MS 101: Algorithms

1
MS 101 Algorithms

• Instructor
• Neelima Gupta
• ngupta_at_cs.du.ac.in

2
Network Flows
3
DOMAIN

• Depict flow of a material (liquid, gas, current)
  from source to sink.
• Source Material generator.
• Sink Final destination to receive material.
• Flow Rate of flow of material.
• Conduit Medium for material transfer (pipe,
  wires).
• Capacity Maximum rate at which material can flow
  through conduit.

4
Goal

• To solve the Maximum Flow problem
• Compute greatest rate at which material can flow
  from source to sink subject to specified
  constraints.

5
TECHNIQUE

• Flow Network
• A directed graph depicting source, conduit, sink
  modeling the flow.
• Each directed edge represents conduit.
• Vertices are conduit conjunctions through which
  material flows but does not collect .
• At a vertex , Rate of inflow rate of outflow.

6
NOTATIONS AND CONSTRAINTS

• Flow Network G(V,E)
• Edge (u,v) ? E
• Capacity of edge c(u, v) 0
• If (u,v) does not belong to E the c(u,v) 0
• Source s
• Sink t
• Graph connected ie. E V - 1

7

• Flow in G a real valued function
• f V X V ? R satisfying constraints
• Capacity Constraint For all u,v ? V, f(u,v)
  c(u,v)
• Flow from one vertex to another must not
  exceed given capacity.
• Skew Symmetry For all u,v ? V, f(u,v) -f(v,u)
• Flow from a vertex u to a vertex is the
  negative of the flow in the reverse direction.
• Flow Conservation For all u ? V- s,t
• Sf(u,v) 0
• v ? V

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• f(u,v) Flow from u to v.
• It can be either positive , negative or zero.
• Value of flow f Sf(u,v) 0 Total flow
  out of source. v ? V
• Total flow out of a vertex V ? u, v 0
• By Skew symmetry Sf(u,v) 0
• 
  u ? V
• for all v ? V- s,t ie.. total flow into
  vertex 0.
• Total net flow at a vertex
• Total positive flow leaving a vertex Total
  positive flow entering a vertex .
• So by Flow Conservation property total net
  flow at a vertex 0

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Flow Network
SINGLE SOURCE SINGLE SINK
14
1
2
19
18
T
S
5
4
15
20
3
4
16
S
SOURCE

SINK


I
J
K
K is Capacity
T
10
MULTIPLE SOURCE MULTIPLE SINK
10
s1
12
3
15
s2
t1
5
8
6
14
s3
20
t2
7
13
s4
11
18
t3
2
s5
11
To solve multiple source multiple sink problem

• Add a supersource s and directed edge (s,si) with
  c(s,si) 8 for each i1,2,.,m
• Add a supersink t and directed edge (ti,t) with
  c(ti,t) 8 for each i 1,2,.,n

12
s1
10
8
3
12
t1
s2
15
5
8
8
6
8
T
8
S
8
t2
s3
20
14
8
8
13
7
8
t3
s4
18
11
2
s5
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Ford Fulkerson Method

• Instructor Ms Neelima Gupta

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• Ford_Fulkerson(G,s,t)
• 1. initialize flow f to 0
• 2. while there exists augmenting path p
• 3. do augment f along p
• 4. return f

15
RESIDUAL NETWORK

• RESIDUAL CAPACITY OF (u,v) cf
• The amount of additional flow that can be
  pushed from u to v before exceeding c(u,v) .
• cf c(u,v) f(u,v)
• Eg - c(u,v)10, f(u,v)5 then flow can
  be increased by 5 units ie.. residual capacity
  5.
• RESIDUAL NETWORK OF G INDUCED BY f , Gf
• Gf (V,Ef)
• where Ef (u,v) ? V X V cf(u,v) gt 0
• ie. Each edge of the residual network or
  residual edge can admit flow gt 0.

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AUGMENTING PATHS

• AUGMENTING PATH ( P)
• Path from s to t in residual network Gf .
• The augmenting path admits additional
  positive flow from u to v within capacity
  constraint.
• The maximum amount by which flow can be
  increased on each edge in P is residual capacity
  of P, cf mincf (u,v) (u,v) is on P.
• Example if cf (u,v)3 then the flow through each
  edge of p can be increased by 3 without violating
  capacity constraint.

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ALGORITHM

• FORD_FULKERSON(G,s,t)
• for each edge (u,v) ? EG
• do fu,v? 0
• fv,u? 0
• while there exists a path from s to t in Gf
• do cf (p)?minc(u,v) (u,v) is in p
• for each edge (u,v) in p
• do fu,v? fu,v cf (p)
• fu,v? - fu,v

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• DESCRIPTION OF ALGORITHM
• 1-3 initialise flow f to 0.
• 4-8 finds augmenting path p in residual network
  and augments flow f along p by residual capacity
  cf (p).
• When no augmenting path exists flow f is maximum.

