Mixing
Binary self-rotation
$begingroup$
Given a binary 3D array, for each layer, cyclically rotate up each of its columns as many steps as indicated by the binary encoding of the columns of the layer above it, and then cyclically rotate left each of its rows as many steps as indicated by the binary encoding of the rows of the layer below it.
There will always be at least three layers. The top layer's columns and the bottom layer's rows should not be rotated.
Walk-through
Lets start with the small 4-layer, 2-row, 3-column array:
[[[1,0,1],
[1,0,0]],
[[1,0,1],
[0,1,1]],
[[0,1,1],
[1,1,1]],
[[1,1,0],
[1,1,1]]]
The first step is evaluating the numbers encoded in binary by the columns and rows of each layer:
3 0 2
5 [[[1,0,1],
4 [1,0,0]],
2 1 3
5 [[1,0,1],
3 [0,1,1]],
1 3 3
3 [[0,1,1],
7 [1,1,1]],
3 3 1
6 [[1,1,0],
7 [1,1,1]]]
The first layer, [[1,0,1],[1,0,0]] will not have its columns rotated, but its rows will be cyclically rotated left 5 steps and 3 step respectively, thus becoming [[1,1,0],[1,0,0]].
The second layer, [[1,0,1],[0,1,1]], will have its columns cyclically rotated up 3, 0, and 2 steps respectively, giving [[0,0,1],[1,1,1]], and then the rows are cyclically rotated left 3 and 7 steps respectively, with no visible change.
The third layer, [[0,1,1],[1,1,1]] rotated up 2, 1, and 3 steps stays the same, and neither does rotating left 6 and 7 steps do anything.
Finally, the fourth layer, [[1,1,0],[1,1,1]] rotated up 1, 3, and 3 steps is [[1,1,1],[1,1,0]], but its rows are not rotated afterwards, as it is the last layer.
Putting all the layers together again, gives us the binary self-rotated 3D array:
[[[1,1,0],
[1,0,0]],
[[0,0,1],
[1,1,1]],
[[0,1,1],
[1,1,1]],
[[1,1,1],
[1,1,0]]]
Example cases:
[[[1,0,1],[1,0,0]],[[1,0,1],[0,1,1]],[[0,1,1],[1,1,1]],[[1,1,0],[1,1,1]]] gives[[[1,1,0],[1,0,0]],[[0,0,1],[1,1,1]],[[0,1,1],[1,1,1]],[[1,1,1],[1,1,0]]]
[[[1]],[[1]],[[0]]] gives[[[1]],[[1]],[[0]]]
[[[1,0,1],[1,0,1],[1,0,1]],[[0,0,1],[0,0,1],[0,0,1]],[[1,0,0],[1,0,1],[0,0,1]]] gives[[[0,1,1],[0,1,1],[0,1,1]],[[0,1,0],[1,0,0],[0,1,0]],[[1,0,1],[1,0,1],[0,0,0]]]
code-golf matrix cellular-automata 3d
asked Jan 22 at 9:00
AdámAdám
28.9k276204
$endgroup$
add a comment |
$begingroup$
Given a binary 3D array, for each layer, cyclically rotate up each of its columns as many steps as indicated by the binary encoding of the columns of the layer above it, and then cyclically rotate left each of its rows as many steps as indicated by the binary encoding of the rows of the layer below it.
There will always be at least three layers. The top layer's columns and the bottom layer's rows should not be rotated.
Walk-through
Lets start with the small 4-layer, 2-row, 3-column array:
[[[1,0,1],
[1,0,0]],
[[1,0,1],
[0,1,1]],
[[0,1,1],
[1,1,1]],
[[1,1,0],
[1,1,1]]]
The first step is evaluating the numbers encoded in binary by the columns and rows of each layer:
3 0 2
5 [[[1,0,1],
4 [1,0,0]],
2 1 3
5 [[1,0,1],
3 [0,1,1]],
1 3 3
3 [[0,1,1],
7 [1,1,1]],
3 3 1
6 [[1,1,0],
7 [1,1,1]]]
The first layer, [[1,0,1],[1,0,0]] will not have its columns rotated, but its rows will be cyclically rotated left 5 steps and 3 step respectively, thus becoming [[1,1,0],[1,0,0]].
The second layer, [[1,0,1],[0,1,1]], will have its columns cyclically rotated up 3, 0, and 2 steps respectively, giving [[0,0,1],[1,1,1]], and then the rows are cyclically rotated left 3 and 7 steps respectively, with no visible change.
The third layer, [[0,1,1],[1,1,1]] rotated up 2, 1, and 3 steps stays the same, and neither does rotating left 6 and 7 steps do anything.
Finally, the fourth layer, [[1,1,0],[1,1,1]] rotated up 1, 3, and 3 steps is [[1,1,1],[1,1,0]], but its rows are not rotated afterwards, as it is the last layer.
Putting all the layers together again, gives us the binary self-rotated 3D array:
[[[1,1,0],
[1,0,0]],
[[0,0,1],
[1,1,1]],
[[0,1,1],
[1,1,1]],
[[1,1,1],
[1,1,0]]]
Example cases:
[[[1,0,1],[1,0,0]],[[1,0,1],[0,1,1]],[[0,1,1],[1,1,1]],[[1,1,0],[1,1,1]]] gives[[[1,1,0],[1,0,0]],[[0,0,1],[1,1,1]],[[0,1,1],[1,1,1]],[[1,1,1],[1,1,0]]]
[[[1]],[[1]],[[0]]] gives[[[1]],[[1]],[[0]]]
[[[1,0,1],[1,0,1],[1,0,1]],[[0,0,1],[0,0,1],[0,0,1]],[[1,0,0],[1,0,1],[0,0,1]]] gives[[[0,1,1],[0,1,1],[0,1,1]],[[0,1,0],[1,0,0],[0,1,0]],[[1,0,1],[1,0,1],[0,0,0]]]
code-golf matrix cellular-automata 3d
asked Jan 22 at 9:00
AdámAdám
28.9k276204
$endgroup$
add a comment |
$begingroup$
Given a binary 3D array, for each layer, cyclically rotate up each of its columns as many steps as indicated by the binary encoding of the columns of the layer above it, and then cyclically rotate left each of its rows as many steps as indicated by the binary encoding of the rows of the layer below it.
