Question

Local search for minimum spanning trees. Consider the set of all spanning trees (not just minimum ones) of a weighted, connected, undirected graph$\mathit{G}\mathbf{=}\mathbf{(}\mathbf{V}\mathbf{,}\mathbf{E}\mathbf{)}\mathbf{.}$

Recall from Section 5.1 that adding an edge to a spanning tree$\mathit{T}$ creates an unique cycle, and subsequently removing any other edge$\mathbf{e}\mathbf{\text{'}}\mathbf{\ne }\mathbf{e}$ from this cycle gives back a different spanning tree$\mathit{T}\mathbf{\text{'}}$ . We will say that differ by a single edge swap $\mathbf{(}\mathbf{e}\mathbf{,}\mathbf{e}\mathbf{\text{'}}\mathbf{)}$and that they are neighbors.

a) Show that it is possible to move from any spanning tree $\mathit{T}$to any other spanning tree $\mathit{T}\mathbf{\text{'}}$ by performing a series of edge-swaps, that is, by moving from neighbor to neighbor. At most how many edge-swaps are needed?

(b) Show that if$\mathit{T}\mathbf{\text{'}}$ is an MST, then it is possible to choose these swaps so that the costs of the spanning trees encountered along the way are nonincreasing. In other words, if the sequence of spanning trees encountered is

$\mathbf{T}\mathbf{=}{\mathbf{T}}_{\mathbf{0}}\mathbf{\to }{\mathbf{T}}_{\mathbf{1}}\mathbf{\to }{\mathbf{T}}_{\mathbf{2}}\mathbf{\to }\mathbf{·}\mathbf{·}\mathbf{·}\mathbf{\to }{\mathbf{T}}_{\mathbf{k}}\mathbf{=}\mathbf{T}\mathbf{\text{'}}\mathbf{,}$

then $\mathbf{cost}\mathbf{\left(}{\mathbf{T}}_{\mathbf{i}\mathbf{+}\mathbf{1}}\mathbf{\right)}\mathbf{\le }\mathbf{cost}\mathbf{\left(}{\mathbf{T}}_{\mathbf{i}}\mathbf{\right)}\mathbf{for}\mathbf{all}\mathbf{i}\mathbf{<}\mathbf{k}\mathbf{.}$

(c) Consider the following local search algorithm which is given as input an undirected graph $\mathit{G}$.

Let $\mathit{T}$be any spanning tree of $\mathit{G}$

while there is an edge-swap$\mathbf{(}\mathbf{e}\mathbf{,}\mathbf{e}\mathbf{\text{'}}\mathbf{)}$ which reduces cost($\mathit{T}$):

$\begin{array}{l}\mathbf{}\mathbf{T}\mathbf{\leftarrow }\mathbf{T}\mathbf{+}\mathbf{e}\mathbf{-}\mathbf{e}\mathbf{\text{'}}\\ \mathbf{return}\mathbf{T}\end{array}$

Show that this procedure always returns a minimum spanning tree. At most how many iterations does it take ?

Short answer

a). yes, it is possible to move from any spanning tree $T$to any other spanning tree $T\text{'}$ by performing a series of edge-swaps, that is, by moving from neighbour to neighbour.

b).The statement given in the second part is proved belowz.

c). This procedure returns a minimum spanning tree and at most four iterations it will take.

Step-by-step solution

Step 1: Local search for minimal spanning tree.

A minimal spanning tree is indeed an edge-weighted graph with a weight that is equal to which is less than the value about any other spanning tree.A subgraph is contain all vertices of graph.A subgraph should be contain (V-1) edges without any cycleThe minimum spanning tree is an edge-weighted graph whose weight is not more than the weight of any other spanning tree. Minimum spanning tree cost is defined as the number of edges for a minimal spanning tree may be calculated using the formula below,

$Numberofedges=Numberofvertices–1$

In Equation, replace $"numberofvertices=8"$with the value.

$\begin{array}{c}Numberofedges=8–1\\ =7\end{array}$

To find the minimum distance in a spanning tree. We are using some algorithm.And the cost of minimum spanning tree is calculated by two algorithm these are as follow:

1). Kruskal’s algorithm

2). Prim’s algorithm

Step 2: Edge-swaps for minimum spanning trees.

a)

Here, a new edge-swap heuristic for generating spanning trees with a minimum number of branch vertices, that is the vertices of degree greater than two.to move from any spanning tree $T$to any other spanning tree $T\text{'}$ by performing a series of edge-swaps, that is, by moving from neighbor to neighbor. And therefore all four edges are pseudo-critical. All pairs are distinct. Use the Kruskal algorithm to find the minimum spanning tree by sorting the edges and picking edges from ones with smaller weights. Use a disjoint set to avoid adding redundant edges that result in a cycle.

And hence yes, it is possible to move from any spanning tree to $T$any other spanning tree $T\text{'}$ by performing a series of edge-swaps, that is, by moving from neighbor to neighbour.

