Negative cycles

What if pred(s) ever changes?

If pred(s) changes after InitSSSP, it would signify that the algorithm had taken a path through the input graph already that had led it back to s, meaning that there was some path s -> u-> u_1 ... -> u_k -> s, and the edge u_k -> s would have to be tense. This would mean that the cycle was negative, since s starts with a distance of zero. Thus, the input graph contains a negative cycle that travels back through s.

Logic review

• Converses: \(P \Rightarrow Q\) is not the same statement as \(Q \Rightarrow P\).
• Contrapositives: \(P \Rightarrow Q\) is logically equivalent to \(\neg Q \Rightarrow \neg P\).
  • Sometimes it is easier to prove the contrapositive of the statement you are trying to prove.

Necessary condition for no negative cycles

Prove that, if the input graph has no negative cycles, then \(P \cup X\) is always a dag.

• Suppose \(P \cup X\) is not a dag at some point, so it contains a cycle.
  • We can assume that the first time this happens, \(P\) does not contain the cycle.
• This means that adding tense edges to \(P\) would result in a cycle.
• But the only way these edges could be tense would be if they improved the distance on some node, which would mean the input graph has a negative cycle.

Sufficient condition for no negative cycles

Prove that, if \(P \cup X\) is always a dag, then the input graph has no negative cycles.

• Suppose that the input graph has a negative cycle.
• Apply MetaSSSP, and start by relaxing edges to form a path to this negative cycle from \(s\).
• Then relax the edges in this cycle. Since it is a negative cycle, you will encounter a vertex again through a tense edge. So at this point, \(P \cup X\) is not a dag.

Flow in a directed graph

• There is a start vertex \(s\) and a terminal vertex \(t\).
• Every vertex is reachable from \(s\), and \(t\) is reachable from every vertex.
  • There are paths \(s \rightsquigarrow v \rightsquigarrow t\) for all \(v\).
• Each edge has a capacity measuring how much flow it can carry.
• The total flow into a vertex must equal the total flow out.
  • Analogies: Water pipes, Freeways
• Maximum flow problem: Given a flow network, find the maximum flow from \(s\) to \(t\).

Applications: Optimizing transportation, scheduling, distributed systems, image processing, baseball elimination, etc.

Flows: Definitions

Let \(G\) be a directed graph where each edge \(e\) has capacity \(c(e)\).

• An \((s,t)\)-flow is an assignment of values \(f(e)\) to each edge \(e\) of \(G\) such that the total flow into each vertex \(v\) equals the total flow out (except, possibly, at \(s\) and \(t\)).

\[ \sum_a f(a\rightarrow v) = \sum_z f(v \rightarrow z) \] - The value \(|f|\) of a flow \(f\) is the total net flow out of \(s\) (or equivalently, into \(t\).)

\[ |f| = \sum_z f(s \rightarrow z) - \sum_a f(a \rightarrow s) \]

• A flow is feasible if it obeys the capacity limits on each edge \(e\): \(0 \leq f(e) \leq c(e)\). See Figure 10.2.

Table Groups

Improve the flow?

Can you increase the flow in Figure 10.2, keeping it feasible? If so, try to make its value as big as possible.

Cuts: Definitions

Let \(G\) be a directed graph where each edge \(e\) has capacity \(c(e)\).

• An \((s,t)\)-cut is a partition of the vertices into two sets \(S\) and \(T\), such that \(s\in S\) and \(t\in T\).
• The capacity \(\lVert S, T \rVert\) of a cut \(S,T\) is the sum of the capacities of the edges that go from \(S\) to \(T\).
  • Cutting these edges would eliminate the flow from \(s\) to \(t\).
  • See Figure 10.2.

Find a cut

1. Find a cut in the graph in Figure 10.2, and try to make its capacity as small as possible.

2. Try to find a flow whose value is bigger than the capacity of the cut that you found.

Flows can’t exceed cuts

Lemma. The value of any feasible \((s,t)\)-flow is less than or equal to the capacity of any \((S,T)\)-cut.

Proof.

• The value of the flow is the net flow out of \(s\).
• The net flow out of \(s\) is the sum of the net flows through all the vertices in \(S\) (since all the others have a net of zero).
• This remains true if we remove the edges that go from \(S\) to \(S\) (zero sum).
• The only edges left go from \(S\) to \(T\).
  • So the total flow from \(S\) to \(T\) can’t be more than the capacity.
  • The max happens when all the edges across are saturated, and all the edges back are avoided.

Maxflow-Mincut Theorem

• Lemma. The value of any feasible \((s,t)\)-flow is less than or equal to the capacity of any \((S,T)\)-cut.
• Corollary. If you can find a cut and a flow such that the capacity of the cut equals the value of the flow, you have found a maximum flow.
• Theorem. You can always find such a cut and flow.
  • (Proof next week.)