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

• First studied in Soviet Union by the National
  Commissariat of Transportation in early (19)30s
  by A.N. Tolstoi
• Solution for ship-ping cargo via the
  railroad system

10 sources, 68 destinations, 155 links
3
Introduction

• Maximum flow problem
• Use the concept of distance
• Distance is defined by arc length
• Methods so far use unit length function
• GoldbergRao use adaptive binary length function
  length threshold is set relative to an estimate
  of residual flow

4
O(mn) barrier

• O(mn) - a natural barrier for maximum flow
  algorithms
• that output explicit flow decomposition (the
  total path length is T(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

5
O(mn) barrier (cont.)
6
Beyond the O(mn) barrier

• Dinics algorithm for unit-capacity networks
• Goldberg-Rao algorithm for general networks

7
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

8
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) gt 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

9
Dinics algorithm for unit-capacity networks
10
Dinics 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

11
Dinics algorithm for unit capacity networks
(cont.)

• Algorithm Dinic
• while ds a (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

12
Dinics 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(am)

13
Dinics 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 a
  levels

14
Dinics 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 a
• Proof. Case 1 a m½
• At least m½ levels and m arcs ? there is a pair
  of adjacent levels Vk and Vk1 which have at
  most m½ arcs between them
• The cut has capacity at
  most m½ ? maximum residual flow is m½

15
Dinics algorithm for unit capacity networks
Analysis (cont.)

• Proof. Case 2 a 2n ?
• There exists a pair of adjacent levels Vk and
  Vk1 , each of which has at most n ? nodes
• Assume the contrary at least half levels must
  contain more than n ? nodes
• Then the total number of nodes more than
• a/2n ? n, which is a contradiction
• Thus the cut has capacity
  at most n ? a

16
Dinics 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 a, the maximum number of
  iterations in phase II is bounded by a
• One iteration of phase II is O(m)
• Thus, the second phase takes total O(am)
• Hence, Dinics algorithm for unit capacity
  networks solves the maximum flow problem in
• time

17
Goldberg-Rao algorithm
18
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) gt 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

19
Length Functions (cont.)

• Claim 2 dl is the biggest distance labeling
• Proof (assume the contrary)
• d(v) gtdl(v) for some v
• w is the last node on a shortest v-t path G that
  satisfies d(w) gtdl(w)
• x follows w (w ?t, hence x exists)
• dl(w) lt d(w) l(w,x) d(x) l(w,x) dl(x)
• v-w concatenated with w-t is shorter than G
• Contradiction to selection of G

20
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 Sauf(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) gt 0)
• To get a tighter bound, keep Volf,l small by
  choosing a binary length function that is zero
  for large residual capacity arcs

21
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 ? 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 ? that insures small number of
  non-blocking augmentations

22
Confused?!

• Bold edges strongly connected components
• Edges with X are deleted for having a blocking
  flow and no more capacity

23
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 lt 1 (integral capacities)
• Initial estimate is F S(s,w)u(s,w) nU

24
The Skeleton of Goldberg-Rao algorithm

• while F 1 do
• Compute ?
• while the residual capacity gt 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 ? or a blocking flow
• Augment the current flow, extend it to original
  graph and recompute the residual graph
• Update F

phase
update step
25
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

26
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) gt 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

27
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

28
Extending the flow inside contracted components

• Claim 4 Given a flow of value at most ?, 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

29
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 ? flow is routed to and from the root
  using one arc, which residual capacity is at
  least 2?

30
Extending the flow inside contracted components
(cont.)

• Example

0.5?
0.5?
0.5?
?
?
?
0.5?
0.75?
0.5?
-0.75 ?
?
0.25?
?
R
-0.25 ?
?
0.25?
31
Time bounds

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

32
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)
  

33
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 gt 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 Vj1 2n /dl(s)
• Since all arcs from Vj1 to Vj have length 1,
  Lemma 6 holds

34
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(a) blocking flow
  augmentations in a phase
• Proof Case 1 a m½
• Using Lemma 7 after 4m½ blocking flow update
  steps dl(s) 4m½
• M 2?. Thus, together with Lemma 5

35
Bound of number of blocking augmentations (cont.)

• Proof. Case 2 a n ?
• Using Lemma 7 after 4n ? blocking flow update
  steps dl(s) 4n ?
• M 2?. Thus, together with Lemma 6
• Conclusion Each phase of the algorithm
  terminates in O(a) update steps

36
Finding a flow in one update step

• We use GoldbergTarjan algorithm for computing a
  blocking flow X in each update step in
  O(m?log(n2/m)) time
• If X gt ?, return extra flow to s in O(m) time
• Place X ? units of excess at t
• Process vertices in reverse topological order
• For current vertex reduce flow on the incoming
  arcs to eliminate its excess

37
Time Bounds Summary

• Theorem 9 The maximum flow problem can be solved
  in O(a?m?log(n2/m)?logU) time
• Proof
• ?gtU all arcs are length one
• Each update step finds a blocking flow and
  increases dl(s) by at least one
• After a steps dl(s) a ?F aU
• After O(a) steps F a ? aU ? ?U
• Total time till ? U is O(a?m?log(n2/m))

F uf(S,T) mM /dl(s) mM /a aM aU
38
Time Bounds Summary (cont.)

