CODEWITHSUNDEEP
prolog
Alex Cutajar
Alex Cutajar

Reputation: 107

Solve in prolog resulting in false?

I have tried using a code similar to this link:

Solving a textual logic puzzle in Prolog - Find birthday and month

The problem I'm trying to solve is this(Telephone Conversations). http://www.cis.upenn.edu/~matuszek/cis554-2012/Assignments/prolog-01-logic-puzzle.html

My code: 

dated(Date):-
member(Date,[1928,1929,1932,1935]).
exchanged(Exchange):-
member(Exchange,[al,be,pe,sl]).

solve(X):-
X=[[gertie,Exchange1,Date1],
   [herbert,Exchange2,Date2],
   [miriam,Exchange3,Date3],
   [wallace,Exchange4,Date4]],

exchanged(Exchange1), exchanged(Exchange2), exchanged(Exchange3), exchanged(Exchange4),
Exchange1 \== Exchange2, Exchange1 \== Exchange3, Exchange1 \== Exchange4,
Exchange2 \== Exchange1, Exchange2 \== Exchange3, Exchange2 \== Exchange4,
Exchange3 \== Exchange1, Exchange3 \== Exchange2, Exchange3 \== Exchange4,
Exchange4 \== Exchange1, Exchange4 \== Exchange2, Exchange4 \== Exchange3,

dated(Date1), dated(Date2), dated(Date3), dated(Date4),
Date1 \== Date2, Date1 \== Date3, Date1 \== Date4,
Date2 \== Date1, Date2 \== Date3, Date2 \== Date4,
Date3 \== Date1, Date3 \== Date2, Date3 \== Date4,
Date4 \== Date1, Date4 \== Date2, Date4 \== Date3,

%Herbet's first exchange wasn't for BE
Exchange2 \== be,

%The Person whose first exchange was SL wasn't Getie or Herbert
Exchange1 \== sl,
Exchange2 \== sl,

%The person whose first exchange was BE didn't get the phone in 1935
member([_,be, \+1935], X),

%The person who got the first phone in 1932 didn't have an exchange for AL or BE
member([_, \+al, 1932], X),
member([_, \+be, 1932],X),

%The person who got the first phone in 1928 had an exchange for PE
member([_,pe,1929], X),

%Wallace first exchange was AL
Exchange4 == al.

My problem is this:

?- solve(X).
 false.

Upvotes: 0

Views: 172

Answers (2)

joel76
joel76

Reputation: 5645

Instead of

exchanged(Exchange1), exchanged(Exchange2), exchanged(Exchange3), exchanged(Exchange4),
Exchange1 \== Exchange2, Exchange1 \== Exchange3, Exchange1 \== Exchange4,
Exchange2 \== Exchange1, Exchange2 \== Exchange3, Exchange2 \== Exchange4,
Exchange3 \== Exchange1, Exchange3 \== Exchange2, Exchange3 \== Exchange4,
Exchange4 \== Exchange1, Exchange4 \== Exchange2, Exchange4 \== Exchange3,

dated(Date1), dated(Date2), dated(Date3), dated(Date4),
Date1 \== Date2, Date1 \== Date3, Date1 \== Date4,
Date2 \== Date1, Date2 \== Date3, Date2 \== Date4,
Date3 \== Date1, Date3 \== Date2, Date3 \== Date4,
Date4 \== Date1, Date4 \== Date2, Date4 \== Date3,

you can write

permutation([al,be,pe,sl], [Exchange1, Exchange2, Exchange3, Exchange4]),
permutation([1928,1929,1932,1935], [Date1, Date2, Date3, Date4]),

Upvotes: 0

user1812457
user1812457

Reputation:

So your problem is, your solve predicate doesn't find any solutions. This means, one of the prerequisites for finding a solution fails for all possible paths in the solution tree.

Did you actually try to search which one it is? Surely not, otherwise you would have noticed that this:

member([_,be,\+1935],X)

always fails. Why? What is \+/1? "\+ :Goal is true if Goal cannot be proven". In other words, you cannot use \+ for matching. Instead, you could write:

\+ member([_,be,1935),X).

So with all the corrections:

?- solve(X).
X = [[gertie, be, 1928], [herbert, pe, 1929], [miriam, sl, 1932], [wallace, al, 1935]] ;
false.

Assuming that the rest of the program is correct.

It is really bad to use stackoverflow as an alternative to debugging your code.

Upvotes: 1

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