genetic algorithms (ga) n a class of probabilistic search algorithms. n inspired by natural genetics...

Genetic Algorithms (GA)

A class of probabilistic search algorithms. Inspired by natural genetics and biological

evolution. Uses concept of “survival of fittest”. GA originally developed by John Holland

(1975)

GA –general concepts Iterative procedure(iterative

improvement) Produces a series of “generations

of populations” one per iteration Each member of a population

represents a feasible solution,called a chromosome.

bagchromosomes

Combine and die(mating)

p11

2

3

K

m

evolution

Chromosome-a feasible solution

Selection crossover mutation

Pk

P3

The population during iteration of GA is denoted by the “bag”

Pi={X1i,X2i,……..,Xni}.

Xmi denotes the m th member of the population in i th iteration and n is the size of the population.

Pi;is a bag ,not a set,hence can contain repeated solutions, i.e. all entries need not be unique.

Initial population P1,may be created randomly or by a deterministic (constructive) heuristic.

Cost(x) is the cost of solution x.This function is user supplied.

Assume cost(x)>=0 V X. Let fitness of (x) =1/1+cost(x), thus V x,

0<=fitness(x)<=1. Average-fitness(Pi)=Σ fitness(xj

i)

Best-fitness(Pi) =max{fitness(xji)}

Going from P1 to P2 to P3…..

is called “evolution”. Goal:ensure that it is highly likely that

average-fitness(Pi)>average-fitness(Pj) for i>j. We go from Pi to Pi+1 via an evolution function that

usually consists of three components called operators.

Operators Selection: select some members of current

population to move to the next generation.Try to select “highly fit “ solutions.

Use an (imaginary) roulette wheel technique. Fitness of .37 => 37%

21n

3

4

Area is

Proportional to fitness value

ExampleChromosome fitness

A .21

B .14

C .10

D .02

E .33

F .20

-----------

1.00

A

21% D 2%

10%

F

20% E

33%

B

14%

C

1) When wheel is turned and eventually stops,probability of stopping at A is 0.21

2) The function select-roulette(f) spins roulette wheel by a random degree(uniform between 0 and 2) and returns a chromosome – the one pointed to by the arrow head

Crossover Crossover operators combine features of

highly fit members of a population to create new members.

First select two unique parents(chromosome)

Say Xa and Xb using select-roulette(). Divide parents into two parts; Xa < Xa1 , Xa2> Xb < Xb1 , Xb2> Exchange to get < Xa1 , Xb2> < Xb1 , Xa2>

This is a mating process of highly fit members.

Example: Often bit strings are used to

represent a chromosome.

Xa;100011;1110 ya;100011;1010

Xb;101010;1010 yb;101010;1110

Mutation Mutation operators randomly alters

structure of chromosome. Pm - user supplied

probability .Select a chromosome with probability,Pm usually;.05<Pm<0.2.

Mutation used to introduce new features into a population.

Inversion:

Third operator of GA Like Mutation, it also operates on a single

chromosome. To laterally invert the order of alleles

between two randomly chosen points on chromosome.

Example: Given String: 0 1 2: 3 4 5 6 7: 8

b i d:e f g c h: a

Will Become: b i d:h c g f e: a

Representation

Often a chromosome is encoded as a bit stream. The actual representation might be a structure,such as a floorplan, placement or route(wire).

By operating on bitstream one can generate a bit stream that doesn’t correspond to a legal structure.

Thus non bit string representations are often used including graphs and lists.

Solving the logic partitioning problem via a GA

Problem formulation K chips Each has area Amax Each has thermal capacity Hmax Each has Pm capacity Pmax Design represented by a graph G=(V,E) Each mode v V represents a gate and

has areaA(v),heat dissipation H(v). Each hyperedge e is a subset of modes of

V.(e represents a signal.)

Partition V into disjoint subsets V1,V2,…..,Vk so that

For i=1,2,3,….kPi ={e E | (v,w) e such that

v Vi and w Vi}| <= Pmax Ai = Σ A(v) <= Amax v Vi

Hi = Σ H(v) <= Hmax v Vi

See algorithm1Encoding:Use double link list to tie together one chip to next chip,and elements(gates)within a chip.

