Imposing edges in Minimum Spanning Tree

Imposing edges in Minimum Spanning Tree

Abstract

We are interested in the consequences of imposing edges in a minimum spanning tree. We prove that the sum of the replacement costs in of the imposed edges is a lower bounds of the additional costs. More precisely if r-cost is the replacement cost of the edge , we prove that if we impose a set of nontree edges of then r-cost cost, where is the set of imposed edges and a minimum spanning tree containing all the edges of .

1 Preliminaries

1.1 Graph Theory

A tree is a connected and acyclic graph. A tree is a spanning tree of if and . The edges of are the tree edges of and the edges of are the nontree edges of . A minimum weighted spanning tree (mst) of is a tree whose sum of the cost of the edges it contains is minimum.

We recall the Optimality Conditions of a mst:

Theorem 1

  • [Path Optimality Condition] A spanning tree is a minimum spanning tree if and only if it satisfies the following path optimality conditions: for every nontree edge of , cost cost for every edge contained in the path in connecting nodes and .

  • [Cut Optimality Condition] A spanning tree is a minimum spanning tree if and only if it satisfies the following cut optimality conditions: for every tree edge of , cost cost for every edge contained in the cut formed by deleting edge from .

We will call -tree, a tree which must contain the edge and denote it by .

Property 1

Let be a graph, be an edge of . We compute a minimum spanning -tree of by merging first the nodes and and then by computing a mst.

For the sake of clarity we will consider that is a minimum spanning tree of . The replacement edge of a nontree edge is defined as follows:

Proposition 1 ([dooms07])

Let be a nontree edge of that we want to impose, and be the edge that is not imposed with the maximum cost contained in the path in connecting nodes and . Then, the tree corresponding to the tree in which the edge has been replaced by the edge is a minimum spanning -tree of .

Proof: If the edge is added to the tree then a cycle is created and the Path Optimality Condition implies that the edge of the cycle having the largest cost must be removed. Since a minimum -tree is wanted, the edge that must be removed is because it has the largest cost. Thus a tree is obtained. This tree satisfies the Path Optimality Condition for all the nontree edges because does. also satisifies the path optimality condition for . ∎
Note that it is possible that an edge has no replacement edge, because it closes a path of implied edges. In this case, we will consider that the replacement cost of this edge is infinite.

Notation 1

  • the edges of the simple path from to in the minimum spanning tree .

  • r-edge is the replacement edge of the edge in the minimum spanning tree .

  • r-cost is the replacement cost of the edge . It is defined by cost - costr-edge.

Figure 1: Imposition of the nontree edge and in . is the left graph and the right graph.
Proposition 2

: r-cost r-cost

Proof: two cases must be considered depending on whether r-edge belongs to or not.

1) r-edge

In this case, r-edger-edge so the replacement cost is not changed.

2) r-edge

is computed by applying the replacement operation from : the edge r-edge is removed and is added. Since r-edge then the path from to in is different from because r-edge. Without loss of generality we assume that can reach in when r-edge is removed from . The path can be split into three parts: , and . The edge cannot be a replacement edge because it is imposed in the spanning tree. Thus the replacement edge is either in or in .

can also be split into two parts (that can be empty): and where is the node in and in whose removal in disconnects and (See Fig.1)1. Clearly, we have r-edge, because these edges belong to and the replacement edges have the largest cost. Similarly we have r-edge. In addition r-edger-edge because r-edge. So, every edge in
has a cost that is less than or equal to r-edge.

A similar reasoning can be applied to . can also be split into two parts (that can be empty): and where is the node in and in whose removal in disconnects and (See Fig. 1). We have r-edge, because these edges belong to and the replacement edges have the largest cost. We also have r-edge. In addition r-edger-edge because r-edge. Thus, every edge in has a cost that is less than or equal to r-edge.

Hence, r-edge(r-edge so the replacement cost in is less than or equal to the replacement cost in . ∎
We can now define the wanted proposition:

Proposition 3

Let be an mst and a set of nontree edges of . Then, r-cost r-cost.

Proof By induction. This is true for one edge. We assume it is true for edges. From Proposition 2 we have r-cost r-cost. In addition we have r-cost r-cost. So the proposition holds.∎
This means that we have the final proposition:

Proposition 4

Let be an mst and a set of nontree edges of . Then, costr-cost cost.

References

Footnotes

  1. In fact, there are three possibilities: either there is a path from to through , or a path from to though , or a fork having and as extremities with in the center and path from to . We consider only the latter case which is more general.
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