4.2 All Pairs Shortest Path (Floyd-Warshall) - Dynamic Programming

Floyd-Warshall All Pairs Shortest Path Problem
Dynamic Programming
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[@mr.aravindreddy]

Sir At 6:39 I think for A^1[4,2] the value is 5 but not 8.

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[@tausifahmad2007]

sir i think at 6:39 for A-1[4,2] there will be 5 not 8

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Yes

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Thank you so much for making these videos! It's crazy how much better I understand Floyd-Warshall now

[@nickzervas4937]

When comparing the Floyd Warshall method with the Djiktra method , i think there are 3 important things you should have pointed out : 1) The Djiktra method can be achieved with less complexity , especially with sparse graphs . By using binary heap as a priority queue we achieve O(n*m*logn) and with a fibonacci heap we achieve O( (m+nlogn)*n). 2) The main advantage of this method compared to Djiktra is that it can be implemented in a graph even if we have edges of negative weights ( as long as there are no negatives circles ) . Djiktra is highly limited by that aspect . 3) The best algorithm for this method which benefits both from 1,2 is : Johnsons . It can be achieved in O( (m+nlogn)*n) even if we have negative weights ( no negative cycles )

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[@josepaul2725]

Correction A1 matrix (4,2) = 5

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Right A1 is wrong

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in video - at 8:36 AND 8:38 they changed the value of A'[4][2]=5 , don't get confuse...

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[@kunalchoudhary7477]

correction A^1[4,2]=5.

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I was looking for this comment.I thought i made a mistake taking 5...ty

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@@eurus7509 same here

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[@siddharthmanumusic]

Thank you for the beautiful explanation. One observation: Isn't the time complexity of Dijkstra to find all pairs shortest path same as that of Floyd Warshall? I think the only advantage of FW is that it can also find it for negative weights plus it's easier to implement - no priority queue or extra space complexity.

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Sir, There is a mistake in finding A1 A1=0 3 α 7 8 0 2 15 5 8 0 1 2 5 α 0 Can u check it sir....

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[@sanjeebbaitha96]

till 8:31 there was a mistake 4,2=8 instead of 8.....but corrected nice video

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Sanjeeb Baitha yeah i noticed also

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@13:55 isn't the time complexity again n^3 which we were trying to avoid initially??

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yes it is i think the constant term in this asymptotic n3 is lower than simply using djikstra several times

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@@techwithwhiteboard3483 even my point is also the same...time complexity as of using dijkstra also n times .@abdul Bari sir plz solve this doubt..thanks for wonderful explanation

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We were not trying to reduce time. We just saw a different approach

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6:40,Correction A1 matrix (4,2) = 5

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he wasted my time for this mistake and even confused me

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@@nusratfans39 always check for comments when u face such issue

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@@_outcyrptolist true dat

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Dijkstra takes O(ElogV). So, if we calculate using Dijkstra it would be O(n^3*logn)

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For a dense graph E=O(V^2), So, worst case time complexity will be O(ElogV*V)=O(EVlogV)=O(n^3*logn)..

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If Complexity is O(n^3) using DP also ,then we could have used Dijkstra's also then what is the benefit of using DP

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There are multiples ways to Solve every problem.

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Dijkstra wont work for negative edges

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[@AMITKUMAR-te9bn]

What About Complexity? Is it still N^3.

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Time O(n3) Space O(n2)

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