Get product of the list items by using recursion in Prolog?
up vote
0
down vote
favorite
How do you get a product of the list items by using recursion?
If I ask:
product([s(0), s(s(0)), s(s(0))], S).
The result should be:
S = s(s(s(s(0)))).
But I geep geting wrong results. Or no results.
I tried:
product(, 0).
product(, Res).
product([H1, H2|T], Res) :- T=, mul(H1, H2, Res), product(T, Res).
product([H|T], Res) :- mul(H, Res, X), product(T, X).
mul is multiplication and it works fine.
If I use trace i can see that it finds th result but than it failes for some reason.
Call: (10) product(, s(s(s(s(0))))) ? creep
Fail: (10) product(, s(s(s(s(0))))) ? creep
Any idea anyone?
prolog successor-arithmetics
edited Nov 11 at 12:35
false
10.9k769141
asked Nov 10 at 22:49
Ishmael Black
102
add a comment |
up vote
0
down vote
favorite
How do you get a product of the list items by using recursion?
If I ask:
product([s(0), s(s(0)), s(s(0))], S).
The result should be:
S = s(s(s(s(0)))).
But I geep geting wrong results. Or no results.
I tried:
product(, 0).
product(, Res).
product([H1, H2|T], Res) :- T=, mul(H1, H2, Res), product(T, Res).
product([H|T], Res) :- mul(H, Res, X), product(T, X).
mul is multiplication and it works fine.
If I use trace i can see that it finds th result but than it failes for some reason.
Call: (10) product(, s(s(s(s(0))))) ? creep
Fail: (10) product(, s(s(s(s(0))))) ? creep
Any idea anyone?
prolog successor-arithmetics
edited Nov 11 at 12:35
false
10.9k769141
asked Nov 10 at 22:49
Ishmael Black
102
product(, Res).says that the multiplication of no elements is anything you want. That doesn't make sense.
– lurker
Nov 10 at 23:44
What are you trying to achieve withmul(H, Res, X)?
– lurker
Nov 10 at 23:45
"mul is multiplication and it works fine" - it doesn't work fine for us unless you post the code. You should always provide a Minimal, Complete, and Verifiable example.
– Enigmativity
Nov 11 at 0:09
This is a problem:mul(H1, H2, Res), product(T, Res).. This is likely to fail unless theResin the first expression is unifiable theResin the second.
– lurker
Nov 11 at 0:53
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
How do you get a product of the list items by using recursion?
If I ask:
product([s(0), s(s(0)), s(s(0))], S).
The result should be:
S = s(s(s(s(0)))).
But I geep geting wrong results. Or no results.
I tried:
product(, 0).
product(, Res).
product([H1, H2|T], Res) :- T=, mul(H1, H2, Res), product(T, Res).
product([H|T], Res) :- mul(H, Res, X), product(T, X).
mul is multiplication and it works fine.
If I use trace i can see that it finds th result but than it failes for some reason.
Call: (10) product(, s(s(s(s(0))))) ? creep
Fail: (10) product(, s(s(s(s(0))))) ? creep
Any idea anyone?
prolog successor-arithmetics
edited Nov 11 at 12:35
false
10.9k769141
asked Nov 10 at 22:49
Ishmael Black
102
How do you get a product of the list items by using recursion?
If I ask:
product([s(0), s(s(0)), s(s(0))], S).
The result should be:
S = s(s(s(s(0)))).
But I geep geting wrong results. Or no results.
I tried:
product(, 0).
product(, Res).
product([H1, H2|T], Res) :- T=, mul(H1, H2, Res), product(T, Res).
product([H|T], Res) :- mul(H, Res, X), product(T, X).
mul is multiplication and it works fine.
If I use trace i can see that it finds th result but than it failes for some reason.
Call: (10) product(, s(s(s(s(0))))) ? creep
Fail: (10) product(, s(s(s(s(0))))) ? creep
Any idea anyone?
prolog successor-arithmetics
prolog successor-arithmetics
edited Nov 11 at 12:35
false
10.9k769141
asked Nov 10 at 22:49
Ishmael Black
102
edited Nov 11 at 12:35
false
10.9k769141
asked Nov 10 at 22:49
Ishmael Black
102
edited Nov 11 at 12:35
false
10.9k769141
edited Nov 11 at 12:35
false
10.9k769141
edited Nov 11 at 12:35
false
10.9k769141
10.9k769141
asked Nov 10 at 22:49
Ishmael Black
102
asked Nov 10 at 22:49
Ishmael Black
102
asked Nov 10 at 22:49
Ishmael Black
102
102
product(, Res).says that the multiplication of no elements is anything you want. That doesn't make sense.
– lurker
Nov 10 at 23:44
What are you trying to achieve withmul(H, Res, X)?
– lurker
Nov 10 at 23:45
"mul is multiplication and it works fine" - it doesn't work fine for us unless you post the code. You should always provide a Minimal, Complete, and Verifiable example.
– Enigmativity
Nov 11 at 0:09
This is a problem:mul(H1, H2, Res), product(T, Res).. This is likely to fail unless theResin the first expression is unifiable theResin the second.
