Instructor
Neelima Gupta
[email protected]
Edited by Divya Gaur(39, MCS
'09)
Thanks to: Bhavya(9), Deepika(10),
Deepika Bisht(11) (MCS '09)
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.
Thanks to: Bhavya(9),
Deepika(10), Deepika Bisht(11)
(MCS '09)
Edited by
Divya
Gaur(39,
MCS '09)
Goal
 To solve the Maximum Flow problem
Compute greatest rate at which material can flow from
source to sink subject to specified constraints.
Thanks to: Bhavya(9),
Deepika(10), Deepika Bisht(11)
(MCS '09)
Edited by
Divya
Gaur(39,
MCS '09)
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.
Thanks to: Bhavya(9),
Deepika(10), Deepika Bisht(11)
(MCS '09)
Edited by
Divya
Gaur(39,
MCS '09)
NOTATIONS AND
CONSTRAINTS
 Flow Network : G(V,E)
 Edge : (u,v) є 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
Thanks to: Bhavya(9),
Deepika(10), Deepika Bisht(11)
(MCS '09)
Edited by
Divya
Gaur(39,
MCS '09)
 Flow in G a real valued function
f: V X V → R satisfying constraints:
 Capacity Constraint: For all u,v є V, f(u,v) ≤
c(u,v)
Flow from one vertex to another must not exceed
given capacity.
 Skew Symmetry: For all u,v є 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 є V- {s,t}
Σf(u,v) = 0
vєV
Thanks to: Bhavya(9),
Deepika(10), Deepika Bisht(11)
(MCS '09)
Edited by
Divya
Gaur(39,
MCS '09)
 f(u,v) : Flow from u to v.
It can be either positive , negative or zero.
Value of flow |f| =Σf(u,v) = 0 : Total flow out of
source.
vєV
 Total flow out of a vertex V ≠ u, v =0
By Skew symmetry Σf(u,v) = 0
uєV
for all v є 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
Thanks to: Bhavya(9),
Deepika(10), Deepika Bisht(11)
(MCS '09)
Edited by
Divya
Gaur(39,
MCS '09)
Flow Network
SINGLE SOURCE SINGLE SINK
14
1
2
4
S
5
3
T
4
16
S
: SOURCE
TThanks: to:SINK
Bhavya(9),
Deepika(10), Deepika Bisht(11)
(MCS '09)
I
K
J
: K is Capacity
Edited by
Divya
Gaur(39,
MCS '09)
MULTIPLE SOURCE MULTIPLE SINK
s1
s2
t1
s3
t2
s4
t3
s5
Thanks to: Bhavya(9),
Deepika(10), Deepika Bisht(11)
(MCS '09)
Edited by
Divya
Gaur(39,
MCS '09)
To solve multiple source multiple sink
problem
 Add a supersource s and directed edge (s,si) with c(s,si)
= ∞ for each i=1,2,….,m
 Add a supersink t and directed edge (ti,t) with c(ti,t) =
∞ for each i =1,2,….,n
Thanks to: Bhavya(9),
Deepika(10), Deepika Bisht(11)
(MCS '09)
Edited by
Divya
Gaur(39,
MCS '09)
s1
t1
s2
T
S
s3
s4
s5
Thanks to: Bhavya(9),
Deepika(10), Deepika Bisht(11)
(MCS '09)
t2
t3
Edited by
Divya
Gaur(39,
MCS '09)
Instructor: Ms Neelima Gupta
Edited by Divya Gaur(39, MCS
'09)
Thanks to: Bhavya(9), Deepika(10),
Deepika Bisht(11) (MCS '09)
 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
Thanks to: Bhavya(9),
Deepika(10), Deepika Bisht(11)
(MCS '09)
Edited by
Divya
Gaur(39,
MCS '09)
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) є V X V: cf(u,v) > 0 }
ie. Each edge of the residual network or residual edge can
Edited by
admit
flow
>
0.
Thanks to: Bhavya(9),
Divya
Deepika(10), Deepika Bisht(11)
(MCS '09)
Gaur(39,
MCS '09)
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 = min{cf (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.
Thanks to: Bhavya(9),
Deepika(10), Deepika Bisht(11)
(MCS '09)
Edited by
Divya
Gaur(39,
MCS '09)
ALGORITHM
FORD_FULKERSON(G,s,t)
1. for each edge (u,v) є E[G]
2.
do f[u,v]← 0
3.
f[v,u]← 0
4. while there exists a path from s to t in Gf
5.
do cf (p)←min{c(u,v) :(u,v) is in p}
6.
for each edge (u,v) in p
7.
do f[u,v]← f[u,v] + cf (p)
8.
f[u,v]← - f[u,v]
Thanks to: Bhavya(9),
Deepika(10), Deepika Bisht(11)
(MCS '09)
Edited by
Divya
Gaur(39,
MCS '09)
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.