• Proof (cont.)
• 1?U Number of phases is O(logU), since each
  phase decreases twice
• Total time for these phases is O(a?m?log(n2/m)?log
  U)
• When ?1, F a, the algorithm ends in O(a) update
  steps

39
Proof of Lemma 7

• In every iteration of a blocking flow, dl(s)
  increases by at least 1
• Denote
• l and dl - initial length and distance functions
• l and dl - 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) lt dl (s)

40
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) gtdl(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)

41
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)

42
Proof of Lemma 7 (cont.)

• We need to show dl(s) lt dl (s)
• Let G be an s-t path in Gf
• By Claim 10
• cost(v,w) dl(w) -dl(v) l(v,w) 0
• l (G) 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 (G) dl(s) c (G)

43
Proof of Lemma 7 (cont.)

• dl (s) l (G) dl(s) c (G)
• Enough to show that along every s-t path G in Gf
  there is an arc (v,w) satisfying c(v,w) gt 0

44
Proof of Lemma 7 (cont.)

• dl(v) gtdl(w)
• dl(v) dl(w) l(v,w)
• lt dl(v) l(v,w) ?
• l(v,w) gt 0 ?
• dl(v) dl(w) l(v,w)

• We routed a blocking flow ? G 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 Gf but in Gf (means that (w,v) is
  in A(f, l, dl) ?dl(w) dl(v) l(v,w) ? dl(w)
  dl(v))

45
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) and l(v,w) 0
• Case 1 - (v,w) in Gf
• 1l(v,w) gtl(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

46
Bad Example

• dl(w) dl(v), l(w,v) 0 and l(v,w) 0
• ? uf(v,w) lt 2?
• uf(w,v) gt 2?
• uf (v,w) uf(v,w) 2? uf(v,w) 2?
• l(v,w) 0

w
v
2?- uf(v,w)
47
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 2? in
  order to be able to route a flow of ? in flow
  extension (remember example)

48
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 ? blocking flow update steps instead
  of 4n ?

49
Length function l

• (v,w) is a special arc if
• 2? uf(v,w) lt 3?
• d(v) d(w)
• uf(w,v) 3?
• 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)

50
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) 3?
• l(v,w) 0 ? uf (v,w) 3?
• (w,v) is in A(f, l, dl) ? pushing a flow (at
  most ?) increased uf (v,w)
• ? uf(v,w) 2?
• ? Either (v,w) is a special arc before
  augmentation or uf(v,w) 3?
• ? (v,w) is in A(f, l, dl) - contradiction

51
Example
while F 1 do Compute ?, l, dl while
(S,T) gt 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
• a minm½, n ? 4
• ? F /a 8

52
Example (cont.)
while F 1 do Compute ?, l, dl while
(S,T) gt 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. ? 8
53
Example (cont.)
while F 1 do Compute ?, l, dl while
(S,T) gt 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. ? 8 (S,T) 15
54
Example (cont.)
while F 1 do Compute ?, l, dl while
(S,T) gt 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. ? 8
1
8/8
2
1
Af
1
8/12
8/16
1
0
3
0/7
0/4
1
1
1
3
2
0/10
55
Example (cont.)
while F 1 do Compute ?, l, dl while
(S,T) gt 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. ? 8 (S,T) 7 ? F 7, ? 2
56
Example (cont.)
while F 1 do Compute ?, l, dl while
(S,T) gt 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 7. ? 2
57
Example (cont.)
while F 1 do Compute ?, l, dl while
(S,T) gt 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 7. ? 2 (S,T) 5
58
Example (cont.)
while F 1 do Compute ?, l, dl while
(S,T) gt 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 7. ? 2
59
Example (cont.)
while F 1 do Compute ?, l, dl while
(S,T) gt 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
0
F 7. ? 2 (S,T) 3 ? F 3, ? 1
60
Example (cont.)
while F 1 do Compute ?, l, dl while
(S,T) gt 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 3. ? 1
8
61
Example (cont.)
while F 1 do Compute ?, l, dl while
(S,T) gt 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 3. ? 1 (S,T) 2
62
Example (cont.)
while F 1 do Compute ?, l, dl while
(S,T) gt 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 3. ? 1
63
Example (cont.)
while F 1 do Compute ?, l, dl while
(S,T) gt 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 3. ? 1 (S,T) 1 ? F 1, ? 1
64
Example (cont.)
while F 1 do Compute ?, l, dl while
(S,T) gt 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 1. ? 1
65
Example (cont.)

