Mixing
How do I make a primitive recursive function that does division?
$begingroup$
I am trying to define a primitive recursive function that does division. I looked at this answer but it seems wrong to me, because according to Wikipedia:
The primitive recursive functions are among the number-theoretic functions, which are functions from the natural numbers (nonnegative integers) {0, 1, 2, ...} to the natural numbers
So the inequality x−t⋅y≥0 will always be true and the function will always keep adding +1. The function given in the answers seems right but only assuming that I have negative numbers. Now how could I build a PRF with just natural numbers?
EDIT: I found a way to either make a division that always rounds up or always rounds down. But I haven't found one yet that always does the correct thing. So far:
Div(x,y,0) = 0
Div(x,y,S(m) = A(Div(x,y,m),V(D(x,M(y,S(m)))))
where S(m) is successor, A is addition, V is 0 if 0 and 1 otherwise, D is subtraction and M is multiplication.
Now the above always rounds down and the next one always rounds up:
Div(x,y,0) = 0
Div(x,y,S(m) = A(Div(x,y,m),V(D(x,M(y,m))))
functions
edited Nov 10 '17 at 5:27
J. M. is not a mathematician
61.5k5152290
asked May 21 '17 at 8:52
HakaishinHakaishin
1166
$endgroup$
add a comment |
$begingroup$
I am trying to define a primitive recursive function that does division. I looked at this answer but it seems wrong to me, because according to Wikipedia:
The primitive recursive functions are among the number-theoretic functions, which are functions from the natural numbers (nonnegative integers) {0, 1, 2, ...} to the natural numbers
So the inequality x−t⋅y≥0 will always be true and the function will always keep adding +1. The function given in the answers seems right but only assuming that I have negative numbers. Now how could I build a PRF with just natural numbers?
EDIT: I found a way to either make a division that always rounds up or always rounds down. But I haven't found one yet that always does the correct thing. So far:
Div(x,y,0) = 0
Div(x,y,S(m) = A(Div(x,y,m),V(D(x,M(y,S(m)))))
where S(m) is successor, A is addition, V is 0 if 0 and 1 otherwise, D is subtraction and M is multiplication.
Now the above always rounds down and the next one always rounds up:
Div(x,y,0) = 0
Div(x,y,S(m) = A(Div(x,y,m),V(D(x,M(y,m))))
functions
edited Nov 10 '17 at 5:27
J. M. is not a mathematician
61.5k5152290
asked May 21 '17 at 8:52
HakaishinHakaishin
1166
$endgroup$
add a comment |
$begingroup$
I am trying to define a primitive recursive function that does division. I looked at this answer but it seems wrong to me, because according to Wikipedia:
The primitive recursive functions are among the number-theoretic functions, which are functions from the natural numbers (nonnegative integers) {0, 1, 2, ...} to the natural numbers
So the inequality x−t⋅y≥0 will always be true and the function will always keep adding +1. The function given in the answers seems right but only assuming that I have negative numbers. Now how could I build a PRF with just natural numbers?
EDIT: I found a way to either make a division that always rounds up or always rounds down. But I haven't found one yet that always does the correct thing. So far:
Div(x,y,0) = 0
Div(x,y,S(m) = A(Div(x,y,m),V(D(x,M(y,S(m)))))
where S(m) is successor, A is addition, V is 0 if 0 and 1 otherwise, D is subtraction and M is multiplication.
Now the above always rounds down and the next one always rounds up:
Div(x,y,0) = 0
Div(x,y,S(m) = A(Div(x,y,m),V(D(x,M(y,m))))
functions
edited Nov 10 '17 at 5:27
J. M. is not a mathematician
61.5k5152290
asked May 21 '17 at 8:52
HakaishinHakaishin
1166
$endgroup$
I am trying to define a primitive recursive function that does division. I looked at this answer but it seems wrong to me, because according to Wikipedia:
The primitive recursive functions are among the number-theoretic functions, which are functions from the natural numbers (nonnegative integers) {0, 1, 2, ...} to the natural numbers
So the inequality x−t⋅y≥0 will always be true and the function will always keep adding +1. The function given in the answers seems right but only assuming that I have negative numbers. Now how could I build a PRF with just natural numbers?
