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Partitioning

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Outline

What is Partitioning

Partitioning Example

Partitioning Theory

Partitioning Algorithms

Goal

Understand partitioning problem

Understand partitioning algorithms

Divide a design into smaller pieces

based on a set of constraints

Constraints

amount of design in a partition

number of nets to/from partition

number of partitions allowed

balance between partition sizes

weight nets crossing partition boundaries

Applications

divide large design into multiple packages

place circuit components on a chip or board

Constraints

12 transistors per package

7 pins per package

few packages as possible

nets of equal weight

Bounds

30 xistors => >=3 packages

21 terminals => <=3 packages

6

4

6

4

4

6

12 xistors

7 pins

12 xistors

5 pins

6

4

6

4

4

6

6 xistors

4 pins

12 xistors

7 pins

8 xistors

5 pins

6

4

10 xistors

6 pins

6

4

etc.

4

6

Partitioning Set Formulation

V a set of nodes (components)

each node r having area a(r)

X a set of nodes (terminals) external to V

S = (S1, S2, ..., SN) a set of subsets of V U X

Si correspond to nets

partition V into disjoint subsets V1, V2,..., Vk such that

area of nodes in Vi<= Ai

number of sets in S which have nodes external and internal to Viis <= Ti, (pin count)

Other formulations

graphs - weighted nets, edges

connection matrix - eigenvectors

Direct

seed each module, grow based on constraints

Group Migration

randomly place components, then move between partitions

Metric Allocation

goodness metric for each component pair, partition to minimize metric

Simulated Annealing

shuffle components among partitions to minimize cost function, permit uphill moves to get around local minima

place a component r from V into each partition Vi

pick relatively independent components

for each remaining component r in V {

for each Vi with area of nodes <= Ai and number of sets in S which have nodes external and internal to Vi <= Ti,

compute cost of placing r in Vi

place r in lowest-cost Vi

}

Complexity

O(V*k)

assumes placement cost computation is O(1)

Issues

sensitive to initial seeding

sensitive to component examination order

gets stuck in local minimum

D

G

C

F

E

B

A

A

D

A

G

E

D

F

B

C

G, F

A

E

D

B

C

E

C

A

A

B

E

E

D

D

B

1. Partition nodes into groups A and B

2. For every a in A, and every b in B {

compute change in terminal counts Da and Db that occur if a and b are swapped.}

set queue to empty and i = 1.

3. Select from all pairs (a, b) the pair (ai, bi) that gives most reduction in total terminal count when swapped.

add to queue

save the improvement in terminal count as gi

4. Remove ai from A and bi from B, recalculate Da and Db.

if A and B not empty, i++, go to 3.

5. Find k such that G = sum of g1 to gk is a minimum.

swap a1,...ak and b1,...bk.

if G < 0 and k > 0, go to 2, else stop.

D

G

C

F

E

B

A

Initial Partition, Cutset = 6

Da Db

Da Db

(A,E) 0 -3

(A,F) -1 0

(A,G) -1 0

(C,E) +1 -3

(C,F) 0 0

(C,G) 0 0

(D,E) -2 -3

(D,F) -2 +1

(D,G) -2 +1

(B,E) +1 -3

(B,F) 0 0

(B,G) 0 0

G

Queue

1: (D,E) -5

2: (A,F) +1 (-1,+3)

...