• Final flow

66
Concluding Remarks

• These results can be extended to a wide class of
  length functions (not binary) to obtain same time
  bounds
• Open issues
• Can the flow decomposition bound be improved upon
  by a strongly polynomial algorithm? For example,
  O(am(logn)O(1))
• Can these results be extended to obtain better
  bounds for the minimum-cost flow problem?

67
Appendix

• Unless you had enough ?

68
Computing dl in linear time

• We use Dials algorithm (1969)
• Init
• Compute GT
• Mark t as reached but not scanned, Distt0
• Mark rest of the nodes as not reached,
  DistvINF, v ?t
• Step 1
• Choose a reached but not scanned node v with
  minimal Dist
• If no such node compute G back and FINISH, else
  go to Step 2

69
Dials algorithm (cont.)

• Step 2
• For each (v,w)
• if Dist(w ) gt Dist(v ) l(v,w), then
• Mark w as reached but not scanned
• Dist(w ) Dist(v ) l(v,w)
• Add (v,w) to the output
• Remove all arcs (u,w) from the output, u ? w
• Mark v as reached and scanned
• Go to Step 1

70
Dials algorithm Example
1
node status Dist
t RnS 0
s nR INF
a nR INF
b nR INF
c nR INF
d nR INF
G
0
0
1
0
1
0
1
1
0
GT
1
71
Dials algorithm Example (cont.)
node status Dist
t RnS 0
s nR INF
a nR INF
b nR INF
c nR INF
d nR INF
1
Dist(b) gt Dist(t) l(t,b)
0
RnS
distance nodes
0
1
b
72
Dials algorithm Example (cont.)
node status Dist
t RnS 0
s nR INF
a nR INF
b RnS 0
c nR INF
d nR INF
RS
1
Dist(d) gt Dist(t) l(t,d)
distance nodes
0 b
1
1
RnS
d
73
Dials algorithm Example (cont.)
node status Dist
t RS 0
s nR INF
a nR INF
b RnS 0
c nR INF
d RnS 1
RnS
1
Dist(a) gt Dist(b) l(b,a)
distance nodes
0 b
1 d
a
74
Dials algorithm Example (cont.)
node status Dist
t RS 0
s nR INF
a RnS 1
b RnS 0
c nR INF
d RnS 1
Dist(d) gt Dist(b) l(b,d)
RS
distance nodes
0
1 d a
d
0
75
Dials algorithm Example (cont.)
node status Dist
t RS 0
s nR INF
a RnS 1
b RS 0
c nR INF
d RnS 0
Dist(c) gt Dist(d) l(d,c)
distance nodes
0 d
1 a
0
RnS
c
RS
76
Dials algorithm Example (cont.)
node status Dist
t RS 0
s nR INF
a RnS 1
b RS 0
c RnS 0
d RS 0
1
RnS
Dist(s) gt Dist(c) l(c,s)
distance nodes
0 c
1 a
s
77
Dials algorithm Example (cont.)
node status Dist
t RS 0
s RnS 1
a RnS 1
b RS 0
c RnS 0
d RS 0
0
Dist(a) gt Dist(c) l(c,a)
distance nodes
0
1 a s
RS
a
78
Dials algorithm Example (cont.)
node status Dist
t RS 0
s RnS 1
a RnS 0
b RS 0
c RS 0
d RS 0
0
RS
Dist(s) gt Dist(a) l(a,s)
distance nodes
0 a
1 s
s
79
Dials algorithm Example (cont.)
node status Dist
t RS 0
s RnS 0
a RS 0
b RS 0
c RS 0
d RS 0
RS
Dist(s) gt Dist(a) l(a,s)
distance nodes
0 s
1
80
Dials algorithm Example (cont.)
Final result
node status Dist
t RS 0
s RS 0
a RS 0
b RS 0
c RS 0
d RS 0
81
Dials algorithm Running Time

• Build GT O(nm)
• Each arc is examined once O(m)
• By keeping bins for possible distances, perform
  Step 1 in linear time O(nU)
• Total O(m nU)

82
References

• Beyond the Flow Decomposition Barrier, A.V.
  Goldberg, Satish Rao
• A Brief History Of Max-Flow, R. Khandekar,
  K.Talwar
• The maximum flow problem, I. Hudson
• Shortest-path forest with topological ordering,
  R.B. Dial