Thanks to: Bhavya(9),
Deepika(10), Deepika Bisht(11)
(MCS '09)
Edited by
Divya
Gaur(39,
MCS '09)
EXECUTION OF BASIC FORDFULKERSON ALGORITHM
14
1
2
4
S
T
5
3
4
16
S
: SOURCE
TThanks: to:SINK
Bhavya(9),
Deepika(10), Deepika Bisht(11)
(MCS '09)
I
K
J
: K is Capacity
Edited by
Divya
Gaur(39,
MCS '09)
PATH CHOSEN :S
1
3
4
14
1
4
S
T
2
5
3
T
4
16
MINIMUM CAPACITY IN THE PATH = 4
Thanks to: Bhavya(9),
Deepika(10), Deepika Bisht(11)
(MCS '09)
Edited by
Divya
Gaur(39,
MCS '09)
C / F : Capacity / Flow
RESIDUAL GRAPH
14
1
S
4
0
2
5
T
12 / 4
3
4
4
a/b
a = original capacity - minimum capacity of the selected
Edited by
path
Thankscto: Bhavya(9),
Divya
b
=
minimum
capacity
of
the
selected
path
Deepika(10), Deepika Bisht(11)
Gaur(39,
(MCS '09)
MCS '09)
c=b
PATH CHOSEN:S
1
14
1
0
S
2
4
T
2
5
T
12 / 4
3
4
MINIMUM CAPACITY IN THE PATH = 5
Thanks to: Bhavya(9),
Deepika(10), Deepika Bisht(11)
(MCS '09)
Edited by
Divya
Gaur(39,
MCS '09)
C / F : Capacity / Flow
RESIDUAL GRAPH
5
1
2
9/5
5
S
0/ 5
0/4
T
12 / 4
3
Thanks to: Bhavya(9),
Deepika(10), Deepika Bisht(11)
(MCS '09)
4
Edited by
Divya
Gaur(39,
MCS '09)
PATH CHOSEN : S
1
S
1
9/5
2
T
2
0/4
0/ 5
T
12 / 4
3
4
MINIMUM CAPACITY IN THE PATH = 9
Thanks to: Bhavya(9),
Deepika(10), Deepika Bisht(11)
(MCS '09)
Edited by
Divya
Gaur(39,
MCS '09)
C / F : Capacity / Flow
RESIDUAL GRAPH
14
1
2
0/ 14
S
0/ 5
0/4
T
12 / 4
3
Thanks to: Bhavya(9),
Deepika(10), Deepika Bisht(11)
(MCS '09)
4
Edited by
Divya
Gaur(39,
MCS '09)
PATH CHOSEN :S
1
S
3
0 / 14
4
T
2
0/4
0/ 5
T
12 / 4
3
4
MINIMUM CAPACITY IN THE PATH = 6
Thanks to: Bhavya(9),
Deepika(10), Deepika Bisht(11)
(MCS '09)
Edited by
Divya
Gaur(39,
MCS '09)
C / F : Capacity / Flow
RESIDUAL GRAPH
1
S
0 / 14
0/4
2
0/ 5
T
6 / 10
3
4
10
Thanks to: Bhavya(9),
Deepika(10), Deepika Bisht(11)
(MCS '09)
Edited by
Divya
Gaur(39,
MCS '09)
PATH CHOSEN: S
(BACKTRACKING)
1
S
3
4
T
2
0 / 14
5
0/4
2
0/ 5
T
6 / 10
3
4
MINIMUM CAPACITY IN THE PATH = 5
Thanks to: Bhavya(9),
Deepika(10), Deepika Bisht(11)
(MCS '09)
Edited by
Divya
Gaur(39,
MCS '09)
C / F : Capacity / Flow
RESIDUAL GRAPH
1
S
2
0 / 14
0/4
T
0
5
1 / 15
3
4
15 / 0
Now the graph has no augmenting path , therefore the flow shown
is the maximum flow.
Flow :- 14+15 = 29
ThanksMaximum
to: Bhavya(9),
Deepika(10), Deepika Bisht(11)
(MCS '09)
Edited by
Divya
Gaur(39,
MCS '09)
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
T=V - S such that s є S and t є 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.
