MAX-FLOW

Question 1

How a flow function is proved to be legal?

Answer 1

Question 2

State Max-Flow problem as a linear programming problem

Answer 2

Question 3

We can use Max-Flow problem to determine whether there are at least two totally distinct ways to get from source point to a target point. How?

Answer 3

Question 4

Which edges of G are found in the residual graph of G?

Answer 4

• Only the edges that can admit more flow, forward and backward.*
• if c(u,v)-f(u,v)>0 then (u,v) appears.*
• if f(u,v)>0 then (v,u) appears.*

Question 5

Answer 5

• Page 718 in the book.*
• The proof is based on the fact that when we look at a specific vertex u,V1 ={v |(u,v)} andV2 = {v |(v,u)} , because we disallow antiparallalism, V1 and V2 don’t share vertices.*

Question 6

What is the residual capacity of p?

Answer 6

It is the maximum amount by which we can increase the flow on each edge in the augmenting path p.

Question 7

What is it?

Answer 7

It is the flow on each edge on the path p in the augmenting path. Note: every edge which is not contained in p has a flow of 0.

Question 8

f is a flowm the net flow f(S,T) across the cut (S,T) is defined to be:

Answer 8

Question 9

What is the capacity of a cut (S,T)?

Answer 9

Question 10

What is the definition of a minimum-cut?

Answer 10

A minimum-cut of a network is a cut whose capacity is minimum over all cuts of the network.

Question 11

Question 12

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Question 13

Prove 1->2

Answer 13

Question 14

Prove 2 to 3

Answer 14

Question 15

Prove 3 to 1

Answer 15

Question 16

For an arbitrarily chosen augmenting paths, does FF always eventually terminates?

Answer 16

• No.*
• The FF method might fail to terminate only if edge capacities are irrational numbers.*
• If the capacities are rational numbers, we can apply an appropriate scaling transformation to make them all intergral.*
• Since with integral, |f| must be increased with at least 1, finding an augmenting path will occur at most |f| times, and since searching a path costs O(|V|+|E’|)=O(|E|) - using BFS, we have a run-time of O(|E|*|f|)*

Question 17

Answer 17

Question 18

• There’s a repeating question about having new limits; limit on the vertrics/multiple sources which produces exactly pi units of flow and such.*
• What is the solution for this cases?*

Answer 18

• The solution is simple to create a new edge which resembles the new limit/condition. This edge’s capacity is the limit and we force the current to move throw it, thus enforcing the condition.*
• If we wish for an exact condition - we can further check whether the maximal current equals the capacity given by the new condition. If so, this condition can be maintained.*

Question 19

• Note:*
• Since capacities are non-negative, an augmenting path is definitely a simple path. why? because we push the stream “forwards” and not “backwards”.*

Question 20

Answer 20

Question 21

Reread the answer because you didn’t understand it at the first time.

Answer 21

Question 22

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Question 23

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Question 24

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Question 25

Prove

Answer 25

Question 26

What is its dual?

Answer 26

Question 27

Answer 27