DiGraph: nearest node that connects all paths

Question

Some background

I am analyzing a function control flowchart that basically maps incoming data to outgoing data. Most blocks are similar to the following:

if (input_variable == SPECIFIC_CONSTANT) { output_variable = TRUE; } else { output_variable = FALSE; } 

A typical control flow graph for such code is as follows

 digraph G { 2 -> 3 -> 5; 2 -> 4 -> 5; } 

Graph image

where the execution of 3 and 4 is due to the value of input_variable , but 5 is independent.

Question

Given a directed graph and the beginning of a node, how to find the nearest node, so that any path from the beginning of a node goes through that node?

Example: Given this graph , how to find 6 starting at 2 or 12 starting at 8 ?

Can the Lowest Common Ancestor algorithm be canceled and will it be effective? how

 for each node in graph: ancestors = node.get_all_ancestors() lca = find_lowest_common_ancestor(ancestors) junction_node[lca] = node get_junction_point(node): return junction_node[node] 

My programming language is Python, and I just opened NetworkX, but any algorithm would be appreciated. I'm not used to graph theory, and I think I skipped the main glossary to find what I'm looking for.

Thank you for your help!

Answer 1

Not the most effective solution, but here is something you need to get started:

Make DFS, then calculate the intersection of all paths (nodes that exist in each path). Then, among these nodes, find the one that will be closest to the beginning in each path:

 >>> paths [] >>> def dfs(G, s, path): ... if s not in G or not G[s]: ... paths.append(path) ... else: ... for t in G[s]: ... dfs({k:[_t for _t in v if _t!=t] for k,v in G.items()}, t, path+[t]) ... >>> dfs(G, 2, []) >>> paths [[3, 4, 6, 7, 12], [3, 4, 6, 8, 9, 10, 12], [3, 4, 6, 8, 9, 12], [3, 4, 6, 8, 11, 12], [3, 5, 6, 7, 12], [3, 5, 6, 8, 9, 10, 12], [3, 5, 6, 8, 9, 12], [3, 5, 6, 8, 11, 12], [4, 6, 7, 12], [4, 6, 8, 9, 10, 12], [4, 6, 8, 9, 12], [4, 6, 8, 11, 12]] >>> for path in paths: print(path) ... [3, 4, 6, 7, 12] [3, 4, 6, 8, 9, 10, 12] [3, 4, 6, 8, 9, 12] [3, 4, 6, 8, 11, 12] [3, 5, 6, 7, 12] [3, 5, 6, 8, 9, 10, 12] [3, 5, 6, 8, 9, 12] [3, 5, 6, 8, 11, 12] [4, 6, 7, 12] [4, 6, 8, 9, 10, 12] [4, 6, 8, 9, 12] [4, 6, 8, 11, 12] >>> nodes = [set(L) for L in paths] >>> commons = functools.reduce(set.intersection, nodes) >>> commons {12, 6} >>> min(commons, key=lambda v: max(L.index(v) for L in paths)) 6 

Now notice how 6 displayed in index 2 in some paths and in index 1 in some other paths. If there was a node (say, x) that appeared in index 1 on the paths where 6 is displayed at index 2, and at index 2, where 6 is displayed at index 1, this would be a connection that this algorithm would interrupt arbitrarily. Depending on your needs, you can determine how best to deal with this case.

Answer 2

You can do something like this:

For each node, find a list of all its ancestors and descendants. If size (ancestors) + size (descendants) + 1 is equal to the size of the network, this is a candidate. Now find a node with at least one ancestor and the maximum number of children.

The list of ancestors and descendants can be easily calculated. Let me know if you are not sure and I will expand my answer.

Answer 3

Reading all the solutions you propose, I came up with an idea. I give my first node the amount of 1. Recursively, all children get an equal share of this amount. In turn, they send this amount down. If the child receives a total of 1 (starting amount), then this is the “connection point”. Here is my implementation (open for comment !!).

