• retagged by
921 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
1,570 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
239
239 views
GO Classes asked Oct 16, 2024
239 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
720
720 views
Vaishnavi01 asked Nov 19, 2018
720 views