There will always be at least three layers. The top layer's columns and the bottom layer's rows should not be rotated.
Walk-through
Lets start with the small 4-layer, 2-row, 3-column array:
[[[1,0,1],
[1,0,0]],
[[1,0,1],
[0,1,1]],
[[0,1,1],
[1,1,1]],
[[1,1,0],
[1,1,1]]]
The first step is evaluating the numbers encoded in binary by the columns and rows of each layer:
3 0 2
5 [[[1,0,1],
4 [1,0,0]],
2 1 3
5 [[1,0,1],
3 [0,1,1]],
1 3 3
3 [[0,1,1],
7 [1,1,1]],
3 3 1
6 [[1,1,0],
7 [1,1,1]]]
The first layer, [[1,0,1],[1,0,0]] will not have its columns rotated, but its rows will be cyclically rotated left 5 steps and 3 step respectively, thus becoming [[1,1,0],[1,0,0]].
The second layer, [[1,0,1],[0,1,1]], will have its columns cyclically rotated up 3, 0, and 2 steps respectively, giving [[0,0,1],[1,1,1]], and then the rows are cyclically rotated left 3 and 7 steps respectively, with no visible change.
The third layer, [[0,1,1],[1,1,1]] rotated up 2, 1, and 3 steps stays the same, and neither does rotating left 6 and 7 steps do anything.
Finally, the fourth layer, [[1,1,0],[1,1,1]] rotated up 1, 3, and 3 steps is [[1,1,1],[1,1,0]], but its rows are not rotated afterwards, as it is the last layer.
Putting all the layers together again, gives us the binary self-rotated 3D array:
[[[1,1,0],
[1,0,0]],
[[0,0,1],
[1,1,1]],
[[0,1,1],
[1,1,1]],
[[1,1,1],
[1,1,0]]]
Example cases:
[[[1,0,1],[1,0,0]],[[1,0,1],[0,1,1]],[[0,1,1],[1,1,1]],[[1,1,0],[1,1,1]]] gives[[[1,1,0],[1,0,0]],[[0,0,1],[1,1,1]],[[0,1,1],[1,1,1]],[[1,1,1],[1,1,0]]]
[[[1]],[[1]],[[0]]] gives[[[1]],[[1]],[[0]]]
[[[1,0,1],[1,0,1],[1,0,1]],[[0,0,1],[0,0,1],[0,0,1]],[[1,0,0],[1,0,1],[0,0,1]]] gives[[[0,1,1],[0,1,1],[0,1,1]],[[0,1,0],[1,0,0],[0,1,0]],[[1,0,1],[1,0,1],[0,0,0]]]
code-golf matrix cellular-automata 3d
asked Jan 22 at 9:00
AdámAdám
28.9k276204
$endgroup$
Given a binary 3D array, for each layer, cyclically rotate up each of its columns as many steps as indicated by the binary encoding of the columns of the layer above it, and then cyclically rotate left each of its rows as many steps as indicated by the binary encoding of the rows of the layer below it.
There will always be at least three layers. The top layer's columns and the bottom layer's rows should not be rotated.
Walk-through
Lets start with the small 4-layer, 2-row, 3-column array:
[[[1,0,1],
[1,0,0]],
[[1,0,1],
[0,1,1]],
[[0,1,1],
[1,1,1]],
[[1,1,0],
[1,1,1]]]
The first step is evaluating the numbers encoded in binary by the columns and rows of each layer:
3 0 2
5 [[[1,0,1],
4 [1,0,0]],
2 1 3
5 [[1,0,1],
3 [0,1,1]],
1 3 3
3 [[0,1,1],
7 [1,1,1]],
3 3 1
6 [[1,1,0],
7 [1,1,1]]]
The first layer, [[1,0,1],[1,0,0]] will not have its columns rotated, but its rows will be cyclically rotated left 5 steps and 3 step respectively, thus becoming [[1,1,0],[1,0,0]].
The second layer, [[1,0,1],[0,1,1]], will have its columns cyclically rotated up 3, 0, and 2 steps respectively, giving [[0,0,1],[1,1,1]], and then the rows are cyclically rotated left 3 and 7 steps respectively, with no visible change.
The third layer, [[0,1,1],[1,1,1]] rotated up 2, 1, and 3 steps stays the same, and neither does rotating left 6 and 7 steps do anything.
Finally, the fourth layer, [[1,1,0],[1,1,1]] rotated up 1, 3, and 3 steps is [[1,1,1],[1,1,0]], but its rows are not rotated afterwards, as it is the last layer.
Putting all the layers together again, gives us the binary self-rotated 3D array:
[[[1,1,0],
[1,0,0]],
[[0,0,1],
[1,1,1]],
[[0,1,1],
[1,1,1]],
[[1,1,1],
[1,1,0]]]
Example cases:
[[[1,0,1],[1,0,0]],[[1,0,1],[0,1,1]],[[0,1,1],[1,1,1]],[[1,1,0],[1,1,1]]] gives[[[1,1,0],[1,0,0]],[[0,0,1],[1,1,1]],[[0,1,1],[1,1,1]],[[1,1,1],[1,1,0]]]
[[[1]],[[1]],[[0]]] gives[[[1]],[[1]],[[0]]]
[[[1,0,1],[1,0,1],[1,0,1]],[[0,0,1],[0,0,1],[0,0,1]],[[1,0,0],[1,0,1],[0,0,1]]] gives[[[0,1,1],[0,1,1],[0,1,1]],[[0,1,0],[1,0,0],[0,1,0]],[[1,0,1],[1,0,1],[0,0,0]]]
code-golf matrix cellular-automata 3d
code-golf matrix cellular-automata 3d
asked Jan 22 at 9:00
AdámAdám
28.9k276204
asked Jan 22 at 9:00
AdámAdám
28.9k276204
asked Jan 22 at 9:00
AdámAdám
28.9k276204
asked Jan 22 at 9:00
AdámAdám
28.9k276204
asked Jan 22 at 9:00
AdámAdám
28.9k276204
28.9k276204
add a comment |
add a comment |
6 Answers
6
active
oldest
votes
$begingroup$
Jelly, 18 17 bytes
ṙ""Ḅ}
Z€çŻṖ$$Z€çḊ
Try it online!