EDIT: I found a way to either make a division that always rounds up or always rounds down. But I haven't found one yet that always does the correct thing. So far:
Div(x,y,0) = 0
Div(x,y,S(m) = A(Div(x,y,m),V(D(x,M(y,S(m)))))
where S(m) is successor, A is addition, V is 0 if 0 and 1 otherwise, D is subtraction and M is multiplication.
Now the above always rounds down and the next one always rounds up:
Div(x,y,0) = 0
Div(x,y,S(m) = A(Div(x,y,m),V(D(x,M(y,m))))
functions
functions
edited Nov 10 '17 at 5:27
J. M. is not a mathematician
61.5k5152290
asked May 21 '17 at 8:52
HakaishinHakaishin
1166
edited Nov 10 '17 at 5:27
J. M. is not a mathematician
61.5k5152290
asked May 21 '17 at 8:52
HakaishinHakaishin
1166
edited Nov 10 '17 at 5:27
J. M. is not a mathematician
61.5k5152290
edited Nov 10 '17 at 5:27
J. M. is not a mathematician
61.5k5152290
edited Nov 10 '17 at 5:27
J. M. is not a mathematician
61.5k5152290
61.5k5152290
asked May 21 '17 at 8:52
HakaishinHakaishin
1166
asked May 21 '17 at 8:52
HakaishinHakaishin
1166
asked May 21 '17 at 8:52
HakaishinHakaishin
1166
1166
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Definition by cases is a valid, derived principle of definition for primitive recursive functions. So is subtraction, and so is equality. I will therefore use them freely.
Moreover, it is a good idea to define not just integer division $d(m,n)$ but also the remainder function $r(m,n)$. One can then write
begin{align}
r(0,n) & = 0 \
r(m+1,n) & = begin{cases} 0 & text{if}; n-1 = r(m,n) \ r(m,n) + 1 &text{otherwise} end{cases}
end{align}
and define integer division by
begin{align*}
d(0,n) & = 0 \
d(m+1,n) & = begin{cases} d(m,n) + 1 & text{if}; r(m,n) = n-1 \
d(m,n) & text{otherwise} end{cases}
end{align*}
edited Apr 1 '18 at 9:04
Community♦
1
answered May 25 '17 at 12:07
Hans HüttelHans Hüttel
3,3572921
$endgroup$
add a comment |
$begingroup$
If y has to divide x such that x>y,
then, initially assuming i=0;
x/y=f(x,y,0);
f(x,y,i)=f(x-y,y,i+1)
quotient =i and remainder =x-y (finally, such that when
y>(x-y)) ,
the recursive step has to be stopped.
Example 1:
25/4=f(25,4,0)
=f(21,4,1)
=f(17,4,2)
=f(13,4,3)
=f(9,4,4)
=f(5,4,5)
=f(1,4,6)
Hence quotient=i=6 and reaminder=x-y=1.
i.e, 25=6*4+1!
Example 2:
30/8=f(30,8,0)
=f(22,8,1)
=f(14,8,2)
=f(6,8,3)
Since 6<8, we stop here.
Therefore 30=3*8+6!!
answered May 21 '17 at 10:15
BabuBabu
1116
$endgroup$
$begingroup$
I am not sure this really answers my question. I know how to divide recursively. But I want to define a primitive recursive function with the form A(x,y,0) = something and A(x,y,i+1) = something depending on A
$endgroup$
– Hakaishin
May 21 '17 at 10:42
add a comment |
$begingroup$
I found a solution to my problem. The main issue was to figure out if $a - b geq 0.$ One has to use a little trick to figure this out. Below is the function I used for it and the final primitive recursive function that does division.
$a-b geq 0$
$BOE(a,0) = 1$
$BOE(a,S(m)) = A(V(D(a,S(m))),E(a,S(m)))$
with $A$ being Addition, $V(0) = 0$ else $1$, $D$ being subtraction, $E$ being equals, $S$ being successor and $M$ being multiplication.
$$ Div(x,y,0) = 0$$
$$Div(x,y,S(m)) = A(A(Div(x,y,m),V(D(x,M(y,S(m))))),E(D(x,M(m,y)),D(M(m,y),x)))$$
edited Jan 28 at 11:04
E.Nole
199114
answered May 25 '17 at 9:15
HakaishinHakaishin
1166
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Definition by cases is a valid, derived principle of definition for primitive recursive functions. So is subtraction, and so is equality. I will therefore use them freely.