Thanks to: Bhavya(9),
Deepika(10), Deepika Bisht(11)
(MCS '09)
Edited by
Divya
Gaur(39,
MCS '09)
CUT
1
S
2
0 / 14
0/4
T
0
5
1 / 15
3
Thanks to: Bhavya(9),
Deepika(10), Deepika Bisht(11)
(MCS '09)
4
Edited by
Divya
Gaur(39,
MCS '09)
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 cut’s capacity is less
than C*’s 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.
Thanks to: Bhavya(9),
Deepika(10), Deepika Bisht(11)
(MCS '09)
Edited by
Divya
Gaur(39,
MCS '09)
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*|).
Thanks to: Bhavya(9),
Deepika(10), Deepika Bisht(11)
(MCS '09)
Edited by
Divya
Gaur(39,
MCS '09)
A DEEPER SIGHT INTO ANALYSIS
 f* can be very big as it depends on capacities.
 f* could be as big as Σc(u,v).
 Since it depends on the values of capacities so it is not
polynomial in input size which is Σlog ci + E + V
where ci’s could be large.
Thanks to: Bhavya(9),
Deepika(10), Deepika Bisht(11)
(MCS '09)
Edited by
Divya
Gaur(39,
MCS '09)
 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*|.
200
v1
1
s
200
Thanks to: Bhavya(9),
Deepika(10), Deepika Bisht(11)
(MCS '09)
200
v2
t
200
Edited by
Divya
Gaur(39,
MCS '09)
 Here max flow = 400.
 But if the first augmenting path is
s->v1->v2->t, flow would be one and then choosing s-> v2-> v1- >t in even numbered iterations and s->u->v->t in odd
ones increases flow by just one in each iteration(total of
400 iterations) and perform a total of 400 augmentations.
Thanks to: Bhavya(9),
Deepika(10), Deepika Bisht(11)
(MCS '09)
Edited by
Divya
Gaur(39,
MCS '09)
Thanks to: Swati Vohra (29, MCS '09)
Edmonds Karp Algorithm
u
s
t
v
 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.
Thanks to: Swati Vohra (29,
MCS '09)
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 )
Thanks to: Swati Vohra (29,
MCS '09)
f1
s
f
u
u
v
 As f is the shortest distance from s to u so
f <= f1 and
$f(s,u) <= $f1(s,u) V u.
Thanks to: Swati Vohra (29,
MCS '09)
t
Critical Edge
s
u
v
t
 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.
Thanks to: Swati Vohra (29,
MCS '09)
When (u,v) is not present in
augmenting path
s
u
t
v
Thanks to: Swati Vohra (29,
MCS '09)
For having (u,v) back in augmenting
path
s
u
t
v
 edge (v,u) must be present on augmenting path
 (v,u) need not to be critical
Thanks to: Swati Vohra (29,
MCS '09)
 So,
$f1(s,u) = $f1(s,v) +1
&
$f(s,v) = $f(s,u) +1
$f(s,v) <= $f1(s,v)
Therefore, $f1(s,u) = $f1(s,v) +1
>=($f(s,v) +1)
= $f(s,u) +2
Thanks to: Swati Vohra (29,
MCS '09)
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 can’t
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.
Thanks to: Swati Vohra (29,
MCS '09)
 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(VE^2).
Thanks to: Swati Vohra (29,
MCS '09)
Thanks to: Geetanjali Shivas (13,MCS '09)
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
t1
t2
.
.
.
.
.
.
tm
Thanks to: Geetanjali Shivas
(13,MCS '09)
C
c1
c2
.c3
.
.
.
.
.
cn
 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.
c1
t1
c2
t2
Thanks to: Geetanjali Shivas
(13,MCS '09)
c3
 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.
1
t
 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.
Thanks to: Geetanjali Shivas
(13,MCS '09)
The corresponding flow network
1
1
s
t1
1
1
1
t2
Thanks to: Geetanjali Shivas
(13,MCS '09)
c1
1
c2
1
1
1
c3
t
 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.
Thanks to: Geetanjali Shivas
(13,MCS '09)
 Select the path s-t1-c1-t.
1
0/1
0/1
t1
s
1
t2
Thanks to: Geetanjali Shivas
(13,MCS '09)
c1
1
1
1
c2
c3
0/1
1
1
t
 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
0/1
s
t1
0/1
Thanks to: Geetanjali Shivas
(13,MCS '09)
0/1
0/1
t2
c1
1
c2
c3
0/1
t
0/1
1
 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.
Thanks to: Geetanjali Shivas
(13,MCS '09)