Give You are given a graph$\mathit{G}\mathbf{=}\mathbf{(}\mathit{V}\mathbf{,}\mathit{E}\mathbf{)}$with positive edge weights, and a minimum spanning tree $\mathit{T}\mathbf{=}\mathbf{(}\mathit{V}\mathbf{,}\mathit{E}\mathbf{)}$with respect to these weights; you may assume GandTare given as adjacency lists. Now suppose the weight of a particular edge $\mathit{e}\mathbf{\notin }\mathit{E}\mathbf{\text{'}}$is modified fromw(e)to a new value w'(e). You wish to quickly update the minimum spanning tree T to reflect this change, without recomputing the entire tree from scratch. There are four cases. In each case give a linear-time algorithm for updating the tree.

(a) $\mathit{e}\mathbf{\notin }\mathit{E}\mathbf{\text{'}}$and $\mathit{w}\mathbf{\text{'}}\mathbf{\left(}\mathit{e}\mathbf{\right)}\mathbf{>}\mathit{w}\mathbf{\left(}\mathit{e}\mathbf{\right)}$ .

(b) role="math" localid="1658907878059" $\mathit{e}\mathbf{\notin }\mathit{E}\mathbf{\text{'}}$and $\mathit{w}\mathbf{\text{'}}\mathbf{\left(}\mathit{e}\mathbf{\right)}\mathbf{>}\mathit{w}\mathbf{\left(}\mathit{e}\mathbf{\right)}$ .

(c) role="math" localid="1658907882667" $\mathit{e}\mathbf{\notin }\mathit{E}\mathbf{\text{'}}$and $\mathit{w}\mathbf{\text{'}}\mathbf{\left(}\mathit{e}\mathbf{\right)}\mathbf{>}\mathit{w}\mathbf{\left(}\mathit{e}\mathbf{\right)}$ .

(d) role="math" localid="1658907887400" $\mathit{e}\mathbf{\notin }\mathit{E}\mathbf{\text{'}}$and $\mathit{w}\mathbf{\text{'}}\mathbf{\left(}\mathit{e}\mathbf{\right)}\mathbf{>}\mathit{w}\mathbf{\left(}\mathit{e}\mathbf{\right)}$ .

A minimal spanning tree (MST as a minimum spanning tree) is a subgroup of something like a tree that has the shortest packets from the source to all of its vertex points. A graph G = (V,E) Some minimum spanning tree is shown, along with edge weights that are positive. $T=(V,E)$ such as relation to all of these weights. Take a look at the clustering lists. Graph G and T are supplied. After this step suppose that a particular edge’s weight is altered from w(e) to w'(e) .

a).

Algorithm:

Input:

Graph: G = (V,E)

minimum spanning tree: T = (V,E)

Procedure is given as

Declare T and G as adjacency list

Declare e ,

if (e is in $\left[e\right]$ )

Modify $w\left(e\right)$to $w\text{'}=e$.

Else If (e is not in [E])

Update w (e) > w'(e)

Explanation of the given equation is as follows:

Consider the minimum spanning tree when the weight of one edge $e\notin E\text{'}$is decreased. Assume that $e=(u,v)$ .Add edge e to T tree that will create a distinct cycle, that can be found applying breadth first search and not considering weights. And this will take $O\left(\left|T\right|\right)=O\left(\left|V\right|\right)$ time. And remove the cycle's maximum weight edge $O\left(\left|V\right|\right)$.

In this way linear time algorithm for updating a tree when $e\notin E\text{'}$and w(e) > w'(e) is given.

(b).

Algorithm:

Input:

Graph: G = (V,E)

minimum spanning tree: T = (V,E)

Procedure is given as

Declare T and G as adjacency list

Declare e ,

if ( is in $\left[e\right]$ )

Modify w(e) to w'(e) .

Else If (e is not in [E])

Making change w(e) to w'(e) .

Else If (e is not in [E])

Update w'(e)<w(e)<p=" ">

Update T with w'(e) .

End.

Explanation:

Take into account the MST while one of the edge's weights is increased, as well as the outcomes of a Kruskal method.$e\notin E$. That outcome is just the same as when edge demand is higher and also the minimum value gets chosen by the algorithm. (This suggests that the edges' weights are distinct.) Whenever the tree is updated in this fashion, a linear time approach is used. $e\notin E$ and w(e) < w'(e) is given.

(c)

Algorithm:

Input:

Graph: G = (V,E)

minimum spanning tree: T = (V,E)

Procedure is given as

Declare T and G as adjacency list

Declare e ,

if ( is in $\left[e\right]$)

Update w'(e) <w (e) <p=" ">

Update $T_u$ with w' = (e)

Update $T_v$with w'(e)

end

Explanation:

Analyse how minimum spanning t until one of the edges is heavier than the other $e\notin E\text{'}$. And it is increased Let e = (u,v) and just let the subtrees that were created by deleting them e be $T_v$and $T_u$. With BFS (breadth first search )(ignoring weights of edges) , It really is possible to detect whichever vertices are all in the localid="1658916013612" $T_u$ and which are in $T_v$ in time localid="1658917474694" $O\left(\left|V\right|+\left|E\right|\right)$Assume each node is marked with its membership.

Each edge is checked and edge e' with having one endpoint $T_u$ and $T_v$having the other only are kept This is possible to be done in time $O\left(\left|V\right|\right)$ . Then the complete runtime is $O\left(\left|V\right|+\left|E\right|\right)$. Whenever the tree is updated throughout this fashion, a linear time approach is used. $e\notin E$and w(e) > w'(e) is given.

(d).

Algorithm:

Input:

Graph: G = (V,E)

minimum spanning tree: T = (V,E)

Procedure is given as

Declare T and G as adjacency list

Declare e ,

wedge cost W ;

decrement

when (w - d ), new cost T'

If (e is in [E])

Update w'(e)<w(e)<p=" "style=" box-sizing-box;"></w(e)<>

Update $T_u$ with w’(e) with T'

Update $T_v$ with w’(e) with T'

End

Explanation:

Whenever one of the edge's weights changes, think about the MST. $e\in E\text{'}$ is reduced. Assume that T is the tree, and W is the weight of the tree. Once the value of the edge gets reduced by d, their actual cost is ( W - d ). Describe the subtrees of T as $T_u$ and $T_v$ , when edge e is deleted. Assuming T somehow doesn't stay an MST, its edge e is required for any T' to continue an MST. But both $T_u$ and $T_v$ Otherwise, T may not even be basic to begin with if they weren't MSTs with each respective node sets. As a result, the weight $T_v$ can’t be lesser ( W - d ). Whenever a tree is updated in this fashion, a linear time approach is used $e\notin E\text{'}$and w(e) > w'(e) .