Moreover, it is a good idea to define not just integer division $d(m,n)$ but also the remainder function $r(m,n)$. One can then write
begin{align}
r(0,n) & = 0 \
r(m+1,n) & = begin{cases} 0 & text{if}; n-1 = r(m,n) \ r(m,n) + 1 &text{otherwise} end{cases}
end{align}
and define integer division by
begin{align*}
d(0,n) & = 0 \
d(m+1,n) & = begin{cases} d(m,n) + 1 & text{if}; r(m,n) = n-1 \
d(m,n) & text{otherwise} end{cases}
end{align*}
edited Apr 1 '18 at 9:04
Community♦
1
answered May 25 '17 at 12:07
Hans HüttelHans Hüttel
3,3572921
$endgroup$
add a comment |
$begingroup$
Definition by cases is a valid, derived principle of definition for primitive recursive functions. So is subtraction, and so is equality. I will therefore use them freely.
Moreover, it is a good idea to define not just integer division $d(m,n)$ but also the remainder function $r(m,n)$. One can then write
begin{align}
r(0,n) & = 0 \
r(m+1,n) & = begin{cases} 0 & text{if}; n-1 = r(m,n) \ r(m,n) + 1 &text{otherwise} end{cases}
end{align}
and define integer division by
begin{align*}
d(0,n) & = 0 \
d(m+1,n) & = begin{cases} d(m,n) + 1 & text{if}; r(m,n) = n-1 \
d(m,n) & text{otherwise} end{cases}
end{align*}
edited Apr 1 '18 at 9:04
Community♦
1
answered May 25 '17 at 12:07
Hans HüttelHans Hüttel
3,3572921
$endgroup$
add a comment |
$begingroup$
Definition by cases is a valid, derived principle of definition for primitive recursive functions. So is subtraction, and so is equality. I will therefore use them freely.
Moreover, it is a good idea to define not just integer division $d(m,n)$ but also the remainder function $r(m,n)$. One can then write
begin{align}
r(0,n) & = 0 \
r(m+1,n) & = begin{cases} 0 & text{if}; n-1 = r(m,n) \ r(m,n) + 1 &text{otherwise} end{cases}
end{align}
and define integer division by
begin{align*}
d(0,n) & = 0 \
d(m+1,n) & = begin{cases} d(m,n) + 1 & text{if}; r(m,n) = n-1 \
d(m,n) & text{otherwise} end{cases}
end{align*}
edited Apr 1 '18 at 9:04
Community♦
1
answered May 25 '17 at 12:07
Hans HüttelHans Hüttel
3,3572921
$endgroup$
Definition by cases is a valid, derived principle of definition for primitive recursive functions. So is subtraction, and so is equality. I will therefore use them freely.
Moreover, it is a good idea to define not just integer division $d(m,n)$ but also the remainder function $r(m,n)$. One can then write
begin{align}
r(0,n) & = 0 \
r(m+1,n) & = begin{cases} 0 & text{if}; n-1 = r(m,n) \ r(m,n) + 1 &text{otherwise} end{cases}
end{align}
and define integer division by
begin{align*}
d(0,n) & = 0 \
d(m+1,n) & = begin{cases} d(m,n) + 1 & text{if}; r(m,n) = n-1 \
d(m,n) & text{otherwise} end{cases}
end{align*}
edited Apr 1 '18 at 9:04
Community♦
1
answered May 25 '17 at 12:07
Hans HüttelHans Hüttel
3,3572921
edited Apr 1 '18 at 9:04
Community♦
1
edited Apr 1 '18 at 9:04
Community♦
1
edited Apr 1 '18 at 9:04
Community♦
1
1
answered May 25 '17 at 12:07
Hans HüttelHans Hüttel
3,3572921
answered May 25 '17 at 12:07
Hans HüttelHans Hüttel
3,3572921
answered May 25 '17 at 12:07
Hans HüttelHans Hüttel
3,3572921
3,3572921
add a comment |
add a comment |
$begingroup$
If y has to divide x such that x>y,
then, initially assuming i=0;
x/y=f(x,y,0);
f(x,y,i)=f(x-y,y,i+1)
quotient =i and remainder =x-y (finally, such that when
y>(x-y)) ,
the recursive step has to be stopped.
