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the unix(argv([Word|Constraints])), line makes it look like it's intended to run from the command prompt, instead of in the SWI-Prolog interface. If you replace that with getWord(Word), the program should prompt you for words instead. Then just start up SWI-Prolog, type in['filename.pl']. at the...

Assuming you're representing the sequences as lists of r,g, and b (since prolog interprets capitals as variables), and you want to accept an empty sequence, you want something like: gbr(g,b,r). accept([]). accept([X]). accept([X,Y]):-X /== Y. accept([X,Y,Z|Xs]):- X /== Y, not gbr(X,Y,Z)...

You seem to be missing rules for 3 and 4 cards in a suit. You should assign a value of 0 to those cases. Otherwise, prolog can't unify hands containing those numbers of cards, so it will say 'No'.

General hint: make sure you dig recursion Hint on previous-to: It's obviously true if X is the head of a list and Y is a member of the list. So you must first figure out a member predicate, then reduce Z to the point where X is the head. Hint on number-occurences: number-occurences(X,[],0)...

You probably want to start with some simpler predicates you can use in constructing the more complex ones, such as: / member(X,Ys) returns true if X is a member of Ys member(X,[X|_Xs]). member(X,[_Y|Ys]):-member(X,Ys). / select(X,Ys,Zs) returns true if Zs is Ys with the first occurence of X...

You should throw in a recursive step. i.e. add: oad(X,Y,D) :- distance(X,Y,D). {base case} oad(X,Y,D) :- distance(X,Z,D1), oad(Z,Y,D2), D is D1+D2.