Breif description of Kernighan-Lin graph partitioning method, applications,pros and cons of the method

1. Kernighan-Lin
Method

2. What is it?
• Kernighan-Lin is a method of partitioning a
graph containing nodes and vertices into
separate subsets that are connected together
in an optimal manner. Since a graph can be
used to represent an electrical network
containing blocks, the Kernighan-Lin
algorithm can be extended to partitioning
circuits into sub-circuits.

3. Application
• Important application to VLSI circuits. Used
to find minimal numbers of connections
between partitions to improve speed or
decrease power consumption.

4. • Kernighan-Lin is an iterative algorithm. This
means that the graph/circuit may already be
partitioned, but application of Kernighan-Lin
will try to improve or optimize the partition.
Kernighan-Lin is iterative as opposed to
constructive.
• Kernighan-Lin is a greedy algorithm. This
means the algorithm will make changes if
there is a benefit right away without
consideration to other possible ways of
obtaining an optimal solution.

5. • Kernighan-Lin is a deterministic algorithm
because the same result will be achieved
every time the algorithm is applied. The
same result will be the same number of nets
crossing the bisection, but not necessarily the
same nets.
• One bisection or cut is made to the partition
only, so partitioning using Kernighan-Lin will
result in only two partitions. Partitions must
be equal size.

6. Steps…
• Draw a line separating your graph into two halves
(partitions) with an equal number of vertices
(blocks or cells) in each partition.
• Count the number of edges that cross the line.
This number is called your net cut and the goal is
to decrease this number. In terms of VLSI
circuits, you will be decreasing the number of
connections between blocks, which could increase
the speed of the circuit, decrease power
consumption, and depict other desirable results.

7. • Find the edge cost of all vertices in the graph.
Finding edge cost is done by finding the number
of connections each vertex has within its own
partition and subtracting that from the number of
connections each vertex has with vertices in the
other partition.
• Determine the maximum gain by swapping any
two nodes. The gain equation is given below:
G = D1 + D2 – 2C12

8. • Swap the two nodes with the maximum gain.
Note that if all node pairing gains have been
calculated and the maximum gain is zero or
negative, the nodes with the highest gain
should still be swapped.
• Subtract the gain from the original net cut to
get the new net cut.
• Fix the nodes that were just swapped in place
• Repeat steps until the maximum gain is zero
or negative

9. Advantage
• Algorithm is Robust.

10. Disadvantages
• Results are random because the algorithm starts
with a random partition
• Computationally intensive which makes the
algorithm slow
• Only two partitions are created
• Partitions have to be equal in size so the
algorithm does not attempt to find optimal
partition sizes when they may (and probably do)
exist
• Does not allow cells to remain fixed in place when
they may need to be for timing or other reasons

11. • Does not solve problems with weighted edges
very well
• Solution largely dependant on the first swap

12. Solved Example

13. • After initial partitioning

14. Step 1:
Initial
Partition
A={ 2, 3, 4 } B={ 1, 5, 6 }
Step 2: Compute D
Partition A Ea Ia Compute D Partition B Eb Ib Compute D2
2 1 2 -1 1 1 0 1
3 0 1 -1 5 1 1 0
4 2 1 1 6 1 1 0

15. Step 3: Computing
Gain
Possible
Pairs
C(a,b)
Gain
G21
1 -2
G25 0 -1
G26 0 -1
G31 0 0
G35 0 -1
G36 0 -1
G41 0 2
G45 1 -1
G46 1 -1
Largest G value G41 = +2
A' = A' - { 4 } = { 2, 3 }
B' = B' = { 1 } = { 5, 6 }

16. Iteration
2:
D values connected to ( 4, 1) are
2 in A'
5, 6 in B'
Step 1:
Computing
new D
A' B' CA'4 CA'1 CB'4 CB'1
2 5 1 1 1 0
3 6 0 0 1 0
D'2 -1
D'5 -2
D'6 -2

17. Computin
g Gain
Step 2:
Possible
Pairs C(a,b) Gain
G25 0 -4
G26 0 -2
G35 0 -3
G36 0 -3
Since all Gain values are equal, we
arbitrarily choose any one,
A' = A' - { 3 } = { 2 }
B' = B' - { 6 } = { 5 }
we
choose G36

18. Iteration 3:
( 3 , 6 ) connected to 2
in A'
and 5 in B'
Step 1:
Computin
g new D
A' B' CA'3 CA'6 CB'3 CB'6
2 5 1 0 0 1
D''2 0
D''5 0
Therefore last pair (2,5) and
gain is 1

19. Determining K
Step 5: G1 = +2
G2 = -3 i.e G1 + G2 = -1
G3 = 1 G1 + G2 + G3 = 0

20. References:
• COMPARATIVE STUDY OF CIRCUIT PARTITIONING
ALGORITHMS Zoltan Baruch1, Octavian Creţ2, Kalman
Pusztai3 .
• Network decomposition using Kernighan Lin strategy aided
harmony search algorithm G.A. Ezhilarasi,K.S.Swarup.

21. Thank You 