Example 1:
25/4=f(25,4,0)
=f(21,4,1)
=f(17,4,2)
=f(13,4,3)
=f(9,4,4)
=f(5,4,5)
=f(1,4,6)
Hence quotient=i=6 and reaminder=x-y=1.
i.e, 25=6*4+1!
Example 2:
30/8=f(30,8,0)
=f(22,8,1)
=f(14,8,2)
=f(6,8,3)
Since 6<8, we stop here.
Therefore 30=3*8+6!!
answered May 21 '17 at 10:15
BabuBabu
1116
$endgroup$
$begingroup$
I am not sure this really answers my question. I know how to divide recursively. But I want to define a primitive recursive function with the form A(x,y,0) = something and A(x,y,i+1) = something depending on A
$endgroup$
– Hakaishin
May 21 '17 at 10:42
add a comment |
$begingroup$
If y has to divide x such that x>y,
then, initially assuming i=0;
x/y=f(x,y,0);
f(x,y,i)=f(x-y,y,i+1)
quotient =i and remainder =x-y (finally, such that when
y>(x-y)) ,
the recursive step has to be stopped.
Example 1:
25/4=f(25,4,0)
=f(21,4,1)
=f(17,4,2)
=f(13,4,3)
=f(9,4,4)
=f(5,4,5)
=f(1,4,6)
Hence quotient=i=6 and reaminder=x-y=1.
i.e, 25=6*4+1!
Example 2:
30/8=f(30,8,0)
=f(22,8,1)
=f(14,8,2)
=f(6,8,3)
Since 6<8, we stop here.
Therefore 30=3*8+6!!
answered May 21 '17 at 10:15
BabuBabu
1116
$endgroup$
$begingroup$
I am not sure this really answers my question. I know how to divide recursively. But I want to define a primitive recursive function with the form A(x,y,0) = something and A(x,y,i+1) = something depending on A
$endgroup$
– Hakaishin
May 21 '17 at 10:42
add a comment |
$begingroup$
If y has to divide x such that x>y,
then, initially assuming i=0;
x/y=f(x,y,0);
f(x,y,i)=f(x-y,y,i+1)
quotient =i and remainder =x-y (finally, such that when
y>(x-y)) ,
the recursive step has to be stopped.
Example 1:
25/4=f(25,4,0)
=f(21,4,1)
=f(17,4,2)
=f(13,4,3)
=f(9,4,4)
=f(5,4,5)
=f(1,4,6)
Hence quotient=i=6 and reaminder=x-y=1.
i.e, 25=6*4+1!
Example 2:
30/8=f(30,8,0)
=f(22,8,1)
=f(14,8,2)
=f(6,8,3)
Since 6<8, we stop here.
Therefore 30=3*8+6!!
answered May 21 '17 at 10:15
BabuBabu
1116
$endgroup$
If y has to divide x such that x>y,
then, initially assuming i=0;
x/y=f(x,y,0);
f(x,y,i)=f(x-y,y,i+1)
quotient =i and remainder =x-y (finally, such that when
y>(x-y)) ,
the recursive step has to be stopped.
Example 1:
25/4=f(25,4,0)
=f(21,4,1)
=f(17,4,2)
=f(13,4,3)
=f(9,4,4)
=f(5,4,5)
=f(1,4,6)
Hence quotient=i=6 and reaminder=x-y=1.
i.e, 25=6*4+1!
Example 2:
30/8=f(30,8,0)
=f(22,8,1)
=f(14,8,2)
=f(6,8,3)
Since 6<8, we stop here.
Therefore 30=3*8+6!!
answered May 21 '17 at 10:15
BabuBabu
1116
answered May 21 '17 at 10:15
BabuBabu
1116
answered May 21 '17 at 10:15
BabuBabu
1116
answered May 21 '17 at 10:15
BabuBabu
1116
1116
$begingroup$
I am not sure this really answers my question. I know how to divide recursively. But I want to define a primitive recursive function with the form A(x,y,0) = something and A(x,y,i+1) = something depending on A
$endgroup$
– Hakaishin
May 21 '17 at 10:42
add a comment |
$begingroup$
I am not sure this really answers my question. I know how to divide recursively. But I want to define a primitive recursive function with the form A(x,y,0) = something and A(x,y,i+1) = something depending on A
$endgroup$
– Hakaishin
May 21 '17 at 10:42
$begingroup$
I am not sure this really answers my question. I know how to divide recursively. But I want to define a primitive recursive function with the form A(x,y,0) = something and A(x,y,i+1) = something depending on A
$endgroup$
– Hakaishin
May 21 '17 at 10:42
$begingroup$
I am not sure this really answers my question. I know how to divide recursively. But I want to define a primitive recursive function with the form A(x,y,0) = something and A(x,y,i+1) = something depending on A
$endgroup$
– Hakaishin
May 21 '17 at 10:42
add a comment |
$begingroup$
I found a solution to my problem. The main issue was to figure out if $a - b geq 0.$ One has to use a little trick to figure this out. Below is the function I used for it and the final primitive recursive function that does division.
