04/19/23 1

VLSI Physical Design Automation

Prof. David Pan

dpan@ece.utexas.edu

Office: ACES 5.434

Detailed Routing (I)

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After Global Routing: Detailed Routing

The routing regions are divided into channels and switchboxes.

So only need to consider the channel routing problem and the switchbox routing problem.

AA

BB

304/19/23

Channel Routing for Different Styles

• For Gate-array design, channel widths are fixed. The goal is to finish routing of all the nets.

• For Standard-cell and Full-custom design, channels are expandable. The goal is to route all nets using the minimum channel width.

• We will consider the case when the channels are expandable.

404/19/23

Channel Ordering

AA

BB

The width of A is not known untilThe width of A is not known untilA is routed, we must route A first.A is routed, we must route A first.

AA

BB

CC DDWhat should be the routingWhat should be the routingorder for this example?order for this example?

504/19/23

Channel Ordering

CC DD

BB AA

No feasible No feasible channel order!channel order!

CC DD

BB AA

Need to use switchboxNeed to use switchbox

1. Fix the terminals between A & B1. Fix the terminals between A & B2. Route B, C, then D (channel)2. Route B, C, then D (channel)3. Route A (switchbox)3. Route A (switchbox)

604/19/23

Routing Grid Models

• Grid-based model is the most commonly used.• We will focus on it in this course.

Grid-based RoutingGrid-based Routing Gridless RoutingGridless Routing

704/19/23

Channel Routing Terminology

Upper boundaryUpper boundary

Lower boundaryLower boundary

TracksTracks

TerminalsTerminalsViaVia

TrunksTrunks BranchesBranches

DoglegDogleg

804/19/23

Routing Layer Models

HV modelHV modelVH modelVH model

HVH modelHVH modelVHV modelVHV model

Layer 1Layer 1Layer 2Layer 2Layer 3Layer 3ViaVia

1 layer1 layer

2 layers2 layers

3 layers3 layers

904/19/23

Channel Routing Problem

• Input: – Two vectors of the same length to represent the pins on two

sides of the channel.– Number of layers and layer model used.

• Output:– Connect pins of the same net together.– Minimize the channel width.– Minimize the number of vias.

1004/19/23

Channel Routing Problem

11 33 00 00 22 11 11 00

33 00 11 22 00 33 00 00

Example: (13002110) (30120300)where 0 = no terminal

1104/19/23

Constraint Graphs

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Vertical constraint graphVertical constraint graph

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22

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Horizontal constraint graphHorizontal constraint graph

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3344

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11 2233

55 4466

1204/19/23

Lower Bound on Channel Width00 11 66 11 22 33 55

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11 2233

55 4466

LocalLocaldensitydensity 11 33 44 44 44 44 22

Channel density =Channel density =Maximum local densityMaximum local density

Lower bound = 4Lower bound = 4

Lower bound on channel width = Channel density

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Left-edge Channel Routing Algorithm

“Wire Routing by Optimizing ChannelAssignment within Large Apertures”, A. Hashimoto and J. Stevens, DAC 1971, pages 155-169.

1404/19/23

Features of Left-edge Algorithm

• Assumptions:– One horizontal routing layer– No vertical constraint, e.g., VHV model

• Always gives a solution with channel width equal channel density, i.e., optimal solution.

00 00 00 22 00 11 00

00 11 00 00 00 22 00

2211

Vertical constraintVertical constraintmay occur here.may occur here.

1504/19/23

Left-edge Algorithm

1. Sort the horizontal segments of the nets in increasing order of their left end points.

2. Place them one by one greedily on the bottommost available track.

1604/19/23

Channel Density

Local density at column Cld(C) = # nets split by column C

Channel Densityd = max ld( C ) all C

Each net spans over an interval

Horizontal Constraint Graph(HCG)node : netedge: two intervals intersect

Size of max clique in HCG= channel density

A lower bound:# tracks channel density

a a

a

c

a

bd

ef

ld(x)

e

a

dc

b

f

1704/19/23

Thm: If the density of a set of intervals is d, then they can be packed into d tracks.Proof: I1=(a,b) I2=(c,d)Define: I1<I2 iff b<c or I1=I2

reflective: I1<I1

anti-symmetric: I1<I2, I2<I1 I1=I2

transitive: I1<I2, I2<I3 I1<I3

The interval set with binary relation < forms a partially ordered set (POSET)!!