How?
ṙ""Ḅ} - Link 1, rotation helper: 3d matrix to rotate, 3d matrix of rotation instructions
} - use the right argument for:
Ḅ - un-binary (vectorises) - get the rotation amounts as a 2d matrix
" - zip with:
" - zip with:
ṙ - rotate (the current row) left by (the current amount)
Z€çŻṖ$ $Z€çḊ - Main Link: 3d matrix, M
Z€ - transpose €ach (layer of M)
$ - last two links as a monad:
$ - last two links as a monad:
Ż - prepend a zero
Ṗ - pop (i.e. remove the tail)
ç - call the last Link as a dyad (i.e. f(Z€ result, ŻṖ$ result) )
Z€ - transpose €ach (layer of that)
Ḋ - dequeue (i.e. remove the head layer of M)
ç - call the last Link as a dyad (i.e. f(Z€çŻṖ$$Z€ result, Ḋ result) )
Note: $$ (or possibly $$ ... $$?) seems to mess up the code-block formatting (but only once posted, not in the preview), so I added a space to make my life easier.
answered Jan 22 at 13:27
Jonathan AllanJonathan Allan
52.7k535170
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Python 2, 220 211 209 185 176 174 164 161 159 bytes
lambda m:map(R,z(map(R,z(m,['']+[z(*l)for l in m])),m[1:]+['']))
R=lambda(l,L):map(lambda r,i:r[i:]+r[:i or 0],z(*l),[int(`b`[1::3],2)%len(b)for b in L])
z=zip
Try it online!
-2 bytes, thanks to Jonathan Allan
answered Jan 22 at 14:08
TFeldTFeld
15.7k21248
$endgroup$
$begingroup$
Since you handleNoneduring the slicing for the rotation I believe both of the['0']can become[].
$endgroup$
– Jonathan Allan
Jan 22 at 17:29
$begingroup$
@JonathanAllan Thanks :)
$endgroup$
– TFeld
Jan 23 at 8:01
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APL+WIN, 53 39 bytes
Many thanks to Adám for saving 14 bytes
(1 0↓⍉2⊥⍉m⍪0)⌽(¯1 0↓2⊥2 1 3⍉0⍪m)⊖[2]m←⎕
Try it online! Courtesy of Dyalog Classic
Prompts for input of a 3d array of the form:
4 2 3⍴1 0 1 1 0 0 1 0 1 0 1 1 0 1 1 1 1 1 1 1 0 1 1 1
which yields:
1 0 1
1 0 0
1 0 1
0 1 1
0 1 1
1 1 1
1 1 0
1 1 1
Explanation:
m←⎕ Prompt for input
(¯1 0↓2⊥2 1 3⍉0⍪m) Calculate column rotations
(1 0↓⍉2⊥⍉m⍪0) Calculate row rotations
(...)⌽(...)⊖[2]m Apply column and row rotation and output resulting 3d array:
1 1 0
1 0 0
0 0 1
1 1 1
0 1 1
1 1 1
1 1 1
1 1 0
answered Jan 22 at 14:49
GrahamGraham
2,50678
$endgroup$
$begingroup$
Instead of enclosing and using¨, just process the entire array at once. Try it online!
$endgroup$
– Adám
Jan 22 at 17:08
$begingroup$
@Adám Many thanks. I do not know why I over thought this one and went the nested route :( Getting old?
$endgroup$
– Graham
Jan 22 at 18:44
add a comment |
$begingroup$
R, 226 216 205 bytes
-21 bytes thanks to digEmAll
function(a,L=`for`){d=dim(b<-a)
r=function(a,n,l=sum(a|1))a[(1:l+sum(n*2^(sum(n|1):1-1))-1)%%l+1]
L(i,I<-2:d[3],L(j,1:d,b[j,,i]<-r(b[j,,i],a[j,,i-1])))
L(i,I-1,L(k,1:d[2],b[,k,i]<-r(b[,k,i],a[,k,i+1])))
b}
Try it online!
answered Feb 23 at 1:11
ASCII-onlyASCII-only
4,3591338
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05AB1E, 41 39 bytes
εNĀiø¹N<èøJC‚øε`._}ø}N¹g<Êi¹N>èJC‚øε`._
This feels way too long.. Can definitely be golfed some more.
Try it online or verify all test cases.
Explanation:
ε # Map each layer in the (implicit) input to:
# (`N` is the layer-index of this map)
NĀi # If it is not the first layer:
ø # Zip/transpose the current layer; swapping rows/columns
¹N<è # Get the `N-1`'th layer of the input
ø # Zip/transpose; swapping rows/columns
J # Join all inner lists (the columns) together
C # And convert it from binary to integer
‚ # Pair it with the current layer's columns we're mapping
ø # Zip/transpose; to pair each integer with a layer's columns
ε } # Map over these pairs:
` # Push both values of the pair separately to the stack
._ # Rotate the column the integer amount of times
ø # Zip/transpose the rows/columns of the current layer back
} # Close the if-statement
N¹g<Êi # If this is not the last layer (layer-index-1 != amount_of_layers):
¹N>è # Get the `N+1`'th layer of the input
J # Join all inner lists (the rows) together
C # And convert it from binary to integer
‚ # Pair it with the current layer's rows we're mapping
ø # Zip/transpose; to pair each integer with a layer's rows
ε # Map over these pairs:
` # Push both values of the pair separately to the stack
._ # Rotate the row the integer amount of times
# (implicitly output the result after the layer-mapping is done)
answered Jan 22 at 9:56
Kevin CruijssenKevin Cruijssen
40.1k563206
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$begingroup$
Wolfram Language (Mathematica), 138 131 125 123 bytes
t=Map@Thread
m=MapThread[r=RotateLeft,#,2]&
b=(a=ArrayPad)[Map@Fold[#+##&]/@#,1]~r~#2~a~-1&
g=m@{t@m@{t@#,t@#~b~-1},#~b~1}&
Try it online!
