First time here? Checkout the FAQ!
+4 votes

asked in Algorithms by Active (1.3k points)   | 85 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.


answered by Veteran (22.1k points)  
selected by
Nice explanation @Gabbar..
+2 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)

answered by Boss (7.7k points)  

Top Users Mar 2017
  1. rude

    4768 Points

  2. sh!va

    3054 Points

  3. Rahul Jain25

    2920 Points

  4. Kapil

    2728 Points

  5. Debashish Deka

    2602 Points

  6. 2018

    1574 Points

  7. Vignesh Sekar

    1422 Points

  8. Akriti sood

    1378 Points

  9. Bikram

    1342 Points

  10. Sanjay Sharma

    1126 Points

Monthly Topper: Rs. 500 gift card

21,517 questions
26,844 answers
23,181 users