Multi layer channel routing complexity and algorithms
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Multi-Layer Channel Routing Complexity and Algorithms. Rajat K. Pal. Chapter 4. A General Framework for Track Assignment in Multi-Layer Channel Routing. Presented By- Sumaya Kazary Std. ID: 0409052009. TAH Basics.

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Multi layer channel routing complexity and algorithms

Multi-Layer Channel RoutingComplexity and Algorithms

Rajat K. Pal


Chapter 4
Chapter 4

A General Framework for Track Assignment in

Multi-Layer Channel Routing

  • Presented By-

    • Sumaya Kazary

    • Std. ID: 0409052009


Tah basics
TAH Basics

  • The Algorithm TAH assigns intervals to tracks from top to bottom in the presence of Vertical constraints.

  • In the first iteration, the algorithm assigns a set of non-overlapping intervals to the top most track. Then it delete the nets corresponding to these intervals from the channel.

  • In the second iteration, it assigns a set of non-overlapping intervals to the second track.

  • The iterative process continues till all the nets are assigned to tracks in the channel.


Some notations a few definitions

6

3

5

4

1

7

2

I3

I1

I2

I7

I5

I4

I6

Some Notations & a few definitions

  • dmax→channel density.

  • vmax→the length of the longest path in the VCG.

  • idvi→ indegree of vertex vi in the VCG.

  • dpvi→length of the longest path from a source to vi.

  • htvi→ length of the longest path from a sink vertex to vi.

  • spvi→span of a net ni.

  • zdvi→ zonal density of a vertex v.


For the following channel

TOP: 3 1 3 0 0 5 6 0 3 0 0

BOTTOM: 1 2 4 2 4 1 5 7 0 7 6

The routing solution using TAH framework for VH routing is

3

1

3

0

0

5

6

0

0

0

3

1

2

4

2

4

1

5

7

7

6

0



I3

I1

I2

I7

I5

I4

I6

0

3

1

3

0

3

0

0

5

6

0

1

2

4

1

0

6

4

5

7

7

2

  • The Channel contains total 7 nets.

  • Density of this channel is dmax = 4.


6

1

2

3

5

3

7

4

1

7

4

6

2

5

VCG

HNCG(G*)

Vmax =4


  • In the TAH algorithm ,for each iteration, certain no. of intervals are assigned in a track from top to bottom.

  • In each iteration ,it computes a clique such that there is no overlapped interval is present & assign it to a track.

  • At the beginning, it compares the channel density (dmax)and the longest path in the VCG(vmax).

  • Two cases may arise:-

    • dmax>vmax

    • dmax≤vmax


Executing the algorithm steps s1 3 6 7

6 intervals are assigned in a track from top to bottom.

3

5

4

1

7

2

Executing the Algorithm Steps:S1={3,6,7}

  • Iteration-1As, dmax=vmax=4. So case II

    and S3={6} [As S3={vi| htvi=vmax}]

    Step Efor vertex 6 < od6,1,sp6> is <1,1,4>.

    So ,

    Step Flimitvi=max(dmax,vmax)-htvi+1

    wt:<f(t),ht,zd,sp>

    limit3=4-3+1=2>t .So,f(1)3=(1-1+1)÷(2-1+1)=1/2.

    limit6=4-4+1=1=t. So,f(1)6=1-1+1=1.

    limit7=4-1+1=4>t. So,f(1)7=(1-1+1)÷(4-1+1)=1/4.

    Now, ht3=3,ht6=4 ,ht7=1; zd3=4,zd6=3,zd7=3 and

    sp3=8,sp6=4, sp7=2 So, 3:<1/2,3,4,8>

    6:<1,4,3,4> 7: <1/4,1,3,2>

C1={6}.

So,C2={6}.


  • Step G intervals are assigned in a track from top to bottom.In step E, span usage (spvi) by a net was maximized, now it is checked with minimizing, so the weights are <odvi, 1, -spvi>.In this case,6:<1, 1, -4> .

    So, again let’s pick C3 = {6}.

    Step H similar to step F, so, C4 = {6}

    So net 6 is assigned to track 1.

    Now, HNCG & VCG are rearranged excluding node 6 and its adjacent edges.


0 intervals are assigned in a track from top to bottom.

3

1

3

0

3

0

0

5

6

0

I3

I1

I2

I7

I5

I4

1

2

4

1

0

6

4

5

7

7

2

Density of this channel is dmax = 4.


1 intervals are assigned in a track from top to bottom.

2

3

5

3

7

4

1

7

4

5

2

VCG

HNCG(G*)

Vmax =3


Executing the algorithm steps s1 3 5 7

3 intervals are assigned in a track from top to bottom.

5

4

1

7

2

Executing the Algorithm Steps:S1={3,5,7}

  • Iteration-2Now , dmax>vmax. So case I. and S2={1,2,3,4}

    Step ANow, S1 S2={3}

    for vertex 3 < cn3,1,sp3> is <1,1,8>.

    So ,

    Step Blimitvi=max(dmax,vmax)-htvi+1 wt:<f(t),zd,ht,sp>

    limit3=4-3+1=2=t .So,f(2)3=2-2+1=1. zd3=4, ht3=3, sp3=8

    limit5=4-3+1=2=t. So,f(2)5=2-2+1=1. zd5=3, ht5=3, sp5=1

    limit7=4-1+1=4>t. So,f(2)7=(2-1+1)÷(4-1+1)=1/2. zd7=2, ht7=1, sp7=2

    So, 3:<1,4,3,8>; 5:<1,3,3,1>7: <1/2,2,1,2>

C1={3}.

So,C2={3}.


  • Step C intervals are assigned in a track from top to bottom.In step A, span usage (spvi) by a net was maximized, now it is checked with minimizing, so the weights are <cnvi, 1, -spvi>.In this case,3:<1, 1, -8> .

    So, again let’s pick C3 = {3}.

    Step D Similar to step F, so, C4 = {3}

    So net 3 is assigned to track 2.

    Now, again HNCG & VCG are re-arranged excluding node 3 and its adjacent edges. And iteration 3 is proceed.


0 intervals are assigned in a track from top to bottom.

3

1

3

0

3

0

0

5

6

0

I1

I2

I7

I5

I4

1

2

4

1

0

6

4

5

7

7

2

Density of this channel is dmax = 3.


1 intervals are assigned in a track from top to bottom.

2

5

7

4

1

7

4

2

5

VCG

HNCG(G*)

Vmax =3

Now again dmax=vmax i.e.,case II


Now the final routing solution using TAH framework for VH routing is

3

1

3

0

0

5

6

0

0

0

3

1

2

4

2

4

1

5

7

7

6

0


Time complexity
Time complexity routing is

  • For initial part ,time complexity--

    • To compute the HNCG & G* is O(n+e).

    • To compute the VCG is O(n).

    • To compute the span of every interval is O(n).

    • htvi and dpvi is computed by 1 scan of VCG needs O(n).

      So,for initial part, it takes time O(n+e).

  • Each iteration part takes O(n+e).

  • If total track=k, then total time complexity of the algorithm is O(k(n+e)).


Thank You routing is


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