leftarrow\
color{grey}downarrow&smallcolor{grey}nwarrow&smallcolor{grey}nearrow&4\
smallcolor{grey}nearrow&1&color{grey}uparrow&2
end{matrix}$$



There's no cheese anymore around the Hungry Mouse, so it stops there.



The challenge



Given the initial cheese configuration, your code must print or return the sum of the remaining piles once the Hungry Mouse has stopped eating them.



For the above example, the expected answer is $12$.



Rules




  • Because the size of the input matrix is fixed, you may take it as either a 2D array or a one-dimensional array.

  • Each value from $1$ to $16$ is guaranteed to appear exactly once.

  • This is code-golf.


Test cases



[ [ 4,  3,  2,  1], [ 5,  6,  7,  8], [12, 11, 10,  9], [13, 14, 15, 16] ] --> 0
[ [ 8, 1, 9, 14], [11, 6, 5, 16], [13, 15, 2, 7], [10, 3, 12, 4] ] --> 0
[ [ 1, 2, 3, 4], [ 5, 6, 7, 8], [ 9, 10, 11, 12], [13, 14, 15, 16] ] --> 1
[ [10, 15, 14, 11], [ 9, 3, 1, 7], [13, 5, 12, 6], [ 2, 8, 4, 16] ] --> 3
[ [ 3, 7, 10, 5], [ 6, 8, 12, 13], [15, 9, 11, 4], [14, 1, 16, 2] ] --> 12
[ [ 8, 9, 3, 6], [13, 11, 7, 15], [12, 10, 16, 2], [ 4, 14, 1, 5] ] --> 34
[ [ 8, 11, 12, 9], [14, 5, 10, 16], [ 7, 3, 1, 6], [13, 4, 2, 15] ] --> 51
[ [13, 14, 1, 2], [16, 15, 3, 4], [ 5, 6, 7, 8], [ 9, 10, 11, 12] ] --> 78
[ [ 9, 10, 11, 12], [ 1, 2, 4, 13], [ 7, 8, 5, 14], [ 3, 16, 6, 15] ] --> 102
[ [ 9, 10, 11, 12], [ 1, 2, 7, 13], [ 6, 16, 4, 14], [ 3, 8, 5, 15] ] --> 103









share|improve this question




















  • 27




    +1 for that mouse character
    – Luis Mendo
    2 days ago








  • 2




    ...make that 103: [[9, 10, 11, 12], [1, 2, 7, 13], [6, 16, 4, 14], [3, 8, 5, 15]]
    – Jonathan Allan
    2 days ago








  • 7




    What a nicely written challenge! I'll keep it in mind for the best-of nominations.
    – xnor
    2 days ago






  • 6




    After misreading I was a little sad that this was not a hungry moose.
    – akozi
    2 days ago






  • 1




    @akozi Sandboxed challenge "The Hungry Moose": codegolf.meta.stackexchange.com/a/17123/39328
    – lirtosiast
    yesterday















up vote
66
down vote

favorite
7









up vote
66
down vote

favorite
7






7





Sixteen piles of cheese are put on a 4x4 square. They're labeled from $1$ to $16$. The smallest pile is $1$ and the biggest one is $16$.



The Hungry Mouse is so hungry that it always goes straight to the biggest pile (i.e. $16$) and eats it right away.



After that, it goes to the biggest neighboring pile and quickly eats that one as well. (Yeah ... It's really hungry.) And so on until there's no neighboring pile anymore.



A pile may have up to 8 neighbors (horizontally, vertically and diagonally). There's no wrap-around.



Example



We start with the following piles of cheese:



$$begin{matrix}
3&7&10&5\
6&8&12&13\
15&9&11&4\
14&1&16&2
end{matrix}$$



The Hungry Mouse first eats $16$, and then its biggest neighbor pile, which is $11$.



$$begin{matrix}
3&7&10&5\
6&8&12&13\
15&9&🐭&4\
14&1&color{grey}uparrow&2
end{matrix}$$



Its next moves are $13$, $12$, $10$, $8$, $15$, $14$, $9$, $6$, $7$ and $3$ in this exact order.



$$begin{matrix}
🐭&color{grey}leftarrow&smallcolor{grey}swarrow&5\
smallcolor{grey}nearrow&smallcolor{grey}swarrow&color{grey}uparrow&color{grey}leftarrow\
color{grey}downarrow&smallcolor{grey}nwarrow&smallcolor{grey}nearrow&4\
smallcolor{grey}nearrow&1&color{grey}uparrow&2
end{matrix}$$



There's no cheese anymore around the Hungry Mouse, so it stops there.



The challenge



Given the initial cheese configuration, your code must print or return the sum of the remaining piles once the Hungry Mouse has stopped eating them.



For the above example, the expected answer is $12$.



Rules




  • Because the size of the input matrix is fixed, you may take it as either a 2D array or a one-dimensional array.

  • Each value from $1$ to $16$ is guaranteed to appear exactly once.

  • This is code-golf.


Test cases



[ [ 4,  3,  2,  1], [ 5,  6,  7,  8], [12, 11, 10,  9], [13, 14, 15, 16] ] --> 0
[ [ 8, 1, 9, 14], [11, 6, 5, 16], [13, 15, 2, 7], [10, 3, 12, 4] ] --> 0
[ [ 1, 2, 3, 4], [ 5, 6, 7, 8], [ 9, 10, 11, 12], [13, 14, 15, 16] ] --> 1
[ [10, 15, 14, 11], [ 9, 3, 1, 7], [13, 5, 12, 6], [ 2, 8, 4, 16] ] --> 3
[ [ 3, 7, 10, 5], [ 6, 8, 12, 13], [15, 9, 11, 4], [14, 1, 16, 2] ] --> 12
[ [ 8, 9, 3, 6], [13, 11, 7, 15], [12, 10, 16, 2], [ 4, 14, 1, 5] ] --> 34
[ [ 8, 11, 12, 9], [14, 5, 10, 16], [ 7, 3, 1, 6], [13, 4, 2, 15] ] --> 51
[ [13, 14, 1, 2], [16, 15, 3, 4], [ 5, 6, 7, 8], [ 9, 10, 11, 12] ] --> 78
[ [ 9, 10, 11, 12], [ 1, 2, 4, 13], [ 7, 8, 5, 14], [ 3, 16, 6, 15] ] --> 102
[ [ 9, 10, 11, 12], [ 1, 2, 7, 13], [ 6, 16, 4, 14], [ 3, 8, 5, 15] ] --> 103









share|improve this question















Sixteen piles of cheese are put on a 4x4 square. They're labeled from $1$ to $16$. The smallest pile is $1$ and the biggest one is $16$.



The Hungry Mouse is so hungry that it always goes straight to the biggest pile (i.e. $16$) and eats it right away.



After that, it goes to the biggest neighboring pile and quickly eats that one as well. (Yeah ... It's really hungry.) And so on until there's no neighboring pile anymore.



A pile may have up to 8 neighbors (horizontally, vertically and diagonally). There's no wrap-around.



Example



We start with the following piles of cheese:



$$begin{matrix}
3&7&10&5\
6&8&12&13\
15&9&11&4\
14&1&16&2
end{matrix}$$



The Hungry Mouse first eats $16$, and then its biggest neighbor pile, which is $11$.



$$begin{matrix}
3&7&10&5\
6&8&12&13\
15&9&🐭&4\
14&1&color{grey}uparrow&2
end{matrix}$$



Its next moves are $13$, $12$, $10$, $8$, $15$, $14$, $9$, $6$, $7$ and $3$ in this exact order.



$$begin{matrix}
🐭&color{grey}leftarrow&smallcolor{grey}swarrow&5\
smallcolor{grey}nearrow&smallcolor{grey}swarrow&color{grey}uparrow&color{grey}leftarrow\
color{grey}downarrow&smallcolor{grey}nwarrow&smallcolor{grey}nearrow&4\
smallcolor{grey}nearrow&1&color{grey}uparrow&2
end{matrix}$$



There's no cheese anymore around the Hungry Mouse, so it stops there.



The challenge



Given the initial cheese configuration, your code must print or return the sum of the remaining piles once the Hungry Mouse has stopped eating them.



For the above example, the expected answer is $12$.



Rules




  • Because the size of the input matrix is fixed, you may take it as either a 2D array or a one-dimensional array.

  • Each value from $1$ to $16$ is guaranteed to appear exactly once.

  • This is code-golf.


Test cases



[ [ 4,  3,  2,  1], [ 5,  6,  7,  8], [12, 11, 10,  9], [13, 14, 15, 16] ] --> 0
[ [ 8, 1, 9, 14], [11, 6, 5, 16], [13, 15, 2, 7], [10, 3, 12, 4] ] --> 0
[ [ 1, 2, 3, 4], [ 5, 6, 7, 8], [ 9, 10, 11, 12], [13, 14, 15, 16] ] --> 1
[ [10, 15, 14, 11], [ 9, 3, 1, 7], [13, 5, 12, 6], [ 2, 8, 4, 16] ] --> 3
[ [ 3, 7, 10, 5], [ 6, 8, 12, 13], [15, 9, 11, 4], [14, 1, 16, 2] ] --> 12
[ [ 8, 9, 3, 6], [13, 11, 7, 15], [12, 10, 16, 2], [ 4, 14, 1, 5] ] --> 34
[ [ 8, 11, 12, 9], [14, 5, 10, 16], [ 7, 3, 1, 6], [13, 4, 2, 15] ] --> 51
[ [13, 14, 1, 2], [16, 15, 3, 4], [ 5, 6, 7, 8], [ 9, 10, 11, 12] ] --> 78
[ [ 9, 10, 11, 12], [ 1, 2, 4, 13], [ 7, 8, 5, 14], [ 3, 16, 6, 15] ] --> 102
[ [ 9, 10, 11, 12], [ 1, 2, 7, 13], [ 6, 16, 4, 14], [ 3, 8, 5, 15] ] --> 103






code-golf matrix






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edited 2 days ago

























asked 2 days ago









Arnauld

69.4k586293




69.4k586293








  • 27




    +1 for that mouse character
    – Luis Mendo
    2 days ago








  • 2




    ...make that 103: [[9, 10, 11, 12], [1, 2, 7, 13], [6, 16, 4, 14], [3, 8, 5, 15]]
    – Jonathan Allan
    2 days ago








  • 7




    What a nicely written challenge! I'll keep it in mind for the best-of nominations.
    – xnor
    2 days ago






  • 6




    After misreading I was a little sad that this was not a hungry moose.
    – akozi
    2 days ago






  • 1




    @akozi Sandboxed challenge "The Hungry Moose": codegolf.meta.stackexchange.com/a/17123/39328
    – lirtosiast
    yesterday
















  • 27




    +1 for that mouse character
    – Luis Mendo
    2 days ago








  • 2




    ...make that 103: [[9, 10, 11, 12], [1, 2, 7, 13], [6, 16, 4, 14], [3, 8, 5, 15]]
    – Jonathan Allan
    2 days ago








  • 7




    What a nicely written challenge! I'll keep it in mind for the best-of nominations.
    – xnor
    2 days ago






  • 6




    After misreading I was a little sad that this was not a hungry moose.
    – akozi
    2 days ago






  • 1




    @akozi Sandboxed challenge "The Hungry Moose": codegolf.meta.stackexchange.com/a/17123/39328
    – lirtosiast
    yesterday










27




27




+1 for that mouse character
– Luis Mendo
2 days ago






+1 for that mouse character
– Luis Mendo
2 days ago






2




2




...make that 103: [[9, 10, 11, 12], [1, 2, 7, 13], [6, 16, 4, 14], [3, 8, 5, 15]]
– Jonathan Allan
2 days ago






...make that 103: [[9, 10, 11, 12], [1, 2, 7, 13], [6, 16, 4, 14], [3, 8, 5, 15]]
– Jonathan Allan
2 days ago






7




7




What a nicely written challenge! I'll keep it in mind for the best-of nominations.
– xnor
2 days ago




What a nicely written challenge! I'll keep it in mind for the best-of nominations.
– xnor
2 days ago




6




6




After misreading I was a little sad that this was not a hungry moose.
– akozi
2 days ago




After misreading I was a little sad that this was not a hungry moose.
– akozi
2 days ago




1




1




@akozi Sandboxed challenge "The Hungry Moose": codegolf.meta.stackexchange.com/a/17123/39328
– lirtosiast
yesterday






@akozi Sandboxed challenge "The Hungry Moose": codegolf.meta.stackexchange.com/a/17123/39328
– lirtosiast
yesterday












14 Answers
14






active

oldest

votes

















up vote
11
down vote














Python 2, 133 130 bytes





a=input();m=16
for i in range(m):a[i*5:i*5]=0,
while m:i=a.index(m);a[i]=0;m=max(a[i+x]for x in[-6,-5,-4,-1,1,4,5,6])
print sum(a)


Try it online!



Takes a flattened list of 16 elements.



How it works



a=input();m=16

# Add zero padding on each row, and enough zeroes at the end to avoid index error
for i in range(m):a[i*5:i*5]=0,

# m == maximum element found in last iteration
# i == index of last eaten element
# eaten elements of `a` are reset to 0
while m:i=a.index(m);a[i]=0;m=max(a[i+x]for x in[-6,-5,-4,-1,1,4,5,6])
print sum(a)





share|improve this answer























  • The adjacent-cell expression a[i+x]for x in[-6,-5,-4,-1,1,4,5,6] can be shortened to a[i+j+j/3*2-6]for j in range(9) (the zero entry is harmless). Python 3 can surely do shorter by hardcoding a length-8 bytestring, but Python 2 might still be better overall.
    – xnor
    2 days ago






  • 1




    Though your zero padding loop is clever, it looks like it's shorter to take a 2D list: a=[0]*5 for r in input():a=r+[0]+a. Perhaps there's a yet shorter string slicing solution that doesn't require iterating.
    – xnor
    2 days ago


















up vote
8
down vote














MATL, 50 49 47 bytes



16:HZ^!"2G@m1ZIm~]v16eXK68E16b"Ky0)Y)fyX-X>h]s-


Input is a matrix, using ; as row separator.



Try it online! Or verify all test cases.



Explanation



16:HZ^!  % Cartesian power of [1 2 ... 16] with exponent 2, transpose. Gives a 
% 2-row matrix with 1st column [1; 1], 2nd [1; 2], ..., last [16; 16]
" % For each column, say [k; j]
2 % Push 2
G@m % Push input matrix, then current column [k; j], then check membership.
% This gives a 4×4 matrix that contains 1 for entries of the input that
% contain k or j
1ZI % Connected components (based on 8-neighbourhood) of nonzero entries.
% This gives a 4×4 matrix with each connected component labeled with
% values 1, 2, ... respectively
m~ % True if 2 is not present in this matrix. That means there is only
% one connected component; that is, k and j are neighbours in the
% input matrix, or k=j
] % End
v16e % The stack now has 256 values. Concatenate them into a vector and
% reshape as a 16×16 matrix. This matrix describes neighbourhood: entry
% (k,j) is 1 if values k and j are neighbours in the input or if k=j
XK % Copy into clipboard K
68E % Push 68 times 2, that is, 136, which is 1+2+...+16
16 % Push 16. This is the initial value eaten by the mouse. New values will
% be appended to create a vector of eaten values
b % Bubble up the 16×16 matrix to the top of the stack
" % For each column. This just executes the loop 16 times
K % Push neighbourhood matrix from clipboard K
y % Copy from below: pushes a copy of the vector of eaten values
0) % Get last value. This is the most recent eaten value
Y) % Get that row of the neighbourhood matrix
f % Indices of nonzeros. This gives a vector of neighbours of the last
% eaten value
y % Copy from below: pushes a copy of the vector of eaten values
X- % Set difference (may give an empty result)
X> % Maximum value. This is the new eaten value (maximum neighbour not
% already eaten). May be empty, if all neighbours are already eaten
h % Concatenate to vector of eaten values
] % End
s % Sum of vector of all eaten values
- % Subtract from 136. Implicitly display





share|improve this answer























  • Idk MatLab, but can you save a little if you push -136 instead of +136?
    – Titus
    yesterday










  • @Titus Hm I don't see how
    – Luis Mendo
    yesterday










  • or the other way round: I thought instead of 1) push 136 2) push each eaten value 3) sum up eaten values 4) subtract from 136 -> 1) push 136 2) push negative of eaten value 3) sum up stack. But as it obviously is only one byte each; it´s probably no gain.
    – Titus
    yesterday










  • @Titus Ah, yes, I think that uses the same number of bytes. Also, I need each eaten value (not its negative) for the set difference; negating would have to be done at the end
    – Luis Mendo
    yesterday




















up vote
7
down vote














Python 2, 111 bytes





i=x=a=input()
while x:x,i=max((y,j)for j,y in enumerate(a)if i>or 2>i/4-j/4>-2<i%4-j%4<2);a[i]=0
print sum(a)


Try it online!



Method and test cases adapted from Bubbler. Takes a flat list on STDIN.



The code checks whether two flat indices i and j represent touching cells by checking that both row different i/4-j/4 and column difference i%4-j%4 are strictly between -2 and 2. The first pass instead has this check automatically succeed so that the largest entry is found disregarding adjacency.






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    up vote
    6
    down vote













    PHP, 177 174 171 bytes



    for($v=16;$v;$u+=$v=max($p%4-1?max($a[$p-5],$a[$p-1],$a[$p+3]):0,$a[$p-4],$a[$p+4],$p%4?max($a[$p-3],$a[$p+1],$a[$p+5]):0))$a[$p=array_search($v,$a=&$argv)]=0;echo 120-$u;


    Run with -nr, provide matrix elements as arguments or try it online.






    share|improve this answer






























      up vote
      3
      down vote














      Charcoal, 47 bytes



      EA⭆ι§αλ≔QθW›θA«≔⌕KAθθJ﹪θ⁴÷θ⁴≔⌈KMθA»≔ΣEKA⌕αιθ⎚Iθ


      Try it online! Link is to verbose version of code. Explanation:



      EA⭆ι§αλ


      Convert the input numbers into alphabetic characters (A=0 .. Q=16) and print them as a 4x4 grid.



      ≔Qθ


      Start by eating the Q, i.e. 16.



      W›θA«


      Repeat while there is something to eat.



      ≔⌕KAθθ


      Find where the pile is. This is a linear view in row-major order.



      J﹪θ⁴÷θ⁴


      Convert to co-ordinates and jump to that location.



      ≔⌈KMθ


      Find the largest adjacent pile.






      Eat the current pile.



      ≔ΣEKA⌕αιθ


      Convert the piles back to integers and take the sum.



