Find longest path less than or equal to given value of an acyclic, directed graph in Python

Question

So let's say I have an acyclic, directed graph G with nodes a0, a1, a2, b0, b1, b2, c0, c1, c2 and I know both the outgoing distance for each node to any of it's neighbors. What algorithm can I use to find the longest path between any two nodes that is less than a given length?

EDIT: I have looked at the Wikipedia page for the Longest path problem but I don't know whether to use the algorithm outlined in the 'Acyclic graphs and critical paths' section or the one in the 'Parameterized complexity' section. I'd have problems with either case. In the former, the algorithm requires you to have the incoming distance (I have the outgoing distance) and to be honest, the description in the latter goes over my head as far as implementation.

EDIT2: My completed graph is as follows

graph = {
             "a0": set(["a1", "b0])
             "a1": set(["a2", "b1"])
             "a2": set(["b2"])
             "b0": set(["b1", "c0"])
             "b1": set(["b2", "c1"])
             "b2": set(["c2"])
             "c0": set(["c1"])
             "c1": set(["c2"])
             "c2": set([])
}

I also have the outgoing distance in a separate dictionary e.g.

edge_distance = {
                 "a0": 0
                 "a1": 4
                 "a2": 12
                 "b0": 2
                 "b1": 1
                 "b2": 4
                 "c0": 7
                 "c1": 2
                 "c2": 1
}

Technically the only node for which I have the incoming distance is c2

The path I want to find is a0 to c2

EDIT3: My graph, toposorted:

   c2 b2 c1 a2 b1 c0 a1 b0 a0

Answer 1

'''

This meets the problem spec as given, except that it shows all distances between two points. It is trivial to choose the largest one that is below some threshold.

It is not at all optimized, and is not general, in that the problem spec as given has a single distance for every edge leaving any particular node. Nonetheless, it should be easy to modify to make more general.

Also, the major variables are global, but this is easily remedied as well.

NOTE: I am not sure the problem spec is quite correct, because the distance out of a0 is given as 0, but the distance out of c2 (which doesn't go connect anywhere) is given as 1. In any case the scenario (and the example code) is quite contrived, because it is highly unlikely that all edges leaving a given node would have the same length in real life.

'''

if 1:

import collections

# Data given in problem description

# Keys are starting nodes, values are destination nodes.

graph = {
            "a0": set(["a1", "b0"]),
            "a1": set(["a2", "b1"]),
            "a2": set(["b2"]),
            "b0": set(["b1", "c0"]),
            "b1": set(["b2", "c1"]),
            "b2": set(["c2"]),
            "c0": set(["c1"]),
            "c1": set(["c2"]),
            "c2": set([]),
}

# Length of every outbound edge from the node

edge_distance = {
                "a0": 0,
                "a1": 4,
                "a2": 12,
                "b0": 2,
                "b1": 1,
                "b2": 4,
                "c0": 7,
                "c1": 2,
                "c2": 1,
                }

def sort_graph():
    ''' Develop a sorted list using Kahn's algorithm
    '''
    # Number of incoming vertices to each node
    incoming = collections.defaultdict(int)

    #All nodes with vertices exiting them
    startnodes = set()

    # All nodes with vertices entering them
    endnodes = set()

    for start in graph:
        for end in graph[start]:
            startnodes.add(start)
            endnodes.add(end)
            incoming[end] += 1

    ordered = []
    startnodes -= endnodes
    while startnodes:
        start = startnodes.pop()
        ordered.append(start)
        for end in graph[start]:
            incoming[end] -= 1
            if not incoming[end]:
                startnodes.add(end)
    if sum(incoming.values()):
        raise ValueError("Graph has at least one cycle")

    return ordered

ordered = sort_graph()

def calc_dists(start):
    ''' Returns a dictionary containing all possible
        distances from a given start node to each other
        node, expressed as a set of possible distances
        for each target node.  The set will be empty
        if the target node is not reachable from the
        start node.
    '''
    dist_to_node = collections.defaultdict(set)
    dist_to_node[start].add(0)
    for start in ordered:
        cumulative = dist_to_node[start]
        dist = edge_distance[start]
        for end in graph[start]:
            dist_to_node[end].update(dist + prev for prev in cumulative)
    return dist_to_node

# Show all possible distances between a0 and c2
print(calc_dists('a0')['c2'])