Algorithm 1 (Genetic Algorithm)

n : population sizePs : selection percentagePc: crossover percentage/*100 – Ps*/Pm: mutation probability thresholdPk: population bag belonging to iteration k

GA (n,Ps,Pm)

begin best_fitness 0;best_solution null;Po create_initial_population(n);i0;/* generation counter*/ repeat for each member xPi do fitness1/(1+cost(x)); if (fitness > best_fitness) then best_solution x; best_fitness fitness; end if; end for; if (user-specified exit condition is satisfied) then exit else Pi+1evolve(Pi,n,Ps,Pm); ii+1 ; end if;forever;end;

function evolve(p,n,Ps,Pm); bs : bag of solutions selected bc : bag of solutions crossed over bm : the bag bs bc after mutation begin bs select(p,n,Ps); bc crossover(p,n-|bs|); bm mutate(bs bc,n,Pm); return(bm); end;

function select(p,n,Ps) begin bs {}; i 0; while(i<Ps *n) do x select_roulette(p); bs bs {x}; ii+1; end while;return(bs) end;

function mutate(p,n,Pm)

begin

bm {};

for each member xp

if random(0,1) Pm then

bm bm{mutate(x)};

else

bm bm {x};

end if;

end for;

return(bm);

end;

Figure 1: Representation scheme for a structure

Chip0

R15R14R6R5R1

R11R0

R12R7R2

R4R3

R10R8

chip1

R13R9chip4

chip3

chip2

Ri:components of a design(gates)

CrossoverFigure 2 shows that emphasizing a simple crossover

can produce an illegal(illogical) child(partition). For example chip 3 of child one has two R12 entries. In addition R14 occurs in both chip 1 and chip3. There are two ways to avoid illogic structures.

1. Use a cost function that penalizes illogical structures so they die a natural death within a few generations.

2. Use a crossover operator that does not produce illogical structures.

We will use the “uniform crossover” operator that is in the second category.

We split two parents into distinct parts and exchange the parts to form two child structures.

Illegal structures produced by simple crossover

chip0

chip1

chip3

chip2

chip1

chip0

chip2

chip1

chip0

chip3

chip2

chip0

chip3

chip3

chip2

chip1

R4R15

R6R4

R13R1

R15R14R11R9R3R2R0

R8R5

R11

R1

R3R2R0

R3R8R5

R12R10R7

R9R5R4

R2R1R0

R15R7R6R4

R11R10R9R13

R10R9R8R7R6

R15R14R13R12R12R10R14R13R12

R8R7R6

R14

R3

R5

R2R1

R11

R0

Parent 1 Child 1

Parent 2 Child 2

Uniform Crossover For each gate i in the design,let Cx(i) denote the

chip that contains gate i in structure (chromosome X). The children are constructed as follows.

Cy1(i) = Cx1(i) if T(i) =1

Cx2(i) otherwise

Cy2(i) = Cx1(i) if T(i)=0

Cx2(i) otherwise

Thus child Y1 is produced by assigning the i th gate to the same structure as in the parent X1(X2) when T(i)=1(T(i)=0).

Child Y2 is produced by reversing the roles of the parents

Parent 1 Parent 2

chip0 R0 R4

R9

R11R5

R6

R10R8R2R0

R7R6R5R4R3

R11R9R1

R7R3R0

R10R6R4R2R1

R11R9R5R9R7R6R5R3

R11R4R1R0

R8

R2

R8R7R3

R1

R10R8R2

chip1

chip0

chip2

chip1

chip0

chip2

chip1

chip0

chip2

chip2

chip1

R10

Child 1 Child 2

0 1 1 1 0 1 0 1 1 0 1 0

Template for crossover 0 1 2 3 4 5 6 7 8 9 10 11

Example of uniform crossover

“0’s” “1’s” Parent 1 Parent 2

chip0 R0 R4

R9

R11R5

R6

R10R8R2R0

R7R6R5R4R3

R11R9R1

R7R3R0

R10R6R4R2R1

R11R9R5R9R7R6R5R3

R11R4R1R0

R8

R2

R8R7R3

R1

R10R8R2

chip1

chip0

chip2

chip1

chip0

chip2

chip1

chip0

chip2

chip2

chip1

R10

Child 1 Child 2

0 1 1 1 0 1 0 1 1 0 1 0


0 0

“0’s” Parent 1 parent 2

chip0 R0 R4

R9

R11R5

R6

R10R8R2R0

R7R6R5R4R3

R11R9R1

R7R3R0

R10R6R4R2R1

R11R9R5R9R7R6R5R3

R11R4R1R0

R8

R2

R8R7R3

R1

R10R8R2

chip1

chip0

chip2

chip1

chip0

chip2

chip1

chip0

chip2

chip2

chip1

R10

Child 1 Child 2

0 1 1 1 0 1 0 1 1 0 1 0


00 0


chip0 R0 R4

R9

R11R5

R6

R10R8R2R0

R7R6R5R4R3

R11R9R1

R7R3R0

R10R6R4R2R1

R11R9R5R9R7R6R5R3

R11R4R1R0

R8

R2

R8R7R3

R1

R10R8R2

chip1

chip0

chip2

chip1

chip0

chip2

chip1

chip0

chip2

chip2

chip1

R10

Child 1 Child 2

0 1 1 1 0 1 0 1 1 0 1 0

Template for crossover0 1 2 3 4 5 6 7 8 9 10 11

1 111


chip0 R0 R4

R9

R11R5

R6

R10R8R2R0

R7R6R5R4R3

R11R9R1

R7R3R0

R10R6R4R2R1

R11R9R5R9R7R6R5R3

R11R4R1R0

R8

R2

R8R7R3

R1

R10R8R2

chip1

chip0

chip2

chip1

chip0

chip2

chip1

chip0

chip2

chip2

chip1

R10

Child 1 Child 2

0 1 1 1 0 1 0 1 1 0 1 0


111

T-a binary string whose length is the number of elements to be partitioned.T is also called a template.