      ⎚Iθ


      Clear the canvas and output the result.






      share|improve this answer




























        up vote
        3
        down vote














        R, 128 124 bytes





        r=rbind(0,cbind(0,matrix(scan(),4,4),0),0)
        m=which(r==16)
        while(r[m]){r[m]=0
        m=which(r==max(r[m+c(-7:-5,-1,1,5:7)]))}
        sum(r)


        Try it online!



        TIO link is slightly different, I am still trying to figure out how to make it work.



        I do feel like I can golf a lot more out of this. But this works for now.



        It creates a 4x4 matrix (which helped me to visualize things), pads it with 0's, then begins from 16 and searches it's surrounding "piles" for the next largest, and so forth.



        Upon conclusion, it does output a warning, but it is of no consequence and does not change the result.



        EDIT: -4 bytes by compressing the initialization of the matrix into 1 line






        share|improve this answer






























          up vote
          2
          down vote













          JavaScript, 122 bytes



          I took more than a couple of wrong turns on this one and now I've run out of time for further golfing but at least it's working. Will revisit tomorrow (or, knowing me, on the train home this evening!), if I can find a minute.



          a=>(g=n=>n?g([-6,-5,-4,-1,1,4,5,6].map(x=>n=a[x+=i]>n?a[x]:n,a[i=a.indexOf(n)]=n=0)|n)-n:120)(16,a=a.flatMap(x=>[...x,0]))


          Try it online






          share|improve this answer

















          • 1




            +1 for flatMap() :p
            – Arnauld
            2 days ago










          • :D I think this is the first time I've used it for golf! Out of interest (and to give me a target when I come back to this), what was your score when you tried it?
            – Shaggy
            2 days ago






          • 1




            I didn't try to golf it. Unless I have a specific trick in mind, I usually just write clean code to test my challenges.
            – Arnauld
            2 days ago










          • Ok, you've convinced me to try. :) I'm currently at <s>116</s> 114.
            – Arnauld
            2 days ago












          • 98 bytes after a good night's sleep. (Sorry about the multiple notifications.)
            – Arnauld
            yesterday


















          up vote
          2
          down vote














          Jelly,  31 30  29 bytes



          ³œiⱮZIỊȦ
          ⁴ṖŒPŒ!€Ẏ⁴;ⱮṢÇƇṪ
          FḟÇS


          Since the method is far too slow to run within 60s with the mouse starting on 16 this starts her off at 9 and limits her ability such that she is only able to eat 9s or less Try it online! (thus here she eats 9, 2, 7, 4, 8, 6, 3 leaving 97).



          How?



          ³œiⱮZIỊȦ - Link 1, isSatisfactory?: list of integers, possiblePileChoice
          ³ - (using a left argument of) program's 3rd command line argument (M)
          Ɱ - map across (possiblePileChoice) with:
          œi - first multi-dimensional index of (the item) in (M)
          Z - transpose the resulting list of [row, column] values
          I - get the incremental differences
          Ị - insignificant? (vectorises an abs(v) <= 1 test)
          Ȧ - any and all? (0 if any 0s are present in the flattened result [or if it's empty])

          ⁴ṖŒPŒ!€Ẏ⁴;ⱮṢÇƇṪ - Link 2, getChosenPileList: list of lists of integers, M
          ⁴ - literal 16
          Ṗ - pop -> [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15]
          ŒP - power-set -> [,[1],[2],...,[1,2],[1,3],...,[2,3,7],...,[1,2,3,4,5,6,7,8,9,10,11,12,13,14,15]]
          € - for each:
          Œ! - all permutations
          Ẏ - tighten (to a single list of all these individual permutations)
          ⁴ - (using a left argument of) literal 16
          Ɱ - map across it with:
          ; - concatenate (put a 16 at the beginning of each one)
          Ṣ - sort the resulting list of lists
          Ƈ - filter keep those for which this is truthy:
          Ç - call last Link as a monad (i.e. isSatisfactory(possiblePileChoice)
          Ṫ - tail (get the right-most, i.e. the maximal satisfactory one)

          FḟÇS - Main Link: list of lists of integers, M
          F - flatten M
          Ç - call last Link (2) as a monad (i.e. get getChosenPileList(M))
          ḟ - filter discard (the resulting values) from (the flattened M)
          S - sum





          share|improve this answer























          • Ah yeah, power-set is not enough!
            – Jonathan Allan
            2 days ago






          • 2




            @Arnauld - finally got a little time to golf :D This should work, but will be (way) too slow for running at TIO with the test case you used before.
            – Jonathan Allan
            2 days ago


















          up vote
          2
          down vote













          SAS, 236 219 bytes



          Input on punch cards, one line per grid (space-separated), output printed to the log.



          This challenge is slightly complicated by some limitations of arrays in SAS:




          • There is no way to return the row and column indexes of a matching element from multidimensional data-step array - you have to treat the array as 1-d and then work them out for yourself.

          • If you go out of bounds, SAS throws an error and halts processing rather than returning null / zero.


          Updates:




          • Removed infile cards; statement (-13)

          • Used wildcard a: for array definition rather than a1-a16 (-4)


          Golfed:



          data;input a1-a16;array a[4,4]a:;p=16;t=136;do while(p);m=whichn(p,of a:);t=t-p;j=mod(m-1,4)+1;i=ceil(m/4);a[i,j]=0;p=0;do k=max(1,i-1)to min(i+1,4);do l=max(1,j-1)to min(j+1,4);p=max(p,a[k,l]);end;end;end;put t;cards;
          <insert punch cards here>
          ;


          Ungolfed:



          data;                /*Produce a dataset using automatic naming*/
          input a1-a16; /*Read 16 variables*/
          array a[4,4] a:; /*Assign to a 4x4 array*/
          p=16; /*Initial pile to look for*/
          t=136; /*Total cheese to decrement*/
          do while(p); /*Stop if there are no piles available with size > 0*/
          m=whichn(p,of a:); /*Find array element containing current pile size*/
          t=t-p; /*Decrement total cheese*/
          j=mod(m-1,4)+1; /*Get column number*/
          i=ceil(m/4); /*Get row number*/
          a[i,j]=0; /*Eat the current pile*/
          /*Find the size of the largest adjacent pile*/
          p=0;
          do k=max(1,i-1)to min(i+1,4);
          do l=max(1,j-1)to min(j+1,4);
          p=max(p,a[k,l]);
          end;
          end;
          end;
          put t; /*Print total remaining cheese to log*/
          /*Start of punch card input*/
          cards;
          4 3 2 1 5 6 7 8 12 11 10 9 13 14 15 16
          8 1 9 14 11 6 5 16 13 15 2 7 10 3 12 4
          1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
          10 15 14 11 9 3 1 7 13 5 12 6 2 8 4 16
          3 7 10 5 6 8 12 13 15 9 11 4 14 1 16 2
          8 9 3 6 13 11 7 15 12 10 16 2 4 14 1 5
          8 11 12 9 14 5 10 16 7 3 1 6 13 4 2 15
          13 14 1 2 16 15 3 4 5 6 7 8 9 10 11 12
          9 10 11 12 1 2 4 13 7 8 5 14 3 16 6 15
          9 10 11 12 1 2 7 13 6 16 4 14 3 8 5 15
          ; /*End of punch card input*/
          /*Implicit run;*/





          share|improve this answer























          • +1 for use of punch cards in PPCG :)
            – GNiklasch
            2 days ago


















          up vote
          2
          down vote













          Java 10, 272 271 bytes





          m->{int r,c,R=4,M=1,x,y,X,Y;for(r=c=X=Y=0;R-->0;)for(C=4;C-->0;)if(m[R][C]>15)m[r=R][c=C]=0;for(;M!=0;m[r=X][c=Y]=0)for(M=-1,C=9;C-->0;)try{if((R=m[x=C<3?r-1:C>5?r+1:r][y=C%3<1?c-1:C%3>1?c+1:c])>M){M=R;X=x;Y=y;}}catch(Exception e){}for(var Z:m)for(int z:Z)M+=z;return M;}


          -1 byte indirectly thanks to @Serverfrog.



          The cells are checked the same as in my answer for the All the single eights challenge.



          Try it online.



          Explanation:



          m->{                       // Method with integer-matrix parameter and integer return
          int r, // Row-coordinate for the largest number
          c, // Column-coordinate for the largest number
          R=4,C, // Row and column indices (later reused as temp integers)
          M=1, // Largest number the mouse just ate, starting at 1
          x,y,X,Y; // Temp integers
          for(r=c=X=Y=0; // Start `r`, `c`, `X`, and `Y` at 0
          R-->0;) // Loop `R` in the range (4, 0]:
          for(C=4;C-->0;) // Inner loop `C` in the range (4, 0]:
          if(m[R][C]>15) // If the current cell is 16:
          m[r=R][c=C] // Set `r,c` to this coordinate
          =0; // And empty this cell
          for(;M!=0; // Loop as long as the largest number isn't 0:
          ; // After every iteration:
          m[r=X][c=Y] // Change the `r,c` coordinates,
          =0) // And empty this cell
          for(M=-1, // Reset `M` to -1
          C=9;C-->0;) // Inner loop `C` in the range (9, 0]:
          try{if((R= // Set `R` to:
          m[x=C<3? // If `C` is 0, 1, or 2:
          r-1 // Look at the previous row
          :C>5? // Else-if `C` is 6, 7, or 8:
          r+1 // Look at the next row
          : // Else (`C` is 3, 4, or 5):
          r] // Look at the current row
          [y=C%3<1? // If `C` is 0, 3, or 6:
          c-1 // Look at the previous column
          :C%3>1? // Else-if `C` is 2, 5, or 8:
          c+1 // Look at the next column
          : // Else (`C` is 1, 4, or 7):
          c]) // Look at the current column
          >M){ // And if the number in this cell is larger than `M`
          M=R; // Change `M` to this number
          X=x;Y=y;} // And change the `X,Y` coordinate to this cell
          }catch(Exception e){}
          // Catch and ignore ArrayIndexOutOfBoundsExceptions
          // (try-catch saves bytes in comparison to if-checks)
          for(var Z:m) // Then loop over all rows of the matrix:
          for(int z:Z) // Inner loop over all columns of the matrix:
          M+=z; // And sum them all together in `M` (which was 0)
          return M;} // Then return this sum as result





          share|improve this answer























          • could you not to int r=c=X=Y=0,R=4,M=1,x,y; ?
            – Serverfrog
            6 hours ago










          • @Serverfrog I'm afraid that's not possible when declaring the variables in Java. Your suggestion did give me an idea to save a byte though, by using int r,c,R=4,M=1,x,y,X,Y;for(r=c=X=Y=0;, so thanks. :)
            – Kevin Cruijssen
            3 hours ago




















          up vote
          1
          down vote













          J, 82 bytes



          g=.](]*{:@[~:])]_1}~[:>./]{~((,-)1 5 6 7)+]i.{:
          [:+/[:(g^:_)16,~[:,0,~0,0,0,.~0,.]


          Try it online!



          I plan to golf this more tomorrow, and perhaps write a more J-ish solution similar to this one, but I figured I'd try the flattened approach since I hadn't done that before.






          share|improve this answer























          • Do you really need the leftmost ] in g?
            – Galen Ivanov
            2 days ago






          • 1




            Thanks Galen, you're right. It's the least of the issues with this code :) I have a much better solution which I'll implement when I have time.
            – Jonah
            yesterday


















          up vote
          1
          down vote














          Red, 277 bytes



          func[a][k: 16 until[t:(index? find load form a k)- 1
          p: do rejoin[t / 4 + 1"x"t % 4 + 1]a/(p/1)/(p/2): 0
          m: 0 foreach d[-1 0x-1 1x-1 -1x0 1x0 -1x1 0x1 1][j: p + d
          if all[j/1 > 0 j/1 < 5 j/2 > 0 j/2 < 5 m < t: a/(j/1)/(j/2)][m: t]]0 = k: m]s: 0
          foreach n load form a[s: s + n]s]


          Try it online!



          It's really long solution and I'm not happy with it, but I spent so much time fixing it to work in TIO (apparently there are many differences between the Win and Linux stable versions of Red), so I post it anyway...



          More readable:



          f: func [ a ] [
          k: 16
          until [
          t: (index? find load form a n) - 1
          p: do rejoin [ t / 4 + 1 "x" t % 4 + 1 ]
          a/(p/1)/(p/2): 0
          m: 0
          foreach d [ -1 0x-1 1x-1 -1x0 1x0 -1x1 0x1 1 ] [
          j: p + d
          if all[ j/1 > 0
          j/1 < 5
          j/2 > 0
          j/2 < 5
          m < t: a/(j/1)/(j/2)
          ] [ m: t ]
          ]
          0 = k: m
          ]
          s: 0
          foreach n load form a [ s: s + n ]
          s
          ]





          share|improve this answer




























            up vote
            1
            down vote













            Not my best work. There's some definite improvements to be done, some probably fundamental to the algorithm used -- I'm sure it can be improved by using only an int, but I couldn't figure out how to efficiently enumerate neighbors that way. I'd love to see a PowerShell solution that uses only a single dimensional array!




            PowerShell Core, 348 bytes





            Function F($o){$t=120;$a=@{-1=,0*4;4=,0*4};0..3|%{$a[$_]=[int](-join$o[(3+18*$_)..(3+18*$_+13)]-split',')+,0};$m=16;while($m-gt0){0..3|%{$i=$_;0..3|%{if($a[$i][$_]-eq$m){$r=$i;$c=$_}}};$m=($a[$r-1][$c-1],$a[$r-1][$c],$a[$r-1][$c+1],$a[$r][$c+1],$a[$r][$c-1],$a[$r+1][$c-1],$a[$r+1][$c],$a[$r+1][$c+1]|Measure -Max).Maximum;$t-=$m;$a[$r][$c]=0}$t}


            Try it online!





            More readable version:



            Function F($o){
            $t=120;
            $a=@{-1=,0*4;4=,0*4};
            0..3|%{$a[$_]=[int](-join$o[(3+18*$_)..(3+18*$_+13)]-split',')+,0};
            $m=16;
            while($m-gt0){
            0..3|%{$i=$_;0..3|%{if($a[$i][$_]-eq$m){$r=$i;$c=$_}}};
            $m=($a[$r-1][$c-1],$a[$r-1][$c],$a[$r-1][$c+1],$a[$r][$c+1],$a[$r][$c-1],$a[$r+1][$c-1],$a[$r+1][$c],$a[$r+1][$c+1]|Measure -Max).Maximum;
            $t-=$m;
            $a[$r][$c]=0
            }
            $t
            }





            share|improve this answer




























              up vote
              1
              down vote













              Powershell, 143 141 136 130 122 121 bytes





              $a=,0*5+($args|%{$_+0})
              for($n=16;$i=$a.IndexOf($n)){$a[$i]=0
              $n=(-1,1+-6..-4+4..6|%{$a[$i+$_]}|sort)[-1]}$a|%{$s+=$_}
              $s


              Less golfed test script:



              $f = {

              $a=,0*5+($args|%{$_+0})
              for($n=16;$i=$a.IndexOf($n)){
              $a[$i]=0
              $n=(-1,1+-6..-4+4..6|%{$a[$i+$_]}|sort)[-1]
              }
              $a|%{$s+=$_}
              $s

              }

              @(
              ,( 0 , ( 4, 3, 2, 1), ( 5, 6, 7, 8), (12, 11, 10, 9), (13, 14, 15, 16) )
              ,( 0 , ( 8, 1, 9, 14), (11, 6, 5, 16), (13, 15, 2, 7), (10, 3, 12, 4) )
              ,( 1 , ( 1, 2, 3, 4), ( 5, 6, 7, 8), ( 9, 10, 11, 12), (13, 14, 15, 16) )
              ,( 3 , (10, 15, 14, 11), ( 9, 3, 1, 7), (13, 5, 12, 6), ( 2, 8, 4, 16) )
              ,( 12 , ( 3, 7, 10, 5), ( 6, 8, 12, 13), (15, 9, 11, 4), (14, 1, 16, 2) )
              ,( 34 , ( 8, 9, 3, 6), (13, 11, 7, 15), (12, 10, 16, 2), ( 4, 14, 1, 5) )
              ,( 51 , ( 8, 11, 12, 9), (14, 5, 10, 16), ( 7, 3, 1, 6), (13, 4, 2, 15) )
              ,( 78 , (13, 14, 1, 2), (16, 15, 3, 4), ( 5, 6, 7, 8), ( 9, 10, 11, 12) )
              ,( 102, ( 9, 10, 11, 12), ( 1, 2, 4, 13), ( 7, 8, 5, 14), ( 3, 16, 6, 15) )
              ,( 103, ( 9, 10, 11, 12), ( 1, 2, 7, 13), ( 6, 16, 4, 14), ( 3, 8, 5, 15) )
              ) | % {
              $expected, $a = $_
              $result = &$f @a
              "$($result-eq$expected): $result"
              }


              Output:



              True: 0
              True: 0
              True: 1
              True: 3
              True: 12
              True: 34
              True: 51
              True: 78
              True: 102
              True: 103


              Explanation:



              First, add top and bottom borders of 0 and make a single dimensional array:





              0 0 0 0 0
              # # # # 0
              # # # # 0
              # # # # 0
              # # # # 0



              0 0 0 0 0 # # # # 0 # # # # 0 # # # # 0 # # # # 0


              Powershell returns $null if you try to get the value behind the end of the array.



              Second, loop biggest neighbor pile started from 16 to non-zero-maximum. And nullify it (The Hungry Mouse eats it).





              for($n=16;$i=$a.IndexOf($n)){
              $a[$i]=0
              $n=(-1,1+-6..-4+4..6|%{$a[$i+$_]}|sort)[-1]
              }


              Third, sum of the remaining piles.






              share|improve this answer























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                14 Answers
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                up vote
                11
                down vote














                Python 2, 133 130 bytes





                a=input();m=16
                for i in range(m):a[i*5:i*5]=0,
                while m:i=a.index(m);a[i]=0;m=max(a[i+x]for x in[-6,-5,-4,-1,1,4,5,6])
                print sum(a)


                Try it online!



                Takes a flattened list of 16 elements.



                How it works



                a=input();m=16

                # Add zero padding on each row, and enough zeroes at the end to avoid index error
                for i in range(m):a[i*5:i*5]=0,

                # m == maximum element found in last iteration
                # i == index of last eaten element
                # eaten elements of `a` are reset to 0
                while m:i=a.index(m);a[i]=0;m=max(a[i+x]for x in[-6,-5,-4,-1,1,4,5,6])
                print sum(a)





                share|improve this answer























                • The adjacent-cell expression a[i+x]for x in[-6,-5,-4,-1,1,4,5,6] can be shortened to a[i+j+j/3*2-6]for j in range(9) (the zero entry is harmless). Python 3 can surely do shorter by hardcoding a length-8 bytestring, but Python 2 might still be better overall.
                  – xnor
                  2 days ago






                • 1




                  Though your zero padding loop is clever, it looks like it's shorter to take a 2D list: a=[0]*5 for r in input():a=r+[0]+a. Perhaps there's a yet shorter string slicing solution that doesn't require iterating.
                  – xnor
                  2 days ago















                up vote
                11
                down vote














                Python 2, 133 130 bytes





                a=input();m=16
                for i in range(m):a[i*5:i*5]=0,
                while m:i=a.index(m);a[i]=0;m=max(a[i+x]for x in[-6,-5,-4,-1,1,4,5,6])
                print sum(a)


                Try it online!



