Partitioning. 1 st homework: Deadline: 13 Esfand 2

Partitioning

• 1st homework: Deadline: 13 Esfand

2

Partitioning

• PROBLEM: Large complex systems:Break up into smaller subsystemsCalled partitioningRequirements for a good partition

− Original functionality of the system is not changed− Minimize the interconnection between subsystems− Process should be simple and efficient

3

Partitioning Issues

Each subsystem can be designed independently

− → speeding up the design process

4

60 IPs

260 =32000 years 6 x 210 = 5 ms

10 IPs 10 IPs

10 IPs 10 IPs

10 IPs 10 IPs

Partitioning Issues

Decomposition scheme has to minimize the interconnections between subsystems.

Decomposition may be carried out hierarchically until each subsystem is of manageable size.

5

Circuit:

Cut ca: four external connections

1

2

4

5

3

6

7 8

5

6

48

7 23

1

56

48

7 2

3 1

Cut ca

Cut cb

Block A Block B Block A Block B

Cut cb: two external connections

Introduction

6

Circuit Example

• Start with a circuit of 48 logic gates:

Partition 1 – 4 lines 15 gates

Partition 2 – 4 lines 16 gates

Partition 3 – 17 gates

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System Hierarchy

8

Levels of Partitioning

System Level Partitioning

Board Level Partitioning

Chip Level Partitioning

System

PCBs

Chips

Subcircuits/ Blocks

9

System Level Partitioning

• Objectives:Optimize the system performance.Minimize the no. of boards.

• Constraints:Board size is fixed.Terminal count on each board is fixed.

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System

PCBs

Board Level Partitioning

• Objectives:Minimize the interchip connections.Optimize the performance and reliability

by minimizing the no. of chips.• Constraints:

Chip sizeTerminal count on a chip Multi-FPGA Design

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PCBs

Chips

Chip Level Partitioning

• Objectives:Minimizing the no. of nets cut by the

partitioning.Avoid cutting critical nets by the partitioning.

• Constraints:Terminal counts

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Chips

Subcircuits/ Blocks

Graph Partitoining: Terminology

5 6

4

2

1

3 3

2

4

5 61

Graph G2: Nodes 1, 2, 6.

Graph G1: Nodes 3, 4, 5.

Collection of cut edges Cut set: (1,3), (2,3), (5,6),

Block (Partition)

Cells

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Problem Formulation

شامل گره هايhypergraph G = (V, E)يک • V = {v1, …, vn}و

•hyperedge هايE = {e1, …, em} که هر( ei V ،)

V:را طوري تقسيم کنيد که

Vi Vj = , ij

ViVj = V

•Cut.مرز بخشها :

•Cut size تعداد يالهاي قطع کنندة :cut. 15

Objective Functions

pi

16

Constraints

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Constraints

TSV (Through Silicon Vias) in 3D chips

Ababei, Mogal and Kia Bazargan, “Three-dimensional Place and Route for FPGAs,” TCAD 2005.

www.micromagic.com

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Constraints: Area

• Bounded-Size Partitioning:

Aimin ≤ Area(Vi) ≤ Ai

max

• Balanced Partitioning:

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Scope of the Problem

[©Rutenbar]

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Multi-Terminal Nets

:two-terminal netsمدلسازي بر حسب •

1 )complete graph:

اشکال: تعداد زياد يالها غير واقعي •است.

.netحسن: تقارن •21

Multi-Terminal Nets

(2 Spanning tree:

netحسن: در نهايت، يک درخت براي •کافي است.

اشکال: از قبل نمي دانيم کدام درخت •بهترين است.

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مسائل خاص هر سبک طراحي

• Full-custom:

:cut size محدوديت •

d

pT ii

امpartition iمحيط نهايي Pitch size

باشد.hierarchical مي تواند •

مساحت کل = مجموع مساحت اجزا + مجموع مساحت •routing.مساحتهاي تلف شده +

ثابتها را cutsizeاگر

کم کنيم، به احتمال زياد کم

مي شود.

کيفيت placement آن

را تعيين مي کند.

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مسائل خاص هر سبک طراحي

• Standard Cell، Gate Array و FPGA:

شود که partition هدف: طوري مدار •cut sizeحداقل شود

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تقسيم بندي الگوريتمها

Constructive Iterative

Improvement

Partitioning Algorithms

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تقسيم بندي الگوريتمها

DeterministicStochastic

Partitioning Algorithms

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تقسيم بندي الگوريتمها

Simulation-Based

Group Migration

Partitioning Algorithms

Multi-Level .…

.…

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Example• Given this set of components• Divide it into 3 parts

4 logic cells per segment− Lots of solutions but only 1 optimal

A B C D

E F G H

I J K L

1 1

4

4

2

2 2

6 6

6

5 5 10

10 11

11

12

123

3

9

99

8 88

7

7 7

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Solution

• Minimum number of external connections:

A B

C L

D F

H I

E G

J K

1

2 2 6

6

4 4 55

10

11

9

3

8

12 7

PROBLEM: How to find this solution?