Map[Thread]is equivalent toTranspose[a, {1,3,2}], which transposes the columns and rows.
Fold[#+##&]is shorter thanIntegerDigits[#,2]for converting from binary.
answered Jan 27 at 1:50
lirtosiastlirtosiast
18.2k438109
$endgroup$
add a comment |
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6 Answers
6
active
oldest
votes
6 Answers
6
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
Jelly, 18 17 bytes
ṙ""Ḅ}
Z€çŻṖ$$Z€çḊ
Try it online!
How?
ṙ""Ḅ} - Link 1, rotation helper: 3d matrix to rotate, 3d matrix of rotation instructions
} - use the right argument for:
Ḅ - un-binary (vectorises) - get the rotation amounts as a 2d matrix
" - zip with:
" - zip with:
ṙ - rotate (the current row) left by (the current amount)
Z€çŻṖ$ $Z€çḊ - Main Link: 3d matrix, M
Z€ - transpose €ach (layer of M)
$ - last two links as a monad:
$ - last two links as a monad:
Ż - prepend a zero
Ṗ - pop (i.e. remove the tail)
ç - call the last Link as a dyad (i.e. f(Z€ result, ŻṖ$ result) )
Z€ - transpose €ach (layer of that)
Ḋ - dequeue (i.e. remove the head layer of M)
ç - call the last Link as a dyad (i.e. f(Z€çŻṖ$$Z€ result, Ḋ result) )
Note: $$ (or possibly $$ ... $$?) seems to mess up the code-block formatting (but only once posted, not in the preview), so I added a space to make my life easier.
answered Jan 22 at 13:27
Jonathan AllanJonathan Allan
52.7k535170
$endgroup$
add a comment |
$begingroup$
Jelly, 18 17 bytes
ṙ""Ḅ}
Z€çŻṖ$$Z€çḊ
Try it online!
How?
ṙ""Ḅ} - Link 1, rotation helper: 3d matrix to rotate, 3d matrix of rotation instructions
} - use the right argument for:
Ḅ - un-binary (vectorises) - get the rotation amounts as a 2d matrix
" - zip with:
" - zip with:
ṙ - rotate (the current row) left by (the current amount)
Z€çŻṖ$ $Z€çḊ - Main Link: 3d matrix, M
Z€ - transpose €ach (layer of M)
$ - last two links as a monad:
$ - last two links as a monad:
Ż - prepend a zero
Ṗ - pop (i.e. remove the tail)
ç - call the last Link as a dyad (i.e. f(Z€ result, ŻṖ$ result) )
Z€ - transpose €ach (layer of that)
Ḋ - dequeue (i.e. remove the head layer of M)
ç - call the last Link as a dyad (i.e. f(Z€çŻṖ$$Z€ result, Ḋ result) )
Note: $$ (or possibly $$ ... $$?) seems to mess up the code-block formatting (but only once posted, not in the preview), so I added a space to make my life easier.
answered Jan 22 at 13:27
Jonathan AllanJonathan Allan
52.7k535170
$endgroup$
add a comment |
$begingroup$
Jelly, 18 17 bytes
ṙ""Ḅ}
Z€çŻṖ$$Z€çḊ
Try it online!
How?
ṙ""Ḅ} - Link 1, rotation helper: 3d matrix to rotate, 3d matrix of rotation instructions
} - use the right argument for:
Ḅ - un-binary (vectorises) - get the rotation amounts as a 2d matrix
" - zip with:
" - zip with:
ṙ - rotate (the current row) left by (the current amount)
Z€çŻṖ$ $Z€çḊ - Main Link: 3d matrix, M
Z€ - transpose €ach (layer of M)
$ - last two links as a monad:
$ - last two links as a monad:
Ż - prepend a zero
Ṗ - pop (i.e. remove the tail)
ç - call the last Link as a dyad (i.e. f(Z€ result, ŻṖ$ result) )
Z€ - transpose €ach (layer of that)
Ḋ - dequeue (i.e. remove the head layer of M)
ç - call the last Link as a dyad (i.e. f(Z€çŻṖ$$Z€ result, Ḋ result) )
Note: $$ (or possibly $$ ... $$?) seems to mess up the code-block formatting (but only once posted, not in the preview), so I added a space to make my life easier.
answered Jan 22 at 13:27
Jonathan AllanJonathan Allan
52.7k535170
$endgroup$
Jelly, 18 17 bytes
ṙ""Ḅ}
Z€çŻṖ$$Z€çḊ
Try it online!
How?
ṙ""Ḅ} - Link 1, rotation helper: 3d matrix to rotate, 3d matrix of rotation instructions
} - use the right argument for:
Ḅ - un-binary (vectorises) - get the rotation amounts as a 2d matrix
" - zip with:
" - zip with:
ṙ - rotate (the current row) left by (the current amount)
Z€çŻṖ$ $Z€çḊ - Main Link: 3d matrix, M
Z€ - transpose €ach (layer of M)
$ - last two links as a monad:
$ - last two links as a monad:
Ż - prepend a zero
Ṗ - pop (i.e. remove the tail)
ç - call the last Link as a dyad (i.e. f(Z€ result, ŻṖ$ result) )
Z€ - transpose €ach (layer of that)
Ḋ - dequeue (i.e. remove the head layer of M)
ç - call the last Link as a dyad (i.e. f(Z€çŻṖ$$Z€ result, Ḋ result) )
Note: $$ (or possibly $$ ... $$?) seems to mess up the code-block formatting (but only once posted, not in the preview), so I added a space to make my life easier.
answered Jan 22 at 13:27
Jonathan AllanJonathan Allan
52.7k535170
edited Jan 22 at 18:14
answered Jan 22 at 13:27
Jonathan AllanJonathan Allan
52.7k535170
answered Jan 22 at 13:27
Jonathan AllanJonathan Allan
52.7k535170
answered Jan 22 at 13:27
Jonathan AllanJonathan Allan
52.7k535170
52.7k535170
add a comment |
add a comment |
$begingroup$
Python 2, 220 211 209 185 176 174 164 161 159 bytes
lambda m:map(R,z(map(R,z(m,['']+[z(*l)for l in m])),m[1:]+['']))
R=lambda(l,L):map(lambda r,i:r[i:]+r[:i or 0],z(*l),[int(`b`[1::3],2)%len(b)for b in L])
z=zip
Try it online!