                Takes a flattened list of 16 elements.



                How it works



                a=input();m=16

                # Add zero padding on each row, and enough zeroes at the end to avoid index error
                for i in range(m):a[i*5:i*5]=0,

                # m == maximum element found in last iteration
                # i == index of last eaten element
                # eaten elements of `a` are reset to 0
                while m:i=a.index(m);a[i]=0;m=max(a[i+x]for x in[-6,-5,-4,-1,1,4,5,6])
                print sum(a)





                share|improve this answer























                • The adjacent-cell expression a[i+x]for x in[-6,-5,-4,-1,1,4,5,6] can be shortened to a[i+j+j/3*2-6]for j in range(9) (the zero entry is harmless). Python 3 can surely do shorter by hardcoding a length-8 bytestring, but Python 2 might still be better overall.
                  – xnor
                  2 days ago






                • 1




                  Though your zero padding loop is clever, it looks like it's shorter to take a 2D list: a=[0]*5 for r in input():a=r+[0]+a. Perhaps there's a yet shorter string slicing solution that doesn't require iterating.
                  – xnor
                  2 days ago













                up vote
                11
                down vote










                up vote
                11
                down vote










                Python 2, 133 130 bytes





                a=input();m=16
                for i in range(m):a[i*5:i*5]=0,
                while m:i=a.index(m);a[i]=0;m=max(a[i+x]for x in[-6,-5,-4,-1,1,4,5,6])
                print sum(a)


                Try it online!



                Takes a flattened list of 16 elements.



                How it works



                a=input();m=16

                # Add zero padding on each row, and enough zeroes at the end to avoid index error
                for i in range(m):a[i*5:i*5]=0,

                # m == maximum element found in last iteration
                # i == index of last eaten element
                # eaten elements of `a` are reset to 0
                while m:i=a.index(m);a[i]=0;m=max(a[i+x]for x in[-6,-5,-4,-1,1,4,5,6])
                print sum(a)





                share|improve this answer















                Python 2, 133 130 bytes





                a=input();m=16
                for i in range(m):a[i*5:i*5]=0,
                while m:i=a.index(m);a[i]=0;m=max(a[i+x]for x in[-6,-5,-4,-1,1,4,5,6])
                print sum(a)


                Try it online!



                Takes a flattened list of 16 elements.



                How it works



                a=input();m=16

                # Add zero padding on each row, and enough zeroes at the end to avoid index error
                for i in range(m):a[i*5:i*5]=0,

                # m == maximum element found in last iteration
                # i == index of last eaten element
                # eaten elements of `a` are reset to 0
                while m:i=a.index(m);a[i]=0;m=max(a[i+x]for x in[-6,-5,-4,-1,1,4,5,6])
                print sum(a)






                share|improve this answer














                share|improve this answer



                share|improve this answer








                edited 2 days ago

























                answered 2 days ago









                Bubbler

                5,729756




                5,729756












                • The adjacent-cell expression a[i+x]for x in[-6,-5,-4,-1,1,4,5,6] can be shortened to a[i+j+j/3*2-6]for j in range(9) (the zero entry is harmless). Python 3 can surely do shorter by hardcoding a length-8 bytestring, but Python 2 might still be better overall.
                  – xnor
                  2 days ago






                • 1




                  Though your zero padding loop is clever, it looks like it's shorter to take a 2D list: a=[0]*5 for r in input():a=r+[0]+a. Perhaps there's a yet shorter string slicing solution that doesn't require iterating.
                  – xnor
                  2 days ago


















                • The adjacent-cell expression a[i+x]for x in[-6,-5,-4,-1,1,4,5,6] can be shortened to a[i+j+j/3*2-6]for j in range(9) (the zero entry is harmless). Python 3 can surely do shorter by hardcoding a length-8 bytestring, but Python 2 might still be better overall.
                  – xnor
                  2 days ago






                • 1




                  Though your zero padding loop is clever, it looks like it's shorter to take a 2D list: a=[0]*5 for r in input():a=r+[0]+a. Perhaps there's a yet shorter string slicing solution that doesn't require iterating.
                  – xnor
                  2 days ago
















                The adjacent-cell expression a[i+x]for x in[-6,-5,-4,-1,1,4,5,6] can be shortened to a[i+j+j/3*2-6]for j in range(9) (the zero entry is harmless). Python 3 can surely do shorter by hardcoding a length-8 bytestring, but Python 2 might still be better overall.
                – xnor
                2 days ago




                The adjacent-cell expression a[i+x]for x in[-6,-5,-4,-1,1,4,5,6] can be shortened to a[i+j+j/3*2-6]for j in range(9) (the zero entry is harmless). Python 3 can surely do shorter by hardcoding a length-8 bytestring, but Python 2 might still be better overall.
                – xnor
                2 days ago




                1




                1




                Though your zero padding loop is clever, it looks like it's shorter to take a 2D list: a=[0]*5 for r in input():a=r+[0]+a. Perhaps there's a yet shorter string slicing solution that doesn't require iterating.
                – xnor
                2 days ago




                Though your zero padding loop is clever, it looks like it's shorter to take a 2D list: a=[0]*5 for r in input():a=r+[0]+a. Perhaps there's a yet shorter string slicing solution that doesn't require iterating.
                – xnor
                2 days ago










                up vote
                8
                down vote














                MATL, 50 49 47 bytes



                16:HZ^!"2G@m1ZIm~]v16eXK68E16b"Ky0)Y)fyX-X>h]s-


                Input is a matrix, using ; as row separator.



                Try it online! Or verify all test cases.



                Explanation



                16:HZ^!  % Cartesian power of [1 2 ... 16] with exponent 2, transpose. Gives a 
                % 2-row matrix with 1st column [1; 1], 2nd [1; 2], ..., last [16; 16]
                " % For each column, say [k; j]
                2 % Push 2
                G@m % Push input matrix, then current column [k; j], then check membership.
                % This gives a 4×4 matrix that contains 1 for entries of the input that
                % contain k or j
                1ZI % Connected components (based on 8-neighbourhood) of nonzero entries.
                % This gives a 4×4 matrix with each connected component labeled with
                % values 1, 2, ... respectively
                m~ % True if 2 is not present in this matrix. That means there is only
                % one connected component; that is, k and j are neighbours in the
                % input matrix, or k=j
                ] % End
                v16e % The stack now has 256 values. Concatenate them into a vector and
                % reshape as a 16×16 matrix. This matrix describes neighbourhood: entry
                % (k,j) is 1 if values k and j are neighbours in the input or if k=j
                XK % Copy into clipboard K
                68E % Push 68 times 2, that is, 136, which is 1+2+...+16
                16 % Push 16. This is the initial value eaten by the mouse. New values will
                % be appended to create a vector of eaten values
                b % Bubble up the 16×16 matrix to the top of the stack
                " % For each column. This just executes the loop 16 times
                K % Push neighbourhood matrix from clipboard K
                y % Copy from below: pushes a copy of the vector of eaten values
                0) % Get last value. This is the most recent eaten value
                Y) % Get that row of the neighbourhood matrix
                f % Indices of nonzeros. This gives a vector of neighbours of the last
                % eaten value
                y % Copy from below: pushes a copy of the vector of eaten values
                X- % Set difference (may give an empty result)
                X> % Maximum value. This is the new eaten value (maximum neighbour not
                % already eaten). May be empty, if all neighbours are already eaten
                h % Concatenate to vector of eaten values
                ] % End
                s % Sum of vector of all eaten values
                - % Subtract from 136. Implicitly display





                share|improve this answer























                • Idk MatLab, but can you save a little if you push -136 instead of +136?
                  – Titus
                  yesterday










                • @Titus Hm I don't see how
                  – Luis Mendo
                  yesterday










                • or the other way round: I thought instead of 1) push 136 2) push each eaten value 3) sum up eaten values 4) subtract from 136 -> 1) push 136 2) push negative of eaten value 3) sum up stack. But as it obviously is only one byte each; it´s probably no gain.
                  – Titus
                  yesterday










                • @Titus Ah, yes, I think that uses the same number of bytes. Also, I need each eaten value (not its negative) for the set difference; negating would have to be done at the end
                  – Luis Mendo
                  yesterday

















                up vote
                8
                down vote














                MATL, 50 49 47 bytes



                16:HZ^!"2G@m1ZIm~]v16eXK68E16b"Ky0)Y)fyX-X>h]s-


                Input is a matrix, using ; as row separator.



                Try it online! Or verify all test cases.



                Explanation



                16:HZ^!  % Cartesian power of [1 2 ... 16] with exponent 2, transpose. Gives a 
                % 2-row matrix with 1st column [1; 1], 2nd [1; 2], ..., last [16; 16]
                " % For each column, say [k; j]
                2 % Push 2
                G@m % Push input matrix, then current column [k; j], then check membership.
                % This gives a 4×4 matrix that contains 1 for entries of the input that
                % contain k or j
                1ZI % Connected components (based on 8-neighbourhood) of nonzero entries.
                % This gives a 4×4 matrix with each connected component labeled with
                % values 1, 2, ... respectively
                m~ % True if 2 is not present in this matrix. That means there is only
                % one connected component; that is, k and j are neighbours in the
                % input matrix, or k=j
                ] % End
                v16e % The stack now has 256 values. Concatenate them into a vector and
                % reshape as a 16×16 matrix. This matrix describes neighbourhood: entry
                % (k,j) is 1 if values k and j are neighbours in the input or if k=j
                XK % Copy into clipboard K
                68E % Push 68 times 2, that is, 136, which is 1+2+...+16
                16 % Push 16. This is the initial value eaten by the mouse. New values will
                % be appended to create a vector of eaten values
                b % Bubble up the 16×16 matrix to the top of the stack
                " % For each column. This just executes the loop 16 times
                K % Push neighbourhood matrix from clipboard K
                y % Copy from below: pushes a copy of the vector of eaten values
                0) % Get last value. This is the most recent eaten value
                Y) % Get that row of the neighbourhood matrix
                f % Indices of nonzeros. This gives a vector of neighbours of the last
                % eaten value
                y % Copy from below: pushes a copy of the vector of eaten values
                X- % Set difference (may give an empty result)
                X> % Maximum value. This is the new eaten value (maximum neighbour not
                % already eaten). May be empty, if all neighbours are already eaten
                h % Concatenate to vector of eaten values
                ] % End
                s % Sum of vector of all eaten values
                - % Subtract from 136. Implicitly display





                share|improve this answer























                • Idk MatLab, but can you save a little if you push -136 instead of +136?
                  – Titus
                  yesterday










                • @Titus Hm I don't see how
                  – Luis Mendo
                  yesterday










                • or the other way round: I thought instead of 1) push 136 2) push each eaten value 3) sum up eaten values 4) subtract from 136 -> 1) push 136 2) push negative of eaten value 3) sum up stack. But as it obviously is only one byte each; it´s probably no gain.
                  – Titus
                  yesterday










                • @Titus Ah, yes, I think that uses the same number of bytes. Also, I need each eaten value (not its negative) for the set difference; negating would have to be done at the end
                  – Luis Mendo
                  yesterday















                up vote
                8
                down vote










                up vote
                8
                down vote










                MATL, 50 49 47 bytes



                16:HZ^!"2G@m1ZIm~]v16eXK68E16b"Ky0)Y)fyX-X>h]s-


                Input is a matrix, using ; as row separator.



                Try it online! Or verify all test cases.



                Explanation



                16:HZ^!  % Cartesian power of [1 2 ... 16] with exponent 2, transpose. Gives a 
                % 2-row matrix with 1st column [1; 1], 2nd [1; 2], ..., last [16; 16]
                " % For each column, say [k; j]
                2 % Push 2
                G@m % Push input matrix, then current column [k; j], then check membership.
                % This gives a 4×4 matrix that contains 1 for entries of the input that
                % contain k or j
                1ZI % Connected components (based on 8-neighbourhood) of nonzero entries.
                % This gives a 4×4 matrix with each connected component labeled with
                % values 1, 2, ... respectively
                m~ % True if 2 is not present in this matrix. That means there is only
                % one connected component; that is, k and j are neighbours in the
                % input matrix, or k=j
                ] % End
                v16e % The stack now has 256 values. Concatenate them into a vector and
                % reshape as a 16×16 matrix. This matrix describes neighbourhood: entry
                % (k,j) is 1 if values k and j are neighbours in the input or if k=j
                XK % Copy into clipboard K
                68E % Push 68 times 2, that is, 136, which is 1+2+...+16
                16 % Push 16. This is the initial value eaten by the mouse. New values will
                % be appended to create a vector of eaten values
                b % Bubble up the 16×16 matrix to the top of the stack
                " % For each column. This just executes the loop 16 times
                K % Push neighbourhood matrix from clipboard K
                y % Copy from below: pushes a copy of the vector of eaten values
                0) % Get last value. This is the most recent eaten value
                Y) % Get that row of the neighbourhood matrix
                f % Indices of nonzeros. This gives a vector of neighbours of the last
                % eaten value
                y % Copy from below: pushes a copy of the vector of eaten values
                X- % Set difference (may give an empty result)
                X> % Maximum value. This is the new eaten value (maximum neighbour not
                % already eaten). May be empty, if all neighbours are already eaten
                h % Concatenate to vector of eaten values
                ] % End
                s % Sum of vector of all eaten values
                - % Subtract from 136. Implicitly display





                share|improve this answer















                MATL, 50 49 47 bytes



                16:HZ^!"2G@m1ZIm~]v16eXK68E16b"Ky0)Y)fyX-X>h]s-


                Input is a matrix, using ; as row separator.



                Try it online! Or verify all test cases.



                Explanation



                16:HZ^!  % Cartesian power of [1 2 ... 16] with exponent 2, transpose. Gives a 
                % 2-row matrix with 1st column [1; 1], 2nd [1; 2], ..., last [16; 16]
                " % For each column, say [k; j]
                2 % Push 2
                G@m % Push input matrix, then current column [k; j], then check membership.
                % This gives a 4×4 matrix that contains 1 for entries of the input that
                % contain k or j
                1ZI % Connected components (based on 8-neighbourhood) of nonzero entries.
                % This gives a 4×4 matrix with each connected component labeled with
                % values 1, 2, ... respectively
                m~ % True if 2 is not present in this matrix. That means there is only
                % one connected component; that is, k and j are neighbours in the
                % input matrix, or k=j
                ] % End
                v16e % The stack now has 256 values. Concatenate them into a vector and
                % reshape as a 16×16 matrix. This matrix describes neighbourhood: entry
                % (k,j) is 1 if values k and j are neighbours in the input or if k=j
                XK % Copy into clipboard K
                68E % Push 68 times 2, that is, 136, which is 1+2+...+16
                16 % Push 16. This is the initial value eaten by the mouse. New values will
                % be appended to create a vector of eaten values
                b % Bubble up the 16×16 matrix to the top of the stack
                " % For each column. This just executes the loop 16 times
                K % Push neighbourhood matrix from clipboard K
                y % Copy from below: pushes a copy of the vector of eaten values
                0) % Get last value. This is the most recent eaten value
                Y) % Get that row of the neighbourhood matrix
                f % Indices of nonzeros. This gives a vector of neighbours of the last
                % eaten value
                y % Copy from below: pushes a copy of the vector of eaten values
                X- % Set difference (may give an empty result)
                X> % Maximum value. This is the new eaten value (maximum neighbour not
                % already eaten). May be empty, if all neighbours are already eaten
                h % Concatenate to vector of eaten values
                ] % End
                s % Sum of vector of all eaten values
                - % Subtract from 136. Implicitly display






                share|improve this answer














                share|improve this answer



                share|improve this answer








                edited 2 days ago

























                answered 2 days ago









                Luis Mendo

                73.7k885289




                73.7k885289












                • Idk MatLab, but can you save a little if you push -136 instead of +136?
                  – Titus
                  yesterday










                • @Titus Hm I don't see how
                  – Luis Mendo
                  yesterday










                • or the other way round: I thought instead of 1) push 136 2) push each eaten value 3) sum up eaten values 4) subtract from 136 -> 1) push 136 2) push negative of eaten value 3) sum up stack. But as it obviously is only one byte each; it´s probably no gain.
                  – Titus
                  yesterday










                • @Titus Ah, yes, I think that uses the same number of bytes. Also, I need each eaten value (not its negative) for the set difference; negating would have to be done at the end
                  – Luis Mendo
                  yesterday




















                • Idk MatLab, but can you save a little if you push -136 instead of +136?
                  – Titus
                  yesterday










                • @Titus Hm I don't see how
                  – Luis Mendo
                  yesterday










                • or the other way round: I thought instead of 1) push 136 2) push each eaten value 3) sum up eaten values 4) subtract from 136 -> 1) push 136 2) push negative of eaten value 3) sum up stack. But as it obviously is only one byte each; it´s probably no gain.
                  – Titus
                  yesterday










                • @Titus Ah, yes, I think that uses the same number of bytes. Also, I need each eaten value (not its negative) for the set difference; negating would have to be done at the end
                  – Luis Mendo
                  yesterday


















                Idk MatLab, but can you save a little if you push -136 instead of +136?
                – Titus
                yesterday




                Idk MatLab, but can you save a little if you push -136 instead of +136?
                – Titus
                yesterday












                @Titus Hm I don't see how
                – Luis Mendo
                yesterday




                @Titus Hm I don't see how
                – Luis Mendo
                yesterday












                or the other way round: I thought instead of 1) push 136 2) push each eaten value 3) sum up eaten values 4) subtract from 136 -> 1) push 136 2) push negative of eaten value 3) sum up stack. But as it obviously is only one byte each; it´s probably no gain.
                – Titus
                yesterday




                or the other way round: I thought instead of 1) push 136 2) push each eaten value 3) sum up eaten values 4) subtract from 136 -> 1) push 136 2) push negative of eaten value 3) sum up stack. But as it obviously is only one byte each; it´s probably no gain.
                – Titus
                yesterday












                @Titus Ah, yes, I think that uses the same number of bytes. Also, I need each eaten value (not its negative) for the set difference; negating would have to be done at the end
                – Luis Mendo
                yesterday






                @Titus Ah, yes, I think that uses the same number of bytes. Also, I need each eaten value (not its negative) for the set difference; negating would have to be done at the end
                – Luis Mendo
                yesterday












                up vote
                7
                down vote














                Python 2, 111 bytes





                i=x=a=input()
                while x:x,i=max((y,j)for j,y in enumerate(a)if i>or 2>i/4-j/4>-2<i%4-j%4<2);a[i]=0
                print sum(a)


                Try it online!



