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CS473-Algorithms I

CS473-Algorithms I. Lecture ? Network Flows Finding Max Flow. FORD-FULKERSON METHOD / ALGORITHM. • iterative algorithm: start with initial flow f = [ 0 ] with | f| = 0 at each iteration, increase | f | by finding an augmenting path

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CS473-Algorithms I

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  1. CS473-Algorithms I Lecture ? Network Flows Finding Max Flow Lecture ?

  2. FORD-FULKERSON METHOD / ALGORITHM • iterative algorithm: start with initial flow f =[0] with |f| = 0 • at each iteration, increase | f | by finding an augmenting path • repeat this process until no augmenting path can be found • by max-flow min-cut thm: upon termination this process yields a max flow FORD-FULKERSON-METHOD(G, s, t) initialize flow f to 0 while  an augmenting path p do augment flow f along path p returnf Lecture ?

  3. FORD-FULKERSON METHOD / ALGORITHM • basic Ford-Fulkerson Algorithm: data structures • note(u,v)  Ef only if (u,v)  E or (v,u)  E • maintain an adj-list representation of directed graph G' = (V', E'), where • E' = {(u,v): (u,v)  E or (v,u)  E}, i.e., • for each v  Adj[u] in G' maintain the record • note: G' used to represent both G and Gf , i.e., for any edge (u,v)  E' • c[u,v] > 0  (u,v)  E and cf[u,v] > 0  (u,v)  Ef v f(u,v) c(u,v) cf (u,v) Lecture ?

  4. FORD-FULKERSON METHOD / ALGORITHM • COMPUTE-GF (G', f) • for each edge (u,v)  E'do • ifc[u,v] – f [u,v] > 0 then • cf [u,v] ← c[u,v] – f[u,v] • else • cf[u,v] ← 0 • returnG' • CANCEL (G', u, v) • min← {f [u,v], f [v,u]} • f [u,v] ← f [u,v] – min • f [v,u] ← f [v,u] – min • FORD-FULKERSON (G', s, t) • for each edge (u,v)  E'do • f [u,v] ← 0 • cf [u,v] ← 0 • Gf ← COMPUTE-GF(G', f) • while  an st path p in Gfdo • cf (p) ← min {cf[u,v]: (u,v) p} • for each edge (u,v) pdo • f [u,v] ←f [u,v] + cf(p) • CANCEL(G', u, v) • Gf← COMPUTE-GF(G', f) Lecture ?

  5. FORD-FULKERSON ALGORITHM • augmenting path in Gf is chosen arbitrarily • performance if capacities are integers: O(E |f*|) • while-loop: time to find s t path in Gf = O(E') = O(E) • # of while-loop iterations:≤ |f*|, where f* = max flow • so, running time is good if capacities are integers and |f*| is small Lecture ?

  6. FORD-FULKERSON ALGORITHM Gf0 f0 u u 0/100 0/100 100 100 s 0/1 t s 1 t 0/100 100 0/100 100 v v p = < s, u, v, t > cf (p) = 1 | f1 | = | f0 | + | fp | = 0+1 = 1 Lecture ?

  7. FORD-FULKERSON ALGORITHM Gf1 f1 u u 1/100 0/100 100 99 1 s 1/1 t s 1 t 99 0/100 100 1 1/100 v v p = < s, v, u, t > cf (p) = 1 | f2 | = | f1 | + | fp | = 1+1 = 2 Lecture ?

  8. FORD-FULKERSON ALGORITHM Gf2 f2 u u 1/100 1/100 99 99 1 1 s 0/1 t s 1 t 99 99 1/100 1 1 1/100 v v p = < s, u, v, t > cf (p) = 1 | f3 | = | f2 | + | fp | = 2+1 = 3 Lecture ?

  9. FORD-FULKERSON ALGORITHM Gf3 f3 u u 2/100 1/100 98 99 2 1 s 1/1 t s 1 t 99 98 1/100 2 1 2/100 v v p = < s, v, u, t > cf (p) = 1 | f4 | = | f3 | + | fp | = 3+1 = 4 Lecture ?

  10. FORD-FULKERSON ALGORITHM • choose p = < s, u, v, t > in odd iterations • choose p = < s, v, u, t > in even iterations • 200 augmentations to reach |f*| = 200 • might never terminate for non-integer capacities • efficient algorithms: • augment along max-capacity path in Gf:not mentioned in textbook • augment along breadth-first path in Gf:Edmonds-Karp algorithm O(VE2) Lecture ?

  11. EDMONDS-KARP ALGORITHM • def: δf (s,v) = shortest path distance from s to v in Gf • unit edge weights in Gfδf (s,v) = breadth-first distance from s to v in Gf • L7: v V – {s,t}; δ(s,v) inGf’s increases monotonically with each augmentation • proof:suppose • (i) a flow f on G induces Gf • (ii) fp along an augmenting path in Gf produces f’ = f + fp on G • (iii) f’ on G induces Gf’ • notation:δ(s,v) = δf (s,v) and δ’(s,v) = δf’ (s,v) Lecture ?

