1

Multi-Bend Bus-Driven Floorplanning

Jill H.Y.Law & Evangeline F.Y.Young

The Chinese University of Hong Kong

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Outline

Introduction Background Sequence Pair

Methodology Shape Validation Bus Ordering Floorplan Realization

Experimental Results Conclusion

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Introduction

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Background

Why is bus-driven floorplanning important? Previous work:

Hua Xiang, Xiaoping Tang and Martin D.F. Wong “Bus-Driven Floorplanning” ICCAD 2003

0-bend is not enough for buses going through many blocks

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Problem Formulation

A set of n blocks B = {b0, b1, …, bn-1}

A set of m buses U = {u0, u1, …, um-1}, where each bus is associated with a width ti

Decide the position of blocks, such that Buses go through their blocks Chip area minimized Bus area minimized

Buses can have at most 2-bends

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Problem Formulation

What is meant by “going through” ? Assume block height > bus width

A

B

C

bus i >= ti

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Problem Formulation

AB

E

C

D A

F

J

H

E

DB

C G

I

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Sequence Pair (SP)

A pair of sequences of n elements (…a…b…, …a…b…), a is on the left of b (…a…b…, …b…a…), a is on top of b

Example: (acbde, daceb)

d

a c

e

b

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Methodology

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Overview

While Temp > threshold Apply a move to obtain a new SP Evaluate the floorplan

Shape Validation Bus Ordering Floorplan Realization

Accept or reject according to Cost and Temp

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Steps

Assumption 2 layers for bus routing

Evaluation Shape Validation

0-bend 1-bend 2-bend

Bus Ordering Floorplan Realization

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Shape Validation

Check the bus one by one From the relative position of the blocks, check if a

bus of at most 2 bends can go through all blocks

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Shape Validation – 0 bend

Step 1: Extract the related blocks from the sequence pair Example: For a bus ABE Given a SP (ABCDE, ABCED) (ABE, ABE)

Step 2: Check relative position between blocks Example: For a bus ABC Horizontal bus: (ABCDE, ADEBC) (ABC, ABC) Vertical bus: (CDBEA , ABCDE) (CBA, ABC)

A

BC

A

B

C

(ABCDE, ADEBC) (CDBEA, ABCDE)

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Shape Validation – 1 bend

L-shape bus 1 horizontal component 1 vertical component

How to recognize them? Step 1: Extract the related blocks (X, Y) Step 2: Find the Longest Common

Subsequence (LCS) of (X, Y) Horizontal component

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Shape Validation – 1 bend

Step 3: Check if the remaining blocks in reverse order Example: (ABCDEF, ABCFED) LCSABCD, remaining blocks (DEF, FED)

Step 4: Check if T-shape T-shape also contains one horizontal component and

one vertical component T-shape is kept for later 2-bend checking

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Shape Validation – 1 bend

When will a T-shape be formed?

A

B C

A

B

C

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Shape Validation – 1 bend

Example: (ABCDE, ADCBE) Horizontal Component: ABE Vertical Component: CD

AB

EC

D

AB

E

C

D

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Shape Validation – 2 bend

C-Shapes, Z-Shapes, mirrored Z-Shapes … HVH or VHV Assume HVH (VHV is similar)

How to recognize them? Step 1: Extract the related blocks (X, Y) Step 2: Find the LCS of (X, YR) vertical

Example: (ABCDE, ADCBE) Find the LCS of (ABCDE, EBCDA): BCD

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Shape Validation – 2 bend

Step 3: Put the remaining blocks in different relationships with the vertical component

B

C

D

E

A

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Shape Validation – 2 bend

8 possible relationships

Step 4: Put the blocks into 2 horizontal components

B

C

D

Verticalcomponent

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Shape Validation – 2 bend

Example: (GHABCEFD, EFDCBGHA) (GHABCEFD, AHGBCDFE) vertical component Upper = {A, H, G}, Lower = {D, E, F} A C-shape can be formed

