Beyond the flow decomposition barrier

1. Beyond the flow decomposition barrier Authors: A. V. Goldberg and Satish Rao (1997) Presented by Yana Branzburg Advanced Algorithms Seminar Instructed by Prof. Haim Kaplan Tel Aviv University, June 2007

2. Introduction • Maximum flow problem • Use the concept of distance • Distance is defined by arc length • Methods so far use unit length function • Goldberg&Rao use adaptive binary length function: length threshold is set relative to an estimate of residual flow

3. Ω(mn) barrier • Ω(mn) - a natural barrier for maximum flow algorithms • that output explicit flow decomposition (the total path length is Θ(mn) in the worst case) • that augment flow one path at the time, for each augmenting path, one arc at the time • Although the barrier does not apply to algorithms that use preflows or data structures like dynamic trees, no one beats it

4. Ω(mn) barrier (cont.)

5. Beyond the Ω(mn) barrier • Dinic’s algorithm for unit-capacity networks • Goldberg-Rao algorithm for general networks

6. Background • flow network: • G = (V, E) • source s, sink t • capacity function u: E {1, …, U} • flow function f: E  {1, …, U} • f(a) ≤ u(a) • conservation constraint • flow value is

7. Background (cont.) • residual capacity uf (v,w)= u(v,w) – f(v,w) + f(w,v) • Arc (v,w) is a residual arc if uf (v,w) > 0 • residual flow is the difference between optimal flow f* and current flow f • In blocking flow every directed path s-t contains an arc with residual capacity 0 • Without loss of generality assume: • G has no parallel arcs • if arc (v,w) is in G, but (w,v) is not, add (w,v) to G, define u(w,v)=0 • either f(v,w) = 0 or f(w,v) = 0

8. Dinic’s algorithm for unit-capacity networks

9. Dinic’s algorithm for unit capacity networks • A unit capacity network is one in which all arc have unit capacities • Algorithm of two phases: • Find augmenting paths, until distance between s and t is • Augment the found flow to an optimal flow

10. Dinic’s algorithm for unit capacity networks (cont.) • Algorithm Dinic • while ds ≤ α(ds stands for length of the shortest path from s to t) • Find blocking flow in admissible network • Compute residual network • while there is an s-t path in the residual network • Find an s-t path P • Augment one unit of flow along P • Compute residual network

11. Dinic’s algorithm for unit capacity networks - Analysis • Phase I: • Find a blocking flow: • Find a s-t path • Route one unit of flow • Delete all arcs of the path (all saturated) • Repeat until there is no s-t path • Each arc is looked only once, hence O(m) • Thus the first phase total takes O(αm)

12. Dinic’s algorithm for unit capacity networks – Analysis (cont.) • Phase II: • Blocking flow increases the distance between s and t by at least one • Let the vertices be partitioned into levels • Thus, at the end of phase I, there are at least α levels

13. Dinic’s algorithm for unit capacity networks – Analysis (cont.) • Lemma 1: At the end of phase I, the maximum flow in the residual network is at most α • Proof. Case 1:α = m½ • At least m½ levels and m arcs  there is a pair of adjacent levels Vk and Vk+1 which have at most m½ arcs between them • The cut has capacity at most m½  maximum residual flow is m½

14. Dinic’s algorithm for unit capacity networks – Analysis (cont.) • Proof. Case 2:α = 2n⅔ • There exists a pair of adjacent levels Vk and Vk+1 , each of which has at most n⅓nodes • Assume the contrary: at least half levels must contain more than n⅓ nodes • Then the total number of nodes more than α/2*n⅓≥ n, which is a contradiction • Thus the cut has capacity at most n⅔≤ α

15. Dinic’s algorithm for unit capacity networks – Analysis (cont.) • Each iteration of phase II routes one unit of additional flow • Since after phase I the maximum flow in the residual network ≤ α, the maximum number of iterations in phase II is bounded by α • One iteration of phase II is O(m) • Thus, the second phase takes total O(αm) • Hence, Dinic’s algorithm for unit capacity networks solves the maximum flow problem in time

16. Goldberg-Rao algorithm

17. Length Functions • length function ℓ: E R+ • distance labeling d: V R+ with respect to a length function ℓ if: • d(t) = 0 • for every (v,w) in E : d(v) ≤ d(w) + ℓ(v,w) • admissible arc (v,w) is a residual arc where d(v) > d(w) or d(v) = d(w) and ℓ(v,w) = 0 • In case of binary length function, admissible arc satisfies d(v) = d(w) + ℓ(v,w) • distance function dℓ(v): distance from v to t under ℓ • dℓ(v) is a distance labeling

