# Create a recursive function in prolog

1. Mar 10, 2009

### Wez

can anyone help me out with this plz
i want to create a recursive function in prolog to do the following thing

sum lists(X; Y;Z) holds for lists X = [x1; : : : xn], Y = [y1; : : : yn] and
Z of numbers if and only if Z = [z1; : : : zn] and zi = xi+yi (1  i  n).

2. Mar 11, 2009

### haki

Re: Prolog

This is 'Hello World' level for lists in Prolog.

For starters create a 'function' that sums a list.

sum([1,2,3],X).
X = 6.

You should get the idea.

Do you self study Prolog?

3. Mar 11, 2009

### Wez

Re: Prolog

yeah i do.....i have that one done...but i wanted for two list

e.g [1,2,3,4] and [1,2,3,4] would output [2,4,6,8]

and for the second would be how many elements are repeated

e.g [1,2,3,4] and [1,6,7,4] would output 2

thanks

4. Mar 11, 2009

### csprof2000

Re: Prolog

First, state it in a logical manner. For instance,

Z is the sum of lists X and Y iff Z_i = X_i + Y_i. This is good, but this is not a recursive enough definition to be implemented directly in prolog. Perhaps a better way of saying the same thing would involve mentioning the first elements of the lists X, Y, and Z, and then what has to be true of the lists' tails...

So you would write:

listSum(X, Y, Z) :- < something about the lists' heads > , <something about the lists' tails>.

5. Mar 11, 2009

### haki

Re: Prolog

Are you familiar with the syntax of the lists? namely [H|T]? Notice that you can use [H|T] for your 'return' variable not just for 'incoming' variable this is what I mean:

e.g.

Code (Text):
sum([1,2,3,4,5],Sum,ReverseSteps).
Sum = 15
ReverseSteps = [15, 14, 12, 9, 5]]
you are familiar with this

Code (Text):
sum([],0).
sum([H|T],Sum):-
sum(T,Sum1),
Sum is Sum1 + H.
the only thing I can think you are not familiar is to pass something back in the list like this

Code (Text):
sum1([],0,[]).
sum1([H|T],Sum,[Sum|ReturnList]):-
sum1(T,Sum1,ReturnList),
Sum is Sum1 + H.
the solution to your original problem is similar in concept to this - how to use the return of predicate in the list. In the example it made sense to jump into recursion in first step, it is not so in your problem.

HTH.

6. Mar 11, 2009

### haki

Re: Prolog

Aside. If you are using SWI Prolog, I recommend to use the debugger for tracing the execution - it is not great but it helps.

7. May 27, 2009

### CylonMath

Re: Prolog

Code (Text):
sum(X,Y,Z) :- X == [], write(over).
sum(X,Y,Z) :- [Tmp1 | X_new] = X, [Tmp2 | Y_new] = Y, Tmp is Tmp1 + Tmp2, [ Tmp | Z ]=Z1 , sum(X_new, Y_new, Z1) .