-2 bytes, thanks to Jonathan Allan
answered Jan 22 at 14:08
TFeldTFeld
15.7k21248
$endgroup$
$begingroup$
Since you handleNoneduring the slicing for the rotation I believe both of the['0']can become[].
$endgroup$
– Jonathan Allan
Jan 22 at 17:29
$begingroup$
@JonathanAllan Thanks :)
$endgroup$
– TFeld
Jan 23 at 8:01
add a comment |
$begingroup$
Python 2, 220 211 209 185 176 174 164 161 159 bytes
lambda m:map(R,z(map(R,z(m,['']+[z(*l)for l in m])),m[1:]+['']))
R=lambda(l,L):map(lambda r,i:r[i:]+r[:i or 0],z(*l),[int(`b`[1::3],2)%len(b)for b in L])
z=zip
Try it online!
-2 bytes, thanks to Jonathan Allan
answered Jan 22 at 14:08
TFeldTFeld
15.7k21248
$endgroup$
$begingroup$
Since you handleNoneduring the slicing for the rotation I believe both of the['0']can become[].
$endgroup$
– Jonathan Allan
Jan 22 at 17:29
$begingroup$
@JonathanAllan Thanks :)
$endgroup$
– TFeld
Jan 23 at 8:01
add a comment |
$begingroup$
Python 2, 220 211 209 185 176 174 164 161 159 bytes
lambda m:map(R,z(map(R,z(m,['']+[z(*l)for l in m])),m[1:]+['']))
R=lambda(l,L):map(lambda r,i:r[i:]+r[:i or 0],z(*l),[int(`b`[1::3],2)%len(b)for b in L])
z=zip
Try it online!
-2 bytes, thanks to Jonathan Allan
answered Jan 22 at 14:08
TFeldTFeld
15.7k21248
$endgroup$
Python 2, 220 211 209 185 176 174 164 161 159 bytes
lambda m:map(R,z(map(R,z(m,['']+[z(*l)for l in m])),m[1:]+['']))
R=lambda(l,L):map(lambda r,i:r[i:]+r[:i or 0],z(*l),[int(`b`[1::3],2)%len(b)for b in L])
z=zip
Try it online!
-2 bytes, thanks to Jonathan Allan
answered Jan 22 at 14:08
TFeldTFeld
15.7k21248
edited Jan 23 at 8:01
answered Jan 22 at 14:08
TFeldTFeld
15.7k21248
answered Jan 22 at 14:08
TFeldTFeld
15.7k21248
answered Jan 22 at 14:08
TFeldTFeld
15.7k21248
15.7k21248
$begingroup$
Since you handleNoneduring the slicing for the rotation I believe both of the['0']can become[].
$endgroup$
– Jonathan Allan
Jan 22 at 17:29
$begingroup$
@JonathanAllan Thanks :)
$endgroup$
– TFeld
Jan 23 at 8:01
add a comment |
$begingroup$
Since you handleNoneduring the slicing for the rotation I believe both of the['0']can become[].
$endgroup$
– Jonathan Allan
Jan 22 at 17:29
$begingroup$
@JonathanAllan Thanks :)
$endgroup$
– TFeld
Jan 23 at 8:01
$begingroup$
Since you handle
None during the slicing for the rotation I believe both of the ['0'] can become [].$endgroup$
– Jonathan Allan
Jan 22 at 17:29
$begingroup$
Since you handle
None during the slicing for the rotation I believe both of the ['0'] can become [].$endgroup$
– Jonathan Allan
Jan 22 at 17:29
$begingroup$
@JonathanAllan Thanks :)
$endgroup$
– TFeld
Jan 23 at 8:01
$begingroup$
@JonathanAllan Thanks :)
$endgroup$
– TFeld
Jan 23 at 8:01
add a comment |
$begingroup$
APL+WIN, 53 39 bytes
Many thanks to Adám for saving 14 bytes
(1 0↓⍉2⊥⍉m⍪0)⌽(¯1 0↓2⊥2 1 3⍉0⍪m)⊖[2]m←⎕
Try it online! Courtesy of Dyalog Classic
Prompts for input of a 3d array of the form:
4 2 3⍴1 0 1 1 0 0 1 0 1 0 1 1 0 1 1 1 1 1 1 1 0 1 1 1
which yields:
1 0 1
1 0 0
1 0 1
0 1 1
0 1 1
1 1 1
1 1 0
1 1 1
Explanation:
m←⎕ Prompt for input
(¯1 0↓2⊥2 1 3⍉0⍪m) Calculate column rotations
(1 0↓⍉2⊥⍉m⍪0) Calculate row rotations
(...)⌽(...)⊖[2]m Apply column and row rotation and output resulting 3d array:
1 1 0
1 0 0
0 0 1
1 1 1
0 1 1
1 1 1
1 1 1
1 1 0
answered Jan 22 at 14:49
GrahamGraham
2,50678
$endgroup$
$begingroup$
Instead of enclosing and using¨, just process the entire array at once. Try it online!
$endgroup$
– Adám
Jan 22 at 17:08
$begingroup$
@Adám Many thanks. I do not know why I over thought this one and went the nested route :( Getting old?