18. Length Functions (cont.) • Claim 2:dℓ is the biggest distance labeling • Proof (assume the contrary): • d(v) >dℓ(v) for some v • w is the last node on a shortest v-t path Γthat satisfies d(w) >dℓ(w) • x follows w (w ≠t, hence x exists) • dℓ(w) < d(w) ≤ ℓ(w,x) +d(x) ≤ ℓ(w,x) +dℓ(x) • v-w concatenated with w-t is shorter than Γ • Contradiction to selection of Γ

19. Volume of Network • A residual arc a has • “width” uf(a) • length ℓ(a) • volume volf,ℓ(a)= uf(a) ℓ(a) • network volume Volf,ℓ = Σauf(a) ℓ(a) • The residual flow value is upper bounded by Volf,ℓ∕dℓ(s) (since any path from s to t has length at least dℓ(s) > 0) • To get a tighter bound, keep Volf,ℓ small by choosing a binary length function that is zero for large residual capacity arcs

20. Binary Blocking Flow Algorithm - Idea • Ensure the algorithms terminates after not too many augmentations by showing that every iteration increases dℓ(s) - but we have zero arcs?! • Contract strongly connected components (nodes) formed by zero-length arcs  • Bound the augmenting flow in a node by Δ to be able to reroute the flow in a restored node  • Possible not to find a blocking flow  can not guarantee increase of dℓ(s)  • Choose Δ that insures small number of non-blocking augmentations

21. Stopping Condition of the Algorithm • Define F - an upper bound on the residual flow value • Update F every time our estimate improves • Stop when F < 1 (integral capacities) • Initial estimate is F = Σ(s,w)u(s,w) ≤ nU

22. The Skeleton of Goldberg-Rao algorithm • while F ≥ 1 do • Compute Δ • while the residual capacity > F /2 • Compute length function ℓ, distance function dℓ • Find admissible graph A(f, ℓ, dℓ) • Contract strongly connected components of A • Find a flow of value Δ or a blocking flow • Augment the current flow, extend it to original graph and recompute the residual graph • Update F phase update step

23. Estimate a residual capacity • A residual capacity uf(S,T) of an s-t cut can serve as an upper bound • canonical cut (Sk ,Tk) : • (Sk ,Tk) is an s-t cut • Initially (S,T) =({s },V \{s })is the current cut • Find a canonical cut with the smallest residual capacity • When the capacity of the found cut ≤ F /2, update F to this capacity value

24. Estimate a residual capacity (cont.) • Claim 3: The minimum capacity canonical cut can be found in O(m) time • Proof: • Every arc may cross at most one canonical cut (its length at most 1) • For k =1,…,dℓ(s), initialize uf(Sk,Tk) =0 • For every arc (v,w) : if dℓ(v) > dℓ(w), increase uf(Sk,Tk) by uf(v,w), k = dℓ(v) • Find minimum among all uf(Sk,Tk) • Denote the minimum capacity canonical cut

25. Binary length function • A first estimate of a length function: * this length function will have to be modified for reasons that will come up during time analysis • Given ℓ, next we show how to handle the contraction of zero arcs

26. Extending the flow inside contracted components • Claim 4: Given a flow of value at most Δ, we can correct it to a flow of the same value in the original graph in O(m) time • Proof (for one component): • Choose an arbitrary vertex as a root • Form an in-tree and an out-tree • in-tree: a rooted tree with a path from each node to the root • out-tree: a rooted tree with a path from the root to each node

27. Extending the flow inside contracted components (cont.) • Proof (cont.): • For each vertex compute its flow balance: sum of flow entering minus sum of flow leaving the vertex • Route the positive balances to the root using the in-tree • Route the resulting flow excess from the room to negative balances via out-tree • At most Δ flow is routed to and from the root using one arc, which residual capacity is at least 2Δ

28. in tree out tree Extending the flow inside contracted components (cont.) • Example: 0.5Δ +0.5Δ 0.5Δ +Δ Δ +Δ 0.5Δ 0.75Δ +0.5Δ -0.75 Δ Δ +0.25Δ Δ R -0.25 Δ Δ 0.25Δ

29. Time bounds • Reminder: choice of Δ must insure small number of non blocking augmentations • Denote • Let • Thus, number of non blocking augmentations of value Δ is at most α in each phase • How many blocking augmentations per phase there are? Need to show that O(α)

30. Bound of number of blocking augmentations • Lemma 5: For a 0-1 length function on the residual graph: where M is the maximum residual capacity of a length one arc • Proof: • Every arc crosses at most one of the canonical cuts, for such an arc, ℓ(a) =1 • Total capacity of length one arcs ≤ mM, there are dℓ(s) canonical cuts • Thus, minimum capacity canonical cut ≤ mM / dℓ(s)