– lurker
Nov 11 at 0:53
add a comment |
product(, Res).says that the multiplication of no elements is anything you want. That doesn't make sense.
– lurker
Nov 10 at 23:44
What are you trying to achieve withmul(H, Res, X)?
– lurker
Nov 10 at 23:45
"mul is multiplication and it works fine" - it doesn't work fine for us unless you post the code. You should always provide a Minimal, Complete, and Verifiable example.
– Enigmativity
Nov 11 at 0:09
This is a problem:mul(H1, H2, Res), product(T, Res).. This is likely to fail unless theResin the first expression is unifiable theResin the second.
– lurker
Nov 11 at 0:53
product(, Res). says that the multiplication of no elements is anything you want. That doesn't make sense.– lurker
Nov 10 at 23:44
product(, Res). says that the multiplication of no elements is anything you want. That doesn't make sense.– lurker
Nov 10 at 23:44
What are you trying to achieve with
mul(H, Res, X)?– lurker
Nov 10 at 23:45
What are you trying to achieve with
mul(H, Res, X)?– lurker
Nov 10 at 23:45
"mul is multiplication and it works fine" - it doesn't work fine for us unless you post the code. You should always provide a Minimal, Complete, and Verifiable example.
– Enigmativity
Nov 11 at 0:09
"mul is multiplication and it works fine" - it doesn't work fine for us unless you post the code. You should always provide a Minimal, Complete, and Verifiable example.
– Enigmativity
Nov 11 at 0:09
This is a problem:
mul(H1, H2, Res), product(T, Res).. This is likely to fail unless the Res in the first expression is unifiable the Res in the second.– lurker
Nov 11 at 0:53
This is a problem:
mul(H1, H2, Res), product(T, Res).. This is likely to fail unless the Res in the first expression is unifiable the Res in the second.– lurker
Nov 11 at 0:53
add a comment |
1 Answer
1
active
oldest
votes
up vote
0
down vote
accepted
I think you'll find that this does the trick:
product(, 0).
product([H], H).
product([H1, H2|T], Res) :-
mul(H1, H2, H),
product([H|T], Res).
The first predicate is a trivial base case.
The second is where the list only contains a single element - so that's the answer.
The third is where you have a list of two or more elements - simply perform the mul/3 on the first two and then recursively call product/2. This will eventually match predicate 2 and complete.
Your sample input does yield s(s(s(s(0)))).
Please in the future include the definition for mul/3. I had to search the internet to find one.
%Addition
sum(0,M,M). %the sum of an integer M and 0 is M.
sum(s(N),M,s(K)) :- sum(N,M,K). %The sum of the successor of N and M is the successor of the sum of N and M.
%Multiplication
%Will work for mul(s(s(0)),s(s(0)),X) but not terminate for mul(X,Y,s(s(0)))
mul(0,M,0). %The product of 0 with any integer is 0
mul(s(N),M,P) :-
mul(N,M,K),
sum(K,M,P). %The product of the successor of N and M is the sum of M with the product of M and N. --> (N+1)*M = N*M + M
answered Nov 11 at 0:56
Enigmativity
74.4k764129
Thank you. Sorry I forgot to add mul and sum.
– Ishmael Black
Nov 11 at 12:05
@IshmaelBlack - No worries.
– Enigmativity
Nov 11 at 22:10
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
I think you'll find that this does the trick:
product(, 0).
product([H], H).
product([H1, H2|T], Res) :-
mul(H1, H2, H),
product([H|T], Res).
The first predicate is a trivial base case.
The second is where the list only contains a single element - so that's the answer.
The third is where you have a list of two or more elements - simply perform the mul/3 on the first two and then recursively call product/2. This will eventually match predicate 2 and complete.
Your sample input does yield s(s(s(s(0)))).
Please in the future include the definition for mul/3. I had to search the internet to find one.
%Addition
sum(0,M,M). %the sum of an integer M and 0 is M.
sum(s(N),M,s(K)) :- sum(N,M,K). %The sum of the successor of N and M is the successor of the sum of N and M.
%Multiplication
%Will work for mul(s(s(0)),s(s(0)),X) but not terminate for mul(X,Y,s(s(0)))
mul(0,M,0). %The product of 0 with any integer is 0
mul(s(N),M,P) :-
mul(N,M,K),
sum(K,M,P). %The product of the successor of N and M is the sum of M with the product of M and N. --> (N+1)*M = N*M + M
answered Nov 11 at 0:56
Enigmativity
74.4k764129
Thank you. Sorry I forgot to add mul and sum.
– Ishmael Black
Nov 11 at 12:05
@IshmaelBlack - No worries.
– Enigmativity
Nov 11 at 22:10
add a comment |
up vote
0
down vote
accepted
I think you'll find that this does the trick:
product(, 0).
product([H], H).
product([H1, H2|T], Res) :-
mul(H1, H2, H),
product([H|T], Res).
The first predicate is a trivial base case.
The second is where the list only contains a single element - so that's the answer.
The third is where you have a list of two or more elements - simply perform the mul/3 on the first two and then recursively call product/2. This will eventually match predicate 2 and complete.