$endgroup$
– Graham
Jan 22 at 18:44
add a comment |
$begingroup$
APL+WIN, 53 39 bytes
Many thanks to Adám for saving 14 bytes
(1 0↓⍉2⊥⍉m⍪0)⌽(¯1 0↓2⊥2 1 3⍉0⍪m)⊖[2]m←⎕
Try it online! Courtesy of Dyalog Classic
Prompts for input of a 3d array of the form:
4 2 3⍴1 0 1 1 0 0 1 0 1 0 1 1 0 1 1 1 1 1 1 1 0 1 1 1
which yields:
1 0 1
1 0 0
1 0 1
0 1 1
0 1 1
1 1 1
1 1 0
1 1 1
Explanation:
m←⎕ Prompt for input
(¯1 0↓2⊥2 1 3⍉0⍪m) Calculate column rotations
(1 0↓⍉2⊥⍉m⍪0) Calculate row rotations
(...)⌽(...)⊖[2]m Apply column and row rotation and output resulting 3d array:
1 1 0
1 0 0
0 0 1
1 1 1
0 1 1
1 1 1
1 1 1
1 1 0
answered Jan 22 at 14:49
GrahamGraham
2,50678
$endgroup$
$begingroup$
Instead of enclosing and using¨, just process the entire array at once. Try it online!
$endgroup$
– Adám
Jan 22 at 17:08
$begingroup$
@Adám Many thanks. I do not know why I over thought this one and went the nested route :( Getting old?
$endgroup$
– Graham
Jan 22 at 18:44
add a comment |
$begingroup$
APL+WIN, 53 39 bytes
Many thanks to Adám for saving 14 bytes
(1 0↓⍉2⊥⍉m⍪0)⌽(¯1 0↓2⊥2 1 3⍉0⍪m)⊖[2]m←⎕
Try it online! Courtesy of Dyalog Classic
Prompts for input of a 3d array of the form:
4 2 3⍴1 0 1 1 0 0 1 0 1 0 1 1 0 1 1 1 1 1 1 1 0 1 1 1
which yields:
1 0 1
1 0 0
1 0 1
0 1 1
0 1 1
1 1 1
1 1 0
1 1 1
Explanation:
m←⎕ Prompt for input
(¯1 0↓2⊥2 1 3⍉0⍪m) Calculate column rotations
(1 0↓⍉2⊥⍉m⍪0) Calculate row rotations
(...)⌽(...)⊖[2]m Apply column and row rotation and output resulting 3d array:
1 1 0
1 0 0
0 0 1
1 1 1
0 1 1
1 1 1
1 1 1
1 1 0
answered Jan 22 at 14:49
GrahamGraham
2,50678
$endgroup$
APL+WIN, 53 39 bytes
Many thanks to Adám for saving 14 bytes
(1 0↓⍉2⊥⍉m⍪0)⌽(¯1 0↓2⊥2 1 3⍉0⍪m)⊖[2]m←⎕
Try it online! Courtesy of Dyalog Classic
Prompts for input of a 3d array of the form:
4 2 3⍴1 0 1 1 0 0 1 0 1 0 1 1 0 1 1 1 1 1 1 1 0 1 1 1
which yields:
1 0 1
1 0 0
1 0 1
0 1 1
0 1 1
1 1 1
1 1 0
1 1 1
Explanation:
m←⎕ Prompt for input
(¯1 0↓2⊥2 1 3⍉0⍪m) Calculate column rotations
(1 0↓⍉2⊥⍉m⍪0) Calculate row rotations
(...)⌽(...)⊖[2]m Apply column and row rotation and output resulting 3d array:
1 1 0
1 0 0
0 0 1
1 1 1
0 1 1
1 1 1
1 1 1
1 1 0
answered Jan 22 at 14:49
GrahamGraham
2,50678
edited Jan 22 at 18:56
answered Jan 22 at 14:49
GrahamGraham
2,50678
answered Jan 22 at 14:49
GrahamGraham
2,50678
answered Jan 22 at 14:49
GrahamGraham
2,50678
2,50678
$begingroup$
Instead of enclosing and using¨, just process the entire array at once. Try it online!
$endgroup$
– Adám
Jan 22 at 17:08
$begingroup$
@Adám Many thanks. I do not know why I over thought this one and went the nested route :( Getting old?
$endgroup$
– Graham
Jan 22 at 18:44
add a comment |
$begingroup$
Instead of enclosing and using¨, just process the entire array at once. Try it online!
$endgroup$
– Adám
Jan 22 at 17:08
$begingroup$
@Adám Many thanks. I do not know why I over thought this one and went the nested route :( Getting old?
$endgroup$
– Graham
Jan 22 at 18:44
$begingroup$
Instead of enclosing and using
¨, just process the entire array at once. Try it online!$endgroup$
– Adám
Jan 22 at 17:08
$begingroup$
Instead of enclosing and using
¨, just process the entire array at once. Try it online!$endgroup$
– Adám
Jan 22 at 17:08
$begingroup$
@Adám Many thanks. I do not know why I over thought this one and went the nested route :( Getting old?
$endgroup$
– Graham
Jan 22 at 18:44
$begingroup$
@Adám Many thanks. I do not know why I over thought this one and went the nested route :( Getting old?
$endgroup$
– Graham
Jan 22 at 18:44
add a comment |
$begingroup$
R, 226 216 205 bytes
-21 bytes thanks to digEmAll
function(a,L=`for`){d=dim(b<-a)
r=function(a,n,l=sum(a|1))a[(1:l+sum(n*2^(sum(n|1):1-1))-1)%%l+1]
L(i,I<-2:d[3],L(j,1:d,b[j,,i]<-r(b[j,,i],a[j,,i-1])))
L(i,I-1,L(k,1:d[2],b[,k,i]<-r(b[,k,i],a[,k,i+1])))
b}
Try it online!
answered Feb 23 at 1:11
ASCII-onlyASCII-only
4,3591338
$endgroup$
add a comment |
$begingroup$
R, 226 216 205 bytes
-21 bytes thanks to digEmAll
function(a,L=`for`){d=dim(b<-a)
r=function(a,n,l=sum(a|1))a[(1:l+sum(n*2^(sum(n|1):1-1))-1)%%l+1]
L(i,I<-2:d[3],L(j,1:d,b[j,,i]<-r(b[j,,i],a[j,,i-1])))
L(i,I-1,L(k,1:d[2],b[,k,i]<-r(b[,k,i],a[,k,i+1])))
b}
Try it online!
