1. Multi-Layer Channel Routing Complexity and Algorithms
Rajat K. Pal

2. Multi-Layer Channel Routing
Chapter 4
A General Framework for Track Assignment in
Multi-Layer Channel Routing
Presented By-
Sumaya Kazary
Std. ID:

3. 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.

4. 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 . 7 6 5 2 4 1 3 I2 I7 I5 I1 I3 I4 I6

5. For the following channel TOP: 3 1 3 0 0 5 6 0 3 0 0
BOTTOM:
The routing solution using TAH framework for VH routing is 3 1 3 5 6 3 1 2 4 2 4 1 5 7 7 6

6. The TAH Algorithm

8. Density of this channel is dmax = 4.3 1 3 5 6 3 I2 I7 I5 I1 I3 I4 I6 1 2 4 2 4 1 5 7 7 6
The Channel contains total 7 nets.
Density of this channel is dmax = 4.

9. 7 6 5 2 4 1 3 1 2 3 4 5 7 6
VCG
HNCG(G*)
Vmax =4

10. In the TAH algorithm ,for each iteration, certain no
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

11. Executing the Algorithm Steps: S1={3,6,7}
Iteration-1 As, dmax=vmax=4. So case II
and S3={6} [As S3={vi| htvi=vmax}]
Step E for vertex 6 < od6,1,sp6> is <1,1,4>.
So ,
Step F limitvi=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, :<1/2,3,4,8>
6:<1,4,3,4> 7: <1/4,1,3,2> 7 6 5 2 4 1 3
C1={6}.
So,C2={6}.

12. Step G 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.

13. Density of this channel is dmax = 4.3 1 3 5 6 3 I3 I1 I2 I7 I5 I4 1 2 4 2 4 1 5 7 7 6
Density of this channel is dmax = 4.

14. 1 2 3 4 5 7 7 5 2 4 1 3
VCG
HNCG(G*)
Vmax =3

15. Executing the Algorithm Steps: S1={3,5,7}
Iteration-2 Now , dmax>vmax. So case I. and S2={1,2,3,4}
Step A Now, S1 S2={3}
for vertex 3 < cn3,1,sp3> is <1,1,8>.
So ,
Step B limitvi=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> 7 5 2 4 1 3
C1={3}.
So,C2={3}.

16. Step C 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.

17. Density of this channel is dmax = 3.1 3 5 6 3 I1 I2 I7 I5 I4 1 2 4 2 4 1 5 7 7 6
Density of this channel is dmax = 3.

18. Now again dmax=vmax i.e.,case II1 2 4 5 7 7 5 2 4 1
VCG
HNCG(G*)
Vmax =3
Now again dmax=vmax i.e.,case II

19. Now the final routing solution using TAH framework for VH routing is3 1 3 5 6 3 1 2 4 2 4 1 5 7 7 6

20. Time complexity 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)).

21. Thank You