                Method and test cases adapted from Bubbler. Takes a flat list on STDIN.



                The code checks whether two flat indices i and j represent touching cells by checking that both row different i/4-j/4 and column difference i%4-j%4 are strictly between -2 and 2. The first pass instead has this check automatically succeed so that the largest entry is found disregarding adjacency.






                share|improve this answer

























                  up vote
                  7
                  down vote














                  Python 2, 111 bytes





                  i=x=a=input()
                  while x:x,i=max((y,j)for j,y in enumerate(a)if i>or 2>i/4-j/4>-2<i%4-j%4<2);a[i]=0
                  print sum(a)


                  Try it online!



                  Method and test cases adapted from Bubbler. Takes a flat list on STDIN.



                  The code checks whether two flat indices i and j represent touching cells by checking that both row different i/4-j/4 and column difference i%4-j%4 are strictly between -2 and 2. The first pass instead has this check automatically succeed so that the largest entry is found disregarding adjacency.






                  share|improve this answer























                    up vote
                    7
                    down vote










                    up vote
                    7
                    down vote










                    Python 2, 111 bytes





                    i=x=a=input()
                    while x:x,i=max((y,j)for j,y in enumerate(a)if i>or 2>i/4-j/4>-2<i%4-j%4<2);a[i]=0
                    print sum(a)


                    Try it online!



                    Method and test cases adapted from Bubbler. Takes a flat list on STDIN.



                    The code checks whether two flat indices i and j represent touching cells by checking that both row different i/4-j/4 and column difference i%4-j%4 are strictly between -2 and 2. The first pass instead has this check automatically succeed so that the largest entry is found disregarding adjacency.






                    share|improve this answer













                    Python 2, 111 bytes





                    i=x=a=input()
                    while x:x,i=max((y,j)for j,y in enumerate(a)if i>or 2>i/4-j/4>-2<i%4-j%4<2);a[i]=0
                    print sum(a)


                    Try it online!



                    Method and test cases adapted from Bubbler. Takes a flat list on STDIN.



                    The code checks whether two flat indices i and j represent touching cells by checking that both row different i/4-j/4 and column difference i%4-j%4 are strictly between -2 and 2. The first pass instead has this check automatically succeed so that the largest entry is found disregarding adjacency.







                    share|improve this answer












                    share|improve this answer



                    share|improve this answer










                    answered 2 days ago









                    xnor

                    88.9k18184437




                    88.9k18184437






















                        up vote
                        6
                        down vote













                        PHP, 177 174 171 bytes



                        for($v=16;$v;$u+=$v=max($p%4-1?max($a[$p-5],$a[$p-1],$a[$p+3]):0,$a[$p-4],$a[$p+4],$p%4?max($a[$p-3],$a[$p+1],$a[$p+5]):0))$a[$p=array_search($v,$a=&$argv)]=0;echo 120-$u;


                        Run with -nr, provide matrix elements as arguments or try it online.






                        share|improve this answer



























                          up vote
                          6
                          down vote













                          PHP, 177 174 171 bytes



                          for($v=16;$v;$u+=$v=max($p%4-1?max($a[$p-5],$a[$p-1],$a[$p+3]):0,$a[$p-4],$a[$p+4],$p%4?max($a[$p-3],$a[$p+1],$a[$p+5]):0))$a[$p=array_search($v,$a=&$argv)]=0;echo 120-$u;


                          Run with -nr, provide matrix elements as arguments or try it online.






                          share|improve this answer

























                            up vote
                            6
                            down vote










                            up vote
                            6
                            down vote









                            PHP, 177 174 171 bytes



                            for($v=16;$v;$u+=$v=max($p%4-1?max($a[$p-5],$a[$p-1],$a[$p+3]):0,$a[$p-4],$a[$p+4],$p%4?max($a[$p-3],$a[$p+1],$a[$p+5]):0))$a[$p=array_search($v,$a=&$argv)]=0;echo 120-$u;


                            Run with -nr, provide matrix elements as arguments or try it online.






                            share|improve this answer














                            PHP, 177 174 171 bytes



                            for($v=16;$v;$u+=$v=max($p%4-1?max($a[$p-5],$a[$p-1],$a[$p+3]):0,$a[$p-4],$a[$p+4],$p%4?max($a[$p-3],$a[$p+1],$a[$p+5]):0))$a[$p=array_search($v,$a=&$argv)]=0;echo 120-$u;


                            Run with -nr, provide matrix elements as arguments or try it online.







                            share|improve this answer














                            share|improve this answer



                            share|improve this answer








                            edited 2 days ago

























                            answered 2 days ago









                            Titus

                            12.9k11237




                            12.9k11237






















                                up vote
                                3
                                down vote














                                Charcoal, 47 bytes



                                EA⭆ι§αλ≔QθW›θA«≔⌕KAθθJ﹪θ⁴÷θ⁴≔⌈KMθA»≔ΣEKA⌕αιθ⎚Iθ


                                Try it online! Link is to verbose version of code. Explanation:



                                EA⭆ι§αλ


                                Convert the input numbers into alphabetic characters (A=0 .. Q=16) and print them as a 4x4 grid.



                                ≔Qθ


                                Start by eating the Q, i.e. 16.



                                W›θA«


                                Repeat while there is something to eat.



                                ≔⌕KAθθ


                                Find where the pile is. This is a linear view in row-major order.



                                J﹪θ⁴÷θ⁴


                                Convert to co-ordinates and jump to that location.



                                ≔⌈KMθ


                                Find the largest adjacent pile.






                                Eat the current pile.



                                ≔ΣEKA⌕αιθ


                                Convert the piles back to integers and take the sum.



                                ⎚Iθ


                                Clear the canvas and output the result.






                                share|improve this answer

























                                  up vote
                                  3
                                  down vote














                                  Charcoal, 47 bytes



                                  EA⭆ι§αλ≔QθW›θA«≔⌕KAθθJ﹪θ⁴÷θ⁴≔⌈KMθA»≔ΣEKA⌕αιθ⎚Iθ


                                  Try it online! Link is to verbose version of code. Explanation:



                                  EA⭆ι§αλ


                                  Convert the input numbers into alphabetic characters (A=0 .. Q=16) and print them as a 4x4 grid.



                                  ≔Qθ


                                  Start by eating the Q, i.e. 16.



                                  W›θA«


                                  Repeat while there is something to eat.



                                  ≔⌕KAθθ


                                  Find where the pile is. This is a linear view in row-major order.



                                  J﹪θ⁴÷θ⁴


                                  Convert to co-ordinates and jump to that location.



                                  ≔⌈KMθ


                                  Find the largest adjacent pile.






                                  Eat the current pile.



                                  ≔ΣEKA⌕αιθ


                                  Convert the piles back to integers and take the sum.



                                  ⎚Iθ


                                  Clear the canvas and output the result.






                                  share|improve this answer























                                    up vote
                                    3
                                    down vote










                                    up vote
                                    3
                                    down vote










                                    Charcoal, 47 bytes



                                    EA⭆ι§αλ≔QθW›θA«≔⌕KAθθJ﹪θ⁴÷θ⁴≔⌈KMθA»≔ΣEKA⌕αιθ⎚Iθ


                                    Try it online! Link is to verbose version of code. Explanation:



                                    EA⭆ι§αλ


                                    Convert the input numbers into alphabetic characters (A=0 .. Q=16) and print them as a 4x4 grid.



                                    ≔Qθ


                                    Start by eating the Q, i.e. 16.



                                    W›θA«


                                    Repeat while there is something to eat.



                                    ≔⌕KAθθ


                                    Find where the pile is. This is a linear view in row-major order.



                                    J﹪θ⁴÷θ⁴


                                    Convert to co-ordinates and jump to that location.



                                    ≔⌈KMθ


                                    Find the largest adjacent pile.






                                    Eat the current pile.



                                    ≔ΣEKA⌕αιθ


                                    Convert the piles back to integers and take the sum.



                                    ⎚Iθ


                                    Clear the canvas and output the result.






                                    share|improve this answer













                                    Charcoal, 47 bytes



                                    EA⭆ι§αλ≔QθW›θA«≔⌕KAθθJ﹪θ⁴÷θ⁴≔⌈KMθA»≔ΣEKA⌕αιθ⎚Iθ


                                    Try it online! Link is to verbose version of code. Explanation:



                                    EA⭆ι§αλ


                                    Convert the input numbers into alphabetic characters (A=0 .. Q=16) and print them as a 4x4 grid.



                                    ≔Qθ


                                    Start by eating the Q, i.e. 16.



                                    W›θA«


                                    Repeat while there is something to eat.



                                    ≔⌕KAθθ


                                    Find where the pile is. This is a linear view in row-major order.



                                    J﹪θ⁴÷θ⁴


                                    Convert to co-ordinates and jump to that location.



                                    ≔⌈KMθ


                                    Find the largest adjacent pile.






                                    Eat the current pile.



                                    ≔ΣEKA⌕αιθ


                                    Convert the piles back to integers and take the sum.



                                    ⎚Iθ


                                    Clear the canvas and output the result.







                                    share|improve this answer












                                    share|improve this answer



                                    share|improve this answer










                                    answered 2 days ago









                                    Neil

                                    78.1k744175




                                    78.1k744175






















                                        up vote
                                        3
                                        down vote














                                        R, 128 124 bytes





                                        r=rbind(0,cbind(0,matrix(scan(),4,4),0),0)
                                        m=which(r==16)
                                        while(r[m]){r[m]=0
                                        m=which(r==max(r[m+c(-7:-5,-1,1,5:7)]))}
                                        sum(r)


                                        Try it online!



                                        TIO link is slightly different, I am still trying to figure out how to make it work.



                                        I do feel like I can golf a lot more out of this. But this works for now.



                                        It creates a 4x4 matrix (which helped me to visualize things), pads it with 0's, then begins from 16 and searches it's surrounding "piles" for the next largest, and so forth.



                                        Upon conclusion, it does output a warning, but it is of no consequence and does not change the result.



                                        EDIT: -4 bytes by compressing the initialization of the matrix into 1 line






                                        share|improve this answer



























                                          up vote
                                          3
                                          down vote














                                          R, 128 124 bytes





                                          r=rbind(0,cbind(0,matrix(scan(),4,4),0),0)
                                          m=which(r==16)
                                          while(r[m]){r[m]=0
                                          m=which(r==max(r[m+c(-7:-5,-1,1,5:7)]))}
                                          sum(r)


                                          Try it online!



                                          TIO link is slightly different, I am still trying to figure out how to make it work.



                                          I do feel like I can golf a lot more out of this. But this works for now.



                                          It creates a 4x4 matrix (which helped me to visualize things), pads it with 0's, then begins from 16 and searches it's surrounding "piles" for the next largest, and so forth.



                                          Upon conclusion, it does output a warning, but it is of no consequence and does not change the result.



                                          EDIT: -4 bytes by compressing the initialization of the matrix into 1 line






                                          share|improve this answer

























                                            up vote
                                            3
                                            down vote










                                            up vote
                                            3
                                            down vote










                                            R, 128 124 bytes





                                            r=rbind(0,cbind(0,matrix(scan(),4,4),0),0)
                                            m=which(r==16)
                                            while(r[m]){r[m]=0
                                            m=which(r==max(r[m+c(-7:-5,-1,1,5:7)]))}
                                            sum(r)


                                            Try it online!



                                            TIO link is slightly different, I am still trying to figure out how to make it work.



                                            I do feel like I can golf a lot more out of this. But this works for now.



                                            It creates a 4x4 matrix (which helped me to visualize things), pads it with 0's, then begins from 16 and searches it's surrounding "piles" for the next largest, and so forth.



                                            Upon conclusion, it does output a warning, but it is of no consequence and does not change the result.



                                            EDIT: -4 bytes by compressing the initialization of the matrix into 1 line






                                            share|improve this answer















                                            R, 128 124 bytes





                                            r=rbind(0,cbind(0,matrix(scan(),4,4),0),0)
                                            m=which(r==16)
                                            while(r[m]){r[m]=0
                                            m=which(r==max(r[m+c(-7:-5,-1,1,5:7)]))}
                                            sum(r)


                                            Try it online!



                                            TIO link is slightly different, I am still trying to figure out how to make it work.



                                            I do feel like I can golf a lot more out of this. But this works for now.



                                            It creates a 4x4 matrix (which helped me to visualize things), pads it with 0's, then begins from 16 and searches it's surrounding "piles" for the next largest, and so forth.



                                            Upon conclusion, it does output a warning, but it is of no consequence and does not change the result.



                                            EDIT: -4 bytes by compressing the initialization of the matrix into 1 line







                                            share|improve this answer














                                            share|improve this answer



                                            share|improve this answer








                                            edited yesterday

























                                            answered yesterday









                                            Sumner18

                                            3115




                                            3115






















                                                up vote
                                                2
                                                down vote













                                                JavaScript, 122 bytes



                                                I took more than a couple of wrong turns on this one and now I've run out of time for further golfing but at least it's working. Will revisit tomorrow (or, knowing me, on the train home this evening!), if I can find a minute.



                                                a=>(g=n=>n?g([-6,-5,-4,-1,1,4,5,6].map(x=>n=a[x+=i]>n?a[x]:n,a[i=a.indexOf(n)]=n=0)|n)-n:120)(16,a=a.flatMap(x=>[...x,0]))


                                                Try it online






                                                share|improve this answer

















                                                • 1




                                                  +1 for flatMap() :p
                                                  – Arnauld
                                                  2 days ago










                                                • :D I think this is the first time I've used it for golf! Out of interest (and to give me a target when I come back to this), what was your score when you tried it?
                                                  – Shaggy
                                                  2 days ago






                                                • 1




                                                  I didn't try to golf it. Unless I have a specific trick in mind, I usually just write clean code to test my challenges.
                                                  – Arnauld
                                                  2 days ago










                                                • Ok, you've convinced me to try. :) I'm currently at <s>116</s> 114.
                                                  – Arnauld
                                                  2 days ago












                                                • 98 bytes after a good night's sleep. (Sorry about the multiple notifications.)
                                                  – Arnauld
                                                  yesterday















                                                up vote
                                                2
                                                down vote













                                                JavaScript, 122 bytes



                                                I took more than a couple of wrong turns on this one and now I've run out of time for further golfing but at least it's working. Will revisit tomorrow (or, knowing me, on the train home this evening!), if I can find a minute.



                                                a=>(g=n=>n?g([-6,-5,-4,-1,1,4,5,6].map(x=>n=a[x+=i]>n?a[x]:n,a[i=a.indexOf(n)]=n=0)|n)-n:120)(16,a=a.flatMap(x=>[...x,0]))


                                                Try it online






                                                share|improve this answer

















                                                • 1




                                                  +1 for flatMap() :p
                                                  – Arnauld
                                                  2 days ago










                                                • :D I think this is the first time I've used it for golf! Out of interest (and to give me a target when I come back to this), what was your score when you tried it?
                                                  – Shaggy
                                                  2 days ago






                                                • 1




                                                  I didn't try to golf it. Unless I have a specific trick in mind, I usually just write clean code to test my challenges.
                                                  – Arnauld
                                                  2 days ago










                                                • Ok, you've convinced me to try. :) I'm currently at <s>116</s> 114.
                                                  – Arnauld
                                                  2 days ago












                                                • 98 bytes after a good night's sleep. (Sorry about the multiple notifications.)
                                                  – Arnauld
                                                  yesterday













                                                up vote
                                                2
                                                down vote










                                                up vote
                                                2
                                                down vote









                                                JavaScript, 122 bytes



                                                I took more than a couple of wrong turns on this one and now I've run out of time for further golfing but at least it's working. Will revisit tomorrow (or, knowing me, on the train home this evening!), if I can find a minute.



                                                a=>(g=n=>n?g([-6,-5,-4,-1,1,4,5,6].map(x=>n=a[x+=i]>n?a[x]:n,a[i=a.indexOf(n)]=n=0)|n)-n:120)(16,a=a.flatMap(x=>[...x,0]))


                                                Try it online






                                                share|improve this answer












                                                JavaScript, 122 bytes



                                                I took more than a couple of wrong turns on this one and now I've run out of time for further golfing but at least it's working. Will revisit tomorrow (or, knowing me, on the train home this evening!), if I can find a minute.



                                                a=>(g=n=>n?g([-6,-5,-4,-1,1,4,5,6].map(x=>n=a[x+=i]>n?a[x]:n,a[i=a.indexOf(n)]=n=0)|n)-n:120)(16,a=a.flatMap(x=>[...x,0]))


                                                Try it online







                                                share|improve this answer












                                                share|improve this answer



                                                share|improve this answer










                                                answered 2 days ago









                                                Shaggy

                                                18.2k21663




                                                18.2k21663








                                                • 1




                                                  +1 for flatMap() :p
                                                  – Arnauld
                                                  2 days ago










                                                • :D I think this is the first time I've used it for golf! Out of interest (and to give me a target when I come back to this), what was your score when you tried it?
                                                  – Shaggy
                                                  2 days ago






                                                • 1




                                                  I didn't try to golf it. Unless I have a specific trick in mind, I usually just write clean code to test my challenges.
                                                  – Arnauld
                                                  2 days ago










                                                • Ok, you've convinced me to try. :) I'm currently at <s>116</s> 114.
                                                  – Arnauld
                                                  2 days ago












                                                • 98 bytes after a good night's sleep. (Sorry about the multiple notifications.)
                                                  – Arnauld
                                                  yesterday














                                                • 1




                                                  +1 for flatMap() :p
                                                  – Arnauld
                                                  2 days ago










                                                • :D I think this is the first time I've used it for golf! Out of interest (and to give me a target when I come back to this), what was your score when you tried it?
                                                  – Shaggy
                                                  2 days ago






                                                • 1




                                                  I didn't try to golf it. Unless I have a specific trick in mind, I usually just write clean code to test my challenges.
                                                  – Arnauld
                                                  2 days ago










                                                • Ok, you've convinced me to try. :) I'm currently at <s>116</s> 114.
                                                  – Arnauld
                                                  2 days ago












                                                • 98 bytes after a good night's sleep. (Sorry about the multiple notifications.)
                                                  – Arnauld
                                                  yesterday








                                                1




                                                1




                                                +1 for flatMap() :p
                                                – Arnauld
                                                2 days ago




                                                +1 for flatMap() :p
                                                – Arnauld
                                                2 days ago












                                                :D I think this is the first time I've used it for golf! Out of interest (and to give me a target when I come back to this), what was your score when you tried it?
                                                – Shaggy
                                                2 days ago




                                                :D I think this is the first time I've used it for golf! Out of interest (and to give me a target when I come back to this), what was your score when you tried it?
                                                – Shaggy
                                                2 days ago




                                                1




                                                1




                                                I didn't try to golf it. Unless I have a specific trick in mind, I usually just write clean code to test my challenges.
                                                – Arnauld
                                                2 days ago




                                                I didn't try to golf it. Unless I have a specific trick in mind, I usually just write clean code to test my challenges.
                                                – Arnauld
                                                2 days ago












                                                Ok, you've convinced me to try. :) I'm currently at <s>116</s> 114.
                                                – Arnauld
                                                2 days ago






                                                Ok, you've convinced me to try. :) I'm currently at <s>116</s> 114.
                                                – Arnauld
                                                2 days ago














                                                98 bytes after a good night's sleep. (Sorry about the multiple notifications.)
                                                – Arnauld
                                                yesterday




                                                98 bytes after a good night's sleep. (Sorry about the multiple notifications.)
                                                – Arnauld
                                                yesterday










                                                up vote
                                                2
                                                down vote














                                                Jelly,  31 30  29 bytes



                                                ³œiⱮZIỊȦ
                                                ⁴ṖŒPŒ!€Ẏ⁴;ⱮṢÇƇṪ
                                                FḟÇS


                                                Since the method is far too slow to run within 60s with the mouse starting on 16 this starts her off at 9 and limits her ability such that she is only able to eat 9s or less Try it online! (thus here she eats 9, 2, 7, 4, 8, 6, 3 leaving 97).



