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 rcost is the replacement cost of the edge , we prove that if we impose a set of nontree edges of then rcost 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 Then, the tree corresponding to the tree in which the edge has been replaced by the edge is a minimum spanning tree of .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 .

redge is the replacement edge of the edge in the minimum spanning tree .

rcost is the replacement cost of the edge . It is defined by cost  costredge.
Proposition 2
: rcost rcost
Proof: two cases must be considered depending on whether redge belongs to or not.
1) redge
In this case, redgeredge so the replacement cost is not changed.
2) redge
is computed by applying the replacement operation from : the edge redge is removed and is added. Since redge then the path from to in is different from because redge. Without loss of generality we assume that can reach in when redge 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)
has a cost that is less than or equal to redge.
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 redge, because these edges belong to and the replacement edges have the largest cost. We also have redge. In addition redgeredge because redge. Thus, every edge in has a cost that is less than or equal to redge.
Hence, redge(redge 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, rcost rcost.
Proof
By induction. This is true for one edge. We assume it is true for edges. From Proposition 2 we have
rcost rcost. In addition we have
rcost rcost. 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, costrcost cost.
References
Footnotes
 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.