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CS473-Algorithms I

CS473-Algorithms I. Lecture ? Network Flows Flow Networks & Flows. FLOW NETWORKS & FLOWS. • A flow network: a directed graph G = ( V , E ) Two distinguished vertices : a source s and a sink t Each edge has a nonegative capacity c( u , v )  0

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CS473-Algorithms I

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  1. CS473-Algorithms I Lecture ? Network Flows Flow Networks & Flows Lecture ?

  2. FLOW NETWORKS & FLOWS •A flow network: a directed graph G = ( V , E ) • Two distinguished vertices : a source s and a sink t • Each edge has a nonegative capacity c(u,v)  0 (if (u,v)  E then c(u,v) = 0) • for convenience :  v V – { s,t }, sv t, i.e., every vertex v lies on some path from s to t. Lecture ?

  3. FLOW NETWORKS & FLOWS •A positive flow p on G: a fn p:VxV  R0 satisfying • capacity constraint: 0  p(u,v)  c(u,v),  u,v V • i.e., flow from one vertex to another cannot exceed the capacity • note : p(u,v)> 0  (u,v)  E with c(u,v) > 0 • flow conservation: Kirschoff’s current law • total positive flow leaving a vertex = total positive flow entering the vertex  u  V- { s,t } Lecture ?

  4. FLOW NETWORKS & FLOWS • value of a positive flow: • a sample flow network G and a positive flow p on G: p/c for every edge note: flow  capacity at every edge 2/2 v1 v3 1/3 2/3 note: flow conservation holds at every vertex (except s and t) 1/1 1/3 s t 2/3 1/2 0/1 2/2 1/2 v2 v4 2/3 Lecture ?

  5. FLOW NETWORKS & FLOWS 2/2 v1 v3 1/3 2/3 1/1 1/3 s t 2/3 1/2 0/1 2/2 1/2 v2 v4 2/3 Lecture ?

  6. FLOW NETWORKS & FLOWS • cancellation : can say positive flow either goes from u to v or from v to u, but not both • if not true, can transform by cancellation to be true • e.g.,  3 units of net positive flow from u to v •  • can be obtained by canceling 2 units of flow in each direction • capacity constraint still satisfied : flows only decrease • flow conservation still satisfied : flow-in & flow-out both reduced by the same amount 5/6 u v 2/4 3/6 u v 0/4 Lecture ?

  7. NET FLOW VERSUS POSITIVE FLOW DEFINITIONS • positive flowis more intuitive • net flow brings mathematical simplification: half as many summations to write • A net flow f on G: a fn f:VxV  Rsatisfying • capacity constraint:  u,v V f(u,v)  c(u,v) • skew symmetry:  u,v V f(u,v)= - f(v,u) • thus, f(u,u)= - f(u,u)  f(u,u)=0  net flow from a vertex toitself is 0 • flow conservation:  u V- {s,t}, • total net flow into a vertex is 0 • Nonzero net flow from u to v(u,v)  E, or (v,u)  E, or both. • value of a net flow : |f|= net flow out of the source Lecture ?

  8. NET FLOW VERSUS POSITIVE FLOW DEFINITIONS • equivalence of net flow and positive flow definitions: • define net flow in terms of positive flow: • f(u,v)= p(u,v) - p(v,u) • Given definition of p, this def. of f satisfies (1)capacity constraint, (2)skew symmetry, and (3)flow constraint. Lecture ?

  9. NET FLOW VERSUS POSITIVE FLOW DEFINITIONS • define positive flow in terms of net flow: f(u,v), if f(u,v) > 0 0, if f(u,v)  0 • Given definition of f, this def. of p satisfies (1)capacity constraint, (2)flow constraint. p(u,v)= Lecture ?

  10. NET FLOW VERSUS POSITIVE FLOW DEFINITIONS split summation because of skew symmetry • proof of (2)  flow conservation in net flow model means that •  positive net flow leaving a vertex = positive net flow entering the vertex Lecture ?

