CS473-Algorithms I

1
CS473-Algorithms I

• Lecture ?
• Network Flows
• Finding Max Flow

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 return f
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 ? Adju in G' maintain the record
• note G' used to represent both G and Gf , i.e.,
  for any edge (u,v) ? E'
• cu,v gt 0 ? (u,v) ? E and cf u,v gt 0 ? (u,v) ?
  Ef

v
f(u,v)
c(u,v)
cf (u,v)
4
FORD-FULKERSON METHOD / ALGORITHM
COMPUTE-GF (G', f) for each edge (u,v) ? E'
do if cu,v f u,v gt 0 then cf u,v ?
cu,v fu,v else cf u,v ? 0 return
G' 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 s t path p in
Gf do cf (p) ? min cf u,v (u,v) ? p for
each edge (u,v) ? p do f u,v ? f u,v
cf (p) CANCEL(G', u, v) Gf ?
COMPUTE-GF(G', f)
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

6
FORD-FULKERSON ALGORITHM
Gf0
f0
u
u
0/100
0/100
100
100
s
t
0/1
s
t
1
0/100
100
0/100
100
v
v
p lt s, u, v, t gt cf (p) 1 f1 f0
fp 01 1
7
FORD-FULKERSON ALGORITHM
Gf1
f1
u
u
1/100
0/100
100
99
1
s
t
1/1
s
t
1
99
0/100
100
1
1/100
v
v
p lt s, v, u, t gt cf (p) 1 f2 f1
fp 11 2
8
FORD-FULKERSON ALGORITHM
Gf2
f2
u
u
1/100
1/100
99
99
1
1
s
t
0/1
s
t
1
99
99
1/100
1
1
1/100
v
v
p lt s, u, v, t gt cf (p) 1 f3 f2
fp 21 3
9
FORD-FULKERSON ALGORITHM
Gf3
f3
u
u
2/100
1/100
98
99
2
1
s
t
1/1
s
t
1
99
98
1/100
2
1
2/100
v
v
p lt s, v, u, t gt cf (p) 1 f4 f3
fp 31 4
10
FORD-FULKERSON ALGORITHM

• choose p lt s, u, v, t gt in odd iterations
• choose p lt s, v, u, t gt 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)

11
EDMONDS-KARP ALGORITHM

• def df (s,v) shortest path distance from s to
  v in Gf
• unit edge weights in Gf ? df (s,v)
  breadth-first distance from s to v in Gf
• L7 ? v ? V s,t d (s,v) in Gfs 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 d(s,v) df (s,v) and d(s,v) df
  (s,v)

12
ANALYSIS OF EDMONDS-KARP ALGORITHM

• want to prove d' (s,v) d(s,v) ? v ? V s,t
• for contradiction assume d'(s,v) lt d(s,v) for
  some v
• without loss of generality assume d' (s,v) is
  minimal over all such vertices
• i.e., d' (s,v) lt d' (s,u) ? u ? V s,t ? d'
  (s,u) lt d(s,u)
• thus d' (s,u) lt d' (s,v) ? d' (s,u) d(s,u) (1)

13
ANALYSIS OF EDMONDS-KARP ALGORITHM

• consider a shortest s v path p' s u ? v in
  Gf
• d'(s,u) d'(s,v) 1 by subpath is also a
  shortest path
• d' (s,u) lt d' (s,v) ? by (1), d' (s,u) d(s,u)
  (2)
• consider this edge (u,v) ? Ef ' question was
  (u,v) ? Ef ?
• YES i.e., (u,v) ? Ef ' and (u,v) ? Ef note
  f(u,v) lt c(u,v)
• d(s,v) d(s,u) 1 by triangle inequality
• d' (s,u) 1 by (2)
• d' (s,v) ? d' (s,v) d(s,v)
  contradiction

14
ANALYSIS OF EDMONDS-KARP ALGORITHM

• NO i.e., (u,v) ? Ef ' but (u,v) ? Ef note
  f(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)
• d(s,v) d(s,u) 1 by subpath is also a
  shortest path
• d' (s,u) 1 by (2)
• (d' (s,v) 1) 1 by subpath is
  also a shortest path
• d' (s,v) 2
• lt d'(s,v) ? d'(s,v)
  d(s,v) contradiction

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) (ffp)(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

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
• d(s,v) d(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

17
ANALYSIS OF EDMONDS-KARP ALGORITHM

• assume this event occurs along a path p' s u
  ? v t in Gf ' later
• d'(s,u) d'(s,v) 1 since p' s u ? v
  t is a shortest path
• d(s,v) 1 by L7
• (d(s,u) 1) 1 since p' s u
  ? v t is a shortest path
• d(s,u) 2 ? d' (s,u) d(s,u) 2
• i.e., d(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 d(s,u) V - 2 and d(s,u) never decreases
  by L7
• ?(E) edges in Gfs ? 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)

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

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
  matching M
• def G(V,E) is a bipartite graph if VL?R where
  LnR? such that E(u,v) u ? L and v ? R

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

21
MAXIMUM BIPARTITE MATCHING PROBLEM

• example two matchings M1 M2 on a sample graph
  with M1 2 M2 3

L
R
L
R
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)
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'

23
FINDING A MAXIMUM BIPARTITE MATCHING
1
u1
u1
v1
1
1
v1
1
u2
u2
1
1
1
v2
1
v2
1
u3
s
t
u3
1
1
v3
1
1
v3
1
u4
u4
1
1
v4
v4
1
u5
u5
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) IF M is a matching in G, THEN ? an
  integer-valued f on G' with f M (b) IF
  f is an integer-valued f on G', THEN ? a
  matching M in G with M f

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

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

27
FINDING A MAXIMUM BIPARTITE MATCHING

• proof L8 (b) let f be a integer-valued flow in
  G'
• define M(u,v) u ? L, v ? R, and f (u,v) gt 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

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

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
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
u3
u3
s
t
v3
v3
1
u4
u4
1
v4
v4
1
u5
u5