MultiLayer Channel Routing Complexity and Algorithms

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Multi-Layer Channel RoutingComplexity and
Algorithms

• Rajat K. Pal

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Chapter 4

• A General Framework for Track Assignment in
• Multi-Layer Channel Routing

Presented By- Sumaya Kazary Std. ID
0409052009
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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.

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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 .

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• 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

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The TAH Algorithm
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• The Channel contains total 7 nets.
• Density of this channel is dmax 4.

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VCG
HNCG(G)
Vmax 4
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• 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-
• dmaxgtvmax
• dmaxvmax

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Executing the Algorithm Steps S13,6,7

• Iteration-1 As, dmaxvmax4. So case II
• and S36 As S3vi htvivmax
• Step E for vertex 6 lt od6,1,sp6gt is
  lt1,1,4gt.
• So ,
  
• Step F limitvimax(dmax,vmax)-htvi1
  
  
  
• wtltf(t),ht,zd,spgt
• limit34-312gtt .So,f(1)3(1-11)(2-11)1/2.
• limit64-411t. So,f(1)61-111.
• limit74-114gtt. So,f(1)7(1-11)(4-11)1/4.
• Now, ht33,ht64 ,ht71 zd34,zd63,zd73 and
• sp38,sp64, sp72 So,
  3lt1/2,3,4,8gt
• 6lt1,4,3,4gt 7 lt1/4,1,3,2gt

C16.
So,C26.
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• Step G In step E, span usage (spvi) by a net
  was maximized, now it is checked with minimizing,
  so the weights are ltodvi, 1, -spvigt.In this
  case,6lt1, 1, -4gt .
• So, again lets 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.

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I3
I1
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I7
I5
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Density of this channel is dmax 4.
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VCG
HNCG(G)
Vmax 3
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Executing the Algorithm Steps S13,5,7

• Iteration-2 Now , dmaxgtvmax. So case I. and
  S21,2,3,4
• Step A Now, S1 S23
• for vertex 3 lt cn3,1,sp3gt is lt1,1,8gt.
• So ,
  
• Step B limitvimax(dmax,vmax)-htvi1
  wtltf(t),zd,ht,spgt
• limit34-312t .So,f(2)32-211. zd34,
  ht33, sp38
• limit54-312t. So,f(2)52-211. zd53,
  ht53, sp51
• limit74-114gtt. So,f(2)7(2-11)(4-11)1/2.
  zd72, ht71, sp72
• So, 3lt1,4,3,8gt 5lt1,3,3,1gt 7
  lt1/2,2,1,2gt

C13.
So,C23.
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• Step C In step A, span usage (spvi) by a net
  was maximized, now it is checked with minimizing,
  so the weights are ltcnvi, 1, -spvigt.In this
  case,3lt1, 1, -8gt .
• So, again lets 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.

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Density of this channel is dmax 3.
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VCG
HNCG(G)
Vmax 3
Now again dmaxvmax i.e.,case II
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• Now the final routing solution using TAH
  framework for VH routing is

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Time complexity

• For initial part ,time complexity--
• To compute the HNCG G is O(ne).
• 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(ne).
• Each iteration part takes O(ne).
• If total trackk, then total time complexity of
  the algorithm is O(k(ne)).

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• Thank You