  11. FLOW NETWORKS & MAXIMUM FLOW PROBLEM • maximum flow problem : given a flow network G with source s and sink t • find a flow of maximum value from s to t • flow network with multiple sources and sinks : • a flow network G with m sources {s1, s2,..., sm}= Sm and n sinks {t1, t2,..., tn}= Tn • Max flow problem : find a flow of max value from m sources to n sinks • Can reduce to an ordinary max-flow problem witha single source & a single sink Lecture ?

  12. FLOW NETWORKS & MAXIMUM FLOW PROBLEM • add a supersources and a supersinkt such that • Add a directed edge (s,si) with capacity c(s,si)= for i=1,2,...,m • Add a directed edge (t,ti) with capacity c(t,ti)= for i=1,2,...,n • i.e., V = V  {s,t}; E = E {(s,si) with c(s,si)=:si}{(t,ti) with c(t,ti)=:ti} Lecture ?

  13. FLOW NETWORKS & MAXIMUM FLOW PROBLEM Example: A flow network with multiple sources and sinks 10 10 s1 s1 14  14 6 v1 6 v1 s2 s1   4 6 t1 4 6 t1 s v1 v1 7 7 t  4 4 s3 s1 t1 t1 5 5   v1 7 v1 7 s4 s1 3 3 Lecture ?

  14. NET FLOW VERSUS POSITIVE FLOW DEFINITIONS • Claim : any flow in a multiple-source multiple-sink flow network G = (v, t) • Corresponds to a flow of identical value in G = ( v, t ) • Proof : define f for G interms of G Lecture ?

  15. IMPLICIT SUMMATION NOTATION • functions over sets imply summation; because will write lots of summations. where X,Y  V Example: value of a flow f = Example: flow conservation : Lecture ?

  16. IMPLICIT SUMMATION NOTATION • L1 : let G=(V,E) be a flow network, and let f be a flow on G, then: • (a) For X  V; f(X,X) = 0 • Because each f(u,v) canceled by f(v,u)= – f(u,v) due to skew symmetry. • (b) for X,Y  V; f(X,Y) = – f(Y,X) • Generalization of f(X,X)=0, same reasoning • (c) for X,Y,Z  V, with XY= • f(X Y,Z) = f(X,Z)+f(Y,Z) & f(Z,X Y) = f(Z,X)+f(Z,Y) • Split summation into sum of two summations, one over X, one over Y Lecture ?

  17. IMPLICIT SUMMATION NOTATION f(x,y) X X Y f(y,x) f(X,X) = 0 f(X,Y) = – f(Y,X) XY X f(x,z) Z Y f(y,z) f(XY,Z) = f(X,Z) + f(Y,Z) Lecture ?

  18. IMPLICIT SUMMATION NOTATION • (d) For any X  V – {s,t}; f(X,V) = 0  f(V,X) = – f(X,V) = 0 due to flow conservation • L2: |f| = f(V,t) Lecture ?

  19. CUTS OF FLOW NETWORKS & UPPER BOUND ON MAX FLOW • def: A cut(S, T) of a flow network is a partition of V into S and T = V-S such that s S and t  T • Similar to the cut definition given for MST • Differences: G is a directed graph here & we insist that s S and t  T • def: f(S, T) is the net flow across the cut (S, T) of G for a flow f on G • Add all edges S T and negative of all edges T  S due to skew symmetry • def: c(S, T) is the capacity across the cut (S, T) of G • not like flow because no skew symmetry; just add edges S T (no neg. values) Lecture ?

  20. CUTS OF FLOW NETWORKS & UPPER BOUND ON MAX FLOW Example: (S1, T1) = ({s, a, b}, {c, d, t}) • f(S1, T1) = f(a, c) + f(b, c) + f(b, d) = 12 + (-4) + 11 = 19 • c(S1, T1) = c(a, c) + c(b, d) = 12 + 14 = 26 (S2, T2) = ({s, a}, {b, c, d, t}) • f(S2, T2) = f(s, b) + f(a, b) + f(a, c) = 8 + (-1) + 12 = 19 • c(S2, T2) = c(s, b) + c(a, b) + c(a, c) = 13 + 10 + 12 = 35 12/12 a c 11/16 15/20 1/4 s 0/10 4/9 7/7 t 8/13 4/4 b d 11/14 S2 T2 S1 T1 T1 Lecture ?