  12. ANALYSIS OF EDMONDS-KARP ALGORITHM • want to prove: δ'(s,v) ≥ δ(s,v)  v V – {s,t} • for contradiction: assume δ'(s,v) < δ(s,v) for some v • without loss of generality: assume δ'(s,v) is minimal over all such vertices • i.e., δ'(s,v) < δ'(s,u)  u  V – {s,t} δ'(s,u) < δ(s,u) • thus δ'(s,u) < δ'(s,v) δ'(s,u) ≥ δ(s,u) (1) Lecture ?

  13. ANALYSIS OF EDMONDS-KARP ALGORITHM • consider a shortest sv path p': su → v in Gf’ • δ'(s,u) = δ'(s,v) – 1 by “subpath is also a shortest path” δ'(s,u) < δ'(s,v)  by (1), δ'(s,u) ≥ δ(s,u) (2) • consider this edge (u,v)  Ef': question: was (u,v)  Ef ? • YES:i.e., (u,v)  Ef'and (u,v)  Ef ; notef(u,v) < c(u,v) • δ(s,v) ≤ δ(s,u) + 1 by triangle inequality ≤ δ'(s,u) + 1 by (2) = δ'(s,v)  δ'(s,v) ≥ δ(s,v) contradiction Lecture ?

  14. ANALYSIS OF EDMONDS-KARP ALGORITHM • NO: i.e.,(u,v)  Ef' but (u,v)  Ef ; notef(u,v)=c(u,v) • fp pushes flow back along edge (u,v)  (v,u)  p in Gf • i.e., augmenting path p is of the form p = s v → u t • note: p is a breadth-first path from s to t (Edmond-Karp algorithm) • δ(s,v) =δ(s,u) – 1 by “subpath is also a shortest path” ≤ δ'(s,u) – 1 by (2) = (δ'(s,v) – 1) – 1 by “subpath is also a shortest path” = δ'(s,v) – 2 < δ'(s,v)  δ'(s,v) ≥ δ(s,v) contradiction Lecture ?

  15. ANALYSIS OF EDMONDS-KARP ALGORITHM • what is the bound on the total # of augmentations? • def: an edge (u,v) in Gf is critical on an augmenting path p in Gf if cf (u,v) = cf (p) • at least one edge on an augmenting path must be critical • a critical edge disappears from Gf after the augmentation • f'(u,v) = (f+fp)(u,v) = f (u,v) + fp(u,v) = f (u,v) + (c(u,v) – f (u,v)) = c(u,v)  cf'(u,v) = c'(u,v) – f'(u,v) = c(u,v) – c(u,v) = 0 Lecture ?

  16. ANALYSIS OF EDMONDS-KARP ALGORITHM • Thm:number of flow augmentations is O(EV) • proof: find a bound on the number of times an edge (u,v) becomes critical • let (u,v) becomes critical the first time along path p = s u → v t in Gf • δ(s,v) = δ(s,u) + 1 (1), since p is a shortest path • then (u,v) disappears from the residual network • (u,v) can reappear an another augmenting path • only after net flow from u to v is decreased • this only happens if (v,u) appears on an augmenting path Lecture ?

  17. ANALYSIS OF EDMONDS-KARP ALGORITHM • assume this event occurs along a path p' = s u → v t in Gf ' later • δ'(s,u) =δ'(s,v) + 1 since p' = s u → v t is a shortest path ≥ δ(s,v) + 1 by L7 =(δ(s,u)+ 1) + 1since p' = s u → v t is a shortest path = δ(s,u) + 2  δ'(s,u) ≥ δ(s,u) + 2 • i.e., δ(s,u) increase by ≥ 2 each time (u,v) becomes critical, except the first time • thus each edge (u,v) can become critical at most O(V) times, since • 1 ≤ δ(s,u) ≤ | V | - 2 and δ(s,u) never decreases by L7 • (E) edges in Gf’s # of flow augmentations = O(VE) • running time of Edmonds-Karp algorithm: |E|×O(VE) = O(E2V) • finding an augmenting path in a Gf = breadth-first search = O(E) Lecture ?

  18. MAXIMUM BIPARTITE MATCHING PROBLEM • many combinatorial optimization problems can be reduced to a max-flow problem • maximum bipartite matching problem is a typical example Lecture ?

  19. MAXIMUM BIPARTITE MATCHING PROBLEM • given an undirected graph G = (V, E) • def: a matching is a subset of edges M  E such that  v  V, at most one edge of M is incident to v • def: a vertex v  V is matched by a matching M if some edge M is incident to v, otherwise v is unmatched • def: a maximum matching M* is a matching M of maximum cardinality, i.e., |M*| ≥ |M| for any matchingM • def: G=(V,E) is a bipartite graph if V=LR where L∩R= such that E={(u,v): u  L and v  R} Lecture ?