                                                How?



                                                ³œiⱮZIỊȦ - Link 1, isSatisfactory?: list of integers, possiblePileChoice
                                                ³ - (using a left argument of) program's 3rd command line argument (M)
                                                Ɱ - map across (possiblePileChoice) with:
                                                œi - first multi-dimensional index of (the item) in (M)
                                                Z - transpose the resulting list of [row, column] values
                                                I - get the incremental differences
                                                Ị - insignificant? (vectorises an abs(v) <= 1 test)
                                                Ȧ - any and all? (0 if any 0s are present in the flattened result [or if it's empty])

                                                ⁴ṖŒPŒ!€Ẏ⁴;ⱮṢÇƇṪ - Link 2, getChosenPileList: list of lists of integers, M
                                                ⁴ - literal 16
                                                Ṗ - pop -> [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15]
                                                ŒP - power-set -> [,[1],[2],...,[1,2],[1,3],...,[2,3,7],...,[1,2,3,4,5,6,7,8,9,10,11,12,13,14,15]]
                                                € - for each:
                                                Œ! - all permutations
                                                Ẏ - tighten (to a single list of all these individual permutations)
                                                ⁴ - (using a left argument of) literal 16
                                                Ɱ - map across it with:
                                                ; - concatenate (put a 16 at the beginning of each one)
                                                Ṣ - sort the resulting list of lists
                                                Ƈ - filter keep those for which this is truthy:
                                                Ç - call last Link as a monad (i.e. isSatisfactory(possiblePileChoice)
                                                Ṫ - tail (get the right-most, i.e. the maximal satisfactory one)

                                                FḟÇS - Main Link: list of lists of integers, M
                                                F - flatten M
                                                Ç - call last Link (2) as a monad (i.e. get getChosenPileList(M))
                                                ḟ - filter discard (the resulting values) from (the flattened M)
                                                S - sum





                                                share|improve this answer























                                                • Ah yeah, power-set is not enough!
                                                  – Jonathan Allan
                                                  2 days ago






                                                • 2




                                                  @Arnauld - finally got a little time to golf :D This should work, but will be (way) too slow for running at TIO with the test case you used before.
                                                  – Jonathan Allan
                                                  2 days ago















                                                up vote
                                                2
                                                down vote














                                                Jelly,  31 30  29 bytes



                                                ³œiⱮZIỊȦ
                                                ⁴ṖŒPŒ!€Ẏ⁴;ⱮṢÇƇṪ
                                                FḟÇS


                                                Since the method is far too slow to run within 60s with the mouse starting on 16 this starts her off at 9 and limits her ability such that she is only able to eat 9s or less Try it online! (thus here she eats 9, 2, 7, 4, 8, 6, 3 leaving 97).



                                                How?



                                                ³œiⱮZIỊȦ - Link 1, isSatisfactory?: list of integers, possiblePileChoice
                                                ³ - (using a left argument of) program's 3rd command line argument (M)
                                                Ɱ - map across (possiblePileChoice) with:
                                                œi - first multi-dimensional index of (the item) in (M)
                                                Z - transpose the resulting list of [row, column] values
                                                I - get the incremental differences
                                                Ị - insignificant? (vectorises an abs(v) <= 1 test)
                                                Ȧ - any and all? (0 if any 0s are present in the flattened result [or if it's empty])

                                                ⁴ṖŒPŒ!€Ẏ⁴;ⱮṢÇƇṪ - Link 2, getChosenPileList: list of lists of integers, M
                                                ⁴ - literal 16
                                                Ṗ - pop -> [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15]
                                                ŒP - power-set -> [,[1],[2],...,[1,2],[1,3],...,[2,3,7],...,[1,2,3,4,5,6,7,8,9,10,11,12,13,14,15]]
                                                € - for each:
                                                Œ! - all permutations
                                                Ẏ - tighten (to a single list of all these individual permutations)
                                                ⁴ - (using a left argument of) literal 16
                                                Ɱ - map across it with:
                                                ; - concatenate (put a 16 at the beginning of each one)
                                                Ṣ - sort the resulting list of lists
                                                Ƈ - filter keep those for which this is truthy:
                                                Ç - call last Link as a monad (i.e. isSatisfactory(possiblePileChoice)
                                                Ṫ - tail (get the right-most, i.e. the maximal satisfactory one)

                                                FḟÇS - Main Link: list of lists of integers, M
                                                F - flatten M
                                                Ç - call last Link (2) as a monad (i.e. get getChosenPileList(M))
                                                ḟ - filter discard (the resulting values) from (the flattened M)
                                                S - sum





                                                share|improve this answer























                                                • Ah yeah, power-set is not enough!
                                                  – Jonathan Allan
                                                  2 days ago






                                                • 2




                                                  @Arnauld - finally got a little time to golf :D This should work, but will be (way) too slow for running at TIO with the test case you used before.
                                                  – Jonathan Allan
                                                  2 days ago













                                                up vote
                                                2
                                                down vote










                                                up vote
                                                2
                                                down vote










                                                Jelly,  31 30  29 bytes



                                                ³œiⱮZIỊȦ
                                                ⁴ṖŒPŒ!€Ẏ⁴;ⱮṢÇƇṪ
                                                FḟÇS


                                                Since the method is far too slow to run within 60s with the mouse starting on 16 this starts her off at 9 and limits her ability such that she is only able to eat 9s or less Try it online! (thus here she eats 9, 2, 7, 4, 8, 6, 3 leaving 97).



                                                How?



                                                ³œiⱮZIỊȦ - Link 1, isSatisfactory?: list of integers, possiblePileChoice
                                                ³ - (using a left argument of) program's 3rd command line argument (M)
                                                Ɱ - map across (possiblePileChoice) with:
                                                œi - first multi-dimensional index of (the item) in (M)
                                                Z - transpose the resulting list of [row, column] values
                                                I - get the incremental differences
                                                Ị - insignificant? (vectorises an abs(v) <= 1 test)
                                                Ȧ - any and all? (0 if any 0s are present in the flattened result [or if it's empty])

                                                ⁴ṖŒPŒ!€Ẏ⁴;ⱮṢÇƇṪ - Link 2, getChosenPileList: list of lists of integers, M
                                                ⁴ - literal 16
                                                Ṗ - pop -> [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15]
                                                ŒP - power-set -> [,[1],[2],...,[1,2],[1,3],...,[2,3,7],...,[1,2,3,4,5,6,7,8,9,10,11,12,13,14,15]]
                                                € - for each:
                                                Œ! - all permutations
                                                Ẏ - tighten (to a single list of all these individual permutations)
                                                ⁴ - (using a left argument of) literal 16
                                                Ɱ - map across it with:
                                                ; - concatenate (put a 16 at the beginning of each one)
                                                Ṣ - sort the resulting list of lists
                                                Ƈ - filter keep those for which this is truthy:
                                                Ç - call last Link as a monad (i.e. isSatisfactory(possiblePileChoice)
                                                Ṫ - tail (get the right-most, i.e. the maximal satisfactory one)

                                                FḟÇS - Main Link: list of lists of integers, M
                                                F - flatten M
                                                Ç - call last Link (2) as a monad (i.e. get getChosenPileList(M))
                                                ḟ - filter discard (the resulting values) from (the flattened M)
                                                S - sum





                                                share|improve this answer















                                                Jelly,  31 30  29 bytes



                                                ³œiⱮZIỊȦ
                                                ⁴ṖŒPŒ!€Ẏ⁴;ⱮṢÇƇṪ
                                                FḟÇS


                                                Since the method is far too slow to run within 60s with the mouse starting on 16 this starts her off at 9 and limits her ability such that she is only able to eat 9s or less Try it online! (thus here she eats 9, 2, 7, 4, 8, 6, 3 leaving 97).



                                                How?



                                                ³œiⱮZIỊȦ - Link 1, isSatisfactory?: list of integers, possiblePileChoice
                                                ³ - (using a left argument of) program's 3rd command line argument (M)
                                                Ɱ - map across (possiblePileChoice) with:
                                                œi - first multi-dimensional index of (the item) in (M)
                                                Z - transpose the resulting list of [row, column] values
                                                I - get the incremental differences
                                                Ị - insignificant? (vectorises an abs(v) <= 1 test)
                                                Ȧ - any and all? (0 if any 0s are present in the flattened result [or if it's empty])

                                                ⁴ṖŒPŒ!€Ẏ⁴;ⱮṢÇƇṪ - Link 2, getChosenPileList: list of lists of integers, M
                                                ⁴ - literal 16
                                                Ṗ - pop -> [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15]
                                                ŒP - power-set -> [,[1],[2],...,[1,2],[1,3],...,[2,3,7],...,[1,2,3,4,5,6,7,8,9,10,11,12,13,14,15]]
                                                € - for each:
                                                Œ! - all permutations
                                                Ẏ - tighten (to a single list of all these individual permutations)
                                                ⁴ - (using a left argument of) literal 16
                                                Ɱ - map across it with:
                                                ; - concatenate (put a 16 at the beginning of each one)
                                                Ṣ - sort the resulting list of lists
                                                Ƈ - filter keep those for which this is truthy:
                                                Ç - call last Link as a monad (i.e. isSatisfactory(possiblePileChoice)
                                                Ṫ - tail (get the right-most, i.e. the maximal satisfactory one)

                                                FḟÇS - Main Link: list of lists of integers, M
                                                F - flatten M
                                                Ç - call last Link (2) as a monad (i.e. get getChosenPileList(M))
                                                ḟ - filter discard (the resulting values) from (the flattened M)
                                                S - sum






                                                share|improve this answer














                                                share|improve this answer



                                                share|improve this answer








                                                edited 2 days ago

























                                                answered 2 days ago









                                                Jonathan Allan

                                                50.1k534165




                                                50.1k534165












                                                • Ah yeah, power-set is not enough!
                                                  – Jonathan Allan
                                                  2 days ago






                                                • 2




                                                  @Arnauld - finally got a little time to golf :D This should work, but will be (way) too slow for running at TIO with the test case you used before.
                                                  – Jonathan Allan
                                                  2 days ago


















                                                • Ah yeah, power-set is not enough!
                                                  – Jonathan Allan
                                                  2 days ago






                                                • 2




                                                  @Arnauld - finally got a little time to golf :D This should work, but will be (way) too slow for running at TIO with the test case you used before.
                                                  – Jonathan Allan
                                                  2 days ago
















                                                Ah yeah, power-set is not enough!
                                                – Jonathan Allan
                                                2 days ago




                                                Ah yeah, power-set is not enough!
                                                – Jonathan Allan
                                                2 days ago




                                                2




                                                2




                                                @Arnauld - finally got a little time to golf :D This should work, but will be (way) too slow for running at TIO with the test case you used before.
                                                – Jonathan Allan
                                                2 days ago




                                                @Arnauld - finally got a little time to golf :D This should work, but will be (way) too slow for running at TIO with the test case you used before.
                                                – Jonathan Allan
                                                2 days ago










                                                up vote
                                                2
                                                down vote













                                                SAS, 236 219 bytes



                                                Input on punch cards, one line per grid (space-separated), output printed to the log.



                                                This challenge is slightly complicated by some limitations of arrays in SAS:




                                                • There is no way to return the row and column indexes of a matching element from multidimensional data-step array - you have to treat the array as 1-d and then work them out for yourself.

                                                • If you go out of bounds, SAS throws an error and halts processing rather than returning null / zero.


                                                Updates:




                                                • Removed infile cards; statement (-13)

                                                • Used wildcard a: for array definition rather than a1-a16 (-4)


                                                Golfed:



                                                data;input a1-a16;array a[4,4]a:;p=16;t=136;do while(p);m=whichn(p,of a:);t=t-p;j=mod(m-1,4)+1;i=ceil(m/4);a[i,j]=0;p=0;do k=max(1,i-1)to min(i+1,4);do l=max(1,j-1)to min(j+1,4);p=max(p,a[k,l]);end;end;end;put t;cards;
                                                <insert punch cards here>
                                                ;


                                                Ungolfed:



                                                data;                /*Produce a dataset using automatic naming*/
                                                input a1-a16; /*Read 16 variables*/
                                                array a[4,4] a:; /*Assign to a 4x4 array*/
                                                p=16; /*Initial pile to look for*/
                                                t=136; /*Total cheese to decrement*/
                                                do while(p); /*Stop if there are no piles available with size > 0*/
                                                m=whichn(p,of a:); /*Find array element containing current pile size*/
                                                t=t-p; /*Decrement total cheese*/
                                                j=mod(m-1,4)+1; /*Get column number*/
                                                i=ceil(m/4); /*Get row number*/
                                                a[i,j]=0; /*Eat the current pile*/
                                                /*Find the size of the largest adjacent pile*/
                                                p=0;
                                                do k=max(1,i-1)to min(i+1,4);
                                                do l=max(1,j-1)to min(j+1,4);
                                                p=max(p,a[k,l]);
                                                end;
                                                end;
                                                end;
                                                put t; /*Print total remaining cheese to log*/
                                                /*Start of punch card input*/
                                                cards;
                                                4 3 2 1 5 6 7 8 12 11 10 9 13 14 15 16
                                                8 1 9 14 11 6 5 16 13 15 2 7 10 3 12 4
                                                1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
                                                10 15 14 11 9 3 1 7 13 5 12 6 2 8 4 16
                                                3 7 10 5 6 8 12 13 15 9 11 4 14 1 16 2
                                                8 9 3 6 13 11 7 15 12 10 16 2 4 14 1 5
                                                8 11 12 9 14 5 10 16 7 3 1 6 13 4 2 15
                                                13 14 1 2 16 15 3 4 5 6 7 8 9 10 11 12
                                                9 10 11 12 1 2 4 13 7 8 5 14 3 16 6 15
                                                9 10 11 12 1 2 7 13 6 16 4 14 3 8 5 15
                                                ; /*End of punch card input*/
                                                /*Implicit run;*/





                                                share|improve this answer























                                                • +1 for use of punch cards in PPCG :)
                                                  – GNiklasch
                                                  2 days ago















                                                up vote
                                                2
                                                down vote













                                                SAS, 236 219 bytes



                                                Input on punch cards, one line per grid (space-separated), output printed to the log.



                                                This challenge is slightly complicated by some limitations of arrays in SAS:




                                                • There is no way to return the row and column indexes of a matching element from multidimensional data-step array - you have to treat the array as 1-d and then work them out for yourself.

                                                • If you go out of bounds, SAS throws an error and halts processing rather than returning null / zero.


                                                Updates:




                                                • Removed infile cards; statement (-13)

                                                • Used wildcard a: for array definition rather than a1-a16 (-4)


                                                Golfed:



                                                data;input a1-a16;array a[4,4]a:;p=16;t=136;do while(p);m=whichn(p,of a:);t=t-p;j=mod(m-1,4)+1;i=ceil(m/4);a[i,j]=0;p=0;do k=max(1,i-1)to min(i+1,4);do l=max(1,j-1)to min(j+1,4);p=max(p,a[k,l]);end;end;end;put t;cards;
                                                <insert punch cards here>
                                                ;


                                                Ungolfed:



                                                data;                /*Produce a dataset using automatic naming*/
                                                input a1-a16; /*Read 16 variables*/
                                                array a[4,4] a:; /*Assign to a 4x4 array*/
                                                p=16; /*Initial pile to look for*/
                                                t=136; /*Total cheese to decrement*/
                                                do while(p); /*Stop if there are no piles available with size > 0*/
                                                m=whichn(p,of a:); /*Find array element containing current pile size*/
                                                t=t-p; /*Decrement total cheese*/
                                                j=mod(m-1,4)+1; /*Get column number*/
                                                i=ceil(m/4); /*Get row number*/
                                                a[i,j]=0; /*Eat the current pile*/
                                                /*Find the size of the largest adjacent pile*/
                                                p=0;
                                                do k=max(1,i-1)to min(i+1,4);
                                                do l=max(1,j-1)to min(j+1,4);
                                                p=max(p,a[k,l]);
                                                end;
                                                end;
                                                end;
                                                put t; /*Print total remaining cheese to log*/
                                                /*Start of punch card input*/
                                                cards;
                                                4 3 2 1 5 6 7 8 12 11 10 9 13 14 15 16
                                                8 1 9 14 11 6 5 16 13 15 2 7 10 3 12 4
                                                1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
                                                10 15 14 11 9 3 1 7 13 5 12 6 2 8 4 16
                                                3 7 10 5 6 8 12 13 15 9 11 4 14 1 16 2
                                                8 9 3 6 13 11 7 15 12 10 16 2 4 14 1 5
                                                8 11 12 9 14 5 10 16 7 3 1 6 13 4 2 15
                                                13 14 1 2 16 15 3 4 5 6 7 8 9 10 11 12
                                                9 10 11 12 1 2 4 13 7 8 5 14 3 16 6 15
                                                9 10 11 12 1 2 7 13 6 16 4 14 3 8 5 15
                                                ; /*End of punch card input*/
                                                /*Implicit run;*/





                                                share|improve this answer























                                                • +1 for use of punch cards in PPCG :)
                                                  – GNiklasch
                                                  2 days ago













                                                up vote
                                                2
                                                down vote










                                                up vote
                                                2
                                                down vote









                                                SAS, 236 219 bytes



                                                Input on punch cards, one line per grid (space-separated), output printed to the log.



