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Consider a complete undirected graph with vertex set $\{0, 1, 2, 3, 4\}$. Entry $W_{ij}$ in the matrix $W$ below is the weight of the edge $\{i, j\}$

$W=\begin{pmatrix} 0 & 1 & 8 & 1 & 4 \\ 1 & 0 & 12 & 4 & 9 \\ 8 & 12 & 0 & 7 & 3 \\ 1 & 4 & 7 & 0 & 2 \\ 4 & 9 & 3 & 2 & 0 \end{pmatrix}$

What is the minimum possible weight of a path $P$ from vertex 1 to vertex 2 in this graph such that $P$ contains at most 3 edges?

1. 7
2. 8
3. 9
4. 10
edited | 653 views

Answer is (B) 8. The possible path is: 1 - 0, 0 - 4, 4 - 2.

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Is there any shortcut method for determining the mini possible wt of a path from vertex 1 to 2.

Because if we take the brute force approach it is very time consuming .Moreover, there are chances of error as well.

@vidhi ....we can apply bellman ford algorithms upto 3 iterations

Start from vertex 1 as source

i=1 representing atmost 1 edge

i=2 representing atmost 2 edge

i=3 representing atmost 3 edge

Use Bellman Ford only for vertex 2, to save time.....

We have to apply bellmanford upto 4(5 - 1) iterations right ?

@gshivam

Using bellman ford is fine, but it is very time consuming.

As first we have to convert the undirected graph to a directed one https://cs.stackexchange.com/questions/24768/shortest-path-in-weightedpositive-or-negative-undirected-graph

And then apply Bellman ford from vertex 1 as src.

But, can anyone suggest any other shortcut method??

ravi_ssj4 no.

Given graph is

Path: 1 -> 0 -> 4 -> 2

So the minimum possible weight of a path from vertex 1 to vertex 2 in this graph

=4+1+3

=8

edited ago by