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Constructive Partitioning• Constructive Partitioning:

Uses a set of rules to find a solution− Most straightforward: seed growth (cluster growth)

• Steps:1. Start a new partition: a seed cell2. Among all cells not yet in a partition, select one at a

time.3. Calculate a gain function, g(m):

− the benefit of adding cell m to the current partition− perhaps the number of connections between m and the cells in

the current partition)

4. Add the cell with max gain to the current partition5. If not reach the size limit,

− Then repeat from step 2. − Else start a new partition from step 1.

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EXAMPLE

• Given the 12 logic cell structure (A to L) on the previous slide find a partition into 3 sets with 4 cells per setuse the gain function g(m) defined as:

− let P(m) be the number of nets (not connections) from cell m to the current partition

− let N(m) be the number of nets from m to cells not yet in partitions

− then g(m) = P(m) - N(m)

select cell C as the seed

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Partition Table 1

• Set up a table for bookkeeping to keep track of the gain at each step - label the final partitions X, Y, and ZPass A B C D E F G H I J K L 1

A B C D

E F G H

I J K L

1 1

4

4

2

2 2

6 6

6

5 5 10

10 11

11

12

123

3

9

99

8 88

7

7 7

C

X

0-2 = -2 A

-2

1-2 = -1

-1

D0-2 = -2

-2 -3 -5 -2 -2 -3 -3 -2 -1

L

A B DB D

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Partition Table 2

Pass A B C D E F G H I J K L

A B C D

E F G H

I J K L

1 1

4

4

2

2 2

6 6

6

5 5 10

10 11

11

12

123

3

9

99

8 88

7

7 7

C

L

2 0-2 1-2 0-2 2-4 0-5 1-2 0-2 0-3 0-3 0-2 -2 -2 -2 -2 -5 -1 -2 -3 -3 -2

3 0-2 1-2 0-2 2-3 0-5 0-2 0-3 1-3 1-2 -2 -1 -2 -1 -5 -2 -3 -2 -1

GE

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Partition Table 3

A B C D

E F G H

I J K L

1 1

4

4

2

2 2

6 6

6

5 5 10

10 11

11

12

123

3

9

99

8 88

7

7 7

C

L

GE

Pass A B C D E F G H I J K L 1 1-1 0-2 1-1 X 0-2 2-2 0-3 0-2 0 -2 0 -2 0 -3 -2

F

2 1-1 0-2 1-1 1-1 X 1-3 1-2 0 -2 0 0 -2 -1

I

3 X 1-2 1-1 1-1 1-3 1-2 -1 0 0 -2 -1

A D

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RESULT

• The final partition has 8 external connections (1,2,5,7,8,9,11,12) worse than the other solution (5 connections) perhaps starting with a different seed cell improve the result

A D

I F

B H

J K

C G

E L

1 2

6

4

5

10

11

9

3

8 12

7

1 11

59

8

2

12

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Iterative Partition

• Constructive Partitioning can produce suboptimal results Iterative partitioning algorithm is often

used to improve the final partition− E.g., swapping logic cells or (or groups of logic cells)

• Group migration methods:mostly based on the Kernighan-Lin Algorithm (K-L

Algorithm)divide a graph into two segments (min-cut

problem).

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Iterative Improvement

Constructive Algorithm

Iterative Improvement

no

yes

TerminationCriterion Met?

Return Best-Seen Solution

Problem Instance

Initial Solution

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Kernighan-Lin Algorithm

“An Efficient heuristic Procedure for Partitioning Graphs”, Kernighan and Lin, The Bell System Technical Journal, 49(2):291-307, 1970.

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K-L Algorithm

• Start with a network already split into two partitions, A and B each with m nodes. Define cab to be the

weight of the connectionbetween nodes a and b

cab = 1 if there is a connectionand 0 if there is not

The cut cost is W where:

1

2

3

4

5

6

7

8

9

10

BbAa

W,

abc

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Definitions

• For any node a in partition A, the external edge cost measures the connections from node a to B