                                                This challenge is slightly complicated by some limitations of arrays in SAS:




                                                • There is no way to return the row and column indexes of a matching element from multidimensional data-step array - you have to treat the array as 1-d and then work them out for yourself.

                                                • If you go out of bounds, SAS throws an error and halts processing rather than returning null / zero.


                                                Updates:




                                                • Removed infile cards; statement (-13)

                                                • Used wildcard a: for array definition rather than a1-a16 (-4)


                                                Golfed:



                                                data;input a1-a16;array a[4,4]a:;p=16;t=136;do while(p);m=whichn(p,of a:);t=t-p;j=mod(m-1,4)+1;i=ceil(m/4);a[i,j]=0;p=0;do k=max(1,i-1)to min(i+1,4);do l=max(1,j-1)to min(j+1,4);p=max(p,a[k,l]);end;end;end;put t;cards;
                                                <insert punch cards here>
                                                ;


                                                Ungolfed:



                                                data;                /*Produce a dataset using automatic naming*/
                                                input a1-a16; /*Read 16 variables*/
                                                array a[4,4] a:; /*Assign to a 4x4 array*/
                                                p=16; /*Initial pile to look for*/
                                                t=136; /*Total cheese to decrement*/
                                                do while(p); /*Stop if there are no piles available with size > 0*/
                                                m=whichn(p,of a:); /*Find array element containing current pile size*/
                                                t=t-p; /*Decrement total cheese*/
                                                j=mod(m-1,4)+1; /*Get column number*/
                                                i=ceil(m/4); /*Get row number*/
                                                a[i,j]=0; /*Eat the current pile*/
                                                /*Find the size of the largest adjacent pile*/
                                                p=0;
                                                do k=max(1,i-1)to min(i+1,4);
                                                do l=max(1,j-1)to min(j+1,4);
                                                p=max(p,a[k,l]);
                                                end;
                                                end;
                                                end;
                                                put t; /*Print total remaining cheese to log*/
                                                /*Start of punch card input*/
                                                cards;
                                                4 3 2 1 5 6 7 8 12 11 10 9 13 14 15 16
                                                8 1 9 14 11 6 5 16 13 15 2 7 10 3 12 4
                                                1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
                                                10 15 14 11 9 3 1 7 13 5 12 6 2 8 4 16
                                                3 7 10 5 6 8 12 13 15 9 11 4 14 1 16 2
                                                8 9 3 6 13 11 7 15 12 10 16 2 4 14 1 5
                                                8 11 12 9 14 5 10 16 7 3 1 6 13 4 2 15
                                                13 14 1 2 16 15 3 4 5 6 7 8 9 10 11 12
                                                9 10 11 12 1 2 4 13 7 8 5 14 3 16 6 15
                                                9 10 11 12 1 2 7 13 6 16 4 14 3 8 5 15
                                                ; /*End of punch card input*/
                                                /*Implicit run;*/





                                                share|improve this answer














                                                SAS, 236 219 bytes



                                                Input on punch cards, one line per grid (space-separated), output printed to the log.



                                                This challenge is slightly complicated by some limitations of arrays in SAS:




                                                • There is no way to return the row and column indexes of a matching element from multidimensional data-step array - you have to treat the array as 1-d and then work them out for yourself.

                                                • If you go out of bounds, SAS throws an error and halts processing rather than returning null / zero.


                                                Updates:




                                                • Removed infile cards; statement (-13)

                                                • Used wildcard a: for array definition rather than a1-a16 (-4)


                                                Golfed:



                                                data;input a1-a16;array a[4,4]a:;p=16;t=136;do while(p);m=whichn(p,of a:);t=t-p;j=mod(m-1,4)+1;i=ceil(m/4);a[i,j]=0;p=0;do k=max(1,i-1)to min(i+1,4);do l=max(1,j-1)to min(j+1,4);p=max(p,a[k,l]);end;end;end;put t;cards;
                                                <insert punch cards here>
                                                ;


                                                Ungolfed:



                                                data;                /*Produce a dataset using automatic naming*/
                                                input a1-a16; /*Read 16 variables*/
                                                array a[4,4] a:; /*Assign to a 4x4 array*/
                                                p=16; /*Initial pile to look for*/
                                                t=136; /*Total cheese to decrement*/
                                                do while(p); /*Stop if there are no piles available with size > 0*/
                                                m=whichn(p,of a:); /*Find array element containing current pile size*/
                                                t=t-p; /*Decrement total cheese*/
                                                j=mod(m-1,4)+1; /*Get column number*/
                                                i=ceil(m/4); /*Get row number*/
                                                a[i,j]=0; /*Eat the current pile*/
                                                /*Find the size of the largest adjacent pile*/
                                                p=0;
                                                do k=max(1,i-1)to min(i+1,4);
                                                do l=max(1,j-1)to min(j+1,4);
                                                p=max(p,a[k,l]);
                                                end;
                                                end;
                                                end;
                                                put t; /*Print total remaining cheese to log*/
                                                /*Start of punch card input*/
                                                cards;
                                                4 3 2 1 5 6 7 8 12 11 10 9 13 14 15 16
                                                8 1 9 14 11 6 5 16 13 15 2 7 10 3 12 4
                                                1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
                                                10 15 14 11 9 3 1 7 13 5 12 6 2 8 4 16
                                                3 7 10 5 6 8 12 13 15 9 11 4 14 1 16 2
                                                8 9 3 6 13 11 7 15 12 10 16 2 4 14 1 5
                                                8 11 12 9 14 5 10 16 7 3 1 6 13 4 2 15
                                                13 14 1 2 16 15 3 4 5 6 7 8 9 10 11 12
                                                9 10 11 12 1 2 4 13 7 8 5 14 3 16 6 15
                                                9 10 11 12 1 2 7 13 6 16 4 14 3 8 5 15
                                                ; /*End of punch card input*/
                                                /*Implicit run;*/






                                                share|improve this answer














                                                share|improve this answer



                                                share|improve this answer








                                                edited yesterday

























                                                answered 2 days ago









                                                user3490

                                                76945




                                                76945












                                                • +1 for use of punch cards in PPCG :)
                                                  – GNiklasch
                                                  2 days ago


















                                                • +1 for use of punch cards in PPCG :)
                                                  – GNiklasch
                                                  2 days ago
















                                                +1 for use of punch cards in PPCG :)
                                                – GNiklasch
                                                2 days ago




                                                +1 for use of punch cards in PPCG :)
                                                – GNiklasch
                                                2 days ago










                                                up vote
                                                2
                                                down vote













                                                Java 10, 272 271 bytes





                                                m->{int r,c,R=4,M=1,x,y,X,Y;for(r=c=X=Y=0;R-->0;)for(C=4;C-->0;)if(m[R][C]>15)m[r=R][c=C]=0;for(;M!=0;m[r=X][c=Y]=0)for(M=-1,C=9;C-->0;)try{if((R=m[x=C<3?r-1:C>5?r+1:r][y=C%3<1?c-1:C%3>1?c+1:c])>M){M=R;X=x;Y=y;}}catch(Exception e){}for(var Z:m)for(int z:Z)M+=z;return M;}


                                                -1 byte indirectly thanks to @Serverfrog.



                                                The cells are checked the same as in my answer for the All the single eights challenge.



                                                Try it online.



                                                Explanation:



                                                m->{                       // Method with integer-matrix parameter and integer return
                                                int r, // Row-coordinate for the largest number
                                                c, // Column-coordinate for the largest number
                                                R=4,C, // Row and column indices (later reused as temp integers)
                                                M=1, // Largest number the mouse just ate, starting at 1
                                                x,y,X,Y; // Temp integers
                                                for(r=c=X=Y=0; // Start `r`, `c`, `X`, and `Y` at 0
                                                R-->0;) // Loop `R` in the range (4, 0]:
                                                for(C=4;C-->0;) // Inner loop `C` in the range (4, 0]:
                                                if(m[R][C]>15) // If the current cell is 16:
                                                m[r=R][c=C] // Set `r,c` to this coordinate
                                                =0; // And empty this cell
                                                for(;M!=0; // Loop as long as the largest number isn't 0:
                                                ; // After every iteration:
                                                m[r=X][c=Y] // Change the `r,c` coordinates,
                                                =0) // And empty this cell
                                                for(M=-1, // Reset `M` to -1
                                                C=9;C-->0;) // Inner loop `C` in the range (9, 0]:
                                                try{if((R= // Set `R` to:
                                                m[x=C<3? // If `C` is 0, 1, or 2:
                                                r-1 // Look at the previous row
                                                :C>5? // Else-if `C` is 6, 7, or 8:
                                                r+1 // Look at the next row
                                                : // Else (`C` is 3, 4, or 5):
                                                r] // Look at the current row
                                                [y=C%3<1? // If `C` is 0, 3, or 6:
                                                c-1 // Look at the previous column
                                                :C%3>1? // Else-if `C` is 2, 5, or 8:
                                                c+1 // Look at the next column
                                                : // Else (`C` is 1, 4, or 7):
                                                c]) // Look at the current column
                                                >M){ // And if the number in this cell is larger than `M`
                                                M=R; // Change `M` to this number
                                                X=x;Y=y;} // And change the `X,Y` coordinate to this cell
                                                }catch(Exception e){}
                                                // Catch and ignore ArrayIndexOutOfBoundsExceptions
                                                // (try-catch saves bytes in comparison to if-checks)
                                                for(var Z:m) // Then loop over all rows of the matrix:
                                                for(int z:Z) // Inner loop over all columns of the matrix:
                                                M+=z; // And sum them all together in `M` (which was 0)
                                                return M;} // Then return this sum as result





                                                share|improve this answer























                                                • could you not to int r=c=X=Y=0,R=4,M=1,x,y; ?
                                                  – Serverfrog
                                                  6 hours ago










                                                • @Serverfrog I'm afraid that's not possible when declaring the variables in Java. Your suggestion did give me an idea to save a byte though, by using int r,c,R=4,M=1,x,y,X,Y;for(r=c=X=Y=0;, so thanks. :)
                                                  – Kevin Cruijssen
                                                  3 hours ago

















                                                up vote
                                                2
                                                down vote













                                                Java 10, 272 271 bytes





                                                m->{int r,c,R=4,M=1,x,y,X,Y;for(r=c=X=Y=0;R-->0;)for(C=4;C-->0;)if(m[R][C]>15)m[r=R][c=C]=0;for(;M!=0;m[r=X][c=Y]=0)for(M=-1,C=9;C-->0;)try{if((R=m[x=C<3?r-1:C>5?r+1:r][y=C%3<1?c-1:C%3>1?c+1:c])>M){M=R;X=x;Y=y;}}catch(Exception e){}for(var Z:m)for(int z:Z)M+=z;return M;}


                                                -1 byte indirectly thanks to @Serverfrog.



                                                The cells are checked the same as in my answer for the All the single eights challenge.



                                                Try it online.



                                                Explanation:



                                                m->{                       // Method with integer-matrix parameter and integer return
                                                int r, // Row-coordinate for the largest number
                                                c, // Column-coordinate for the largest number
                                                R=4,C, // Row and column indices (later reused as temp integers)
                                                M=1, // Largest number the mouse just ate, starting at 1
                                                x,y,X,Y; // Temp integers
                                                for(r=c=X=Y=0; // Start `r`, `c`, `X`, and `Y` at 0
                                                R-->0;) // Loop `R` in the range (4, 0]:
                                                for(C=4;C-->0;) // Inner loop `C` in the range (4, 0]:
                                                if(m[R][C]>15) // If the current cell is 16:
                                                m[r=R][c=C] // Set `r,c` to this coordinate
                                                =0; // And empty this cell
                                                for(;M!=0; // Loop as long as the largest number isn't 0:
                                                ; // After every iteration:
                                                m[r=X][c=Y] // Change the `r,c` coordinates,
                                                =0) // And empty this cell
                                                for(M=-1, // Reset `M` to -1
                                                C=9;C-->0;) // Inner loop `C` in the range (9, 0]:
                                                try{if((R= // Set `R` to:
                                                m[x=C<3? // If `C` is 0, 1, or 2:
                                                r-1 // Look at the previous row
                                                :C>5? // Else-if `C` is 6, 7, or 8:
                                                r+1 // Look at the next row
                                                : // Else (`C` is 3, 4, or 5):
                                                r] // Look at the current row
                                                [y=C%3<1? // If `C` is 0, 3, or 6:
                                                c-1 // Look at the previous column
                                                :C%3>1? // Else-if `C` is 2, 5, or 8:
                                                c+1 // Look at the next column
                                                : // Else (`C` is 1, 4, or 7):
                                                c]) // Look at the current column
                                                >M){ // And if the number in this cell is larger than `M`
                                                M=R; // Change `M` to this number
                                                X=x;Y=y;} // And change the `X,Y` coordinate to this cell
                                                }catch(Exception e){}
                                                // Catch and ignore ArrayIndexOutOfBoundsExceptions
                                                // (try-catch saves bytes in comparison to if-checks)
                                                for(var Z:m) // Then loop over all rows of the matrix:
                                                for(int z:Z) // Inner loop over all columns of the matrix:
                                                M+=z; // And sum them all together in `M` (which was 0)
                                                return M;} // Then return this sum as result





                                                share|improve this answer























                                                • could you not to int r=c=X=Y=0,R=4,M=1,x,y; ?
                                                  – Serverfrog
                                                  6 hours ago










                                                • @Serverfrog I'm afraid that's not possible when declaring the variables in Java. Your suggestion did give me an idea to save a byte though, by using int r,c,R=4,M=1,x,y,X,Y;for(r=c=X=Y=0;, so thanks. :)
                                                  – Kevin Cruijssen
                                                  3 hours ago















                                                up vote
                                                2
                                                down vote










                                                up vote
                                                2
                                                down vote









                                                Java 10, 272 271 bytes





                                                m->{int r,c,R=4,M=1,x,y,X,Y;for(r=c=X=Y=0;R-->0;)for(C=4;C-->0;)if(m[R][C]>15)m[r=R][c=C]=0;for(;M!=0;m[r=X][c=Y]=0)for(M=-1,C=9;C-->0;)try{if((R=m[x=C<3?r-1:C>5?r+1:r][y=C%3<1?c-1:C%3>1?c+1:c])>M){M=R;X=x;Y=y;}}catch(Exception e){}for(var Z:m)for(int z:Z)M+=z;return M;}


                                                -1 byte indirectly thanks to @Serverfrog.



                                                The cells are checked the same as in my answer for the All the single eights challenge.



                                                Try it online.



                                                Explanation:



                                                m->{                       // Method with integer-matrix parameter and integer return
                                                int r, // Row-coordinate for the largest number
                                                c, // Column-coordinate for the largest number
                                                R=4,C, // Row and column indices (later reused as temp integers)
                                                M=1, // Largest number the mouse just ate, starting at 1
                                                x,y,X,Y; // Temp integers
                                                for(r=c=X=Y=0; // Start `r`, `c`, `X`, and `Y` at 0
                                                R-->0;) // Loop `R` in the range (4, 0]:
                                                for(C=4;C-->0;) // Inner loop `C` in the range (4, 0]:
                                                if(m[R][C]>15) // If the current cell is 16:
                                                m[r=R][c=C] // Set `r,c` to this coordinate
                                                =0; // And empty this cell
                                                for(;M!=0; // Loop as long as the largest number isn't 0:
                                                ; // After every iteration:
                                                m[r=X][c=Y] // Change the `r,c` coordinates,
                                                =0) // And empty this cell
                                                for(M=-1, // Reset `M` to -1
                                                C=9;C-->0;) // Inner loop `C` in the range (9, 0]:
                                                try{if((R= // Set `R` to:
                                                m[x=C<3? // If `C` is 0, 1, or 2:
                                                r-1 // Look at the previous row
                                                :C>5? // Else-if `C` is 6, 7, or 8:
                                                r+1 // Look at the next row
                                                : // Else (`C` is 3, 4, or 5):
                                                r] // Look at the current row
                                                [y=C%3<1? // If `C` is 0, 3, or 6:
                                                c-1 // Look at the previous column
                                                :C%3>1? // Else-if `C` is 2, 5, or 8:
                                                c+1 // Look at the next column
                                                : // Else (`C` is 1, 4, or 7):
                                                c]) // Look at the current column
                                                >M){ // And if the number in this cell is larger than `M`
                                                M=R; // Change `M` to this number
                                                X=x;Y=y;} // And change the `X,Y` coordinate to this cell
                                                }catch(Exception e){}
                                                // Catch and ignore ArrayIndexOutOfBoundsExceptions
                                                // (try-catch saves bytes in comparison to if-checks)
                                                for(var Z:m) // Then loop over all rows of the matrix:
                                                for(int z:Z) // Inner loop over all columns of the matrix:
                                                M+=z; // And sum them all together in `M` (which was 0)
                                                return M;} // Then return this sum as result





                                                share|improve this answer














                                                Java 10, 272 271 bytes





                                                m->{int r,c,R=4,M=1,x,y,X,Y;for(r=c=X=Y=0;R-->0;)for(C=4;C-->0;)if(m[R][C]>15)m[r=R][c=C]=0;for(;M!=0;m[r=X][c=Y]=0)for(M=-1,C=9;C-->0;)try{if((R=m[x=C<3?r-1:C>5?r+1:r][y=C%3<1?c-1:C%3>1?c+1:c])>M){M=R;X=x;Y=y;}}catch(Exception e){}for(var Z:m)for(int z:Z)M+=z;return M;}


                                                -1 byte indirectly thanks to @Serverfrog.



                                                The cells are checked the same as in my answer for the All the single eights challenge.



                                                Try it online.