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EXECUTION OF BASIC FORD-FULKERSON ALGORITHM
14
1
2
19
18
T
S
5
4
15
20
3
4
16
S
SOURCE

SINK


I
J
K
T
K is Capacity
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• PATH CHOSEN S 1 3 4 T

14
1
2
19
18
T
S
5
4
15
20
3
4
16
MINIMUM CAPACITY IN THE PATH 4
21
RESIDUAL GRAPH
C / F Capacity / Flow
14
1
2
19
14 / 4
4
4
T
S
5
0
11 / 4
4
12 / 4
20
3
4

4

a/b
a original capacity - minimum capacity of the
selected path b minimum capacity of the
selected path c b
c
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PATH CHOSENS 1 2 4 T
14
1
2
19
14 / 4
T
S
5
0
11 / 4
12 / 4
20
3
4

MINIMUM CAPACITY IN THE PATH 5
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RESIDUAL GRAPH
C / F Capacity / Flow
5
1
2
19
9 / 5
9 / 9
9
T
5
S
0/ 5
0/4
6 / 9
9
12 / 4
20
3
4


24
PATH CHOSEN S 1 2 T
1
2
19
9 / 5
9 / 9

T
S
0/ 5
0 / 4
6 / 9
12 / 4
20
3
4

MINIMUM CAPACITY IN THE PATH 9
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RESIDUAL GRAPH
C / F Capacity / Flow
14
1
2
9
0/ 14
0 / 18
10 / 9
18
T
S
0/ 5
0/4
6 / 9
12 / 4
20
3
4


26
PATH CHOSEN S 3 4 T
1
2
10 / 9
0 / 14
0 \ 18

T
S
0/ 5
0 / 4
6 / 9
12 / 4
20
3
4

MINIMUM CAPACITY IN THE PATH 6
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RESIDUAL GRAPH
C / F Capacity / Flow
1
2
10 / 9
0 / 14
0 \ 18

T
S
0/ 5
0 / 4
6
0 / 15
15
6 / 10
14 / 6
3
4
10

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PATH CHOSEN S 3 4 2
T (BACKTRACKING)
1
2
10 / 9
0 / 14
0 \ 18

T
5
S
0/ 5
0 / 4
0 / 15
14 / 6
6 / 10
3
4

MINIMUM CAPACITY IN THE PATH 5
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RESIDUAL GRAPH
C / F Capacity / Flow
1
2
0 / 14
5 / 14
0 \ 18
14

T
S
0
0 / 4
5
11
0 / 15
1 / 15
9 / 11
3
4
15 / 0

Now the graph has no augmenting path ,
therefore the flow shown is the maximum flow.
Maximum Flow - 1415 29
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CUTS OF FLOW NETWORK

• Flow is maximum iff its residual network has no
  augmenting path.
• CUT(s,t) OF FLOW NETWORK G A partition of V
  into S and
• TV - S such that s ? S and t ? T .
• It is a collection of edges separating the
  network into S and T.
• If f is flow ,then the net flow across the cut is
  f(S,T) and capacity is c(S,T).
• A minimum cut of network is cut with minimum
  capacity over all cuts of the network.

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CUT
1
2
0 / 14
5 / 14
0 \ 18
14

T
S
0
0 / 4
5
11
0 / 15
1 / 15
9 / 11
3
4
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OPTIMALITY

• MAX FLOW MIN CUT THEOREM
• 1.f is maximum flow in G
• 2.Gf contains no augmenting paths
• 3.f c(S,T) for some cut(S,T) of G
• If the flow f reported by the algorithm is
  optimal then there exists a cut C on which
  residual capacity of cut 0.
• The initial capacity of C must have been f and
  must have been utilised completely and hence
  residual capacity 0.
• C is a minimum cut since if another cuts
  capacity is less than Cs then it would not be
  possible to send f flow.
• Optimality is achieved since otherwise some
  augmenting path would be present with capacity at
  least one.

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ANALYSIS OF THE ALGORITHM

• Running time of the algorithm is O(f X time to
  compute augmenting path).
• 1-3 take O(E) time.
• While loop is executed at most f times since
  flow value increases by at least one unit in each
  iteration.
• Time to compute an augmenting path using BFS or
  DFS is O(E).
• Each iteration of the while loop hence takes O(E)
  time.
• Hence total running time of FORD FULKERSON is O(E
  f).

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A DEEPER SIGHT INTO ANALYSIS

• f can be very big as it depends on capacities.
• f could be as big as Sc(u,v).
• Since it depends on the values of capacities so
  it is not polynomial in input size which is Slog
  ci E V
• where cis could be large.

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• Running Time depends on how the augmenting path p
  is determined.
• If capacities are small and f is small running
  time is good.
• Example illustrates effect on time with large
  f.