Step 3: The sequence of spanning trees.

b)

if $T\text{'}$is an MST (minimum spanning tree) , then it is possible to choose these swaps so that the costs of the spanning trees encountered along the way are no increasing. In other words, if the sequence of spanning trees encountered is

$T=T_0\to T_1\to T_2\to ···\to T_k=T\text{'},$

Then, $cost\left(T_{i+1}\right)\le cost\left(T_i\right)foralli<k.$

To improve the lazy implementation of Prim's algorithm, we might try to delete ineligible edges from the priority queue, so that the priority queue contains only the crossing edges. But we can eliminate even more edges.

The key is to note that our only interest is in the minimal edge from each non-tree vertex to a tree vertex. When we add a vertex v to the tree, the only possible change with respect to each non-tree vertex w is that adding v brings w closer than before to the tree. In short, we do not need to keep on the priority queue all of the edges from w to vertices tree.

And need to keep track of the minimum-weight edge and check whether the addition of v to the tree necessitates that we update that minimum (because of an edge v-w that has lower weight), which we can do as we process each edge in s adjacency list.

In other words, we maintain on the priority queue just one edge for each non-tree vertex: the shortest edge that connects it to the tree.

Through this its sequence is must be in the form of,$T=T_0\to T_1\to T_2\to ···\to T_k=T\text{'},$

And cost of its ealiar tree $\left(T_{i+1}\right)$is always less than or equal to the cost of its present tree that is$\left(T_i\right)$ . This is follow for all the value of k where it is always less than i. and it is defined in the mathematical way is shown as:

$cost\left(T_{i+1}\right)\le cost\left(T_i\right)foralli<k.$

Step 3: Local search for minimum spanning trees.

c)

Let’s AE is our first edge with weight 1. Then again we have to search the next minimum edge in the tree that is EF which is also with weight 1 after that we can search the weight which is now equal or greater than 1 which is EB and FB, both have edge weight is 2 then we can choose any of them so let’s take FB. As no cycle is formed so include it in our minimum spanning tree. We next consider the smallest weighted edge from remaining edges FG having weight 3. As no cycle is formed, include it in our minimum spanning tree.

We next consider the smallest weighted edge from remaining edges GH having weight 3. As no cycle is formed, include it in our minimum spanning tree. We next consider the smallest weighted edge from remaining edges GC having weight 4. As no cycle is formed, include it in our minimum spanning tree.

Finally consider the smallest weighted edge from remaining edges GD having weight 5. As no cycle is formed, include it in our minimum spanning tree.

minimum spanning tree is now connected containing V-1 edges.

By adding all the weights of the edges we can get the cost of its minimum spanning tree.To find the number of minimum spanning tree by using Kruskal algorithm is in which starts with the node which have minimum weight on its edge and no cycle is formed, include it in our minimum spanning tree. then again search the next minimum weight of its edge by applying these steps simultaneously we get the connected graph which is know as minimum spanning tree.

consider the smallest weighted edge so there we have two options that is AE or EF we can choose any of them edge.So, lets AE is our first edge with weight 1. Then again we have to search the next minimum edge in the tree that is EF which is also with weight 1 after that we can search the weight which is now equal or greater than 1 which is EB and FB, both have edge weight is 2 then we can choose any of them so let’s take FB. As no cycle is formed so include it in our minimum spanning tree. We next consider the smallest weighted edge from remaining edges FG having weight 3. As no cycle is formed, include it in our minimum spanning tree.

We next consider the smallest weighted edge from remaining edges GH having weight 3. As no cycle is formed, include it in our minimum spanning tree. We next consider the smallest weighted edge from remaining edges GC having weight 4. As no cycle is formed, include it in our minimum spanning tree.

Finally consider the smallest weighted edge from remaining edges GD having weight 5. As no cycle is formed, include it in our minimum spanning tree.

considering the smallest weighted edge so there we have two options that is AE or EF we can choose any of them edge.

So, lets AE is our first edge with weight 1. Then again we have to search the next minimum edge in the tree that is EF which is also with weight 1 after that we can search the weight which is now equal or greater than 1 which is EB and FB, both have edge weight is 2 then we can choose any of them so let’s take EB. As no cycle is formed so include it in our minimum spanning tree. We next consider the smallest weighted edge from remaining edges FG having weight 3. As no cycle is formed, include it in our minimum spanning tree.

We next consider the smallest weighted edge from remaining edges GH having weight three. As no cycle is formed, include it in our minimum spanning tree. We next consider the smallest weighted edge from remaining edges GC having weight 4. As no cycle is formed, include it in our minimum spanning tree.

Finally consider the smallest weighted edge from remaining edges GD having weight 5. As no cycle is formed, include it in our minimum spanning tree.

So, the number of minimum spanning trees are: .According to Kruskal’s algorithm the order of the edges added to the MST are:

By sort all the edges in increasing order of their edge weight and pick the smallest one

So, Let’s start our sorting then we have two options that is AE and EF both has weight one. Let’s pick AE first then after EF. now search for next again we have two options that is EB and FB both has weight 2. Let’s choose FB with weight two also check if the new selected edge does create a cycle in the spanning tree or not if not then select that edge. After that select FG with weight three. Then GH with weight three. Then select wedge GC and GD with weight four and five. Then we get our final order of edges in minimum spanning tree.So, the order of the edges added to the minimum spanning tree are AE,EF,FB,FG,GH,GC,GD

Hence, this procedure returns a minimum spanning tree and at most four iterations it will take.