                                                Explanation:



                                                m->{                       // Method with integer-matrix parameter and integer return
                                                int r, // Row-coordinate for the largest number
                                                c, // Column-coordinate for the largest number
                                                R=4,C, // Row and column indices (later reused as temp integers)
                                                M=1, // Largest number the mouse just ate, starting at 1
                                                x,y,X,Y; // Temp integers
                                                for(r=c=X=Y=0; // Start `r`, `c`, `X`, and `Y` at 0
                                                R-->0;) // Loop `R` in the range (4, 0]:
                                                for(C=4;C-->0;) // Inner loop `C` in the range (4, 0]:
                                                if(m[R][C]>15) // If the current cell is 16:
                                                m[r=R][c=C] // Set `r,c` to this coordinate
                                                =0; // And empty this cell
                                                for(;M!=0; // Loop as long as the largest number isn't 0:
                                                ; // After every iteration:
                                                m[r=X][c=Y] // Change the `r,c` coordinates,
                                                =0) // And empty this cell
                                                for(M=-1, // Reset `M` to -1
                                                C=9;C-->0;) // Inner loop `C` in the range (9, 0]:
                                                try{if((R= // Set `R` to:
                                                m[x=C<3? // If `C` is 0, 1, or 2:
                                                r-1 // Look at the previous row
                                                :C>5? // Else-if `C` is 6, 7, or 8:
                                                r+1 // Look at the next row
                                                : // Else (`C` is 3, 4, or 5):
                                                r] // Look at the current row
                                                [y=C%3<1? // If `C` is 0, 3, or 6:
                                                c-1 // Look at the previous column
                                                :C%3>1? // Else-if `C` is 2, 5, or 8:
                                                c+1 // Look at the next column
                                                : // Else (`C` is 1, 4, or 7):
                                                c]) // Look at the current column
                                                >M){ // And if the number in this cell is larger than `M`
                                                M=R; // Change `M` to this number
                                                X=x;Y=y;} // And change the `X,Y` coordinate to this cell
                                                }catch(Exception e){}
                                                // Catch and ignore ArrayIndexOutOfBoundsExceptions
                                                // (try-catch saves bytes in comparison to if-checks)
                                                for(var Z:m) // Then loop over all rows of the matrix:
                                                for(int z:Z) // Inner loop over all columns of the matrix:
                                                M+=z; // And sum them all together in `M` (which was 0)
                                                return M;} // Then return this sum as result






                                                share|improve this answer














                                                share|improve this answer



                                                share|improve this answer








                                                edited 3 hours ago

























                                                answered 2 days ago









                                                Kevin Cruijssen

                                                34.4k554182




                                                34.4k554182












                                                • could you not to int r=c=X=Y=0,R=4,M=1,x,y; ?
                                                  – Serverfrog
                                                  6 hours ago










                                                • @Serverfrog I'm afraid that's not possible when declaring the variables in Java. Your suggestion did give me an idea to save a byte though, by using int r,c,R=4,M=1,x,y,X,Y;for(r=c=X=Y=0;, so thanks. :)
                                                  – Kevin Cruijssen
                                                  3 hours ago




















                                                • could you not to int r=c=X=Y=0,R=4,M=1,x,y; ?
                                                  – Serverfrog
                                                  6 hours ago










                                                • @Serverfrog I'm afraid that's not possible when declaring the variables in Java. Your suggestion did give me an idea to save a byte though, by using int r,c,R=4,M=1,x,y,X,Y;for(r=c=X=Y=0;, so thanks. :)
                                                  – Kevin Cruijssen
                                                  3 hours ago


















                                                could you not to int r=c=X=Y=0,R=4,M=1,x,y; ?
                                                – Serverfrog
                                                6 hours ago




                                                could you not to int r=c=X=Y=0,R=4,M=1,x,y; ?
                                                – Serverfrog
                                                6 hours ago












                                                @Serverfrog I'm afraid that's not possible when declaring the variables in Java. Your suggestion did give me an idea to save a byte though, by using int r,c,R=4,M=1,x,y,X,Y;for(r=c=X=Y=0;, so thanks. :)
                                                – Kevin Cruijssen
                                                3 hours ago






                                                @Serverfrog I'm afraid that's not possible when declaring the variables in Java. Your suggestion did give me an idea to save a byte though, by using int r,c,R=4,M=1,x,y,X,Y;for(r=c=X=Y=0;, so thanks. :)
                                                – Kevin Cruijssen
                                                3 hours ago












                                                up vote
                                                1
                                                down vote













                                                J, 82 bytes



                                                g=.](]*{:@[~:])]_1}~[:>./]{~((,-)1 5 6 7)+]i.{:
                                                [:+/[:(g^:_)16,~[:,0,~0,0,0,.~0,.]


                                                Try it online!



                                                I plan to golf this more tomorrow, and perhaps write a more J-ish solution similar to this one, but I figured I'd try the flattened approach since I hadn't done that before.






                                                share|improve this answer























                                                • Do you really need the leftmost ] in g?
                                                  – Galen Ivanov
                                                  2 days ago






                                                • 1




                                                  Thanks Galen, you're right. It's the least of the issues with this code :) I have a much better solution which I'll implement when I have time.
                                                  – Jonah
                                                  yesterday















                                                up vote
                                                1
                                                down vote













                                                J, 82 bytes



                                                g=.](]*{:@[~:])]_1}~[:>./]{~((,-)1 5 6 7)+]i.{:
                                                [:+/[:(g^:_)16,~[:,0,~0,0,0,.~0,.]


                                                Try it online!



                                                I plan to golf this more tomorrow, and perhaps write a more J-ish solution similar to this one, but I figured I'd try the flattened approach since I hadn't done that before.






                                                share|improve this answer























                                                • Do you really need the leftmost ] in g?
                                                  – Galen Ivanov
                                                  2 days ago






                                                • 1




                                                  Thanks Galen, you're right. It's the least of the issues with this code :) I have a much better solution which I'll implement when I have time.
                                                  – Jonah
                                                  yesterday













                                                up vote
                                                1
                                                down vote










                                                up vote
                                                1
                                                down vote









                                                J, 82 bytes



                                                g=.](]*{:@[~:])]_1}~[:>./]{~((,-)1 5 6 7)+]i.{:
                                                [:+/[:(g^:_)16,~[:,0,~0,0,0,.~0,.]


                                                Try it online!



                                                I plan to golf this more tomorrow, and perhaps write a more J-ish solution similar to this one, but I figured I'd try the flattened approach since I hadn't done that before.






                                                share|improve this answer














                                                J, 82 bytes



                                                g=.](]*{:@[~:])]_1}~[:>./]{~((,-)1 5 6 7)+]i.{:
                                                [:+/[:(g^:_)16,~[:,0,~0,0,0,.~0,.]


                                                Try it online!



                                                I plan to golf this more tomorrow, and perhaps write a more J-ish solution similar to this one, but I figured I'd try the flattened approach since I hadn't done that before.







                                                share|improve this answer














                                                share|improve this answer



                                                share|improve this answer








                                                edited 2 days ago

























                                                answered 2 days ago









                                                Jonah

                                                1,971816




                                                1,971816












                                                • Do you really need the leftmost ] in g?
                                                  – Galen Ivanov
                                                  2 days ago






                                                • 1




                                                  Thanks Galen, you're right. It's the least of the issues with this code :) I have a much better solution which I'll implement when I have time.
                                                  – Jonah
                                                  yesterday


















                                                • Do you really need the leftmost ] in g?
                                                  – Galen Ivanov
                                                  2 days ago






                                                • 1




                                                  Thanks Galen, you're right. It's the least of the issues with this code :) I have a much better solution which I'll implement when I have time.
                                                  – Jonah
                                                  yesterday
















                                                Do you really need the leftmost ] in g?
                                                – Galen Ivanov
                                                2 days ago




                                                Do you really need the leftmost ] in g?
                                                – Galen Ivanov
                                                2 days ago




                                                1




                                                1




                                                Thanks Galen, you're right. It's the least of the issues with this code :) I have a much better solution which I'll implement when I have time.
                                                – Jonah
                                                yesterday




                                                Thanks Galen, you're right. It's the least of the issues with this code :) I have a much better solution which I'll implement when I have time.
                                                – Jonah
                                                yesterday










                                                up vote
                                                1
                                                down vote














                                                Red, 277 bytes



                                                func[a][k: 16 until[t:(index? find load form a k)- 1
                                                p: do rejoin[t / 4 + 1"x"t % 4 + 1]a/(p/1)/(p/2): 0
                                                m: 0 foreach d[-1 0x-1 1x-1 -1x0 1x0 -1x1 0x1 1][j: p + d
                                                if all[j/1 > 0 j/1 < 5 j/2 > 0 j/2 < 5 m < t: a/(j/1)/(j/2)][m: t]]0 = k: m]s: 0
                                                foreach n load form a[s: s + n]s]


                                                Try it online!



                                                It's really long solution and I'm not happy with it, but I spent so much time fixing it to work in TIO (apparently there are many differences between the Win and Linux stable versions of Red), so I post it anyway...



                                                More readable:



                                                f: func [ a ] [
                                                k: 16
                                                until [
                                                t: (index? find load form a n) - 1
                                                p: do rejoin [ t / 4 + 1 "x" t % 4 + 1 ]
                                                a/(p/1)/(p/2): 0
                                                m: 0
                                                foreach d [ -1 0x-1 1x-1 -1x0 1x0 -1x1 0x1 1 ] [
                                                j: p + d
                                                if all[ j/1 > 0
                                                j/1 < 5
                                                j/2 > 0
                                                j/2 < 5
                                                m < t: a/(j/1)/(j/2)
                                                ] [ m: t ]
                                                ]
                                                0 = k: m
                                                ]
                                                s: 0
                                                foreach n load form a [ s: s + n ]
                                                s
                                                ]





                                                share|improve this answer

























                                                  up vote
                                                  1
                                                  down vote














                                                  Red, 277 bytes



                                                  func[a][k: 16 until[t:(index? find load form a k)- 1
                                                  p: do rejoin[t / 4 + 1"x"t % 4 + 1]a/(p/1)/(p/2): 0
                                                  m: 0 foreach d[-1 0x-1 1x-1 -1x0 1x0 -1x1 0x1 1][j: p + d
                                                  if all[j/1 > 0 j/1 < 5 j/2 > 0 j/2 < 5 m < t: a/(j/1)/(j/2)][m: t]]0 = k: m]s: 0
                                                  foreach n load form a[s: s + n]s]


                                                  Try it online!



                                                  It's really long solution and I'm not happy with it, but I spent so much time fixing it to work in TIO (apparently there are many differences between the Win and Linux stable versions of Red), so I post it anyway...



                                                  More readable:



                                                  f: func [ a ] [
                                                  k: 16
                                                  until [
                                                  t: (index? find load form a n) - 1
                                                  p: do rejoin [ t / 4 + 1 "x" t % 4 + 1 ]
                                                  a/(p/1)/(p/2): 0
                                                  m: 0
                                                  foreach d [ -1 0x-1 1x-1 -1x0 1x0 -1x1 0x1 1 ] [
                                                  j: p + d
                                                  if all[ j/1 > 0
                                                  j/1 < 5
                                                  j/2 > 0
                                                  j/2 < 5
                                                  m < t: a/(j/1)/(j/2)
                                                  ] [ m: t ]
                                                  ]
                                                  0 = k: m
                                                  ]
                                                  s: 0
                                                  foreach n load form a [ s: s + n ]
                                                  s
                                                  ]





                                                  share|improve this answer























                                                    up vote
                                                    1
                                                    down vote










                                                    up vote
                                                    1
                                                    down vote










                                                    Red, 277 bytes



                                                    func[a][k: 16 until[t:(index? find load form a k)- 1
                                                    p: do rejoin[t / 4 + 1"x"t % 4 + 1]a/(p/1)/(p/2): 0
                                                    m: 0 foreach d[-1 0x-1 1x-1 -1x0 1x0 -1x1 0x1 1][j: p + d
                                                    if all[j/1 > 0 j/1 < 5 j/2 > 0 j/2 < 5 m < t: a/(j/1)/(j/2)][m: t]]0 = k: m]s: 0
                                                    foreach n load form a[s: s + n]s]


                                                    Try it online!



                                                    It's really long solution and I'm not happy with it, but I spent so much time fixing it to work in TIO (apparently there are many differences between the Win and Linux stable versions of Red), so I post it anyway...



                                                    More readable:



                                                    f: func [ a ] [
                                                    k: 16
                                                    until [
                                                    t: (index? find load form a n) - 1
                                                    p: do rejoin [ t / 4 + 1 "x" t % 4 + 1 ]
                                                    a/(p/1)/(p/2): 0
                                                    m: 0
                                                    foreach d [ -1 0x-1 1x-1 -1x0 1x0 -1x1 0x1 1 ] [
                                                    j: p + d
                                                    if all[ j/1 > 0
                                                    j/1 < 5
                                                    j/2 > 0
                                                    j/2 < 5
                                                    m < t: a/(j/1)/(j/2)
                                                    ] [ m: t ]
                                                    ]
                                                    0 = k: m
                                                    ]
                                                    s: 0
                                                    foreach n load form a [ s: s + n ]
                                                    s
                                                    ]





                                                    share|improve this answer













                                                    Red, 277 bytes



                                                    func[a][k: 16 until[t:(index? find load form a k)- 1
                                                    p: do rejoin[t / 4 + 1"x"t % 4 + 1]a/(p/1)/(p/2): 0
                                                    m: 0 foreach d[-1 0x-1 1x-1 -1x0 1x0 -1x1 0x1 1][j: p + d
                                                    if all[j/1 > 0 j/1 < 5 j/2 > 0 j/2 < 5 m < t: a/(j/1)/(j/2)][m: t]]0 = k: m]s: 0
                                                    foreach n load form a[s: s + n]s]


                                                    Try it online!



                                                    It's really long solution and I'm not happy with it, but I spent so much time fixing it to work in TIO (apparently there are many differences between the Win and Linux stable versions of Red), so I post it anyway...



                                                    More readable:



                                                    f: func [ a ] [
                                                    k: 16
                                                    until [
                                                    t: (index? find load form a n) - 1
                                                    p: do rejoin [ t / 4 + 1 "x" t % 4 + 1 ]
                                                    a/(p/1)/(p/2): 0
                                                    m: 0
                                                    foreach d [ -1 0x-1 1x-1 -1x0 1x0 -1x1 0x1 1 ] [
                                                    j: p + d
                                                    if all[ j/1 > 0
                                                    j/1 < 5
                                                    j/2 > 0
                                                    j/2 < 5
                                                    m < t: a/(j/1)/(j/2)
                                                    ] [ m: t ]
                                                    ]
                                                    0 = k: m
                                                    ]
                                                    s: 0
                                                    foreach n load form a [ s: s + n ]
                                                    s
                                                    ]






                                                    share|improve this answer












                                                    share|improve this answer



                                                    share|improve this answer










                                                    answered 2 days ago









                                                    Galen Ivanov

                                                    5,93711032




                                                    5,93711032






















                                                        up vote
                                                        1
                                                        down vote













                                                        Not my best work. There's some definite improvements to be done, some probably fundamental to the algorithm used -- I'm sure it can be improved by using only an int, but I couldn't figure out how to efficiently enumerate neighbors that way. I'd love to see a PowerShell solution that uses only a single dimensional array!




                                                        PowerShell Core, 348 bytes





                                                        Function F($o){$t=120;$a=@{-1=,0*4;4=,0*4};0..3|%{$a[$_]=[int](-join$o[(3+18*$_)..(3+18*$_+13)]-split',')+,0};$m=16;while($m-gt0){0..3|%{$i=$_;0..3|%{if($a[$i][$_]-eq$m){$r=$i;$c=$_}}};$m=($a[$r-1][$c-1],$a[$r-1][$c],$a[$r-1][$c+1],$a[$r][$c+1],$a[$r][$c-1],$a[$r+1][$c-1],$a[$r+1][$c],$a[$r+1][$c+1]|Measure -Max).Maximum;$t-=$m;$a[$r][$c]=0}$t}


                                                        Try it online!





                                                        More readable version:



                                                        Function F($o){
                                                        $t=120;
                                                        $a=@{-1=,0*4;4=,0*4};
                                                        0..3|%{$a[$_]=[int](-join$o[(3+18*$_)..(3+18*$_+13)]-split',')+,0};
                                                        $m=16;
                                                        while($m-gt0){
                                                        0..3|%{$i=$_;0..3|%{if($a[$i][$_]-eq$m){$r=$i;$c=$_}}};
                                                        $m=($a[$r-1][$c-1],$a[$r-1][$c],$a[$r-1][$c+1],$a[$r][$c+1],$a[$r][$c-1],$a[$r+1][$c-1],$a[$r+1][$c],$a[$r+1][$c+1]|Measure -Max).Maximum;
                                                        $t-=$m;
                                                        $a[$r][$c]=0
                                                        }
                                                        $t
                                                        }





                                                        share|improve this answer

























                                                          up vote
                                                          1
                                                          down vote













                                                          Not my best work. There's some definite improvements to be done, some probably fundamental to the algorithm used -- I'm sure it can be improved by using only an int, but I couldn't figure out how to efficiently enumerate neighbors that way. I'd love to see a PowerShell solution that uses only a single dimensional array!




                                                          PowerShell Core, 348 bytes





                                                          Function F($o){$t=120;$a=@{-1=,0*4;4=,0*4};0..3|%{$a[$_]=[int](-join$o[(3+18*$_)..(3+18*$_+13)]-split',')+,0};$m=16;while($m-gt0){0..3|%{$i=$_;0..3|%{if($a[$i][$_]-eq$m){$r=$i;$c=$_}}};$m=($a[$r-1][$c-1],$a[$r-1][$c],$a[$r-1][$c+1],$a[$r][$c+1],$a[$r][$c-1],$a[$r+1][$c-1],$a[$r+1][$c],$a[$r+1][$c+1]|Measure -Max).Maximum;$t-=$m;$a[$r][$c]=0}$t}


                                                          Try it online!





                                                          More readable version:



                                                          Function F($o){
                                                          $t=120;
                                                          $a=@{-1=,0*4;4=,0*4};
                                                          0..3|%{$a[$_]=[int](-join$o[(3+18*$_)..(3+18*$_+13)]-split',')+,0};
                                                          $m=16;
                                                          while($m-gt0){
                                                          0..3|%{$i=$_;0..3|%{if($a[$i][$_]-eq$m){$r=$i;$c=$_}}};
                                                          $m=($a[$r-1][$c-1],$a[$r-1][$c],$a[$r-1][$c+1],$a[$r][$c+1],$a[$r][$c-1],$a[$r+1][$c-1],$a[$r+1][$c],$a[$r+1][$c+1]|Measure -Max).Maximum;
                                                          $t-=$m;
                                                          $a[$r][$c]=0
                                                          }
                                                          $t
                                                          }





                                                          share|improve this answer























                                                            up vote
                                                            1
                                                            down vote










                                                            up vote
                                                            1
                                                            down vote









                                                            Not my best work. There's some definite improvements to be done, some probably fundamental to the algorithm used -- I'm sure it can be improved by using only an int, but I couldn't figure out how to efficiently enumerate neighbors that way. I'd love to see a PowerShell solution that uses only a single dimensional array!