Az

azcaI In the example I1 = 0 and I3 =2

The internal edge cost measures the internal connections to a

By

aE ayc In the example E1 = 1 and E3 = 0

1

2

3

4

5

6

7

8

9

10

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Definitions

The cost difference is the difference between external and internal edge costs

1

2

3

4

5

6

7

8

9

10

aaa IED

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Gain Function

• Select any node in A and any node in B if we swap these nodes, a and b, we need to measure

the reduction in the cut weight which is the gain g given below:

abb cDDga

2

As a and b may be connected

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K-L Algorithm

• The K-L Algorithm finds a group of node pairs to swap which increases the gain

• STEPS1) Find two nodes ai from A and bi from B so that the gain from

swapping them is maximum2) Next pretend to swap ai and bi (even if the gain is 0 or

negative)3) Lock them.4) Repeat steps 1 and 2 until all the nodes of A and B pretend

swapped. 5) Now we can select which nodes to actually swap. Suppose

we only swap the first n pairs of nodes found in the above steps. The total gain would be Gn (see below), so select the n which maximizes Gn

n

iigG

n

1

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Iterative Process

• If the maximum value of Gn > 0 then the n node pairs are swappedK-L is applied to this new partition

• If the maximum value of Gn ≤ 0 then we stop

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Example

• Given the graph below, its initial partition has 4 external edges Calculate the gain from

swapping all pairs of nodes(there are 25 pairs)

1

2

3

4

5

6

7

8

9

10

Node E I D 1 1 0 1 2 2 0 2 3 0 2 -2 4 1 1 0 5 0 1 -1 6 1 0 1 7 1 1 0 8 2 1 1 9 0 2 -2 10 0 2 -2

Pair g1,6 1 + 1 - 0 = 21,7 1 + 0 - 2 = -11,8 1 + 1 - 0 = 21,9 1 - 2 - 0 = -11,10 1 - 2 - 0 = -12,6 2 + 1 - 2 = 1 . . .

This is the best 45

First Swap

• Swap 1 and 6 (fake)for a gain of 2

Redo the process withthe new graph

6

2

3

4

5

1

7

8

9

10

Node E I D 2 1 1 0 3 0 2 -2 4 1 1 0 5 0 1 -1 7 0 1 -1 8 2 1 1 9 0 2 -2 10 0 2 -2

Pair g2,7 0 - 1 - 0 = -12,8 0 + 1 - 2 = -12,9 0 - 2 - 0 = -22,10 0 - 2 - 0 = -23,7 -2 - 1 - 0 = -3 . . .5,8 -1 + 1 - 0 = 0 . . .

This is the best 46

Second Swap

• Swap 5 and 8 for again of 0

• Continue this process, the third and fourth swaps will also have a gain of 0 and the final swap will have a gain of -2. As the gains from each swap are summed they max out at

the first swap so keep the swap of 1 and 6 and repeat this process on the new partition.

6

2

3

4

8

1

7

5

9

10

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End of First Iteration

Cut Size gi Node Pair i

4 - - - 0

2 2 2 (1, 6) 1

2 2 0 (5, 8) 2

2 2 0 3

4 0 -2 4

ig

• Find maximum and exchange (actually) up to that point.

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Another Case

Cut Size gi Node Pair i

4 - - - 0

2 2 2 (1, 6) 1

3 1 -1 (5, 8) 2

0 4 3 3

2 2 -2 4

ig

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Kernighan & Lin Partitioning

• If sum to m > 0, some gain, so repeat until sum to m=0

gkk

i

1

i1 2 3 m n

[©Newton]

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KL Algorithm

-2cab

-2cab

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Computational Complexity

For 2n nodes

Initial computation of the D’s takes O(n2) For each pass:

For i = 1 to n: Compute the gain for all free pairs takes O(n2)

- After swapping ai and bi in movei, at most (2n-2i) gains of free nodes are updated

- For n moves:

Pair comparison in movei: (n-i+1)2 = O(n2) pairs to choose from: For n pairs comparison:

Si=1,…,n (2n-2i) = O(n2)

Si=1,…,n (n-i+1)2 = O(n3)

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Computational Complexity

Tentative Exchange, Lock, Log take O(1),

total time = # of passes O(n3)# of passes: usually constant (independent of n)

In practice: improvement almost nothing after 4 passes

To speedup pair comparison:

node pairs can be sorted ahead of time

O(n2log n)

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KL Algorithm

• Drawbacks:• Unable to handle hyperedges.

• Partition sizes must be given in advance.

• But can handle unequal sizes

• Unable to handle vertex weights.

• Can assign unit area = gcd (cell_areas)

• edge weight s= infinity

• Time complexity: high = O(n3)

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KL Algorithm

• Advantages:• Simple.

• Performs well for up to ~102 nodes

• can handle constraints as having some blocks together

• (e.g. components of an adder)

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KL Software Package

• A software package : available from the class site. It will allow you to

− enter a connection matrix− set up an initial partition− run KL

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Main Screen

Make orLoadAdjacency matrix

Then selectGIP

Then Run

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Enter a Matrix

Enterthe size andthen theconnectionmatrix

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