A Framework for Track Assignment Presented by: Kaiser Newaj Asif Multi-Layer Routing (Extensions of the TAH Framework)

For channel TOP: BOTTOM: The routing solution using TAH framework for VH routing is

Multi-Layer Routing For multi-layer V i H i, 2 ≤ i < d max and V i H i+1, 2 ≤ i < d max - 1 routing minimum number of tracks t is calculated at first, t = ceil(d max /i) for V i H i t = ceil(d max /(i + 1)) for V i H i+1

For channel TOP: BOTTOM: HCG HNCG d max = 4 For 4 layer routing i = 2, t = ceil(4/2) = 2

For channel TOP: BOTTOM: VCG Stage 1 If VCG is acyclic then nets in t(i-1) tracks of first (i-1) layers are assigned using MCC1 or MCC2

For channel TOP: BOTTOM: VCG Stage 1 Let us make the VCG cyclic Then a cycle is detected and a vertex with maximum adjacency number is deleted, here 3 (or 1), and that vertex is added to a set D. This process continues until the graph becomes acyclic.

For channel TOP: BOTTOM: VC’ Stage 1 D = { } the nets in set D are assigned to t(i-1) tracks with higher priority according to TAH framework, while the rest nets are assigned normal weights.

When D is exhausted the rest of the VC’ are assigned according to MCC1 or MCC2. Net 3 is assigned alone to a track by TAH. Then the sorted intervals and transitively oriented graph G* are as follows: Stage G*

Nets outside D are assigned to extreme layers in this stage. And if D is not empty that is dealt in next stage. In our example i = 2, so, all (i – 1) layers are assigned nets. In the extreme layers the rest of VC’, VC” are assigned according to TAH Stage VC” HNCG

step 1: S1 = { v i | idv i = 0 } = {2, 4, 5, 7} d max = 2 {2,4} > v max (=0) step 2: S2 = {2,4}, each belongs to a maximum independent set step 3: compute clique in S1∩S2 = {2,4} Stage VC” HNCG

step 3: compute clique using weight. cn 2 = 1, cn 4 = 1, sp 2 = 2, sp 4 = 2. Let’s pick net 2. C1 = {2}. step 4: extend C1 to C2 such that C2 is a subset of S1. To do that is computed for vertices in S1. Stage VC” HNCG

step 4: limit vi = max(d max, v max ) – ht vi + 1 limit 4 = limit 5 = limit 7 = 2 – = 3 f(t) vi = (t – dp vi + 1)/(limit vi – dp i + 1) For 1 st track, t = 1 f(1) 4 = f(1) 5 = f(1) 7 = (1 – 0 + 1)/(2 – 0 + 1) = 2/3 Stage VC” HNCG

step 4: zd 4 = 2, zd 5 = 3, zd 7 = 3. ht 4 = ht 4 = ht 4 = 0 sp 4 = 2, sp 5 = 1, sp 7 = 2 Stage VC” HNCG 7: 5: 4: C2 = {2, 7, 5}

step 5: in step 3 span usage (sp vi ) by a net was maximized, now it is checked with minimizing, so the weights are 2: 4:. So, again let’s pick C3 = {2}. step 6: similar to step 4, so, C4 = {2, 7, 5} Stage VC” HNCG

step 7: then the larger clique of C2 and C4 is assigned to track 1. Again the loop begins for track-2, and only remaining vertex 4 is assigned to it. Stage VC” HNCG

step 7: If after stage 1 and stage 2, there exist unassigned tracks then they are assigned iteratively track by track. Tracks in D (the deleted vertex set from cycles) are assigned in the flanked layers as in stage 1 using TAH Framework. Then the remaining nets are of VC”, and assigned to the flanked layers using MCC1 or MCC2, and to extreme layers using TAH Framework Stage 3

Finally,