ee4271 vlsi design
DESCRIPTION
EE4271 VLSI Design. Logic Synthesis. Logic Synthesis. Starts from RTL description in HDL or Boolean expressions Outputs a standard cell netlist. Boolean Function. f: B m Y n B = {‘0’, ‘1’}, Y = {‘0’, ‘1’, ‘-’} The function is incompletely specified, don’t care ‘-’ - PowerPoint PPT PresentationTRANSCRIPT
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EE4271 VLSI Design
Logic Synthesis
Logic SynthesisStarts from RTL description in HDL or Boolean expressionsOutputs a standard cell netlist
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Boolean Functionf: Bm Yn
B = {‘0’, ‘1’}, Y = {‘0’, ‘1’, ‘-’} The function is incompletely specified, don’t care ‘-’
For each output, space of Bm can be partitioned into
on-set: all inputs leading to output ‘1’ off-set: all inputs leading to output ‘0’ dc-set: all inputs leading to output ‘-’
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Points in Input SpaceAssigning ‘1’ or ‘0’ to each of the m Boolean variables x1, x2, …, xm specifies a point in input space Bm
Example (‘1’, ‘0’, ‘1’) in B3
x1=‘1’ Λ x2=‘0’ Λ x3=‘1’ x1•x2•x3
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TerminologyLiteral: a Boolean variable or its complementMinterm: a product of all input variables or their complements – a point in Bm
Cube: a product of input variables or their complements
The fewer number of variables, the bigger space covered
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Boolean Function Representation
A Boolean function can be specified by a sum of mintermsThe expression has a minterm for each point in the on-set
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Implicant and CoverAn implicant is a cube whose points are either in the on-set or the dc-setA prime implicant is an implicant that is not included in any other implicantA set of prime implicants that together cover all points in the on-set (and some or all points of the dc-set) is called a prime cover
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Irredundant CoverAn prime cover is irredundant when none of its prime implicants can be removed from the coverAn irredundant prime cover is minimal when the cover has the minimal number of prime implicants
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Goal of Logic SynthesisFind a minimum irredundant prime coverAn irredundant prime cover is not necessarily a minimum
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Quine-McCluskey Algorithm
Calculate all prime implicants of the union of the on-set and dc-set, omitting prime implicants that only cover points of dc-setFinds the minimum cost cover of all minterms in the on-set by the obtained prime implicants
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Set Covering ProblemGiven a universe U={e1, …, en}, a collection of subsets {S1, …, Sm} where each subset contains some elements in universecost wi is associated with each subset Si
find a subcollection C (cover) such that C covers the entire universe Famous NP-complete problemOn-set U, a prime implicant an S
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Example of Set Cover
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U={1,2,3,4}S1={1,2}, w=2
S2={1,3,4}, w=3
S3={3}, w=1
S4={2,4}, w=2
S5={2,3}, w=2
S6={1,2,4}, w=3
C={S3,S6} is the optimal solution
S3 is redundant in C={S1,S2,S3} , and C={S1,S2} is not optimal
Greedy Algorithm
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C = empty [the cover]E= empty [record elements which have been covered]While there is uncovered element
Find the subset Si which is most cost effective, that is, the Si with smallest w(Si)/|Si-E|. [weight of subset over the elements in the subset but not covered so far]For each e in Si-E, set price(e)=w(Si)/|Si-E|
Put all e in Si to E
Put Si in the current partial cover C
Running Example (Iter 1)
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U={1,2,3,4}S1={1,2}, w=10S2={1,3,4}, w=9S3={3}, w=7
S4={2,4}, w=4
S5={2,3}, w=2
Iteration 1, E = emptyw(S1)/|S1-E|=10/2=5
w(S2)/|S2-E|=9/3=3
w(S3)/|S3-E|=7/1=7
w(S4)/|S4-E|=4/2=2
w(S5)/|S5-E|=2/2=1
Pick S5
E = {2,3}C = {S5}
Running Example (Iter 2)
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U={1,2,3,4}S1={1,2}, w=10S2={1,3,4}, w=9S3={3}, w=7