C

F

removed

(D,E)

B

A

Final Partition, Cutset = 1

- Minimum G=-5 at k=1
- Swap D and E
- Next iteration:
- all pairs positive
- k=0, quit

A

D

G

E

B

C

F

Complexity

O(n2) per iteration in worst case for n nodes

steps 2, 3

converges in only a few iterations

newer versions are O(nlogn)

Application to partitioning

use bisection

divide into two partitions, then split those partitions, etc.

partition area is ignored, partitions remain balanced

subdivide until partition area is small enough

algorithm also called min-cut since it minimizes terminals

Approach

move one cell at a time between partitions A and B

less restrictive than pair moves

no longer need to maintain partition size balance

use special data structures to minimize cell gain updates

only move cells once per pass

Complexity

prove constant number of updates per cell per pass

runtime per pass = O(P) - P pins/terminals

vs. O(n2) for group migration

small (3-5) number of passes to converge

estimate total time is O(PlogP)

more pins than nets

usually more pins than cells

ni - number of cells on neti

si - size (area) of celli

smax - largest cell = max(si)

S - total size of cells = sum(si)

pi - number of pins on celli

pmax - most pins on a cell = max(pi)

P - total number of pins = sum(pi)

C - total number of cells

N - total number of nets

r - fraction of cell area in partition A

CELL - C-entry array, entry is linked list of nets on cell

NET - N-entry array, entry is linked list of cells on net

Cell Gain

gi - reduction in cutset by moving celli

label each cell with its gain

-pi <= gi <= pi

-pmax <= gi <= pmax

Gain Data Structure

BUCKET array of cell gains

one per partition

O(1) access to cells of a given gain

MAXGAIN tracks cells of max gain

remove cell once it has moved

cells move once per pass

gain does not matter after that

+2

+1

0

-1

BUCKET

+pmax

MAX

GAIN

Cell #

Cell #

-pmax

CELL

1

2

C

Minimize cell gain updates

if all cells are updated on each move - O(C2) algorithm

only cells that share a net with a moved cell must be updated

but big net implies many moves and many updates

only cells on critical nets must be updated

Critical nets

moving a cell would change cutstate

cutstate - whether net is cut or not

critical only if net has 0 or 1 cells in A or B

A=0, B=3

critical

A=1, B=2

critical

A=2, B=2

not critical

Control size balance between partitions

otherwise all cells move to one partition

cutset = 0

Balance criterion

rS - smax <= |A| <= rS + smax

permits some “wiggle room” for cells to move

yes

no

Initially place cells randomly into A and B

Compute cell gains

Algorithm for each pass

for all cells in A and B of maximum gain whose move would not cause imbalance

choose one with best balance result - the base cell

if none qualify, quit pass

move to opposite partition and lock (remove from BUCKET)

unlocked cells are free cells

update cell gains and MAXGAIN pointer

Repeat passes until no moves occur

unlock all cells at beginning of pass

F = “from” partition of base cell

T = “to” partition of base cell

FT(n) = # free T cells of net n

FF(n) = # free F cells of net n

LT(n) = # locked T cells of net n

LF(n) = # locked F cells of net n

for each net n on base cell do

if LT(n) == 0

if FT(n) == 0 UpdateGains(NET(n))

else if FT(n) == 1 UpdateGains(NET(n))

FF(n)--

LT(n)++

if LF(n) == 0

if FF(n) == 0 UpdateGains(NET(n))

else if FF(n) == 1 UpdateGains(NET(n))

Initial partition

cutset = 3

si = 1, r = 0.5, 1 <= |A| <= 3

Final partition

cutset = 1

+3

+1

+1

-1

L

-3

L

+1

a

b

c

d

b

a

c

d

- b, c are highest-gain candidates

- choose b, move and lock

- recompute gains for a and b

- no candidates qualify

- quit pass

- second pass

- no candidates qualify

- quit

L

+1

+1

-1

b

a

c

d

-1

-3

-1

+1

- a, c are highest-gain candidates

- choose c, move and lock

- recompute gains for a, d

b

a

c

d

Fast

O(PlogP)

constant factors are small

arrays, pointer access

Space Efficient

5 words/net overhead

4 words/cell overhead

small constant overhead

Suboptimal

gets stuck in local minima

any one move has negative gain

need multiple moves for positive gain

Example

moving a or b results in -3 or -4 gain

moving both a and b results in +1 gain

a

-3

-3

-4

-4

b