VLSI-- S A AHSAN RAJON.ppt

8/7/2019 VLSI-- S A AHSAN RAJON.ppt

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Overview

Area minimization is the key objective of channel routing. Since the problem of minimizing area in two layer and Threelayer channel routing are known to be NP-Hard, severalheuristics for area minimization have been proposed whichgenerate routing solutions for several standard benchmarkchannels within a small number of tracks more than the optimal

number required to route the channels. TAH is an algorithm that computes optimal or nearly optimal

results for several well known benchmark channels. Even though most of these heuristics run in polynomial time, it is

not known whether there exists a polynomial time algorithm that

computes a routing solution for the given instance of channelrouting problem within a constant number of tracks more thanthe optimal number required for that instance.

Such an algorithm is called Absolute Approximation Algorithm.


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Overview [contd.]

There are very few NP-Hard optimization problems that whoseabsolute approximation can be calculated in polynomial time. One problem is that of determining the minimum number of

colors needed to color a planer graph.o Determining if a planer graph is three colorable is NP-Hard.o However all planer graphs are four colorable.

Another problem is the maximum programs stored problem.o Assume that we have n programs and two storage devices, say disks.o Let li be the amount of storage needed to store the i-th program.o Let L be the storage capacity of each disk.o Determining the maximum number of there n programs that can be

stored on the two disks without splitting s program over the disk is

NP-Hard.o However by considering the programs in order if non-decreasing

storage requirement li, we can obtain a polynomial time absolutearea approximation algorithm.


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Overview [contd.]

The NP-Hardness of computing absolute areaapproximation for a problem depicts the degree tohardness of the problem.

The result of NP-Hardness of absolute areaapproximation problem derived in this chapter depict

the channel routing is indeed a very hardcomputational problem.

Computational problems become easier to solve forsimpler inputs and consequently proving intractability

results become harder.


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Overview [contd.]

A natural question is whether the absolute area approximationproblem is NP-Hard for channels with bounded degree nets. A net with a bounded number of terminals is called a bounded

degree net. N{-Hardness of the AAAP for channels with nets having a

maximum of five terminals per net have been proved.

It has been also proved that computing an absolute areaapproximation solution is NP-Hard in the two dogleg routingmodel for channels with two terminal nets under a certainrestriction.

The restriction imposed is that, nets must be assigned to tracks

in pre-specified groups. The motivation for such restriction of groupings of nets in their

assignment to tracks seems from practical design issues in VLSI,where it is preferable to group certain nets into the same trackprovided their spans do not overlap.


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Overview [contd.]

All the problems considered above for the two layerVH routing model remain NP hard even in the threelayer HVH routing model.


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Absolute Approximation for Two LayerNo-Dogleg Routing is NP-hard It is to prove that Absolute Approximation for Two Layer No-Dogleg VH Routing is NP-hard.

Reduction from the well known problem TNVH of computing theminimum number of tracks for a given instance of two terminalnets of the Channel Routing Problem.

Consider the following approximation problem MNVHAA1:o Given a channel specification of multi terminal nets compute a two-

layer no-dogleg routing solution whose number of tracks is at mostone more than the minimum number of tracks required for that given instance.

o In other words, we want to compute a 1-absolute approximatesolution for two layer no-dogleg routing.

o We show that, the problem MNVHAA1 is as hard as the problemTNVH by a polynomial transformation from TNVH ro MNVHAA1.


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Theorem 7.1: MNVHAA1 is NP Hard

Proof:o TO show that MNVHAA1 is NP Hard, we consider the following

reduction from TNVH to MNVHAA1.o We construct an instance I/ of MNVHAA1 from any instance I of TNVH

using a polynomial time transformation.o Let t be the minimum number of tracks required for a given

instance I.o We constyruct the instance I/ in such a manner that the minimum

number of tracks required to route I/ is 2t+1.o Since 2t+1 tracks are sufficient are sufficient for routing I/,the

absolute area approximation question for I/ can be stated ascomputing a routing solution for I/ within 2t+1 tracks.

o We show that, I has a t track solution if and inly if I/ has 2t+2 trackssolution, thereby showing that, MNVHAA1 is as hard as TNVH.


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Theorem 7.1: MNVHAA1 is NP Hard [contd.]

Let the number of nets in I be n and the length of thechannel in I be m. We construct I/ by duplicating the channel

specification of I into two groups A and B, each groupcontaining n nets.