$a-b geq 0$
$BOE(a,0) = 1$
$BOE(a,S(m)) = A(V(D(a,S(m))),E(a,S(m)))$
with $A$ being Addition, $V(0) = 0$ else $1$, $D$ being subtraction, $E$ being equals, $S$ being successor and $M$ being multiplication.
$$ Div(x,y,0) = 0$$
$$Div(x,y,S(m)) = A(A(Div(x,y,m),V(D(x,M(y,S(m))))),E(D(x,M(m,y)),D(M(m,y),x)))$$
edited Jan 28 at 11:04
E.Nole
199114
answered May 25 '17 at 9:15
HakaishinHakaishin
1166
$endgroup$
add a comment |
$begingroup$
I found a solution to my problem. The main issue was to figure out if $a - b geq 0.$ One has to use a little trick to figure this out. Below is the function I used for it and the final primitive recursive function that does division.
$a-b geq 0$
$BOE(a,0) = 1$
$BOE(a,S(m)) = A(V(D(a,S(m))),E(a,S(m)))$
with $A$ being Addition, $V(0) = 0$ else $1$, $D$ being subtraction, $E$ being equals, $S$ being successor and $M$ being multiplication.
$$ Div(x,y,0) = 0$$
$$Div(x,y,S(m)) = A(A(Div(x,y,m),V(D(x,M(y,S(m))))),E(D(x,M(m,y)),D(M(m,y),x)))$$
edited Jan 28 at 11:04
E.Nole
199114
answered May 25 '17 at 9:15
HakaishinHakaishin
1166
$endgroup$
add a comment |
$begingroup$
I found a solution to my problem. The main issue was to figure out if $a - b geq 0.$ One has to use a little trick to figure this out. Below is the function I used for it and the final primitive recursive function that does division.
$a-b geq 0$
$BOE(a,0) = 1$
$BOE(a,S(m)) = A(V(D(a,S(m))),E(a,S(m)))$
with $A$ being Addition, $V(0) = 0$ else $1$, $D$ being subtraction, $E$ being equals, $S$ being successor and $M$ being multiplication.
$$ Div(x,y,0) = 0$$
$$Div(x,y,S(m)) = A(A(Div(x,y,m),V(D(x,M(y,S(m))))),E(D(x,M(m,y)),D(M(m,y),x)))$$
edited Jan 28 at 11:04
E.Nole
199114
answered May 25 '17 at 9:15
HakaishinHakaishin
1166
$endgroup$
I found a solution to my problem. The main issue was to figure out if $a - b geq 0.$ One has to use a little trick to figure this out. Below is the function I used for it and the final primitive recursive function that does division.
$a-b geq 0$
$BOE(a,0) = 1$
$BOE(a,S(m)) = A(V(D(a,S(m))),E(a,S(m)))$
with $A$ being Addition, $V(0) = 0$ else $1$, $D$ being subtraction, $E$ being equals, $S$ being successor and $M$ being multiplication.
$$ Div(x,y,0) = 0$$
$$Div(x,y,S(m)) = A(A(Div(x,y,m),V(D(x,M(y,S(m))))),E(D(x,M(m,y)),D(M(m,y),x)))$$
edited Jan 28 at 11:04
E.Nole
199114
answered May 25 '17 at 9:15
HakaishinHakaishin
1166
edited Jan 28 at 11:04
E.Nole
199114
edited Jan 28 at 11:04
E.Nole
199114
edited Jan 28 at 11:04
E.Nole
199114
199114
answered May 25 '17 at 9:15
HakaishinHakaishin
1166
answered May 25 '17 at 9:15
HakaishinHakaishin
1166
answered May 25 '17 at 9:15
HakaishinHakaishin
1166
1166
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