answered Feb 23 at 1:11
ASCII-onlyASCII-only
4,3591338
$endgroup$
add a comment |
$begingroup$
R, 226 216 205 bytes
-21 bytes thanks to digEmAll
function(a,L=`for`){d=dim(b<-a)
r=function(a,n,l=sum(a|1))a[(1:l+sum(n*2^(sum(n|1):1-1))-1)%%l+1]
L(i,I<-2:d[3],L(j,1:d,b[j,,i]<-r(b[j,,i],a[j,,i-1])))
L(i,I-1,L(k,1:d[2],b[,k,i]<-r(b[,k,i],a[,k,i+1])))
b}
Try it online!
answered Feb 23 at 1:11
ASCII-onlyASCII-only
4,3591338
$endgroup$
R, 226 216 205 bytes
-21 bytes thanks to digEmAll
function(a,L=`for`){d=dim(b<-a)
r=function(a,n,l=sum(a|1))a[(1:l+sum(n*2^(sum(n|1):1-1))-1)%%l+1]
L(i,I<-2:d[3],L(j,1:d,b[j,,i]<-r(b[j,,i],a[j,,i-1])))
L(i,I-1,L(k,1:d[2],b[,k,i]<-r(b[,k,i],a[,k,i+1])))
b}
Try it online!
answered Feb 23 at 1:11
ASCII-onlyASCII-only
4,3591338
edited Feb 24 at 0:22
answered Feb 23 at 1:11
ASCII-onlyASCII-only
4,3591338
answered Feb 23 at 1:11
ASCII-onlyASCII-only
4,3591338
answered Feb 23 at 1:11
ASCII-onlyASCII-only
4,3591338
4,3591338
add a comment |
add a comment |
$begingroup$
05AB1E, 41 39 bytes
εNĀiø¹N<èøJC‚øε`._}ø}N¹g<Êi¹N>èJC‚øε`._
This feels way too long.. Can definitely be golfed some more.
Try it online or verify all test cases.
Explanation:
ε # Map each layer in the (implicit) input to:
# (`N` is the layer-index of this map)
NĀi # If it is not the first layer:
ø # Zip/transpose the current layer; swapping rows/columns
¹N<è # Get the `N-1`'th layer of the input
ø # Zip/transpose; swapping rows/columns
J # Join all inner lists (the columns) together
C # And convert it from binary to integer
‚ # Pair it with the current layer's columns we're mapping
ø # Zip/transpose; to pair each integer with a layer's columns
ε } # Map over these pairs:
` # Push both values of the pair separately to the stack
._ # Rotate the column the integer amount of times
ø # Zip/transpose the rows/columns of the current layer back
} # Close the if-statement
N¹g<Êi # If this is not the last layer (layer-index-1 != amount_of_layers):
¹N>è # Get the `N+1`'th layer of the input
J # Join all inner lists (the rows) together
C # And convert it from binary to integer
‚ # Pair it with the current layer's rows we're mapping
ø # Zip/transpose; to pair each integer with a layer's rows
ε # Map over these pairs:
` # Push both values of the pair separately to the stack
._ # Rotate the row the integer amount of times
# (implicitly output the result after the layer-mapping is done)
answered Jan 22 at 9:56
Kevin CruijssenKevin Cruijssen
40.1k563206
$endgroup$
add a comment |
$begingroup$
05AB1E, 41 39 bytes
εNĀiø¹N<èøJC‚øε`._}ø}N¹g<Êi¹N>èJC‚øε`._
This feels way too long.. Can definitely be golfed some more.
Try it online or verify all test cases.
Explanation:
ε # Map each layer in the (implicit) input to:
# (`N` is the layer-index of this map)
NĀi # If it is not the first layer:
ø # Zip/transpose the current layer; swapping rows/columns
¹N<è # Get the `N-1`'th layer of the input
ø # Zip/transpose; swapping rows/columns
J # Join all inner lists (the columns) together
C # And convert it from binary to integer
‚ # Pair it with the current layer's columns we're mapping
ø # Zip/transpose; to pair each integer with a layer's columns
ε } # Map over these pairs:
` # Push both values of the pair separately to the stack
._ # Rotate the column the integer amount of times
ø # Zip/transpose the rows/columns of the current layer back
} # Close the if-statement
N¹g<Êi # If this is not the last layer (layer-index-1 != amount_of_layers):
¹N>è # Get the `N+1`'th layer of the input
J # Join all inner lists (the rows) together
C # And convert it from binary to integer
‚ # Pair it with the current layer's rows we're mapping
ø # Zip/transpose; to pair each integer with a layer's rows
ε # Map over these pairs:
` # Push both values of the pair separately to the stack
._ # Rotate the row the integer amount of times
# (implicitly output the result after the layer-mapping is done)
answered Jan 22 at 9:56
Kevin CruijssenKevin Cruijssen
40.1k563206
$endgroup$
add a comment |
$begingroup$
05AB1E, 41 39 bytes
εNĀiø¹N<èøJC‚øε`._}ø}N¹g<Êi¹N>èJC‚øε`._
This feels way too long.. Can definitely be golfed some more.
Try it online or verify all test cases.
Explanation:
ε # Map each layer in the (implicit) input to:
# (`N` is the layer-index of this map)
NĀi # If it is not the first layer:
ø # Zip/transpose the current layer; swapping rows/columns
¹N<è # Get the `N-1`'th layer of the input
ø # Zip/transpose; swapping rows/columns
J # Join all inner lists (the columns) together
C # And convert it from binary to integer
‚ # Pair it with the current layer's columns we're mapping
ø # Zip/transpose; to pair each integer with a layer's columns
ε } # Map over these pairs:
` # Push both values of the pair separately to the stack
._ # Rotate the column the integer amount of times
ø # Zip/transpose the rows/columns of the current layer back
} # Close the if-statement
N¹g<Êi # If this is not the last layer (layer-index-1 != amount_of_layers):
¹N>è # Get the `N+1`'th layer of the input
J # Join all inner lists (the rows) together
C # And convert it from binary to integer
‚ # Pair it with the current layer's rows we're mapping
ø # Zip/transpose; to pair each integer with a layer's rows
ε # Map over these pairs:
` # Push both values of the pair separately to the stack
._ # Rotate the row the integer amount of times
# (implicitly output the result after the layer-mapping is done)
answered Jan 22 at 9:56
Kevin CruijssenKevin Cruijssen
40.1k563206
$endgroup$
05AB1E, 41 39 bytes
εNĀiø¹N<èøJC‚øε`._}ø}N¹g<Êi¹N>èJC‚øε`._
This feels way too long.. Can definitely be golfed some more.