I hope that the BFS construct limits the number of nodes visited.

 class Node(object): """ Object representing a node in a graph where we search the junction node that concentrates all paths starting from a start node. ``reference``: Attaches the real object that contains the relationships. ``initial_value``: Initial amount for the node. Typically 1 for the start node, 0 for the others. ``path``: Set of already traversed nodes that reached the node. Used to prune circular dependencies. """ def __init__(self, reference, initial_value, path=set()): self.reference = reference # See dispatch() for explaination self.value_can_dispatch = self.value_has_received = initial_value self.path = path def receive(self, value): """ Give ``value`` to the node. If the node received 1 (or more for security) in total, it will return True. Else it returns False. """ self.value_has_received += value self.value_can_dispatch += value if self.value_has_received >= 1.: return True return False def dispatch(self, children): """ Dispatch the value received to the children. Returns a filtered list of ``children`` where children involved in a circular dependency are removed. If one child signals that it has received a total of 1, the function will stop and return this one child. """ # Filter successors that are in the path used to access this node so to cut # out cycles true_successors = [child for child in children if child not in self.path] # Cut the received value into equal pieces amount = self.value_can_dispatch/len(true_successors) # We transmit only the value received after the last time it was dispatched # because paths may lead to the same node at different iterations (path # through one node may be longer than through another) and thus the same # node can be asked to dispatch to its children more than once. # The total amount of received value is kept in value_has_received because # the node may receive the complete amount in several steps. Thus, part of # its value may have been dispatched before we notice that the node received # the total amount of 1. self.value_can_dispatch = Fraction(0) for child in true_successors: # If child signaled that he received 1, then raise the winner if child.receive(amount): return child return set(true_successors) def touch(self, other_node): """ "Touches" a node with another, notifying that the node is reachable through another path than the known ones. It adds the elements of the new path as ancestors of the node. """ self.path |= other_node.path | {other_node} def make_child(self, reference): """ Creates a child of the node, pointing to reference. The child receives its path from the current node. """ # This is were the algorithm can go mad. If child is accessed through two # paths, the algorithm will only protect recursion into the first # path. If the successors recurse into the second path, we will not detect # it. => We should update the child path when a second path reaches it. return self.__class__(reference, Fraction(0), self.path | {self}) def __repr__(self): return "<{} {}>".format(self.__class__.__name__, self.reference) def find_junction_node(first_reference, get_uid, get_successors, max_iterations=100): """ Find the junction node of all paths starting from ``first_reference`` in a directed graph. ``get_uid`` is a function accepting a reference to a node in your graph and returning a unique identifier for this reference. ``get_successors`` is a function accepting a reference to a node in your graph. It should return a list of references to all its the children nodes. It may return None if the node has no child. ``max_iterations`` limits the number of pass the algorithm use to find the junction node. If reached, the funciton raises a RuntimeError. Returns ``(jp, ln)`` where ``jp`` is a reference to a node in your graph which is the junction node and ``ln`` the list of nodes in the subgraph between the start node and the junction node. """ # Mapping to already created nodes nodes = {} # Initialise first node with an amount of 1 node = Node(first_reference, Fraction(1, 1)) nodes[get_uid(first_reference)] = node # Initialise first iteration of DFS successors = set() successors.add(node) # Max iteration provides security as I'm not sure the algorithm cannot loop for i in range(max_iterations): next_successors = set() # Process one level of nodes for node in successors: # Find successors in data graph sub_references = get_successors(node.reference) # This happens when we reach the end of the graph, node has no children if sub_references is None: continue # Make a list of Node that are children of node children = set() for reference in sub_references: uid = get_uid(reference) # Does it exist? child = nodes.get(uid, None) if not child: child = node.make_child(reference) nodes[uid] = child else: child.touch(node) children.add(child) # Dispatch the value of node equally between its children result = node.dispatch(children) #print("Children of {}: {!r}".format(node, result)) # DEBUG # If one child received a total of 1 from its parents, it is common to # all paths if isinstance(result, Node): return result.reference, [node.reference for node in result.path] # Else, add the filtered list of children to the set of node to process # in the next level else: next_successors |= result successors = next_successors # Reached end of graph by all paths without finding a junction point if len(successors) == 0: return None raise RuntimeError("Max iteration reached")