EE4271 VLSI Design
Logic Synthesis
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Logic Synthesis
Starts from RTL description in HDL or Boolean
expressions
Outputs a standard cell netlist
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Boolean Function
f: 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 Space
Assigning ‘1’ or ‘0’ to each of the m Boolean variables
x1, x2, …, xm specifies a point in input space Bm
Example
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(‘1’, ‘0’, ‘1’) in B3
x1=‘1’ Λ x2=‘0’ Λ x3=‘1’
x1•x2•x3
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Terminology
Literal: a Boolean variable or its complement
Minterm: 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 minterms
The expression has a minterm for each point in the on-set
f  x1  x 2  x 3  x1  x 2  x 3  x1  x 2  x 3  x1  x 2  x 3  x1  x 2  x 3  x1  x 2  x 3
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Implicant and Cover
An implicant is a cube whose points are either in the on-set or
the dc-set
A prime implicant is an implicant that is not included in any other
implicant
A set of prime implicants that together cover all points in the onset (and some or all points of the dc-set) is called a prime cover
f  x1  x 2  x1  x 3  x 2  x 3  x1  x 2  x 2  x 3  x1  x 3
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Irredundant Cover
An prime cover is irredundant when none of its prime implicants
can be removed from the cover
An irredundant prime cover is minimal when the cover has the
minimal number of prime implicants
f  x1  x 3  x 2  x 3  x1  x 2
f  x1  x 2  x 2  x 3  x1  x 3
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Goal of Logic Synthesis
Find a minimum irredundant prime cover
An irredundant prime cover is not necessarily a
minimum
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Quine-McCluskey Algorithm
Calculate all prime implicants of the union of the onset and dc-set, omitting prime implicants that only
cover points of dc-set
Finds the minimum cost cover of all minterms in the
on-set by the obtained prime implicants
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Set Covering Problem
Given a universe U={e1, …, en}, a collection of subsets
{S1, …, Sm} where each subset contains some
elements in universe
cost wi is associated with each subset Si
find a subcollection C (cover) such that C covers the
entire universe
Famous NP-complete problem
On-set U, a prime implicant an S
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Example of Set Cover
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
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Greedy Algorithm
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
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Running Example (Iter 1)
U={1,2,3,4}
S1={1,2}, w=10
S2={1,3,4}, w=9
S3={3}, w=7
S4={2,4}, w=4
S5={2,3}, w=2
Iteration 1, E = empty
w(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}
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Running Example (Iter 2)
U={1,2,3,4}
S1={1,2}, w=10
S2={1,3,4}, w=9
S3={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}
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Running Example (Iter 3)
U={1,2,3,4}
S1={1,2}, w=10
S2={1,3,4}, w=9
S3={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}
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Approximation Ratio
Denote by OPT the weight of the optimal solution
Denote by ALG the weight of the solution of our
algorithm
Our algorithm is an r-approximation to the optimal
solution if for all instances, ALG<= rOPT
r=1 means that our algorithm computes the optimal
solution
We would like to minimize r
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Proof of Approximation - I
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 solution
At 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/|UE| (average weight per element). Otherwise, on average
covering any remaining element needs weight>OPT/|UE|, thus we need a cover with weight>OPT to cover all
remaining elements
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Proof of Approximation - II
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+1
price(ek)<=OPT/(n-k+1)
price(ek) gives the weight of our solution
price(ek)=OPT(1/n+1/(n-1)+…+1)=lnn OPT
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Tight Bound
n elements and a collection of m=n+1 subsets
U={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
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Better solution?
There is no (1-)lnn-approximation for Weighted Set
Cover problem unless NPDTIME(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 - 1
When w(S)=|S| and # elements in any S <= a
constant D, the lnn-approximation algorithm works
However, we can have better approximation ratio
Recall 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 |SiE| (which are uncovered elements so far) at least
|Si||U-E|/OPT >= |U-E|/OPT
We actually pick the subset which covers at least |UE|/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 - 2
Initially we have n uncovered elements
After iteration 1, # uncovered elements <= n – n/OPT
After iteration 2, # uncovered elements <= (n –
n/OPT) – (n – n/OPT)/OPT
After iteration k, # uncovered elements <= n(11/OPT)k
Solve n(1-1/OPT)k<=x to find the smallest k, i.e., #
iterations such that # uncovered elements <= x
Since (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 - 3
Our solution has at most OPT ln(n/x) + x subsets
Since # 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)+OPT
Our algorithm is a ln(D)+1 approximation which is a
constant approximation
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Summary
Primary implicant
Minimum cost irredundant prime cover
Set Cover
Greedy Algorithm
Approximation Ratio
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