Randomly set the bits in T to 0 and 1.

Select two parent structures X1 and X2 for mating.

Let Y1 and Y2 children resulting from crossover.

Mutations

Probabilistically select a member of the population.

Probability Pm of a chromosome being selected.(0.1 Pm 0.2)

For each selected chromosome (partition),a gate is randomly selected and reassigned to another chip.

Example: In figure1; move R14 from chip2 to chip1.

Cost function

(weighted combination of four conflicting cost factors)

cost(x) = Σ Δ Ai (x)/Amax + Σ Δ Hi (x)/Hmax

+ Σ Δ Pi (x)/Pmax + Δ T(x)/Tmax

Where

Δ Ai = 0 if Ai Amax

Ai-Amax otherwise

Same for Δ Hi and Δ Pi

Fitness(x) = 1/1+cosT(x)

Experimental results

Compare GA with Fiduccia-Mattheyses(FM) and Simulated Annealing(SA). First used large values on pin and area constraints using GA and SA algorithms,and tightened(reduced) the constraints until a feasible solution could not be found. For viper SA needed more pins than GA. FM is the fastest of the three algorithms.in one case GA is 44 times slower than FM. GA is usually slower than SA. GA produces better results than SA and FM.

Example

Beh.VHDL

No Of Lines

No of Components

No of Nets

FIFO 65 69 187

Find 65 90 479

TLC 45 49 203

Move Machine

105 49 366

Viper 350 156 658

See figure 4 for variations in fitness values with increasing generations.

We show Best fitness ever: This is the value of the fitness measure of the best fit(least cost) solution seen thus far.This is a non-decreasing function. Maximum fitness of generation: This is the value of fitness measure of the best fit solution in the current generation. Average fitness of generation: This is the average

fitness of all the solutions in the current generation. Average fitness lags best fitness.

Placement via GA G=(V,E) – hypergraph. V- gates E- nets Assume a Gate Array layout structure. Assign(place) each gate to a unique slot(position). S= {S1,S2,….Src} set of slots. Slots are arranged into rows and columns;r-rows,c- columns. r.c > p - where |v| = p is number of gates. For a given placement,its associated length L is given by L = Σ L(e) where L(e) is an estimate of the length of eεE interconnect (wire) needed to route net e. Find an assignment so that L is minimal.

Genetic Encoding and operators

Encoding

Assume placement consists

of a rectangular grid of slots

as shown in the figure on the

next slide.

Encoding Dummy Gates

Row 4

Row 3

Row 2

Row 1

Row 0

C4 C3

C9

D6

C5C12D5

D7C10

C14C13C7C1D3

C8 C18 C2 D4 C6

C11

C15D2C17D1C16

Col 0 Col 1 Col 2 Col 3 Col 4

Chip

Empty Slot Occupied Slot

C1 (a gate) is in row 1 column 1, etc.

Empty slots are filled with dummy gates that have unique names , zero area, zero heat (power), no connections .

For previous placements the encoding is : C16 , D1 , C17 , D2 , C15 , D3 , C 1 etc. (Reading from bottom to top and left to right)

G

CBA

FED

IH End

Start

A B C D E F G H I

Encoding Using for Placement

Cross Over

A B E A BA B C D E

D C E A B D C C D E

Effect of Single Point Cross Over (Illegal Placements)

• A simple one- point crossover operator leads to illegal

placements as shown above.

Parent 1 Child 1

Parent 2 Child 2

Child 1 has 2 A’s and 2 B’s , no C or D .

Child 2 has 2 D’s and 2 C’s no A or B

To resolve this, we employ a cycle crossover operator .

Cycle Crossover

Let P1 and P2 denote the 2 parents & C1 and C2 denote the 2 children .

Let P1 , P2 , C1 , C2 arrays be of size r.c

Child C1 is formed by first copying a number of genes (gate positions ) in P1 into C1 , and then filling the rest of the genes in C1 from P2 .

First , C1[0] P1[0] Let location i in P1contain the gate that

occupies position 0 in P2 Then C1[i]P1[i]. Continue this process . That if we just filled position i in C1 by copying position i in P1 , then we identify the gate in position i in P2 and determine the location j of that gate in P1 and copy that gate to position j in C1 .