Intervals in a track they form a chain Intervals intersecting a common column anti-chain

Dilworth’s theorem (1950): If the max anti-chain of a POSET is d, then the POSET can be partitioned into d chains

Interval PackingI6

ca b d

I4I3

I1I2

I5

I1

I5 I2

I3

I4

I6

1804/19/23

Left-Edge Algorithm for Interval PackingRepeat

create a new track t

Repeat

put leftmost feasible interval to t

until no move feasible interval

until no move interval

Interval are sorted according to their left endpoints

O(nlogn) time algorithm. Greedy algorithm works!

I6

I4I3

I1 I2

I5

I6

I4I3

I1

I2

I5

1904/19/23

Left-edge Algorithm: Example

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1. Sort by left end points.1. Sort by left end points.

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2. Place nets greedily.2. Place nets greedily.

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Vertical Constraint Consideration

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• The Left-edge algorithm ignores vertical constraints.• When there is only one vertical layer, the algorithm will

produce overlapping of vertical wire segments.

2104/19/23

Lower Bound on Channel Width

00 11 66 11 22 33 55

66 33 55 44 00 22 44

11

22

3344

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Lower bound = 3Lower bound = 3

Length of the longestpath in the vertical con-straint graph

2204/19/23

Lower Bound on Channel Width

density Channel

VCG in thepath longest theofLength max

widthchannelon boundLower

2304/19/23

Constrained Left-edge Algorithm

• Consider vertical constraints.• Similar to the Left-edge algorithm.• Modifications: Place a horizontal segment only if it

does not have any unplaced descendants in the vertical constraint graph Gv. Place it on the bottommost available track above all its descendents in Gv.

2404/19/23

Constrained Left-edge Algorithm

00 11 66 11 22 33 55

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Vertical constraint graphVertical constraint graph

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66 33 55 44 00 22 44

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22

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1. Sort the left end points.1. Sort the left end points.

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2. Place nets greedily.2. Place nets greedily.

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55

66

4455

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2504/19/23

Cycles in Vertical Constraint Graph

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22 00 11

Vertical Vertical

constraintconstraint

graphgraph

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22

• If there is cycle in the vertical constraint graph, the channel is not routable.

• Dogleg can solve the problem.11 00 22

22 00 11

2604/19/23

Reduce Channel Width by Dogleg

Even without cycle in the VCG, Dogleg is useful because it can reduce channel width.

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00 22 00 33 33

Without doglegWithout dogleg With doglegWith dogleg

???

2704/19/23

Deutch’s Dogleg Algorithm

• Split each multi-terminal net into several horizontal segments.

• Split only at columns that contain a pin of the net, i.e., using restricted dogleg only.

• Apply the Constrained Left-edge algorithm.

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Restricted doglegRestricted dogleg11 11 00 22 00

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Unrestricted doglegUnrestricted dogleg

2804/19/23

Deutch’s Dogleg Algorithm

VerticalVertical

constraintconstraint

graphgraph

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44

3322

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Routing without dogleg?Routing without dogleg?

Vertical constraint graph after splitting?Vertical constraint graph after splitting?Routing with dogleg?Routing with dogleg?

2904/19/23

Deutch’s Dogleg Algorithm

VerticalVerticalconstraintconstraint

graphgraphafterafter

splittingsplitting

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112a2a 2b2b

3a3a 3b3b44

Without doglegWithout dogleg By Deutch’s Dogleg alg.By Deutch’s Dogleg alg.

???

??????

3004/19/23

Drawbacks of the ConstrainedLeft-edge Algorithm

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11 33VerticalVertical

constraintconstraint

graphgraph

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By Constrained Left- By Constrained Left- edge algorithmedge algorithm

There is a betterThere is a bettersolution...solution...

3104/19/23

What’s wrong with the Constrained Left-edgeWhat’s wrong with the Constrained Left-edgealgorithm?algorithm?

The Constrained Left-edge algorithm does notThe Constrained Left-edge algorithm does nottake care of the vertical and horizontal constraintstake care of the vertical and horizontal constraintstogether optimally.together optimally.

Drawbacks of the ConstrainedLeft-edge Algorithm