Your sample input does yield s(s(s(s(0)))).
Please in the future include the definition for mul/3. I had to search the internet to find one.
%Addition
sum(0,M,M). %the sum of an integer M and 0 is M.
sum(s(N),M,s(K)) :- sum(N,M,K). %The sum of the successor of N and M is the successor of the sum of N and M.
%Multiplication
%Will work for mul(s(s(0)),s(s(0)),X) but not terminate for mul(X,Y,s(s(0)))
mul(0,M,0). %The product of 0 with any integer is 0
mul(s(N),M,P) :-
mul(N,M,K),
sum(K,M,P). %The product of the successor of N and M is the sum of M with the product of M and N. --> (N+1)*M = N*M + M
answered Nov 11 at 0:56
Enigmativity
74.4k764129
Thank you. Sorry I forgot to add mul and sum.
– Ishmael Black
Nov 11 at 12:05
@IshmaelBlack - No worries.
– Enigmativity
Nov 11 at 22:10
add a comment |
up vote
0
down vote
accepted
up vote
0
down vote
accepted
I think you'll find that this does the trick:
product(, 0).
product([H], H).
product([H1, H2|T], Res) :-
mul(H1, H2, H),
product([H|T], Res).
The first predicate is a trivial base case.
The second is where the list only contains a single element - so that's the answer.
The third is where you have a list of two or more elements - simply perform the mul/3 on the first two and then recursively call product/2. This will eventually match predicate 2 and complete.
Your sample input does yield s(s(s(s(0)))).
Please in the future include the definition for mul/3. I had to search the internet to find one.
%Addition
sum(0,M,M). %the sum of an integer M and 0 is M.
sum(s(N),M,s(K)) :- sum(N,M,K). %The sum of the successor of N and M is the successor of the sum of N and M.
%Multiplication
%Will work for mul(s(s(0)),s(s(0)),X) but not terminate for mul(X,Y,s(s(0)))
mul(0,M,0). %The product of 0 with any integer is 0
mul(s(N),M,P) :-
mul(N,M,K),
sum(K,M,P). %The product of the successor of N and M is the sum of M with the product of M and N. --> (N+1)*M = N*M + M
answered Nov 11 at 0:56
Enigmativity
74.4k764129
I think you'll find that this does the trick:
product(, 0).
product([H], H).
product([H1, H2|T], Res) :-
mul(H1, H2, H),
product([H|T], Res).
The first predicate is a trivial base case.
The second is where the list only contains a single element - so that's the answer.
The third is where you have a list of two or more elements - simply perform the mul/3 on the first two and then recursively call product/2. This will eventually match predicate 2 and complete.
Your sample input does yield s(s(s(s(0)))).
Please in the future include the definition for mul/3. I had to search the internet to find one.
%Addition
sum(0,M,M). %the sum of an integer M and 0 is M.
sum(s(N),M,s(K)) :- sum(N,M,K). %The sum of the successor of N and M is the successor of the sum of N and M.
%Multiplication
%Will work for mul(s(s(0)),s(s(0)),X) but not terminate for mul(X,Y,s(s(0)))
mul(0,M,0). %The product of 0 with any integer is 0
mul(s(N),M,P) :-
mul(N,M,K),
sum(K,M,P). %The product of the successor of N and M is the sum of M with the product of M and N. --> (N+1)*M = N*M + M
answered Nov 11 at 0:56
Enigmativity
74.4k764129
answered Nov 11 at 0:56
Enigmativity
74.4k764129
answered Nov 11 at 0:56
Enigmativity
74.4k764129
answered Nov 11 at 0:56
Enigmativity
74.4k764129
74.4k764129
Thank you. Sorry I forgot to add mul and sum.
– Ishmael Black
Nov 11 at 12:05
@IshmaelBlack - No worries.
– Enigmativity
Nov 11 at 22:10
add a comment |
Thank you. Sorry I forgot to add mul and sum.
– Ishmael Black
Nov 11 at 12:05
@IshmaelBlack - No worries.
– Enigmativity
Nov 11 at 22:10
Thank you. Sorry I forgot to add mul and sum.
– Ishmael Black
Nov 11 at 12:05
Thank you. Sorry I forgot to add mul and sum.
– Ishmael Black
Nov 11 at 12:05
@IshmaelBlack - No worries.
– Enigmativity
Nov 11 at 22:10
@IshmaelBlack - No worries.
– Enigmativity
Nov 11 at 22:10
add a comment |
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product(, Res).says that the multiplication of no elements is anything you want. That doesn't make sense.– lurker
Nov 10 at 23:44
What are you trying to achieve with
mul(H, Res, X)?– lurker
Nov 10 at 23:45
"mul is multiplication and it works fine" - it doesn't work fine for us unless you post the code. You should always provide a Minimal, Complete, and Verifiable example.
– Enigmativity
Nov 11 at 0:09
This is a problem:
mul(H1, H2, Res), product(T, Res).. This is likely to fail unless theResin the first expression is unifiable theResin the second.– lurker
Nov 11 at 0:53