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Rewrite Lemma 2 proof #121
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Pseudoflows, active and deficient nodes are defined here: https://en.wikipedia.org/wiki/Flow_network#Flows Perhaps we can also get a CLRS reference. |
Hmm let's leave this for after essential stuff is done. |
CLRS doesn't contain pseudoflows, I will use "Network Flows Theory, Algorithms, and Applications" instead. |
OK On Mon, Oct 31, 2016 at 3:45 PM, Orfeas Stefanos Thyfronitis Litos <
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We have a draft. However it is already a page long and has some holes, noted in parentheses. The proof is placed directly above the previous one, p. 26. Check it out. |
We have to find a reference for the fact that we can construct a valid flow that maintains the outgoing flow from the source from a pseudoflow. |
The above fact doesn't hold in all cases. Think when the outgoing capacity from the source is greater than the incoming capacity to the sink and the pseudoflow is equal to the capacities. |
What you are describing requires a deficient node. The statement holds when no deficient nodes exist. |
So we are speaking only about pseudoflows that are not preflows. |
We have been using the terms deficit and excess to denote the opposite from the literature. I will fix it in our proof. |
OK On Tue, Nov 8, 2016 at 11:36 AM, Orfeas Stefanos Thyfronitis Litos <
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The proof for Lemma 2 can be simplified quite a bit using the following sketch:
Let$\mathcal{H}$ be an execution of the Transitive Game on graph $\mathcal{G}$ and $j$ the convergence turn. We will show that a flow from $A$ to $B$ can be constructed such that $\Sum_{v \in N^+(A)}{x_{Av}} = Loss_A,j$ . First, we construct a pseudoflow $X$ on $G$ as follows:
Prove here that$X$ is a pseudoflow (i.e. flows are limited by capacities). Furthermore $X$ has no deficient nodes (please add proof). Based on the pseudoflow $X$ , we can now construct a feasible flow $X'$ by eliminating all active nodes. Since this does not affect outgoing flows from $A$ we deduce:
This proves the theorem. QED
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