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    <p>Well, to sum it up, you're doing it wrong.</p> <p>In Prolog, when we want to express a relation instead of a function (several results possible instead of one), we don't use <code>findall/3</code> and <code>member/2</code> directly. We rather state what the relation is and then maybe once it's done if we need a list of results we use <code>findall/3</code>.</p> <p>Here what it means is that we want to express the following relation:</p> <blockquote> <p>Take a list of <code>Key-Values</code> and return a list of <code>Key-[Value]</code> where <code>Value</code> is a member of the <code>Values</code> list.</p> </blockquote> <p>We could do so as follows:</p> <pre><code>% The base case: handle the empty list a_pair_list([], []). % The case where the Values list is empty, then the resulting [Value] is [] a_pair_list([Key-[]|List], [Key-[]|Result]) :- a_pair_list(List, Result). % The case where the Values list is not empty, then Value is a member of Values. a_pair_list([Key-[Not|Empty]|List], [Key-[Value]|Result]) :- member(Value, [Not|Empty]), a_pair_list(List, Result). </code></pre> <p>Once this relation is expressed, we can already obtain all the info we wish:</p> <pre><code>?- a_pair_list([1-[a, b], 2-[], 3-[c]], Result). Result = [1-[a], 2-[], 3-[c]] ; Result = [1-[b], 2-[], 3-[c]] ; false. </code></pre> <p>The desired list is now just a fairly straight-forward <code>findall/3</code> call away:</p> <pre><code>all_pairs_lists(Input, Output) :- findall(Result, a_pair_list(Input, Result), Output). </code></pre> <p>The important thing to remember is that it's way better to stay away from extra logical stuff: <code>!/0</code>, <code>findall/3</code>, etc... because it's often leading to less general programs and/or less correct ones. Here since we can express the relation stated above in a pure and clean way, we should. This way we can limit the annoying use of <code>findall/3</code> at the strict minimum.</p>
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    1. COThanks, this really made things clear. However, it still generates the same stack overflow error for large input lists. Five keys with each having a list of 50 values leads to a stack overflow.
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    2. CObut depending on your specific need it might not be useful to return the list at once. You could look into that too and skip findall entirely.
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    3. CObtw 5 keys with 50 values each is 312M lists, it's a lot of lists. Even extending the stack will leave other problems there I guess. You need to refer to my previous comment I think
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