Group A consists of one copy of I having n twoterminals nets in the first m columns of I/ .

The construction of I/ is such that, in any routingsolution for I/ , any net of group A is assigned to a

track above the track to which any net of group B isassigned.

In order to achieve this separation, we add oneadditional net s called the separator net as follows.


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For each net ai of A and bi of B 1 i n, we couldhave introduced vertical constraints (ai, s) and (s, bi)in order to achieve the separation.

However it is sufficient to introduce such verticlconstraints for nets ai of A whose corresponding nets

in I are lsink vertices of A and for nets bi of B whosecorresponding nets in I are source vertices of B.

Let g(h) be the number of such sink(source) nets in I. Therefore, introducing only g+h new columns in the I/

can we can realize the required seperation. So, I/ has 2t+1 nets and 2m+g+h columns. This completes the construction of I/ .


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Now show that, the instance I has a two layer no-dogleg solutionof t tracks if and only if the instance I/ has a two-layer no-doglegrouting solution of 2t+2 tracks.

Suppose there is a t-track two layer no-dogleg routing solutionfor the channel specification of n-two-terminal nets in I.

We show that there is a two layer no dogleg routing solution S for

I/ using 2t+2 tracks. Since I can be routed within t tracks, the nets of group A in I/ can

be assigned within the topmost t tracks. From the construction of I/ we know that, the separator net s

must be assigned below the nets of group A.

So, s can be assigned to the (t+1)th track from the top. The nets of group B in I/ must be must be assigned below s,within the next t tracks.

This gives a 2t+2 track routing solution for I/, where thebottommost track in S is an empty stack.


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Suppose there is a 2t+2 track routing solution S for I/ . We show that, there is a t-track two-layer no dogleg routing solution

for I/ . Note that, according to the construction of I/ , we have the followings-

o Vertices corresponding to the sink nets in group A are immediate ancestorsof the vertex corresponding to the separator net s.

o Similarly the Vertices corresponding to the sourcenets in group B are

immediate descendants of the vertex corresponding to the separator net s So, from the routing solution S for I/ , no net of group B can be

assigned to a track above the track to which the net of group A isassigned.

Moreover in S, all the nets of group A are seperated by the separatornet s from all the nets of group B.

Therefore either the nets of group A or the nets of group B use only ttracks.

Hence we can compute a two layer no dogleg t-track routing solutionfor the channel specification of the n two-terminal nets ion I.


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So far we have proved that, the problem MNVHAA1 is NP-Hard. In similar manner we can show that, the problem MNVHAAK is

also NP Hard. We pose the problem as follows,.

o Given a channel specification of multi terminal nets compute a two-layer no-dogleg routing solution whose number of tracks is at most k

for any fixed K>1 more than the minimum number of tracks requiredfor that given instance.

o In other words, we want to compute a k-absolute approximatesolution for two layer no-dogleg routing.

o We show that, the problem MNVHAA1 is as hard as the problemTNVH by a polynomial transformation from TNVH ro MNVHAAK.


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Theorem 7.2: MNVHAAk is NP Hard

Proof:o TO show that MNVHAAK is NP hard, we use a polynomial time

transformation similar to that of used in the proof of T7.1.o We construct an instance I/ of MNVHAAK for any instance I of

the TNVH by making a k+1 copies of the channel specificationof I and using k additional separator nets.

o As in the proof of T7.1 we use the separator nets to ensurethat, all nets of one one copy are separated from all othernets of another copy in any routing solution of I/ .

o This can be realized by introducing vertical constraintsbetween the sink nets of the i-th copy and the i-th separator

net, and between the i-th separator net and the source netsof the (i+1)-th copy. 1ik of I/ .


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Theorem 7.2: MNVHAAk is NP Hard [contd.]

So, for any routing solution of I/ a net of the i-th copy must beassigned to a track above the track to which a net of the (i+1)-thcopy is assigned.

Assuming that the instance I has g sink nets and h source nets,we therefore require a total of (k+1)m + k(g+h) columnsand(k+1)n+k nets in I/ .

This completes the construction of I/ . As in the proof in T7.1, we can show that, the instance I has a

two layer no-dogleg solution of t tracks if and only if the instanceI/ has a routing solution of (k+1)t-2k tracks.


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THANK YOU

VLSI-- S A AHSAN RAJON.ppt

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