  21. WHY WE NEED NEGATIVE FLOWS • Reasons behind negative flow across a cut : ?/10 5/10 a c a c ?/10 ?/5 5/10 5/5 s s ?/5 t 0/5 t ?/5 5/5 b d ?/10 b d 5/10 ?/10 5/10 S T S T f(S,T) = f(a,c) – f(c,b) + f(b,d) = 5 – 0 + 5 = 10 = |f| no neg. Flow across the cut Lecture ?

  22. WHY WE NEED NEGATIVE FLOWS • Exploit capacity c(c,b) = 5 to induce negative flow across (S,T) • i.e., sacrifice negative flow of 5 for a better utilization of c(a,c) = 10 & c(b,d) = 10 10/10 a c f '(S,T) = f(a,c) – f(c,b) + f(b,d) = (5+5) – (5) + (5+5) = 10 – 5 – 10 = 15 = | f | 10/10 5/5 s 5/5 t 5/5 b d 10/10 10/10 S T Lecture ?

  23. CUTS OF FLOW NETWORKS & UPPER BOUND ON MAX FLOW • L3: For any flow f on G, the net flow across any cut (S, T) of G = |f| • f(S, T) = f(S, V-S) = f(S, V) – f(S, S) by L1(c)= f(S, V) + 0 = f(S, V) by L1(a)= f(s  (S – s), V) = f(s, V) + f(S – s, V) by L1(c)= f(s, V) + 0 = f(s, V) = |f| by L1(d)since S-s  V – {s, t} Lecture ?

  24. CUTS OF FLOW NETWORKS & UPPER BOUND ON MAX FLOW • C1: the value of any flow f in G is bounded above by the capacity of any cut of G • Let (S, T) be any cut of G • |f| = f(S, T) = Lecture ?

  25. RESIDUAL NETWORKS • intuitively: the residual network Gfof a flow network G with a flow f • Consists of edges that can admit more flow • def:given a flow network G = (V, E) with a flow f • residual capacity of (u, v): cf (u, v) = c(u, v) – f(u, v) • cf (u, v): additional flow we can push from u to v without exceeding c(u, v) • residual network of G induced by f is the graph Gf = (V, Ef) with • Strictly positive residual capacities: Ef = {(u, v)VxV: cf (u, v) > 0} Lecture ?

  26. RESIDUAL NETWORKS • recall: if both (u, v) E and (v, u) E then • Transform such that either f(u, v) = 0 or f(v, u) = 0 by cancellation • examples: (1)both (u, v)  E and (v, u)  E cf (u, v) = c(u, v) – f(u, v) = 5 – 3 = 2 > 0 cf (v, u) = c(v, u) – f(v, u) = 1 –(–3) = 4 > 0 3/5 2 u v u v 0/1 4 Lecture ?

  27. RESIDUAL NETWORKS (2)(u, v) E but (v, u)  E and f(u, v) ≥ 0: (v, u) becomes an edge of Ef • |Ef| ≤ 2|E|, since (u,v)Efonly if at least one of (u,v) and (v,u) is in E • note: cf (u, v) + cf (v, u) = c(u, v) + c(v, u) • cf (u, v) = 5 – 5 = 0 ≤ 0× • cf (v, u) = 0 – (– 5) = 5 > 0 √ 5/5 5 u v u v 2 • cf (u, v) = 5 – 3 = 2 > 0 √ • cf (v, u) = 0 – (– 3) = 3 > 0 √ 3/5 u v u v 3 Lecture ?

  28. FLOW SUMS • def:let f1and f2be flows on G flow sumf1+f2: VxV  R  (f1+f2)(u,v) = f1(u,v)+f2(u,v) • L4:flow sum f1+ f2 satisfies (2) skew symmetry & (3) flow conservation • (2)(f1+ f2)(u, v) = f1(u, v) + f2(u, v) = – f1(v, u) – f2(v, u) = – (f1(v, u) + f2(v, u)) = – (f1+ f2)(u, v) • (3) • note:flow sum may violate(1) capacity constraint • f1(u,v), f2(u,v) ≤ c(u,v) may not mean that f1(u,v)+f2(u,v) ≤ c(u,v) Lecture ?