EF D

C

B

GH

AA, H, G

D, E, F

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Shape Validation – 2 bend

Example: (AEBCDF, EDCBFA) (AEBCDF, AFBCDE) vertical component Upper = {A}, Lower = {D}, LowerLeft = {E}, LowerRight =

{F} No valid 2-bend shape can be formed

D

C

B

A

F

E

A

D

E F

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Shape Validation

Extract the related blocks Check if 0-bend Check if 1-bend Assume HVH

Check if 2-bend Assume VHV

Check if 2-bend Mark it invalid if all “no”

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Steps

Evaluation Shape Validation Bus Ordering Floorplan Realization

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Determine Bus Ordering

AB C

DE

The two bus componentshave ordering constraint

A

B C

DE

FG

I

H J

A

BC

D

The two bus componentshave ordering constraintwhich contradict with one

another

A

B

C

D

The two bus componentshave ordering constraintwhich contradict with one

another

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Bus Ordering

Step 1: Split all the buses into 0-bend bus components

Step 2: Build vertical graph and horizontal graph by looking at each pair of bus components Use a node to represent each bus component If bus component a has to be on top of bus component b,

add an edge from node a to node b If there is cycle, at least one bus component in the cycle

has to be removed Horizontal graph can be built in a similar fashion

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Bus Ordering

Step 3: If there are cycles, remove nodes (bus components) to make the graph acyclic Aim at removing the least number of nodes NP-complete Maximum degree heuristic

Step 4: Remove the corresponding bus components in the other graph as well

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Steps

Evaluation Shape Validation Bus Ordering Floorplan Realization

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Floorplan Realization

Realization = obtaining the positions of the blocks and buses

This step is the same as that in Xiang’s work

X

B

C

AX

B

C

A

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Simulated Annealing

Simulated Annealing Framework Moves:

Swap Rotation

Cost Function Cost = A + ‧ B + ‧ I‧ A: chip area, B: total bus area, I: number of invalid bus

Can consider other aspects by adding more terms in the cost function Total wire length Routing congestion

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Summary

While Temp > threshold Apply a move to obtain a new floorplan Evaluate the floorplan

Shape Validation Bus Ordering Floorplan Realization

Accept or reject according to Cost and Temp

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Experimental Results

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Platform

Language Implemented using C ++

Machine Intel Xeon (2.2 GHz) with 1 G memory

MCNC benchmarks

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Experimental Results

The data set used in Xiang’s work* calculated by [(y1 – y0) / y0]*100%

Xiang’s Work Our Work Comparison*

Time (s) Dead space Time (s) Dead space Time Dead space

apte 15 0.72% 30 0.48% +100.00% -33.33%

xerox 15 0.95% 35 0.42% +133.33% -55.79%

hp 33 0.62% 51 0.29% +54.55% -53.23%

ami33-1 11 0.94% 93 1.00% +745.55% +6.38%

ami33-2 92 1.27% 144 1.19% +56.62% -6.30%

ami49-1 16 0.85% 71 0.56% +343.75% -34.12%

ami49-2 302 0.84% 713 0.58% +136.09% -30.95%

ami49-3 285 1.09% 865 0.60% +203.51% -44.95%

Average: +221.65% -31.54%

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Experimental Results In this data set, each bus has to go through 10 – 15 blocks

* calculated by [(y1 – y0) / y0]*100% Xiang’s Work Our Work Comparison*

Time (s) Dead space Time (s) Dead space Time Dead space

ami33-3 86 1.81% 32 1.01% -62.79% -44.20%

ami33-4 >1000 NA 92 1.90% NA NA

ami33-5 > 1000 NA 95 3.80% NA NA

ami49-4 73 19.34% 88 0.63% +20.55% -96.74%

ami49-5 > 1000 NA 261 1.17% NA NA

ami49-6 > 1000 NA 140 2.19% NA NA

Average: 118 1.78%

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Experimental Results

ami49 – 2

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Experimental Results

ami49 – 3

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Experimental Results

ami49 – 6

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Conclusion

Solve the bus-driven floorplanning problem for 0-bend, 1-bend, 2-bend buses

The presence of 1-bend and 2-bend buses is important, especially when the number of blocks that a bus goes through is large