Kernighan-Lin Method

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Kernighan-Lin Method
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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.

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Application

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

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• 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.

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• 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.

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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.

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• 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
•  

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• 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

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Advantage

• Algorithm is Robust.

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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

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

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Solved Example
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• After initial partitioning

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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
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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 Largest G value G41 2

A' A' - 4 2, 3 A' A' - 4 2, 3
B' B' 1 5, 6 B' B' 1 5, 6
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Iteration 2 D values connected to ( 4, 1) are 2 in A' D values connected to ( 4, 1) are 2 in A' 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
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Computing 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, Since all Gain values are equal, we arbitrarily choose any one, Since all Gain values are equal, we arbitrarily choose any one, Since all Gain values are equal, we arbitrarily choose any one, Since all Gain values are equal, we arbitrarily choose any one,

A' A' - 3 2 A' A' - 3 2
B' B' - 6 5 B' B' - 6 5
we choose G36
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Iteration 3 ( 3 , 6 ) connected to 2 in A' ( 3 , 6 ) connected to 2 in A'
and 5 in B'
Step 1 Computing 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
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Determining K
Step 5 G1 2
G2 -3 i.e G1 G2 -1 i.e G1 G2 -1

G3 1 G1 G2 G3 0 G1 G2 G3 0
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References

• COMPARATIVE STUDY OF CIRCUIT PARTITIONING
  ALGORITHMS Zoltan Baruch1, Octavian Cret2, Kalman
  Pusztai3 .
• Network decomposition using Kernighan Lin
  strategy aided harmony search algorithm G.A.
  Ezhilarasi,K.S.Swarup.

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• Thank You ?