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in Algorithms by Active (1.1k points) | 303 views
C, 1st statement false because if distance b/w  s-a is 1, a-b=1, b-t=1 & s-t=4; So shortest distance is s-a-b-t= 3; & if we add 1 in every edge so s-a-b-t=6 & s-t=5 .In this case shortest path is change.          2nd statement is true; take same data now double each edge so new distance s-a-b-t=6 & s-t=8; So shortest distance remain same.

2 Answers

+5 votes
Best answer

By using counter Example:

Here shortest path from s TO t is A to B and B to D. Let's increase edge weight of each edge by 3. Now the shortes path from s To t is A to D that is 7. So S1 is false.

if we multiply edge weights with same number the edge weights are increased in same ratio so No change in path but total weight has been changed.

So ans is option C.

by Boss (25.6k points)
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Nice explanation @Gabbar..
+4 votes

Consider above graph,


Now suppose shortest path from S to A was S-B and then B-A. shortest path length = SB + BA = 5+5 =10

Now if we increase weight of each edge by 1, then new shortest path will be

shortest path length = min (SB + BA, SA)= min(6+6,11) =11

that means shortest path has changed. So S1 is wrong


Now even if we double the edge weight shortest path remains the same

shortest path length = min(SB + BA, SA) = min(10+10,20) = 20

that means shortest path is still S-B and then B-A

So S2 is correct

Hence answer should be C)

by Loyal (9.2k points)

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