Find a minimum weight spanning tree in the undirected graph where there are negative edges

Question

So I need to solve this graph, I have a general idea on how to solve it but if I am doing this wrong please correct me.

So to find the MST I need to perform Kruskals algorithm on the graph

here is my psuedocode for this Kruskals algorithm

Kruskal(V,E) A = null; for each v contains in V make disjoint set(v) Sort E by weight increasingly for each(v1, v2) contains in E if

Kruskal(V,E)


A = null;
for each v contains in V
     make disjoint set(v)
Sort E by weight increasingly
for each(v1, v2) contains in E 
     if Find(v1) is not equal to Find(v2)
          A = A Union {(v1,v2)}
          Union(v1,v2)
Return A

So first thing I do is find the nodes with the shortest distances right?

1)

I assume S to H has the shortest distance since the d(h,s) = -3

so A = {(h,s)}

so now we follow this pattern

2) A = {(h,s),(s,f)}

3) A = {(h,s),(s,f)(s,n)}

4) A = {(h,s)(s,f)(s,n)(f,k)}

5) A = {(h,s)(s,f)(s,n)(f,k)(s,m)} (we skip H to N because a path is already made from h to n which is through s )

6) A = {(h,s)(s,f)(s,n)(f,k)(s,m)(d,b)}

7) A = {(h,s)(s,f)(s,n)(f,k)(s,m)(d,b)(b,m)}

so now since there is a path that connects to all edges we are good right?

but the thing i dont understand is there there are distance[u,v] which is shorter than the path[u,v] through multiple vertices. For example d[d,m] is shorter then p[d,m] which goes through B first. Am I doing something wrong?

Answer 1

You're not doing anything wrong. There's no guarantee that a MST preserves shortest distances between nodes. Eg: the three node complete graph ABC with edge weights 3, 2, 2 (with apologies for my ascii art):

A --- 2 --- B
|           |
2          /
|         /
C----3---/ 

The minimal spanning tree is CAB but the distance between B and C in the original graph is 3, and in the MST 4.