Today Flow review Augmenting paths Ford-Fulkerson Algorithm Intro - PowerPoint PPT Presentation

Today Flow review Augmenting paths Ford-Fulkerson Algorithm Intro to cuts (reason: prove correctness)

Flow Networks s = source, t = sink. c(e) = capacity of edge e Capacity condition: 0 ≤ f(e) ≤ c(e) Conservation condition: for v ∈ V – {s, t}: � ∑ f(e) = ∑ f(e) e into v e out of v 0 5 2 9 0 0 0 4 10 15 15 10 flow = 4 4 0 0 0 t sink s source 3 5 8 6 10 0 4 15 4 0 0 6 10 15 4 4 7 30 0

Flows The value of a flow f is: v(f) = ∑ f(e) e out of s 0 value = 4 5 2 9 0 0 0 4 10 15 15 10 flow = 4 4 0 0 0 t sink s source 3 5 8 6 10 0 4 15 4 0 0 6 10 15 4 4 7 30 0

Flows The value of a flow f is: v(f) = ∑ f(e) e out of s 9 value = 24 5 2 9 0 9 1 0 10 15 15 10 flow = 10 4 4 6 6 t sink s source 3 5 8 6 10 0 9 15 4 10 1 6 10 15 0 4 7 30 10

Maximum Flow Problem Find s-t flow of maximum value. 9 value = 28 5 2 9 0 9 0 1 10 15 15 10 flow = 10 4 5 8 9 t sink s source 3 5 8 6 10 0 10 15 4 13 3 6 10 15 0 4 7 30 13

Towards a Max-Flow Algorithm Key idea: repeatedly choose paths and “augment” the amount of flow on those paths as much as possible until capacities are met

Towards a Max Flow Algorithm Problem: possible to get stuck at a flow that is not maximum, no more paths with excess capacity 1 × 20 0 0 20 10 × s t 30 0 20 10 20 × × Flow value = 0 20 0 0 20 2

Residual Graph Original edge: e = (u, v) ∈ E. u 17 v Flow f(e), capacity c(e). 6 Create two residual edges residual Forward edge capacity e = (u, v) with capacity c(e) - f(e) u v 11 Backward/reverse edge 6 e ’ = (v, u) with capacity f(e) Residual graph: G f = (V , E f ). E f = edges with positive residual capacity E f = {e : f(e) < c(e)} ∪ {e ’ : f(e) > 0}

Augmenting Path Definition: an s-t path P in G f is an augmenting path Idea: use an augmenting path to augment flow in G Increase flow on forward edges Decrease flow on backward edges Definition: let bottleneck(P, f) be the minimum residual capacity (i.e., capacity in G f ) of any edge in P Example on board

Augmenting Path Use path P in G f to to update flow f Augment(f, P) { / / edge on P with least residual capacity b = bottleneck(P, f) foreach e = (u,v) ∈ P { if e is a forward edge f(e) = f(e) + b / / forward edge: increase flow else let e’ = (v, u) f(e’) = f(e’) - b / / backward edge: decrease flow } return f }

Augmenting Path Claim: Let f be a flow and let f’ = Augment(f, P). Then f’ is a flow. Proof idea: verify capacity and conservation conditions 1) Capacity: by design of residual graph 2) Conservation: check that net change at each node is zero Proof sketch on board

Ford-Fulkerson Algorithm Repeat: find an augmenting path, and augment! Ford-Fulkerson(G, s, t) { foreach e ∈ E f(e) = 0 // initially, no flow G f = copy of G // residual graph = original graph while (there exists an s-t path P in G f ) { f = Augment(f, P) // change the flow update G f // build a new residual graph } return f }

Example 0 4 2 4 G: 0 0 0 6 0 8 10 10 2 0 0 0 0 10 s 3 9 5 10 t Flow value = 0

Example 0 4 2 4 × × G 8 0 0 8 0 6 0 8 10 10 2 0 × 0 8 0 0 10 s 3 9 5 10 t × Flow value = 0 8 4 2 4 G f 6 8 10 10 2 10 s 3 9 5 10 t

Example 0 4 2 4 × G 10 8 0 8 × 6 0 8 10 10 2 0 2 × 2 × 0 10 8 0 10 s 3 9 5 10 t × Flow value = 8 10 4 2 4 G f 8 6 8 10 2 2 10 s 3 9 5 2 t 8

