CS Lunch This Week. Special Talk This Week. Soviet Rail Network, Flow Networks. Slides20 - Network Flow Intro.key - December 5, 2016

Transcription

1 CS Lunch This Week Panel on Sudying Engineering a MHC Wednesday, December, : Kendade Special Talk This Week Learning o Exrac Local Evens from he Web John Foley, UMass Thursday, December, :, Carr Sovie Rail Nework, Reference: On he hisory of he ransporaion and maximum flow problems. lexander Schrijver in Mah Programming, :,. Flow Neworks Flow nework. bsracion for maerial flowing hrough he edges. G = (V, E) = direced graph Two disinguished nodes: s = source, =. capaciy Slides - Nework Flow Inro.key - December,

2 Flows n s- flow is a funcion ha saisfies: Capaciy condiion: For each e E: f(e) c(e) Conservaion condiion: For each v V {s, : f(e) = f(e) e ino y e ou of y flow = Flows The value of a flow f is: v(f) = f(e) e ou of s value = flow = Flows The value of a flow f is: v(f) = f(e) e ou of s value = flow = Slides - Nework Flow Inro.key - December,

3 Maximum Flow Problem Find s- flow of maximum value. value = flow = Residual Graph Original edge: e = (u, v) E. Flow f(e), capaciy c(e). u v Residual edges. e = (u, v) wih capaciy c(e) - f(e) ND e R = (v, u) wih capaciy f(e) Residual graph: G f = (V, E f ). Residual edges wih posiive residual capaciy. E f = {e : f(e) < c(e) {e R : f(e) >. u residual capaciy v ugmening Pah lgorihm Ford-Fulkerson(G, s,, c) { foreach e E f(e) // iniially, no flow // iniially, he residual // graph is he original graph while (here exiss an s- pah P in Gf) { f ugmen(f, c, P) // change he flow updae G f // build a new residual graph ugmen(f, c, P) { b boleneck(p) // edge on P wih leas capaciy // forward edge, increase flow in G // reverse edge, decrease flow in G le e be he forward edge corresponding o e R Slides - Nework Flow Inro.key - December,

4 ugmening Pah lgorihm Ford-Fulkerson(G, s,, c) { foreach e E f(e) Cos - while (here exiss an s- pah P in Gf) { f ugmen(f, c, P) updae G f ugmen(f, c, P) { b boleneck(p) le e be he forward edge corresponding o e R ugmening Pah lgorihm Ford-Fulkerson(G, s,, c) { foreach e E f(e) Cos - while (here exiss an s- pah P in Gf) { f ugmen(f, c, P) updae G f ugmen(f, c, P) { b boleneck(p) ieraions le e be he forward edge corresponding o e R ugmening Pah lgorihm Ford-Fulkerson(G, s,, c) { foreach e E f(e) Cos - while (here exiss an s- pah P in Gf) { f ugmen(f, c, P) updae G f ugmen(f, c, P) { b boleneck(p) le e be he forward edge corresponding o e R ieraions Toal = Slides - Nework Flow Inro.key - December,

5 ugmening Pah lgorihm Ford-Fulkerson(G, s,, c) { foreach e E f(e) while (here exiss an s- pah P in Gf) { f ugmen(f, c, P) updae G f ugmen(f, c, P) { b boleneck(p) le e be he forward edge corresponding o e R Cos O(m+n) O(m+n) o find pah, O(C) ieraions, where C = max flow ieraions Toal = - ugmening Pah lgorihm Ford-Fulkerson(G, s,, c) { foreach e E f(e) while (here exiss an s- pah P in Gf) { f ugmen(f, c, P) updae G f ugmen(f, c, P) { b boleneck(p) le e be he forward edge corresponding o e R Cos O(m+n) O(m+n) o find pah, O(C) ieraions, where C = max flow Toal = O(C(m+n)) ieraions Toal = - ugmening Pah lgorihm Ford-Fulkerson(G, s,, c) { foreach e E f(e) while (here exiss an s- pah P in Gf) { f ugmen(f, c, P) updae G f ugmen(f, c, P) { b boleneck(p) le e be he forward edge corresponding o e R Cos O(m+n) O(m+n) o find pah, O(C) ieraions, where C = max flow Toal = O(C(m+n)) ieraions Toal = O(C(m+n)) - a pseudopolynomial - Slides - Nework Flow Inro.key - December,

6 Cus n s- cu is a pariion (, ) of V wih s and. The capaciy of a cu (, ) is: Σ c(e) e ou of capaciy of cu = Cus capaciy of cu = = (Capaciy is sum of weighs on edges leaving.) Flow value lemma. Le f be any flow, and le (, ) be any s- cu. Then, he ne flow sen across he cu is equal o he amoun leaving s. value = Flows and Cus f(e) - f(e) = v(f) e ou of a e ino a Slides - Nework Flow Inro.key - December,

7 Flows and Cus Flow value lemma. Le f be any flow, and le (, ) be any s- cu. Then, he ne flow sen across he cu is equal o he amoun leaving s. value = f(e) - f(e) = v(f) e ou of a e ino a Flow value lemma. Le f be any flow, and le (, ) be any s- cu. Then, he ne flow sen across he cu is equal o he amoun leaving s. value = Flows and Cus f(e) - f(e) = v(f) e ou of a e ino a Slides - Nework Flow Inro.key - December,