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    <p>1) I don't think that you end in an infinite loop, but you let your program explore all ways that two people are connected, which will be a very large number. Since you probably are only interested in whether they are connected at all, you should add cuts at the end of the <code>connected/3</code> clauses that will prevent backtracking once you have successfully determined that two people are connected, e.g.:</p> <pre><code>connected(X,Y,_) :- X == Y,!. connected(X,Y,Visited) :- ancestor(X,Z), notmember(Z,Visited), connected(Z,Y,[Z|Visited]),!. ... </code></pre> <p>2) I didn't get infinitely increasing values for N when I tested your code, but still the <code>distance/3</code> predicate would determine different paths how two people are connected. Depending on the people, the minimum distance would not be the first to be computed. I would change the definition of <code>distance/3</code> to something like this:</p> <pre><code>distance(X,Y,N) :- findall(N0, distance(X,Y,[],N0), Ns), !, minimum(N, Ns). </code></pre> <p>This is reusing your <code>minimum/2</code> predicate. Note that you should add cuts to the first two clauses of this predicate in order to avoid spurious choicepoints.</p> <p>Regarding your additional questions:</p> <p>3) you have to distinguish between looking for the smallest N and looking for people matching degree N:</p> <pre><code>distance(X,Y,N) :- nonground(N),!, findall(N0, distance(X,Y,[],N0), Ns), !, minimum(N, Ns). distance(X,Y,N) :- distance(X,Y,[X],N). </code></pre> <p>Also, you need to remove the cuts that you added to <code>distance/4</code>. This is different from the cuts added to <code>connected/3</code>!</p> <p>There's probably a better way that avoids distinguishing between these two modes, but all I can think of at the moment is using some kind of breadth-first-search (in order to guarantee the minimum degree)...</p> <p>4) I don't get duplicate answers for queries like <code>?- connected(X,victoria).</code> Do you have an example?</p>
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