S4={2,4}, w=4
S5={2,3}, w=2
Iteration 2, E={2,3}, C = {S5}
w(S1)/|S1-E|=10/1=10
w(S2)/|S2-E|=9/2=4.5
w(S3)/|S3-E|=7/0=+infty
w(S4)/|S4-E|=4/1=4
Pick S4
E = {2,3,4}C = {S5,S4}
Running Example (Iter 3)
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U={1,2,3,4}S1={1,2}, w=10S2={1,3,4}, w=9S3={3}, w=7
S4={2,4}, w=4
S5={2,3}, w=2
Iteration 3, E={2,3,4}, C={S5,S4}
w(S1)/|S1-E|=10/1=10
w(S2)/|S2-E|=9/1=9
w(S3)/|S3-E|=7/0=+infty
Pick S2
E = {2,3,4,1}C = {S5,S4,S2}
Approximation RatioDenote by OPT the weight of the optimal solutionDenote by ALG the weight of the solution of our algorithmOur algorithm is an r-approximation to the optimal solution if for all instances, ALG<= rOPTr=1 means that our algorithm computes the optimal solutionWe would like to minimize r
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Proof of Approximation - I
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Order all elements {e1, …, en} in which they are covered by the algorithm, e.g., E = {2,3,4,1}Denote by OPT the optimal solutionAt any iteration, the uncovered (remaining) elements in the universe (i.e., U-E in the beginning of the iteration) can be covered by a cover of weight at most OPT (since OPT covers all elements)There exists one subset Si with w(Si)/|Si-E|<= OPT/|U-E| (average weight per element). Otherwise, on average covering any remaining element needs weight>OPT/|U-E|, thus we need a cover with weight>OPT to cover all remaining elements
Proof of Approximation - II
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Our algorithm picks the Si with smallest w(Si)/|Si-E|, so its w(Si)/|Si-E|<= OPT/|U-E|.
When an element ek is covered, |U-E|<=n-(k-1)=n-k+1price(ek)<=OPT/(n-k+1)
price(ek) gives the weight of our solution
price(ek)=OPT(1/n+1/(n-1)+…+1)=lnn OPT
Tight Bound
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n elements and a collection of m=n+1 subsetsU={1,2,…n}S1={1}, w(S1)=1/n
S2={2}, w(S2)=1/(n-1)
S3={3}, w(S3)=1/(n-2)
…Sn={n}, w(Sn)=1
Sn+1={1,2,…,n}, w(Sn+1)=1.0001
Our solution {S1,S2,…,Sn}, weight = ln n
Optimal solution {Sn+1}, weight = 1.0001
Better solution?There is no (1-)lnn-approximation for Weighted Set Cover problem unless NPDTIME(nloglogn)
Uriel Feige, A threshold of ln n for approximating set cover, Journal of the ACM, Vol. 45, No. 4, pp. 634 - 652, 1998
There is no better approximation algorithm running in polynomial time for the general weighted set cover problem
It is possible to design better (constant) approximation for specific cases
Bounded size
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Unweighted & Size Bounded - I
When w(S)=|S| and # elements in any S <= a constant D, the lnn-approximation algorithm worksHowever, we can have better approximation ratioRecall that in each iteration, we pick the subset with minimum |Si|/|Si-E|.There exists one subset Si with |Si|/|Si-E|<= OPT/|U-E|. We pick the subset with |Si-E| (which are uncovered elements so far) at least |Si||U-E|/OPT >= |U-E|/OPTWe actually pick the subset which covers at least |U-E|/OPT uncovered elements, where |U-E| is # uncovered elements so far (in the beginning of the current iteration)
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Unweighted & Size Bounded - IIInitially we have n uncovered elementsAfter iteration 1, # uncovered elements <= n – n/OPTAfter iteration 2, # uncovered elements <= (n – n/OPT) – (n – n/OPT)/OPTAfter iteration k, # uncovered elements <= n(1-1/OPT)k
Solve n(1-1/OPT)k<=x to find the smallest k, i.e., # iterations such that # uncovered elements <= xSince (1-1/OPT)OPT • k/OPT < (1/e)k/OPT, we have n/(ek/OPT)<=x k>=OPT ln(n/x)Each iteration, one subset is picked and the last x elements need at most x subsets
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Unweighted & Size Bounded - III
Our solution has at most OPT ln(n/x) + x subsetsSince # elements for any subset is bounded by a constant D. OPT>=n/D, i.e., n/OPT<=D. Set x=n/D, so x<=OPT.OPT ln(n/x) + x<=OPT ln(D)+OPTOur algorithm is a ln(D)+1 approximation which is a constant approximation
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Summary
Primary implicantMinimum cost irredundant prime coverSet CoverGreedy AlgorithmApproximation Ratio
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