retagged by
892 views
0 0 votes

Consider the following statement:

$A)$ If all edge weight of a graph are positive then any subset of edges that connect all vertices and has minimum total weight is a tree.

$B)$ Let $p=<V_{0},V_{1},V_{2},........V_{k} >$ be the shortest path from vertex $V_{0}$ to $V_{k}$ for all $i,j$ such that $0\leq i\leq j\leq k$ let $p_{ij}$ be subpath of. $p$ from vertex $V_{i}$ to $V_{j}$. Then $p_{ij}$ be the shortest path from $V_{i}$ to $V_{j}$

Which statement is correct?


My question is , is tree always needed for minimum weight graph??

and what about B)?? Is it just saying each minimum path between $2$ vertices makes total shortest path?? 

1 Answer

Best answer
2 2 votes
For A, assume that any subset of edges that connect all vertices and has minimum total weight is not a tree, i.e it consists a cycle, since it contains a cycle, so you can remove the edge which is creating the cycle and still cover all the vertices , and the total weight also decreased, hence our assumption that the sum of the weight of the subset of edges which we initially chose was minimum is false.

There is a contradiction.

$\therefore$ ,we can say that any subset of edges that connect all vertices and has minimum total weight is a tree.

For B, this is what is the invariant for Dijkstra Algorithm, this is also true. You can see the proof of correctness of Dijkstra algorithm from any book.
selected by
Position:
Show:

Related questions

1 1 vote
1 1 answer
1.6k
1.6k views
Souvik33 asked Dec 19, 2022
1,558 views
If a -ve weight cycle is reachable from source, the Dijkstra's algorithm gets into an infinite loop TRUEFALSE
1 1 vote
1 1 answer
224
224 views
GO Classes asked Oct 16, 2024
224 views
Let $\text{G = (V, E)}$ be a weighted directed graph. The shortest path from a node $s \in \text{V}$ to a node $t \in \text{V}$ will remain unchanged if: (Multiple option...
0 0 votes
0 0 answers
701
701 views
Vaishnavi01 asked Nov 19, 2018
701 views