Example 0 4 2 4 × G 8 0 10 6 × 6 0 8 10 10 2 2 6 × × 0 10 6 2 8 10 s 3 9 5 10 t × Flow value = 10 16 4 2 4 G f 10 6 8 10 2 10 s 3 7 5 t 2 10

Example × 2 0 4 2 4 × 8 G 8 6 10 × 6 6 8 10 10 2 2 0 × 8 6 8 10 10 s 3 9 5 10 t × Flow value = 16 18 4 2 4 Gf 10 6 6 8 4 2 4 s 3 1 5 t 6 8 10

Example × 3 2 4 2 4 × 7 × 9 G 8 8 10 6 6 8 10 10 0 2 × 9 × 9 8 8 10 10 s 3 9 5 10 t × Flow value = 18 2 19 2 2 4 G f 10 8 6 8 2 2 2 s 3 1 5 t 8 8 10

Example 3 4 2 4 G 10 7 9 6 6 8 10 10 0 2 9 9 10 10 s 3 9 5 10 t max flow = 19 3 1 2 4 G f 10 9 6 7 1 2 1 1 s 3 5 t 9 9 10

Termination Assumption. All capacities are positive integers. Invariant. Every flow value f(e) and every residual capacity c f (e) remains an integer throughout the algorithm. Theorem. Let OPT = value of max flow. The algorithm terminates in at most OPT iterations, with OPT ≤ C, the total capacity of the edges leaving the source. Proof?

Running Time? There are at most C augment operations. How long does it take for each? O(m+n) Find a residual path Compute bottleneck capacity O(m) Update flow O(m) Update residual graph O(m) Total running time: O(C(m+n))

Cuts An s-t cut is a partition (A, B) of V with s ∈ A and t ∈ B. The capacity of a cut (A, B) is c(A,B) = Σ c(e) e out of A B 5 2 9 A 10 15 15 10 4 t sink s source 3 5 8 6 10 15 4 6 10 15 capacity of cut = 30 4 7 30

Cuts capacity of cut = 9 + 15 + 8 + 30 = 62 (Capacity is sum of weights on edges leaving A.) B A 5 2 9 10 15 15 10 4 t sink s source 3 5 8 6 10 15 4 6 10 15 4 7 30

Flows and Cuts Flow value lemma. Let f be any flow, and let (A, B) be any s-t cut. Then, the net flow sent across the cut is equal to the amount leaving s. ∑ f(e) - ∑ f(e) = v(f) e out of a e into a B 9 value = 24 5 2 9 A 0 9 1 0 10 10 15 15 10 4 4 6 6 t sink s source 3 5 8 6 10 0 9 15 4 10 1 6 10 15 0 4 7 30 10

Flows and Cuts Flow value lemma. Let f be any flow, and let (A, B) be any s-t cut. Then, the net flow sent across the cut is equal to the amount leaving s. ∑ f(e) - ∑ f(e) = v(f) e out of a e into a 9 B value = 24 A 5 2 9 0 9 1 0 10 10 15 15 10 4 4 6 6 t sink s source 3 5 8 6 10 0 9 15 4 10 1 6 10 15 0 4 7 30 10

Flows and Cuts Flow value lemma. Let f be any flow, and let (A, B) be any s-t cut. Then, the net flow sent across the cut is equal to the amount leaving s. ∑ f(e) - ∑ f(e) = v(f) e out of a e into a 9 value = 24 5 2 9 A 0 9 1 0 10 10 15 15 10 4 B 4 6 6 t sink s source 3 5 8 6 10 0 9 15 4 10 1 6 10 15 0 4 7 30 10

Flows and Cuts Flow value lemma. Let f be any flow, and let (A, B) be any s-t cut. Then ∑ f(e) - ∑ f(e) = v(f). e out of A e into A Proof: v(f) = ∑ f(e) by definition e out of s by flow conservation, all terms = ∑ ( ∑ f(e) - ∑ f(e) ) except v = s are 0 v ∈ A e out of v e into v if both endpoints of e are in A, = ∑ f(e) - ∑ f(e) there will be canceling terms e out of A e into A for that edge

Max-Flow Min-Cut There is a deep connection between flows and cuts in networks Next time, we will prove that Ford-Fulkerson is correct by proving the Max-Flow Min-Cut Theorem