  20. MAXIMUM BIPARTITE MATCHING PROBLEM • applications:job task assignment problem • Assigning a set L of tasks to a set R of machines • (u,v)  E task u  L can be performed on a machine v  R • a max matching provides work for as many machines as possible Lecture ?

  21. MAXIMUM BIPARTITE MATCHING PROBLEM L R L R • example:two matchings M1& M2on a sample graph with |M1| = 2 & |M2| = 3 u1 u1 v1 v1 u2 u2 v2 v2 u3 u3 v3 v3 u4 u4 v4 v4 u5 u5 M1 = {(u1,v1), (u3,v3)} M2 = {(u2,v1), (u3,v2) , (u5,v3)} Lecture ?

  22. FINDING A MAXIMUM BIPARTITE MATCHING • idea: construct a flow network in which flows correspond to matchings • define the corresponding flow network G'=(V',E') for the bipartite graph as • V' = V  {s}  {t} s, t V • E' = {(s,u):  u L}{(u,v): u  L, v  R, (u,v)  E} {(v,t):  v  R} • assign unit capacity to each edge of E' Lecture ?

  23. FINDING A MAXIMUM BIPARTITE MATCHING 1 u1 u1 1 v1 1 v1 1 u2 u2 1 1 1 v2 1 v2 1 t s u3 u3 1 1 1 v3 1 v3 1 u4 u4 1 1 v4 v4 1 u5 u5 Lecture ?

  24. FINDING A MAXIMUM BIPARTITE MATCHING • def: a flow f on a flow network is integer-valued if f(u,v) is integer u,v  V • L8:(a)IFM is a matching in G, THEN an integer-valued f on G' with | f | = |M|(b)IFf is an integer-valued f on G', THEN a matching M in G with | M | = | f | Lecture ?

  25. FINDING A MAXIMUM BIPARTITE MATCHING • proof L8 (a):let M be a matching in G • define the corresponding flow f on G'as • u,vM; f(s,u)=f(u,v)=f(v,t) = 1 & f(u,v) = 0 for all other edges • first show that f is a flow on G': • 1 unit of flow passes thru the path s → u → v → t for each u,v M • these paths are disjoint s t paths, i.e., no common intermediate vertices • f is a flow on G' satisfying capacity constraint, skew symmetry & flow conservation • because f can be obtained by flow augmentation along these s t disjoint paths Lecture ?

  26. FINDING A MAXIMUM BIPARTITE MATCHING • second show that | f | = |M|: • net flow accross the cut ({s}L, R {t}) = | f | by L3 • | f | = f (s L, R t) = f (s, R t) + f (L, R t) = f (s, R t) + f (L, R) + f (L, t) = 0 + f (L, R) + 0; f (s, R t) = f (L, t) since  no such edges = f (L, R) = |M| since f(u,v) = 1 uL, v R & (u,v) M Lecture ?

  27. FINDING A MAXIMUM BIPARTITE MATCHING • proof L8 (b):let f be a integer-valued flow in G' • defineM={(u,v): u  L, v  R, and f (u,v) > 0} • first show that M is a matching in G: i.e., all edges in M are vertex disjoint • let pe(u) / pl(u) = positive net flow entering / leaving vertex u, u  V • each u L has exactly one incoming edge (s,u) with c(s,u)=1  pe(u) ≤ 1 uL Lecture ?

  28. FINDING A MAXIMUM BIPARTITE MATCHING • since f is integer-valued; u L,pe(u) = 1 pl(u) = 1 due to flow conservation u L,pe(u)=1  exactly one vertex vR f(u,v) = 1 to make pl(u) = 1 • thus, at most one edge leaving each vertex u L carries positive flow = 1 • a symmetric argument holds for each vertex v  R • therefore, M is a matching Lecture ?

  29. FINDING A MAXIMUM BIPARTITE MATCHING • second show that |M| = | f |: • |M| = f (L, R) by above def for M since f (u,v) is either 0 or 1 = f (L, V' – s – L – t) = f (L,V') – f (L,s) – f (L,L) – f (L,t) = 0 – f (L,s) – 0 – 0 = – f (L,s) = f (s,L) = | f | due to skew symmetry & then def. f (L,V') = f (L,V') = 0 due to flow cons. f (L,t) = 0 since no edges from L to t Lecture ?

  30. FINDING A MAXIMUM BIPARTITE MATCHING • example: a matching M with |M| = 3 & a f on the corresponding G' with | f | = 3 u1 u1 v1 v1 1 1 u2 u2 1 1 v2 v2 1 1 t u3 s u3 v3 v3 1 1 u4 u4 v4 v4 1 u5 u5 Lecture ?

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