31. Bound of number of blocking augmentations (cont.) For dense graphs, the following Lemma gives better bounds • Lemma 6: For a 0-1 length function on the residual graph: • Proof: • Let • Since , there are at most dℓ(s)/2 values of k where |Vk | > 2n /dℓ(s) • Thus, at least values of k have |Vk | ≤ 2n /dℓ(s) • By the pigeonhole principle, there is j such that |Vj | ≤ 2n /dℓ(s) and |Vj+1 | ≤ 2n /dℓ(s) • Since all arcs from Vj+1 to Vjhave length 1, Lemma 6 holds

32. Bound of number of blocking augmentations (cont.) • Lemma 7: In every iteration of a blocking flow, dℓ(s) increases by at least 1 (prove later) • Lemma 8: There are at most O(α) blocking flow augmentations in a phase • Proof: Case 1:α = m½ • Using Lemma 7: after 4m½ blocking flow update steps dℓ(s) ≥ 4m½ • M ≤ 2Δ. Thus, together with Lemma 5:

33. Bound of number of blocking augmentations(cont.) • Proof. Case 2:α = n⅔ • Using Lemma 7: after 4n ⅔blocking flow update steps dℓ(s) ≥ 4n ⅔ • M ≤ 2Δ. Thus, together with Lemma 6: • Conclusion: Each phase of the algorithm terminates in O(α) update steps

34. Finding a flow in one update step • We use Goldberg&Tarjan algorithm for computing a blocking flow X in each update step in O(mּlog(n2/m)) time • If X > Δ, return extra flow to s in O(m) time: • Place X – Δ units of excess at t • Process vertices in reverse topological order • For current vertex reduce flow on the incoming arcs to eliminate its excess

35. Time Bounds Summary • Theorem 9: The maximum flow problem can be solved in O(αּmּlog(n2/m)ּlogU) time • Proof: • Δ>U : all arcs are length one • Each update step finds a blocking flow and increases dℓ(s) by at least one • After αsteps dℓ(s) ≥αF ≤αU • After O(α) steps F =αΔ≤αU  Δ≤U • Total time till Δ =U is O(αּmּlog(n2/m)) F =uf(S,T) ≤ mM /dℓ(s) ≤ mM /α = αM ≤ αU

36. Time Bounds Summary (cont.) • Proof (cont.): • 1≤Δ≤U :Number of phases is O(logU), since each phase decreases twice • Total time for these phases is O(αּmּlog(n2/m)ּlogU) • When Δ=1, F ≤α, the algorithm ends in O(α) update steps

37. Proof of Lemma 7 • In every iteration of a blocking flow, dℓ(s) increases by at least 1 • Denote: • ℓ anddℓ - initial length and distance functions • ℓ’ anddℓ’ - after finding a blocking flow and recomputing the residual graph • f and f’ - the flows before and after augmentation respectively • Need to prove: dℓ(s) < dℓ’ (s)

38. Proof of Lemma 7 (cont.) • Claim 10:dℓ is a distance labeling with respect to ℓ’ • Proof: • By definition: dℓ(v) ≤ dℓ(w) + ℓ(v,w) • Need to show: dℓ(v) ≤ dℓ(w) + ℓ’(v,w) • If dℓ(v) ≤ dℓ(w) – trivial • dℓ(v) >dℓ(w) (w,v) is not admissible (for admissible arc dℓ(w)=dℓ(v)+ℓ(w,v)) •  uf’ (v,w) ≤ uf(v,w) •  ℓ’(v,w) ≥ ℓ(v,w)

39. Proof of Lemma 7 (cont.) • Claim 11:dℓ(s)≤dℓ’ (s) • Proof: • By Claim 2: for any distance labeling d, d(v) ≤ dℓ(v) • By Claim 10: dℓ is a distance labeling with respect to ℓ’ •  dℓ(s) ≤ dℓ’ (s)

40. Proof of Lemma 7 (cont.) • We need to show dℓ(s) < dℓ’ (s) • Let Γ be an s-t path in Gf’ • By Claim 10: cost(v,w) =dℓ(w) -dℓ(v) +ℓ’(v,w) ≥0 • ℓ’ (Γ) = dℓ(s) –dℓ(v1) +c(s,v1) + dℓ(v1) –dℓ(v2) +c(v1,v2) +…+ dℓ(vk) –dℓ(t) +c(vk,t) = dℓ(s) – dℓ(t) +c (Γ) = dℓ(s) +c (Γ)