                                                            PowerShell Core, 348 bytes





                                                            Function F($o){$t=120;$a=@{-1=,0*4;4=,0*4};0..3|%{$a[$_]=[int](-join$o[(3+18*$_)..(3+18*$_+13)]-split',')+,0};$m=16;while($m-gt0){0..3|%{$i=$_;0..3|%{if($a[$i][$_]-eq$m){$r=$i;$c=$_}}};$m=($a[$r-1][$c-1],$a[$r-1][$c],$a[$r-1][$c+1],$a[$r][$c+1],$a[$r][$c-1],$a[$r+1][$c-1],$a[$r+1][$c],$a[$r+1][$c+1]|Measure -Max).Maximum;$t-=$m;$a[$r][$c]=0}$t}


                                                            Try it online!





                                                            More readable version:



                                                            Function F($o){
                                                            $t=120;
                                                            $a=@{-1=,0*4;4=,0*4};
                                                            0..3|%{$a[$_]=[int](-join$o[(3+18*$_)..(3+18*$_+13)]-split',')+,0};
                                                            $m=16;
                                                            while($m-gt0){
                                                            0..3|%{$i=$_;0..3|%{if($a[$i][$_]-eq$m){$r=$i;$c=$_}}};
                                                            $m=($a[$r-1][$c-1],$a[$r-1][$c],$a[$r-1][$c+1],$a[$r][$c+1],$a[$r][$c-1],$a[$r+1][$c-1],$a[$r+1][$c],$a[$r+1][$c+1]|Measure -Max).Maximum;
                                                            $t-=$m;
                                                            $a[$r][$c]=0
                                                            }
                                                            $t
                                                            }





                                                            share|improve this answer












                                                            Not my best work. There's some definite improvements to be done, some probably fundamental to the algorithm used -- I'm sure it can be improved by using only an int, but I couldn't figure out how to efficiently enumerate neighbors that way. I'd love to see a PowerShell solution that uses only a single dimensional array!




                                                            PowerShell Core, 348 bytes





                                                            Function F($o){$t=120;$a=@{-1=,0*4;4=,0*4};0..3|%{$a[$_]=[int](-join$o[(3+18*$_)..(3+18*$_+13)]-split',')+,0};$m=16;while($m-gt0){0..3|%{$i=$_;0..3|%{if($a[$i][$_]-eq$m){$r=$i;$c=$_}}};$m=($a[$r-1][$c-1],$a[$r-1][$c],$a[$r-1][$c+1],$a[$r][$c+1],$a[$r][$c-1],$a[$r+1][$c-1],$a[$r+1][$c],$a[$r+1][$c+1]|Measure -Max).Maximum;$t-=$m;$a[$r][$c]=0}$t}


                                                            Try it online!





                                                            More readable version:



                                                            Function F($o){
                                                            $t=120;
                                                            $a=@{-1=,0*4;4=,0*4};
                                                            0..3|%{$a[$_]=[int](-join$o[(3+18*$_)..(3+18*$_+13)]-split',')+,0};
                                                            $m=16;
                                                            while($m-gt0){
                                                            0..3|%{$i=$_;0..3|%{if($a[$i][$_]-eq$m){$r=$i;$c=$_}}};
                                                            $m=($a[$r-1][$c-1],$a[$r-1][$c],$a[$r-1][$c+1],$a[$r][$c+1],$a[$r][$c-1],$a[$r+1][$c-1],$a[$r+1][$c],$a[$r+1][$c+1]|Measure -Max).Maximum;
                                                            $t-=$m;
                                                            $a[$r][$c]=0
                                                            }
                                                            $t
                                                            }






                                                            share|improve this answer












                                                            share|improve this answer



                                                            share|improve this answer










                                                            answered 2 days ago









                                                            Jeff Freeman

                                                            20114




                                                            20114






















                                                                up vote
                                                                1
                                                                down vote













                                                                Powershell, 143 141 136 130 122 121 bytes





                                                                $a=,0*5+($args|%{$_+0})
                                                                for($n=16;$i=$a.IndexOf($n)){$a[$i]=0
                                                                $n=(-1,1+-6..-4+4..6|%{$a[$i+$_]}|sort)[-1]}$a|%{$s+=$_}
                                                                $s


                                                                Less golfed test script:



                                                                $f = {

                                                                $a=,0*5+($args|%{$_+0})
                                                                for($n=16;$i=$a.IndexOf($n)){
                                                                $a[$i]=0
                                                                $n=(-1,1+-6..-4+4..6|%{$a[$i+$_]}|sort)[-1]
                                                                }
                                                                $a|%{$s+=$_}
                                                                $s

                                                                }

                                                                @(
                                                                ,( 0 , ( 4, 3, 2, 1), ( 5, 6, 7, 8), (12, 11, 10, 9), (13, 14, 15, 16) )
                                                                ,( 0 , ( 8, 1, 9, 14), (11, 6, 5, 16), (13, 15, 2, 7), (10, 3, 12, 4) )
                                                                ,( 1 , ( 1, 2, 3, 4), ( 5, 6, 7, 8), ( 9, 10, 11, 12), (13, 14, 15, 16) )
                                                                ,( 3 , (10, 15, 14, 11), ( 9, 3, 1, 7), (13, 5, 12, 6), ( 2, 8, 4, 16) )
                                                                ,( 12 , ( 3, 7, 10, 5), ( 6, 8, 12, 13), (15, 9, 11, 4), (14, 1, 16, 2) )
                                                                ,( 34 , ( 8, 9, 3, 6), (13, 11, 7, 15), (12, 10, 16, 2), ( 4, 14, 1, 5) )
                                                                ,( 51 , ( 8, 11, 12, 9), (14, 5, 10, 16), ( 7, 3, 1, 6), (13, 4, 2, 15) )
                                                                ,( 78 , (13, 14, 1, 2), (16, 15, 3, 4), ( 5, 6, 7, 8), ( 9, 10, 11, 12) )
                                                                ,( 102, ( 9, 10, 11, 12), ( 1, 2, 4, 13), ( 7, 8, 5, 14), ( 3, 16, 6, 15) )
                                                                ,( 103, ( 9, 10, 11, 12), ( 1, 2, 7, 13), ( 6, 16, 4, 14), ( 3, 8, 5, 15) )
                                                                ) | % {
                                                                $expected, $a = $_
                                                                $result = &$f @a
                                                                "$($result-eq$expected): $result"
                                                                }


                                                                Output:



                                                                True: 0
                                                                True: 0
                                                                True: 1
                                                                True: 3
                                                                True: 12
                                                                True: 34
                                                                True: 51
                                                                True: 78
                                                                True: 102
                                                                True: 103


                                                                Explanation:



                                                                First, add top and bottom borders of 0 and make a single dimensional array:





                                                                0 0 0 0 0
                                                                # # # # 0
                                                                # # # # 0
                                                                # # # # 0
                                                                # # # # 0



                                                                0 0 0 0 0 # # # # 0 # # # # 0 # # # # 0 # # # # 0


                                                                Powershell returns $null if you try to get the value behind the end of the array.



                                                                Second, loop biggest neighbor pile started from 16 to non-zero-maximum. And nullify it (The Hungry Mouse eats it).





                                                                for($n=16;$i=$a.IndexOf($n)){
                                                                $a[$i]=0
                                                                $n=(-1,1+-6..-4+4..6|%{$a[$i+$_]}|sort)[-1]
                                                                }


                                                                Third, sum of the remaining piles.






                                                                share|improve this answer



























                                                                  up vote
                                                                  1
                                                                  down vote













                                                                  Powershell, 143 141 136 130 122 121 bytes





                                                                  $a=,0*5+($args|%{$_+0})
                                                                  for($n=16;$i=$a.IndexOf($n)){$a[$i]=0
                                                                  $n=(-1,1+-6..-4+4..6|%{$a[$i+$_]}|sort)[-1]}$a|%{$s+=$_}
                                                                  $s


                                                                  Less golfed test script:



                                                                  $f = {

                                                                  $a=,0*5+($args|%{$_+0})
                                                                  for($n=16;$i=$a.IndexOf($n)){
                                                                  $a[$i]=0
                                                                  $n=(-1,1+-6..-4+4..6|%{$a[$i+$_]}|sort)[-1]
                                                                  }
                                                                  $a|%{$s+=$_}
                                                                  $s

                                                                  }

                                                                  @(
                                                                  ,( 0 , ( 4, 3, 2, 1), ( 5, 6, 7, 8), (12, 11, 10, 9), (13, 14, 15, 16) )
                                                                  ,( 0 , ( 8, 1, 9, 14), (11, 6, 5, 16), (13, 15, 2, 7), (10, 3, 12, 4) )
                                                                  ,( 1 , ( 1, 2, 3, 4), ( 5, 6, 7, 8), ( 9, 10, 11, 12), (13, 14, 15, 16) )
                                                                  ,( 3 , (10, 15, 14, 11), ( 9, 3, 1, 7), (13, 5, 12, 6), ( 2, 8, 4, 16) )
                                                                  ,( 12 , ( 3, 7, 10, 5), ( 6, 8, 12, 13), (15, 9, 11, 4), (14, 1, 16, 2) )
                                                                  ,( 34 , ( 8, 9, 3, 6), (13, 11, 7, 15), (12, 10, 16, 2), ( 4, 14, 1, 5) )
                                                                  ,( 51 , ( 8, 11, 12, 9), (14, 5, 10, 16), ( 7, 3, 1, 6), (13, 4, 2, 15) )
                                                                  ,( 78 , (13, 14, 1, 2), (16, 15, 3, 4), ( 5, 6, 7, 8), ( 9, 10, 11, 12) )
                                                                  ,( 102, ( 9, 10, 11, 12), ( 1, 2, 4, 13), ( 7, 8, 5, 14), ( 3, 16, 6, 15) )
                                                                  ,( 103, ( 9, 10, 11, 12), ( 1, 2, 7, 13), ( 6, 16, 4, 14), ( 3, 8, 5, 15) )
                                                                  ) | % {
                                                                  $expected, $a = $_
                                                                  $result = &$f @a
                                                                  "$($result-eq$expected): $result"
                                                                  }


                                                                  Output:



                                                                  True: 0
                                                                  True: 0
                                                                  True: 1
                                                                  True: 3
                                                                  True: 12
                                                                  True: 34
                                                                  True: 51
                                                                  True: 78
                                                                  True: 102
                                                                  True: 103


                                                                  Explanation:



                                                                  First, add top and bottom borders of 0 and make a single dimensional array:





                                                                  0 0 0 0 0
                                                                  # # # # 0
                                                                  # # # # 0
                                                                  # # # # 0
                                                                  # # # # 0



                                                                  0 0 0 0 0 # # # # 0 # # # # 0 # # # # 0 # # # # 0


                                                                  Powershell returns $null if you try to get the value behind the end of the array.



                                                                  Second, loop biggest neighbor pile started from 16 to non-zero-maximum. And nullify it (The Hungry Mouse eats it).





                                                                  for($n=16;$i=$a.IndexOf($n)){
                                                                  $a[$i]=0
                                                                  $n=(-1,1+-6..-4+4..6|%{$a[$i+$_]}|sort)[-1]
                                                                  }


                                                                  Third, sum of the remaining piles.






                                                                  share|improve this answer

























                                                                    up vote
                                                                    1
                                                                    down vote










                                                                    up vote
                                                                    1
                                                                    down vote









                                                                    Powershell, 143 141 136 130 122 121 bytes





                                                                    $a=,0*5+($args|%{$_+0})
                                                                    for($n=16;$i=$a.IndexOf($n)){$a[$i]=0
                                                                    $n=(-1,1+-6..-4+4..6|%{$a[$i+$_]}|sort)[-1]}$a|%{$s+=$_}
                                                                    $s


                                                                    Less golfed test script:



                                                                    $f = {

                                                                    $a=,0*5+($args|%{$_+0})
                                                                    for($n=16;$i=$a.IndexOf($n)){
                                                                    $a[$i]=0
                                                                    $n=(-1,1+-6..-4+4..6|%{$a[$i+$_]}|sort)[-1]
                                                                    }
                                                                    $a|%{$s+=$_}
                                                                    $s

                                                                    }

                                                                    @(
                                                                    ,( 0 , ( 4, 3, 2, 1), ( 5, 6, 7, 8), (12, 11, 10, 9), (13, 14, 15, 16) )
                                                                    ,( 0 , ( 8, 1, 9, 14), (11, 6, 5, 16), (13, 15, 2, 7), (10, 3, 12, 4) )
                                                                    ,( 1 , ( 1, 2, 3, 4), ( 5, 6, 7, 8), ( 9, 10, 11, 12), (13, 14, 15, 16) )
                                                                    ,( 3 , (10, 15, 14, 11), ( 9, 3, 1, 7), (13, 5, 12, 6), ( 2, 8, 4, 16) )
                                                                    ,( 12 , ( 3, 7, 10, 5), ( 6, 8, 12, 13), (15, 9, 11, 4), (14, 1, 16, 2) )
                                                                    ,( 34 , ( 8, 9, 3, 6), (13, 11, 7, 15), (12, 10, 16, 2), ( 4, 14, 1, 5) )
                                                                    ,( 51 , ( 8, 11, 12, 9), (14, 5, 10, 16), ( 7, 3, 1, 6), (13, 4, 2, 15) )
                                                                    ,( 78 , (13, 14, 1, 2), (16, 15, 3, 4), ( 5, 6, 7, 8), ( 9, 10, 11, 12) )
                                                                    ,( 102, ( 9, 10, 11, 12), ( 1, 2, 4, 13), ( 7, 8, 5, 14), ( 3, 16, 6, 15) )
                                                                    ,( 103, ( 9, 10, 11, 12), ( 1, 2, 7, 13), ( 6, 16, 4, 14), ( 3, 8, 5, 15) )
                                                                    ) | % {
                                                                    $expected, $a = $_
                                                                    $result = &$f @a
                                                                    "$($result-eq$expected): $result"
                                                                    }


                                                                    Output:



                                                                    True: 0
                                                                    True: 0
                                                                    True: 1
                                                                    True: 3
                                                                    True: 12
                                                                    True: 34
                                                                    True: 51
                                                                    True: 78
                                                                    True: 102
                                                                    True: 103


                                                                    Explanation:



                                                                    First, add top and bottom borders of 0 and make a single dimensional array:





                                                                    0 0 0 0 0
                                                                    # # # # 0
                                                                    # # # # 0
                                                                    # # # # 0
                                                                    # # # # 0



                                                                    0 0 0 0 0 # # # # 0 # # # # 0 # # # # 0 # # # # 0


                                                                    Powershell returns $null if you try to get the value behind the end of the array.



                                                                    Second, loop biggest neighbor pile started from 16 to non-zero-maximum. And nullify it (The Hungry Mouse eats it).





                                                                    for($n=16;$i=$a.IndexOf($n)){
                                                                    $a[$i]=0
                                                                    $n=(-1,1+-6..-4+4..6|%{$a[$i+$_]}|sort)[-1]
                                                                    }


                                                                    Third, sum of the remaining piles.






                                                                    share|improve this answer














                                                                    Powershell, 143 141 136 130 122 121 bytes





                                                                    $a=,0*5+($args|%{$_+0})
                                                                    for($n=16;$i=$a.IndexOf($n)){$a[$i]=0
                                                                    $n=(-1,1+-6..-4+4..6|%{$a[$i+$_]}|sort)[-1]}$a|%{$s+=$_}
                                                                    $s


                                                                    Less golfed test script:



                                                                    $f = {

                                                                    $a=,0*5+($args|%{$_+0})
                                                                    for($n=16;$i=$a.IndexOf($n)){
                                                                    $a[$i]=0
                                                                    $n=(-1,1+-6..-4+4..6|%{$a[$i+$_]}|sort)[-1]
                                                                    }
                                                                    $a|%{$s+=$_}
                                                                    $s

                                                                    }

                                                                    @(
                                                                    ,( 0 , ( 4, 3, 2, 1), ( 5, 6, 7, 8), (12, 11, 10, 9), (13, 14, 15, 16) )
                                                                    ,( 0 , ( 8, 1, 9, 14), (11, 6, 5, 16), (13, 15, 2, 7), (10, 3, 12, 4) )
                                                                    ,( 1 , ( 1, 2, 3, 4), ( 5, 6, 7, 8), ( 9, 10, 11, 12), (13, 14, 15, 16) )
                                                                    ,( 3 , (10, 15, 14, 11), ( 9, 3, 1, 7), (13, 5, 12, 6), ( 2, 8, 4, 16) )
                                                                    ,( 12 , ( 3, 7, 10, 5), ( 6, 8, 12, 13), (15, 9, 11, 4), (14, 1, 16, 2) )
                                                                    ,( 34 , ( 8, 9, 3, 6), (13, 11, 7, 15), (12, 10, 16, 2), ( 4, 14, 1, 5) )
                                                                    ,( 51 , ( 8, 11, 12, 9), (14, 5, 10, 16), ( 7, 3, 1, 6), (13, 4, 2, 15) )
                                                                    ,( 78 , (13, 14, 1, 2), (16, 15, 3, 4), ( 5, 6, 7, 8), ( 9, 10, 11, 12) )
                                                                    ,( 102, ( 9, 10, 11, 12), ( 1, 2, 4, 13), ( 7, 8, 5, 14), ( 3, 16, 6, 15) )
                                                                    ,( 103, ( 9, 10, 11, 12), ( 1, 2, 7, 13), ( 6, 16, 4, 14), ( 3, 8, 5, 15) )
                                                                    ) | % {
                                                                    $expected, $a = $_
                                                                    $result = &$f @a
                                                                    "$($result-eq$expected): $result"
                                                                    }


                                                                    Output:



                                                                    True: 0
                                                                    True: 0
                                                                    True: 1
                                                                    True: 3
                                                                    True: 12
                                                                    True: 34
                                                                    True: 51
                                                                    True: 78
                                                                    True: 102
                                                                    True: 103


                                                                    Explanation:



                                                                    First, add top and bottom borders of 0 and make a single dimensional array:





                                                                    0 0 0 0 0
                                                                    # # # # 0
                                                                    # # # # 0
                                                                    # # # # 0
                                                                    # # # # 0



                                                                    0 0 0 0 0 # # # # 0 # # # # 0 # # # # 0 # # # # 0


                                                                    Powershell returns $null if you try to get the value behind the end of the array.



                                                                    Second, loop biggest neighbor pile started from 16 to non-zero-maximum. And nullify it (The Hungry Mouse eats it).





                                                                    for($n=16;$i=$a.IndexOf($n)){
                                                                    $a[$i]=0
                                                                    $n=(-1,1+-6..-4+4..6|%{$a[$i+$_]}|sort)[-1]
                                                                    }


                                                                    Third, sum of the remaining piles.







                                                                    share|improve this answer














                                                                    share|improve this answer



                                                                    share|improve this answer








                                                                    edited yesterday

























                                                                    answered yesterday









                                                                    mazzy

                                                                    1,797313




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