Count number of MSTs in a graph

Question

I thought about this problem many hours, but I didn't find any solutions

The problem is :

How many MSTs does the given graph have (MST := Minimum Spanning Tree)

the graph G is an undirected, connected graph

it's Guaranteed that no vertex's degree exceeds 3 :)

prefer C/C++ solutions (can understand like-code-formed Algorithm too)

and please with a low order :) (running time)

UPDATE

first of all finding all MSTs :) O( |E| log |E| )

others where worse :(

Answer 1

You can try with a backtracking. In each step for one of 'leaf' vertices decide how many 'out' edges to use.

function add_some_vertex_edges(leaves, edge):
  if |leaves| == |V|:
    num_of_MSTs += 1
    return
  v = leaves.pop()  # takes one vertex and removes it from leaves
  let v_edge be set of incident edges to v that are not already in edges and that don't make cycle if added to edges
  for each combination of edges from v_edge:
    add_edges(leaves + incident_vertices, edges + edge_combination)

add_some_vertex_edges({v}, {})

Since vertices degree is <= 3, for each leaf one edge is used to 'enter' it, and because of that |v_edge| <= 2, and because of that search tree is narrow.

Worst case running time is O(2^n), but probably in practical cases it is fast. There are examples with worst case number of MSTs. Example that I can think of is with two parallel lines of vertices and connections between them.

O---O---O---O---O---O---O---O---O---
 \ /     \ /     \ /     \ /     \ /
  X       X       X       X       X ...
 / \     / \     / \     / \     / \
O---O---O---O---O---O---O---O---O---

In this example there are exactly 2^(n/2) MSTs. To calculate this, take 2 leftmost vertices. They can be connected to the rest of a graph in 4 ways.

O---O   O   O   O   O   O---O
         \ /     \ /     \ /
          X       X       X
         / \     / \     / \
O---O , O   O , O---O , O   O

For each group of interconnected 4 vertices there are 4=2^2 possibilities to choose from.