41. Proof of Lemma 7 (cont.) • dℓ’ (s) = ℓ’ (Γ) = dℓ(s) + c (Γ) • Enough to show that along every s-t path Γ in Gf’ there is an arc (v,w) satisfying c(v,w) > 0

42. Proof of Lemma 7 (cont.) dℓ(v) >dℓ(w): dℓ(v) ≤ dℓ(w) + ℓ(v,w) < dℓ(v) + ℓ(v,w)  ℓ(v,w) > 0  dℓ(v) = dℓ(w) + ℓ(v,w) • We routed a blocking flow  Γ contains arc (v,w) that is not in A(f, ℓ, dℓ) • dℓ(v) ≤ dℓ(w) either because: • (v,w) in Gf(otherwise (v,w) would be in A(f, ℓ, dℓ)) or • (v,w) not in Gfbut in Gf’ (means that (w,v) is in A(f, ℓ, dℓ) dℓ(w) = dℓ(v) + ℓ(v,w) dℓ(w) ≥dℓ(v))

43. Proof of Lemma 7 (cont.) • Assume the contrary that c(v,w)=dℓ(w) -dℓ(v) +ℓ’(v,w) =0 • dℓ(v) ≤dℓ(w)  dℓ(w) =dℓ(v) andℓ’(v,w) =0 • Case 1 -(v,w) in Gf : • 1=ℓ(v,w) >ℓ’(v,w) =0(w,v) is in A(f, ℓ, dℓ) • Case 2 - (v,w) not in Gf but in Gf’ : • (w,v) is in A(f, ℓ, dℓ) • But dℓ(w) =dℓ(v) ℓ(w,v) = 0

44. Bad Example • dℓ(w) =dℓ(v), ℓ(w,v) =0 and ℓ’(v,w) =0 • Δ ≤ uf(v,w) < 2Δ • uf(w,v) > 2Δ • uf’ (v,w) ≥ uf(v,w) +2Δ – uf(v,w) ≥ 2Δ • ℓ’(v,w) = 0 w v ≥2Δ- uf(v,w)

45. Solution to bad example • Distort the length function, so that such bad arcs get length zero ahead • Might become an inner arc in a strongly connected component • The capacity of such arcs must be at least 2Δ in order to be able to route a flow of Δ in flow extension(remember example…)

46. A correct length function ℓ • Changes due to new ℓ in Lemma 8 (p. 32): • Case 1 - 6m½ blocking flow update steps instead of 4m½ • Case 2 - 5n ⅔blocking flow update steps instead of 4n ⅔

47. Length function ℓ* • (v,w) is a special arc if: • 2Δ ≤ uf(v,w) < 3Δ • d(v) =d(w) • uf(w,v) ≥ 3Δ • Define new length function ℓ* that: • = zero on special arcs • = ℓ on all other arcs • dℓ* = dℓ • The algorithm determines admissible graph A(f, ℓ*, dℓ) instead of A(f, ℓ, dℓ)

48. Proof of Lemma 7 using ℓ* • dℓ(w) =dℓ(v), ℓ(w,v) =0 and ℓ’(v,w) =0 • ℓ(w,v) =0 uf(w,v) ≥ 3Δ • ℓ’(v,w) =0  uf’(v,w) ≥ 3Δ • (w,v) is in A(f, ℓ*, dℓ)  pushing a flow (at most Δ) increased uf’(v,w) •  uf(v,w) ≥ 2Δ •  Either (v,w) is a special arc before augmentation or uf(v,w) ≥ 3Δ •  (v,w) is in A(f, ℓ*, dℓ) - contradiction

49. 12 16 20 9 T 4 S 10 7 4 13 14 Example while F ≥ 1 do Compute Δ,ℓ, dℓ while (S,T) > F /2 Compute ℓ* Find A(f, ℓ*, dℓ) Contract SCC of A Find a flow f* f’ = f + f* Extend f* to G Recompute ℓ,dℓ,Gf’ Update F • n = 6, m = 10 • U = 20 • F = 29 • α = min{m½, n ⅔ } = 4 • Δ = F /α=8

50. 12 1 1 4/12 2 1 2 1 Gf : Af : 1 1 4/16 1 16 20 4/20 1 1 1 9 0 0 4 3 3 10 7 1 1 4 1 1 4/4 13 4/13 1 1 1 1 2 1 2 1 14 4/14 Example (cont.) while F ≥ 1 do Compute Δ,ℓ, dℓ while (S,T) > F /2 Compute ℓ* Find A(f, ℓ*, dℓ) Contract SCC of A Find a flow f* f’ = f + f* Extend f* to G Recompute ℓ,dℓ,Gf’ Update F F = 29.Δ=8