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    <p>This is based on Alex Martelli's answer, but it should work. It depends on the expression <code>source_node.children</code> yielding an iterable that will iterate over all the children of <code>source_node</code>. It also relies on there being a working way for the <code>==</code> operator to compare two nodes to see if they are the same. Using <code>is</code> may be a better choice. Apparently, in the library you're using, the syntax for getting an iterable over all the children is <code>graph[source_node]</code>, so you will need to adjust the code accordingly.</p> <pre><code>def allpaths(source_node, sink_node): if source_node == sink_node: # Handle trivial case return frozenset([(source_node,)]) else: result = set() for new_source in source_node.children: paths = allpaths(new_source, sink_node, memo_dict) for path in paths: path = (source_node,) + path result.add(path) result = frozenset(result) return result </code></pre> <p>My main concern is that this does a depth first search, it will waste effort when there are several paths from the source to a node that's a grandchild, great grandchild, etc all of source, but not necessarily a parent of sink. If it memoized the answer for a given source and sink node it would be possible to avoid the extra effort.</p> <p>Here is an example of how that would work:</p> <pre><code>def allpaths(source_node, sink_node, memo_dict = None): if memo_dict is None: # putting {}, or any other mutable object # as the default argument is wrong memo_dict = dict() if source_node == sink_node: # Don't memoize trivial case return frozenset([(source_node,)]) else: pair = (source_node, sink_node) if pair in memo_dict: # Is answer memoized already? return memo_dict[pair] else: result = set() for new_source in source_node.children: paths = allpaths(new_source, sink_node, memo_dict) for path in paths: path = (source_node,) + path result.add(path) result = frozenset(result) # Memoize answer memo_dict[(source_node, sink_node)] = result return result </code></pre> <p>This also allows you to save the memoization dictionary between invocations so if you need to compute the answer for multiple source and sink nodes you can avoid a lot of extra effort.</p>
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    1. POlist of all paths from source to sink in directed acyclic graph
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    1. COCool. Can you explain to me how you did this?
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    2. CO@Tyler, yeah, I'm trying to remember exactly how it went. :-)
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    3. COAwesome, this would be really useful.
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