Try it online or verify all test cases.
Explanation:
ε # Map each layer in the (implicit) input to:
# (`N` is the layer-index of this map)
NĀi # If it is not the first layer:
ø # Zip/transpose the current layer; swapping rows/columns
¹N<è # Get the `N-1`'th layer of the input
ø # Zip/transpose; swapping rows/columns
J # Join all inner lists (the columns) together
C # And convert it from binary to integer
‚ # Pair it with the current layer's columns we're mapping
ø # Zip/transpose; to pair each integer with a layer's columns
ε } # Map over these pairs:
` # Push both values of the pair separately to the stack
._ # Rotate the column the integer amount of times
ø # Zip/transpose the rows/columns of the current layer back
} # Close the if-statement
N¹g<Êi # If this is not the last layer (layer-index-1 != amount_of_layers):
¹N>è # Get the `N+1`'th layer of the input
J # Join all inner lists (the rows) together
C # And convert it from binary to integer
‚ # Pair it with the current layer's rows we're mapping
ø # Zip/transpose; to pair each integer with a layer's rows
ε # Map over these pairs:
` # Push both values of the pair separately to the stack
._ # Rotate the row the integer amount of times
# (implicitly output the result after the layer-mapping is done)
answered Jan 22 at 9:56
Kevin CruijssenKevin Cruijssen
40.1k563206
edited Jan 22 at 10:10
answered Jan 22 at 9:56
Kevin CruijssenKevin Cruijssen
40.1k563206
answered Jan 22 at 9:56
Kevin CruijssenKevin Cruijssen
40.1k563206
answered Jan 22 at 9:56
Kevin CruijssenKevin Cruijssen
40.1k563206
40.1k563206
add a comment |
add a comment |
$begingroup$
Wolfram Language (Mathematica), 138 131 125 123 bytes
t=Map@Thread
m=MapThread[r=RotateLeft,#,2]&
b=(a=ArrayPad)[Map@Fold[#+##&]/@#,1]~r~#2~a~-1&
g=m@{t@m@{t@#,t@#~b~-1},#~b~1}&
Try it online!
Map[Thread]is equivalent toTranspose[a, {1,3,2}], which transposes the columns and rows.
Fold[#+##&]is shorter thanIntegerDigits[#,2]for converting from binary.
answered Jan 27 at 1:50
lirtosiastlirtosiast
18.2k438109
$endgroup$
add a comment |
$begingroup$
Wolfram Language (Mathematica), 138 131 125 123 bytes
t=Map@Thread
m=MapThread[r=RotateLeft,#,2]&
b=(a=ArrayPad)[Map@Fold[#+##&]/@#,1]~r~#2~a~-1&
g=m@{t@m@{t@#,t@#~b~-1},#~b~1}&
Try it online!
Map[Thread]is equivalent toTranspose[a, {1,3,2}], which transposes the columns and rows.
Fold[#+##&]is shorter thanIntegerDigits[#,2]for converting from binary.
answered Jan 27 at 1:50
lirtosiastlirtosiast
18.2k438109
$endgroup$
add a comment |
$begingroup$
Wolfram Language (Mathematica), 138 131 125 123 bytes
t=Map@Thread
m=MapThread[r=RotateLeft,#,2]&
b=(a=ArrayPad)[Map@Fold[#+##&]/@#,1]~r~#2~a~-1&
g=m@{t@m@{t@#,t@#~b~-1},#~b~1}&
Try it online!
Map[Thread]is equivalent toTranspose[a, {1,3,2}], which transposes the columns and rows.
Fold[#+##&]is shorter thanIntegerDigits[#,2]for converting from binary.
answered Jan 27 at 1:50
lirtosiastlirtosiast
18.2k438109
$endgroup$
Wolfram Language (Mathematica), 138 131 125 123 bytes
t=Map@Thread
m=MapThread[r=RotateLeft,#,2]&
b=(a=ArrayPad)[Map@Fold[#+##&]/@#,1]~r~#2~a~-1&
g=m@{t@m@{t@#,t@#~b~-1},#~b~1}&
Try it online!
Map[Thread]is equivalent toTranspose[a, {1,3,2}], which transposes the columns and rows.
Fold[#+##&]is shorter thanIntegerDigits[#,2]for converting from binary.
answered Jan 27 at 1:50
lirtosiastlirtosiast
18.2k438109
edited Jan 27 at 17:38
answered Jan 27 at 1:50
lirtosiastlirtosiast
18.2k438109
answered Jan 27 at 1:50
lirtosiastlirtosiast
18.2k438109
answered Jan 27 at 1:50
lirtosiastlirtosiast
18.2k438109
18.2k438109
add a comment |
add a comment |
If this is an answer to a challenge…
…Be sure to follow the challenge specification. However, please refrain from exploiting obvious loopholes. Answers abusing any of the standard loopholes are considered invalid. If you think a specification is unclear or underspecified, comment on the question instead.
…Try to optimize your score. For instance, answers to code-golf challenges should attempt to be as short as possible. You can always include a readable version of the code in addition to the competitive one.
Explanations of your answer make it more interesting to read and are very much encouraged.…Include a short header which indicates the language(s) of your code and its score, as defined by the challenge.
More generally…
…Please make sure to answer the question and provide sufficient detail.
…Avoid asking for help, clarification or responding to other answers (use comments instead).
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StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
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Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
