Zain Zahid (bscs)Air university, islamabadzainzahid99@gmail.com

* BELLMAN FORD ALGO (PART-II)

A

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BH

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Distance

Parent

A 0 0B ∞ 0C ∞ 0D ∞ 0E ∞ 0F ∞ 0G ∞ 0H ∞ 0 ∞

0

∞

∞∞

∞

∞∞

2

For demonstration purpose, we would consider the following graph:• Considering A as the source, assign it the distance zero.• For all vertices except A assign a distance infinity.

A

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∞

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A 0 0B 8 ACDEFGH

2

A

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6

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A 0 0B 8 ACDE 6 AFGH

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A

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∞14

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A 0 0B 8 AC 14 BDE 6 AFGH

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A

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∞

∞14

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Distance

Parent

A 0 0B 8 AC 14 BDE 6 AF 9 EGH

2

A

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∞

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∞14

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Distance

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A 0 0B 8 AC 14 BDE 6 AF 9 EG 8 EH

2

A

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BH

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∞14

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Distance

Parent

A 0 0B 8 AC 14 BD 9 GE 6 AF 9 EG 8 EH

2

A

EF

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C

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BH

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1814

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Distance

Parent

A 0 0B 8 AC 14 BD 9 GE 6 AF 9 EG 8 EH 18 C

2

A

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BH

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1814

Vertices

Distance

Parent

A 0 0B 8 AC 14 BD 9 GE 6 AF 9 EG 8 EH 18 C

2

STEP-2 START CHECKING EACH VERTEX AGAIN

B D E F G H remains same

A

EF

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C

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BH

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187

Vertices

Distance

Parent

A 0 0B 8 AC 7 GD 9 GE 6 AF 9 EG 8 EH 18 C

2

Change will be only in C

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Distance

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A 0 0B 8 AC 7 GD 9 GE 6 AF 9 EG 8 EH 11 C

2

STEP-3 START CHECKING EACH VERTEX AGAIN

Change will be only in H

A

EF

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BH

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Vertices

Distance

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A 0 0B 8 AC 7 GD 9 GE 6 AF 9 EG 8 EH 11 C

2

And B C D E F G remains same

A

EF

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Vertices

Distance

Parent

A 0 0B 8 AC 7 GD 9 GE 6 AF 9 EG 8 EH 11 C

2

STEP-4 START CHECKING EACH VERTEX AGAIN

This time there is no change and all vertices remain sameSo we can stop here

A

EF

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BH

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117

Vertices

Distance

Parent

A 0 0B 8 AC 7 GD 9 GE 6 AF 9 EG 8 EH 11 C

2

Shortest Paths are:B=AB=8C=AEGC=7D=AEGD=9E=AE=6F=AEF=9G=AEG=8H=AEGCH=11

*There are two cases in iterations

*We might not notice any change in distance for any vertex in later iterations. It would be good to stop at that point for making the algorithm efficient.

*We might endlessly notice changes in distances of one or the other vertex in later iterations. This will be the case of negative weight cycles. Hence, we need to stop at some point and that would be the after |V| – 1