  29. FLOW SUMS • How a flow on Gf relates to a flow on the original network G? • L5: let f be a flow on G and let f' be a flow on Gf then, flow sumf+f' is a flow on G with value |f+f'| = |f|+|f'| • First verify that f+f'is a flow on G • ie, show that f+f'satisfies (1) capacity constraint, (2) skew symmetry, (3) flow constraint • note that f' is also a flow on G • so, by L4, f+f'already satisfies (2) and (3) on G • thus it remains to show that f+f'satisfies (1) on G • (f+f')(u, v) = f(u, v)+f'(u, v)≤ f(u, v)+cf(u, v) = f(u, v)+(c(u, v) – f(u, v)) = c(u, v) Lecture ?

  30. FLOW SUMS • Second show that |f+f'| = |f|+|f'| • |f+f'| = = = |f|+|f'| Lecture ?

  31. AUGMENTING PATHS • For a flow f on G • augmenting path p is a simple path from s to t in Gf • cf(p): residual capacity of a path p = • i.e., cf(p) = max. amount of the flow we can ship along edges of p on Gf Lecture ?

  32. AUGMENTING PATHS • L6: let f be a flow on G, and let p be an augmenting path in Gf. Let fp be a flow on Gfwith value |fp| = cf (p) > 0 defined as cf(p) if (u,v)  p in Gf fp(u,v) = 0 otherwise then, f’ = f+fpis a flow on G with value |f’| = |f| + |fp| > |f| Lecture ?

  33. AUGMENTING PATHS 3/5 2/6 5/5 2/3 2/4 • example: a single path p on G • can easily say that it is not all of G since flow is not conserved on p • define flow fp on Gfwith cf (p) = min{2, 4, 5, 2, 2, 2} = 2 p on G with f: s t 2 4 2 2 5 p on Gf : s t 3 2 1 2 2/2 2/4 2/2 2/2 2/5 fp on Gf : s t Lecture ?

  34. AUGMENTING PATHS 3/5 2/6 5/5 2/3 2/4 p on G with f: s t • example (cont.): a single path p on G • flow on p in G that results from augmenting along path p: 2/2 2/4 2/2 2/2 2/5 fp on Gf : s t 5/5 4/6 3/5 0/3 4/4 f + fp on p of G: s t +2 +2 -2 -2 +2 Lecture ?

  35. MAX-FLOW MIN-CUT THEOREM • Thm (max-flow min-cut): the following are equivalent for a flow f on G • (1)f is a maximum flow • (2)Gfcontains no augmenting paths • (3) |f| = c(S, T) for some cut (S, T) of G Lecture ?

  36. MAX-FLOW MIN-CUT THEOREM • proof (1) (2): f is a max flow on G • for contradiction, assume that G contains an augmenting path p with cf(p) > 0 • then, by L6 flow sum f+fp is a flow on Gf with value > |f|, contradiction Lecture ?

  37. MAX-FLOW MIN-CUT THEOREM • proof (2) (3): Gfcontains no augmenting path, i.e., Gfcontains no path from s to t • define S = s  {u:  a path from s to u in Gf} • define T = V – S • (S, T) is a cut: s  S & t  S  t  T because  no path from s to t in Gf Lecture ?

  38. MAX-FLOW MIN-CUT THEOREM • proof (2) (3) (cont.): • for each u  S and v  T, we have f(u, v) = c(u, v) • Otherwise cf(u,v) = c(u,v) – f(u,v) > 0  v  S, contradiction u v v u so, by L3, | f | = f(S, T) =c(S,T) s …. …. u v S V-S Lecture ?

  39. MAX-FLOW MIN-CUT THEOREM • proof (3) (1): |f| = c(S,T) for some cut (S,T) of G • already proved by C1 that | f | ≤ c(S, T) for all cuts (S, T) • so, | f | = c(S, T) implies that f is a max flow Lecture ?

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