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    <p>building on the excellently clear code by Daniel Lyons, a breadth first search:</p> <pre class="lang-prolog prettyprint-override"><code>all_paths(Node, Graph, Paths) :- bfs(Graph, [[Node]-[]], R, []), % or dfs(...) maplist( fst, Paths, R). fst(A, A-_). % utility pair(B, A, A-B). % helpers add(LS,H,[H|LS]). % bfs(_G, [], Z, Z). % queue is empty bfs(Graph, [H|Q], [H|R], Z) :- H = Path-Seen, Path = [Node|_], findall( Next, member(r(Node, Next), Graph), NS), flatten_diff( NS, Seen, WS), % working set of nodes maplist( add(Path), WS, PS), % new paths maplist( pair([Node|Seen]), PS, QH), % new addition to the queue %% append( QH, Q, Q2), % DFS append( Q, QH, Q2), % BFS bfs(Graph, Q2, R, Z). </code></pre> <p>(not tested). <code>flatten_diff(A,B,C)</code> should flatten list of lists <code>A</code>, while removing the elements of it that appear also in list <code>B</code>, producing the list <code>C</code> as the result. </p> <hr> <p>As PeterPanter has noticed, Daniel Lyons's code needs a little tweaking, to not exclude the very last node in its resulting paths.</p> <pre class="lang-prolog prettyprint-override"><code>find(Node, Graph, [Node|Path]) :- find(Node, Graph, Path, []). find(_, _, [], _). find(Node, Graph, [AdjNode|Path], Seen) :- member(r(Node, Adjacent), Graph), member(AdjNode, Adjacent), \+ memberchk(AdjNode, Seen), find(AdjNode, Graph, Path, [Node|Seen]). </code></pre> <p>There are no empty paths produced now, and it works as expected:</p> <pre class="lang-prolog prettyprint-override"><code>11 ?- find(a,[r(a,[b,d]),r(b,[a,c,e]),r(c,[b]), r(d,[a,g]), r(e,[b,d,f]),r(f,[e,g]),r(g,[f])], Path). Path = [a] ; Path = [a, b] ; Path = [a, b, c] ; Path = [a, b, e] ; Path = [a, b, e, d] ; Path = [a, b, e, d, g] ; Path = [a, b, e, d, g, f] ; Path = [a, b, e, f] ; Path = [a, b, e, f, g] ; Path = [a, d] ; Path = [a, d, g] ; Path = [a, d, g, f] ; Path = [a, d, g, f, e] ; Path = [a, d, g, f, e, b] ; Path = [a, d, g, f, e, b, c] ; false. </code></pre>
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    1. POFind all possible paths w/o loops in Graph in Prolog
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    1. COThanks! Fortunately I sat down with a friend of mine today and we figured out something by ourselves. However, I'm still trying to modify your code to get the output demanded by the assignment: only terminal paths. I'm trying to implement an additional terminalcheck predicate, which is supposed to check whether a node in Path has only adjacent nodes already in Seen or not. ` %terminalchk([]). %terminalchk([First|Tail], Graph, Seen) :- % member(r(First,Adjacent), Graph), % member(AdjNode, Adjacent), % memberchk(AdjNode, Seen), % terminalchk(Tail).` is my approach, but it won't work...
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    2. CO@PeterPanter A. rep 1: ability to *accept* an answer to your question. :) B. rep 15: ability to *up-vote* a "good" answer. :) C. what is "terminal path"? Is it a path which ends on a node from which there are no more moves possible?
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    3. COa related answer by CapelliC: http://stackoverflow.com/a/16467644/849891 .
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