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.
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)))
).