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    <p>I suspect what's tripping you up here is that instead of getting the data out of the database, you're having to find it from within an explicit data structure. A first crack at this might look like this:</p> <pre><code>find(_, _, []). find(Node, Graph, [Node|Path]) :- member(r(Node,Adjacent), Graph), member(AdjNode, Adjacent), find(AdjNode, Graph, Path). </code></pre> <p>See how I'm using <code>member/2</code> to find data from within the graph. This solution isn't correct though, because it loops. An improvement might be this:</p> <pre><code>find(Node, Graph, Path) :- find(Node, Graph, Path, []). find(_, _, [], _). find(Node, Graph, [Node|Path], Seen) :- member(r(Node, Adjacent), Graph), member(AdjNode, Adjacent), \+ memberchk(AdjNode, Seen), find(AdjNode, Graph, Path, [Node|Seen]). </code></pre> <p>This one is basically the same as the above version except it has a "seen" list to track where it has already been. This still doesn't produce the output you want, but I think it will be enough to get you on the right track.</p> <p><strong>Edit</strong> in response to your edit,</p> <blockquote> <p>For starters, I think I'll need some kind of base case to abort the recursion.</p> </blockquote> <p>Yes. I chose your first case because I don't think you can safely "consume" the graph during traversal. I suppose you could use <code>select/3</code> in lieu of <code>member/2</code> and pass the graph-without-this-node onward. That might be an interesting thing to try (suggestion!).</p> <blockquote> <ul> <li>How do I make the program use the members of the list in the r(...) term as the next Start?</li> </ul> </blockquote> <p>As demonstrated, use <code>member/2</code> to retrieve things from the graph. It's funny, because you used the exact word for the predicate you need. :)</p> <blockquote> <ul> <li>How do I check if a node has already been "visited"/ How can I remove a node from a specific list in r</li> </ul> </blockquote> <p>As demonstrated in my second set of code, you have another parameter for your auxiliary predicate, and use <code>memberchk/3</code> or <code>member/3</code>.</p> <blockquote> <ul> <li>How do I put the found nodes into the Path list? Simply append? Or execute the recursive call with something like [Path|Start]?</li> </ul> </blockquote> <p>I went with the recursive call. <code>append/3</code> would be more expensive.</p> <p><strong>Edit</strong>: Using <code>findall/3</code> per Will's comment, we can find all the paths at once:</p> <pre><code>all_paths(From, Graph, Paths) :- findall(Path, find(From, Graph, Path), Paths). </code></pre> <p>You can invoke this like so:</p> <pre><code>?- all_paths(a, [r(a,[b,d]),r(b,[a,c,e]),r(c,[b]),r(d,[a,e]), r(e,[b,d,f]),r(f,[e,g]),r(g,[f])], AllPaths). </code></pre> <p>I haven't tested that but it should work.</p>
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    1. POFind all possible paths w/o loops in Graph in Prolog
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    1. USDaniel Lyons
    1. USDaniel Lyons
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    1. COThanks! I'll start checking this out as soon as I'm off work ;)
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    2. CONo problem. :) Let me know if I can help more.
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    3. COmy remark wasn't intended as criticism. I just thought you might add a sample `findall` call. I don't understand about the loops - yes it says so, and this is exactly what you code does. :)
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