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• Here max flow 400.
• But if the first augmenting path is
• s-gtv1-gtv2-gtt, flow would be one and then
  choosing s-gt v2--gt v1- gtt in even numbered
  iterations and s-gtu-gtv-gtt in odd ones increases
  flow by just one in each iteration(total of 400
  iterations) and perform a total of 400
  augmentations.

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Edmonds-Karp Algorithm
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Edmonds Karp Algorithm

• Choose the shortest augmenting path from s to t
  in residual network.
• Path is shortest in terms of no. of edges.
• Critical edge - minimum weight edge of an
  augmenting path.

u
t
s
v
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Notations

• Initial graph - G
• Flow in G form s to u f
• Residual graph Gf
• f(u,v) - shortest path distance from u to v in
  Gf
• Flow in residual graph f1 (computed as shortest
  path over Gf )

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f1
u

• As f is the shortest distance from s to u so
• f lt f1 and
• f(s,u) lt f1(s,u) V u.

t
s
u
v
f
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Critical Edge

• If (u,v) is critical edge along augmenting path
  then
• f(s,v) f(s,u) 1 (edge)
• f(s,v) in Gf increases monotonically with each
  flow augmentation.
• Once the flow is augmented, edge (u,v) disappears
  from the Gf.

s
t
u
v
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When (u,v) is not present in augmenting path
u
s
t
v
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For having (u,v) back in augmenting path

• edge (v,u) must be present on augmenting path
• (v,u) need not to be critical

u
s
t
v
44

• So,
• f1(s,u) f1(s,v) 1
• f(s,v) f(s,u) 1
• f(s,v) lt f1(s,v)
• Therefore, f1(s,u) f1(s,v) 1
• gt(f(s,v) 1)
• f(s,u) 2

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Analysis

• In a critical edge (u,v) , Distance of u from s
  initially is at least 0.
• Every time (u,v) edge becomes critical again
  distance of u from s increased by at least 2.
• Intermediate vertices on shortest path from s to
  u cant contain s,u,t.
• (since (u,v) on the critical path means s !
  t).
• So until u becomes unreachable from s its
  distance is at most IVI-2.
• So every (u,v) edge can become critical at most
  (IVI-2)/2 times.

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• Pair of vertices that can have edge in Gf O(E)
• Each edge can become critical O(V) time.
• Total no. of critical edges O(VE).
• Each iteration of Ford Fulkerson can be
  implemented in O(E) time when augmenting path is
  found by BFS so
• total running time of Edmonds Karp O(VE2).

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Maximum bipartite matching
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• Problem- There is a set T of teachers with a set
  C of courses. A teacher can
  teach only some set of courses. Assign 1
  teacher to 1 course.
• T
  C

t1 t2 . . . . . . tm
c1 c2 .c3 . . . . . cn
49

• Each teacher is to be assigned at most one course
  and each course is to be assigned at most one
  teacher.
• This is a bipartite graph
• Let teachers be 2 and courses be 3.

t1 t2
c1 c2 c3
50

• Convert the problem into flow network problem in
  which flows corresponds to matchings.
• Let source s and sink t be new vertices. Draw an
  edge from source s to each vertices in T and from
  each vertices in C to sink t.
• Capacity of each edge from s to T, T to C, C to t
  is 1.
• In the end, each unit of flow will be equivalent
  to a match between a teacher and a course, so
  each edge will be assigned a capacity of 1.

1
t
51

The corresponding flow network
1
c1 c2 c3


1
1
t1 t2
1
t
1
s
1
1
1
1
52

• Now if we assign t1 to c1, then t2 will not get
  any course and c2, c3 are also not assigned to
  any teachers.
• Therefore we select a subset E of edges such
  that no two edges resides on same vertex. We want
  to find the maximum edges.
• Q- Why capacity from s to T cannot be taken
  more than 1?
• Ans- If capacity from s to T is more than 1
  then any vertex in T would be selected more than
  1 time and matched more than once.

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• Select the path s-t1-c1-t.

1
0/1
c1 c2 c3
0/1
t1 t2
0/1
1
1
t
1
s
1
1
1
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• Now t2 is not having any course. But capacity
  from s to t2 is 1. Therefore we do backtracking.
• Now the new path is s-t2-c1-t1-c2-t.

1
1
c1 c2 c3
0/1
0/1
t1 t2
t
0/1
0/1
s
1
0/1
1
0/1
55

• Now t2 is assigned c1 and t1 is assigned to c2.
• Flow from t1to c2 is 1, t2 to c1 is 1.
• But residual capacity from t1 to c2 is 0 and t2
  to c1 is 0.
• Total network flow is 2.
• Therefore teacher t1 is matched with course c2
  and teacher t2 is matched with course c1.