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without using CLP(FD), you should used combinatorial ability of Prolog expressing the problem and the constraints in appropriate way. For instance <pre><code>puzzle(Name) :- L = [[mason, Pos1, Blond1, Handsome1, UnScared1], [alex, Pos2, Blond2, Handsome2, UnScared2], [steve, Pos3, Blond3, Handsome3, UnScared3], [simon, Pos4, Blond4, Handsome4, UnScared4] ], permutation([1,2,3,4], [Pos1,Pos2,Pos3,Pos4]), maplist(yn, [Blond1, Handsome1, UnScared1, Blond2, Handsome2, UnScared2, Blond3, Handsome3, UnScared3, Blond4, Handsome4, UnScared4 ]), ... </code></pre> Each variable (those symbols starting Uppercase!) is an attribute of a person, and can assume a value from the domain. yn/1 it's a service fact, allows those binary values to assume either yes or no: <pre><code>yn(y). yn(n). </code></pre> The constraints can then be expressed in this way (just the first here) <pre><code>... % Two of them who are not blond are standing on either side of Mason. member([mason, I1, _,_,_], L), member([_, I2, n,_,_], L), member([_, I3, n,_,_], L), (I2>I1, I3>I1 ; I2<I1, I3<I1), ... </code></pre> and the solution will be <pre><code> % One of them is blond, handsome, and unscarred. member([Name, _, y, y, y], L). </code></pre> I'm not sure I understand every constraint (in English), indeed my program doesn't find a solution. The program is rather slow, and calls for CLP(FD). Edit your question (add appropriate tag, for instance), if you are interested in a CLP(FD) solution.
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