This process halts when we get to a gate that has already been copied into C1. This completes a cycle .

Subsequently , we fill each unfilled position X in C1 by copying from P2 the gate assignments at X , i.e

C1[X]P2[X].

To form C2 carry out the same process while interchanging the role of P1 and P2

Example

P1 = I H B A G D E C F

P2 = A B C D E F G H I

I = 0 1 2 3 4 5 6 7 8

C1 = B C E G H

C2 = H B G E C A

D F

IFD

I A

Cycle Crossover

C1[0] P1[0] ; so C1[0]=I P2[0] = A . Location of A in P1 is 3 (i=3) C1[i] P1[i] ; So C1[3] = A P2[3] = D . The Location of D in P1 is 5

etc. The cycle ends when we come back to

I . Then C1[1] P2 [1]= B C1[2] P2[2]= C etc. Show that this procedure always leads to

a legal placement

GA in placement (Module Assignment):

b 4 c 7 e g

1 1 3 6 7

a d i f h

3 2 4 8

(a) Design given in a form of a graph

7 8 9 d e f

4 5 6 c b i

1 2 3 a g h (b) Partition Definition (c) One possible assignment

GA In-placement cont. Thus the string (below) represents the solution in

(c) 1 2 3 4 5 6 7 8 9 a g h c b i d e f with the fitness value of

1/85. Consider 2-modules g and f with the Manhattan

distance of 3 (2 vertically and 1 horizontally)The connection between g and f has a weight of 7,

thus for gf we have 7x3=21, doing as above for all assignments (in C) we get 85.

Other possible solutions: b d e f i g c h a with fitness 1/110 i h a g b f c e d with fitness 1/95

Mutation (In Placement Example)

Randomly select two gates (one gate and a dummy gate ) and interchange them .

Interchange two arbitrary rows or columns (massive mutation )

Do a cyclic rotation of the entire placement i.e. P[0] C[1] ,

P[1] C[2], P[2] C[3], …., P[n] C[0].

Evaluation Function

Use P/ 2 measure .

Lest = ½ Σ Pe , i.e. the sum of all the P/2 ‘s for all nets .

Cost = ΔL/Lmax

Where ΔL = 0 if Lest <= L max

Lest - Lmax otherwiseWhere Lmax is user specified

constraint on the maximum wire length . ( a lower bound)

Let L max = 1000

Lest1 = 1100

Lest2 = 1500

ΔL1 = 100

ΔL2 = 500

Fitness1 = 1/ 1+100

~ 1/100 = 0.01

Fitness2 = 1/1+500

~ 1/500 = 0.002

Design No of

Chips No of Slots

No of Nets

Mcc2 37 7 x 7 4767

Viper 10 4 x 3 658

Find 10 4 x 4 479

TLC 9 4 x 3 203

Move Machine

7 3 x 3 366

FIFO 7 3 x 3 187

MCM placement test cases

Design Wire

Length106 microns

Heat Cost Calories

Exec. Time Sec.

Wire Length106 microns

Heat Cost Calories

Exec. Time Sec

Mcc2 3600 92 4337 4100 1323 13874

Viper 21.6 0 303 25.5 66.67 235

Find 117 25 346 120 129 97

Move 6.5 2.7 130 7.6 0 25

FIFO 1.86 10 150 2.19 0 87

TLC 1.66 5.4 104 2.23 38.8 32

Genetic Algorithm Simulated Annealing

Results of placement for GA and SA

Results

GA produced better results than SA .

For one large problem it was even faster than SA

For large problems where convergence is slow GA is better because one can make use of the multiple solutions (chromosomes).

Review of partitioning

Why – what are the constraints and objectives? What entities get partitioned?

PCB,MCM,chips,etc.

Is partitioning done on functional or structural

information? Techniques: Initial construction serial or parallel.

Conjunctions and disjunctions Kernighan and Lin (K-L) Fiduccia-Mattheyses (F-M) Min-Cut (M-C) Simulated Annealing(SA) Genetic Algorithm(GA) Randomness in an effective technique Same constraints are non-monotonic Cut nets not edges See recent directions in netlist partitioning: A survey by”c.albert and B.kahng”,VLSI integration, pp.1- 81,1995

Categories of techniques

1. Move-based approaches2. Geometric representations3. Combinational formulations4. Clustering approaches

Techniques not discussed 1 Hall’s quadratic placement

2 Vector partitioning3 Dynamic programming4 Min delay5 Max-flow min-cut6 Bipartite flow7 Mathematical programming

Functional partitioning

genetic algorithms (ga) n a class of probabilistic search algorithms. n inspired by natural genetics...

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