Karger’s Min-Cut Algorithm Amihood Amir Bar-Ilan University, 2009
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Karger’s Min-Cut Algorithm
Amihood Amir
Bar-Ilan University, 2009
The Problem
INPUT: Undirected Graph G=(V,E)
OUTPUT: A smallest-size set of edges
such that G’=(V,E-E1) is
not connected.
EE 1
Example
A
C
B
D
GE
F
Example
A
C
B
D
GE
F
E1 = BE
Immediate Solution
For every pair of nodes s,t find the min-cut Ms,t
Choose the pair <s,t> with the smallest Ms,t.
Time= O(n² T(min-cut) )
Time for Min-Cut
Ford-Fulkerson: O(|E|flow)
In our case: max flow ≤ n
so: O(n² |E|n) = O(n² n³)
This can be improved
Improvement
We don’t need all pairs.
Why?
Every node is in one side of a cut.
Therefore: Fix node s. Try all other n-1 options as node t.
Time: )( 4nO
Randomized Solution
Repeat Until 2 nodes left:1. choose edge at random2. “contract” edge
When only 2 node left:Take all edges between them as the min-cut.
Time: O(|E|)=O(n²)
Meaning of “contraction”
Make edge ab with its two adjacent nodes a,b into a single node ab.
All edges of a and b will now be edges from ab (creating a multigraph).
Example:A
C
B
D
GE
F
Example:A
C
B
D
GE
Fcontract
Example:A
C
B
D
GE
Fcontract
A
C
B
D
E
FG
Example:A
C
B
D
GE
Fcontract
A
C
B
D
E
FG
contract
Example:A
C
B
D
GE
Fcontract
A
C
B
D
E
FG
contract
A
C
B E
FGD
Is output min-cut?
Not necessarily.
Is it a cut?
- yes.
Is output min-cut?
Not necessarily.
Is it a cut?
- yes.
Could there be an edge
between these sides that
was deleted?
Is output min-cut?
Not necessarily.
Is it a cut?
- yes.
No. The only deleted
edges were within
the contractions.
Is output min-cut?
Not necessarily.
Is it a cut?
- yes.
cut C={e1,…,ec}No. The only deleted
edges were within
the contractions.
When is C a min-cut?
When none of the edges in
C={e1,…,ec}
are chosen to be contracted.
What is the probability of that happening?
Probability calculation
Note: We can assume that degree of every node in graph throughout contraction process ≥ c
Probability calculation
Note: We can assume that degree of every node in graph throughout contraction process ≥ c
Otherwise:
Cut node and get cut of size < c.
Probability calculationProbability that one of the C edges is chosen at first
stage:
Probability that one of the C edges is chosen after stage i ≤:
Probability of not choosing an edge of C at stage i ≥:
nncc
Ec 2
2/
incin
c
2
2/)(
inin
in
221
Probability calculationNote: The algorithm has n-2 stages (at that
point 2 nodes are left).
Probability of not choosing an edge from C at any stage ≥
=
)1)(1)(1()1)(1)(1)(1( )3(2
)4(2
)5(2
32
22
122
nnnnnnnnnn
))()(())()()(( 31
42
53
35
24
132
nn
nn
nn
nn
Probability calculationWhat do we have?
))()(())()()(( 31
42
53
35
24
132
nn
nn
nn
nn
Probability calculationWhat do we have?
Telescope cancellations:
))()(())()()(( 31
42
53
35
24
132
nn
nn
nn
nn
))()(())()()(( 31
42
53
35
24
132
nn
nn
nn
nn
)( 21
)1(2
nOnn
=
ConcludeRun the algorithm O(n²) times and expect to get C with constant probability.
Total time:
No better than Ford-Fulkerson!
Can we do better? )( 4nO
IDEA!The situation:
n² results, one of which is expected to be good. O(n²)time
What if?Could compute:
n² results, one of which is expected to be good.O(log n)time
This is not quite the situation…
But note: The probability of not hitting C at stage 1 is (almost 1). It grows as we run more and more stages.
What happens if we stop in the middle?
The probability that C not touched in the first i steps ≥
n21
)())(())(( 2
2)()1(
)2)(1(211
132
n
innn
ininin
ininin
nn
nn O
So what is the situation?
When is the probability of not hitting C exactly ½ ?
-- When nodes are left.
Because take then
2n
122
n
i
21
2
2
2
2
2
22
2
22
2
122
nnn
nn
n
n nnnn
This means
If we run two times until nodes are left
we expect that one of them still did not touch C
So, for each of these runs, when nodes are left, stop running and recurse!
2
n
2
n
Time:
)()(2)( 2
2nOTnT n
)log( 2 nnO
The closed form is:
Much better!But what is the overall probability?
Probability:
Assume we are in level d of the recursiontree, with the leaves being level 1.
Let Pd(x) be the probability that there is apath from node x at level d that does not touch C.
We have:Pd(x)= ½ (Probability that at least one of
the two recursions does not touchC)
Probability:Recall: Pr(A or B)= Pr(A)+Pr(B)-Pr(A and B)
We have:Pd(x)= ½ (2Pd-1-(Pd-1)²) Where Pd-1 is the
probability that a child of x does not hit C.= Pd-1 – ½ (Pd-1)²
Claim: For d >1, Pd >1/d
Proof: Note that in the interval [0,1) the function x – ½ x² decreases monotonically.
Probability:We have:Pd = Pd-1 – ½ (Pd-1)².
Inductively: Pd-1 >1/(d-1). So, because of monotonicity, Pd-1 – ½ (Pd-1)² >
ddd
d
ddddd
1
)1(
1
)1(
1
1
1
)1(2
1
1
12
Probability:
Conclude:
Plog n(x) > 1/log n
Which means: If we run the algorithm log n times, we get constant
probability. So…
Total time: O(n² log² n)
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