finding longest path in a graph

Question

I am trying to solve a program, where in I have to find the max number of cities connected for a given list of routes.

for eg: if the given route is [['1', '2'], ['2', '4'], ['1', '11'], ['4', '11']] then max cities connected will be 4 constraint is I can't visit a city which I already have visited.

I need ideas, as in how to progress.

For now, What I have thought is if I could be able to create a dictionary with cities as a key and how many other cities its connected to as its value, i get somewhere near to the solution(I hope). for eg: My dictionary will be {'1': ['2', '11'], '4': ['11'], '2': ['4']} for the above given input. I want help to proceed further and guidance if I am missing anything.

Answer 1

You can use a defaultdict to create your "Graph" from your list of edges/paths:

edges = [['1', '2'], ['2', '4'], ['1', '11'], ['4', '11']]

G = defaultdict(list)
for (s,t) in edges:
    G[s].append(t)
    G[t].append(s)

print G.items()

Output:

[
  ('1', ['2', '11']), 
  ('11', ['1', '4']), 
  ('2', ['1', '4']), 
  ('4', ['2', '11'])
]

Note that I added the edges in both directions, since you're working with an undirected graph. So with the edge (a,b), G[a] will include b and G[b] will include a.

From this, you can use an algorithm like depth-first search or breadth-first search to discover all the paths in the graph.

In the following code, I used DFS:

def DFS(G,v,seen=None,path=None):
    if seen is None: seen = []
    if path is None: path = [v]

    seen.append(v)

    paths = []
    for t in G[v]:
        if t not in seen:
            t_path = path + [t]
            paths.append(tuple(t_path))
            paths.extend(DFS(G, t, seen[:], t_path))
    return paths

Which you can use with:

G = defaultdict(list)
for (s,t) in edges:
    G[s].append(t)
    G[t].append(s)

print DFS(G, '1')

Output:

[('1', '2'), ('1', '2', '4'), ('1', '2', '4', '11'), ('1', '11'), ('1', '11', '4'), ('1', '11', '4', '2')]

So the full code, with the final bit that shows the longest path:

from collections import defaultdict

def DFS(G,v,seen=None,path=None):
    if seen is None: seen = []
    if path is None: path = [v]

    seen.append(v)

    paths = []
    for t in G[v]:
        if t not in seen:
            t_path = path + [t]
            paths.append(tuple(t_path))
            paths.extend(DFS(G, t, seen[:], t_path))
    return paths


# Define graph by edges
edges = [['1', '2'], ['2', '4'], ['1', '11'], ['4', '11']]

# Build graph dictionary
G = defaultdict(list)
for (s,t) in edges:
    G[s].append(t)
    G[t].append(s)

# Run DFS, compute metrics
all_paths = DFS(G, '1')
max_len   = max(len(p) for p in all_paths)
max_paths = [p for p in all_paths if len(p) == max_len]

# Output
print("All Paths:")
print(all_paths)
print("Longest Paths:")
for p in max_paths: print("  ", p)
print("Longest Path Length:")
print(max_len)

Output:

All Paths:
   [('1', '2'), ('1', '2', '4'), ('1', '2', '4', '11'), ('1', '11'), ('1', '11', '4'), ('1', '11', '4', '2')]
Longest Paths:
   ('1', '2', '4', '11')
   ('1', '11', '4', '2')
Longest Path Length:
   4

Note, the "starting point" of your search is specified by the second argument to the DFS function, in this case, it's '1'.

---

Update: As discussed in the comments the above code assumes you have a starting point in mind (specifically the code uses the node labelled '1').

A more general method, in the case that you have no such starting point, would be to perform the search starting at every node, and take the overall longest. (Note: In reality, you could be smarter than this)

Changing the line

all_paths = DFS(G, '1')

to

all_paths = [p for ps in [DFS(G, n) for n in set(G)] for p in ps]

would give you the longest path between any two points.

(This is a silly list comprehension, but it allows me to update only a single line. Put more clearly, it's equivalent to the following:

all_paths = []
for node in set(G.keys()):
    for path in DFS(G, node):
        all_paths.append(path)

or

from itertools import chain
all_paths = list(chain.from_iterable(DFS(G, n) for n in set(G)))

).