Generalized Cofactor
Definition 1 Let f, g be completely
specified functions. The generalized
cofactor of f with respect to g is the
incompletely specified function:
co( f , g )  ( f  g , g , f  g )
Definition 2 Let  = (f, d, r) and g be
given. Then
co(, g )  ( f  g , d  g , r  g )
1
Relation with Shannon Cofactor
• Let g=xi . Shannon cofactor is
f xi  f ( x1 ,, ( xi  1), , xn )
• Generalized cofactor with respect to g=xi
is
co( f , x )  ( f  x , x , f  x )
i
i
i
i
• Note that f  xi  f xi  f  xi  xi  f  xi
In fact fxi is the unique cover of co(f, xi )
independent of the variable xi .
2
Shannon Cofactor
on
f  abc  ab c  a b c  a bc
off
Don’t care
a
co( f , a)  ( f  a, a , f  a)
f a  bc  b c
3
Shannon Cofactor (Cont.)
co( f , a)  ( f  a, a , f  a)
f


f a
fa
So
fa  a
f  a  fa  f  a
4
Properties of Generalized Cofactor
co(f,g)
Shannon Cofactor
x  fx  x  fx  f
 f x y  f xy
 f  g y  f y  g y
f    f 
x
x
Generalized Cofactor
f  g  co( f , g )  g  co( f , g )
co(co( f , g ), h)  co( f , gh)
co( f  g , h)  co( f , h)  co( g , h)
co( f , g )  co( f , g )
We will get back to use of generalized cofactor later
5
Quine-McCluskey Summary
Q-M:
1. Generate cover of all primes
G  p1  p2    p3n / n
2. Make G irredundant (in optimum way)
Note: Q-M is exact i.e. it gives an exact minimum
Heuristic Methods:
1.
2.
3.
Generate (somehow) a cover of  using some of
the primes G  pi  pi    pi
1
2
k
Make G irredundant (maybe not optimumly)
Keep best result - try again (i.e. go to 1)
6
ESPRESSO - Heuristic Two-Level
Minimization
ESPRESSO()
{
(F,D,R)  DECODE()
F  EXPAND(F,R)
F  IRREDUNDANT(F,D)
E  ESSENTIAL_PRIMES(F,D)
F  F-E; D  D  E
do{
do{
F  REDUCE(F,D)
F  EXPAND(F,R)
F  IRREDUNDANT(F,D) }
}while fewer terms in F
//LASTGASP
G  REDUCE_GASP(F,D)
G  EXPAND(G,R)
F  IRREDUNDANT(F  G,D)
//LASTGASP
}while G  
F  F  E; D  D-E
LOWER_OUTPUT(F,D)
RAISE_INPUTS(F,R)
error  (Fon  F) or (F  Fold  D)
return (F,error)
7
ESPRESSO Illustrated
REDUCE
Local minimum
EXPAND
IRREDUNDAN T
Local minimum
8
IRREDUNDANT
Problem:
Given a cover of cubes {ci}, find a minimum subset {cik}
that is also a cover, i.e.
f   cik  f  d
Idea 1:
We are going to create a function g(y) and a new set of
variables yi, one for each cube ci. A minterm in the yspace will indicate a subset of the cubes {ci}.
Example 1:
y=011010 = {c2,c3,c5}
9
IRREDUNDANT
Idea 2:
Create g(y) so that it is the function such that:
g(y) = 1 
c
yik 1
ik
is a cover
i.e. if g(yk) = 1 if and only if {ci | yki =1} is a cover.
Note: g(y) is positive unate (monotone increasing) in all
its variables.
10
Example
Example 2:
f  bc  a c  a b  b c
g ( y1 , y2 , y3 , y4 )  y1 y4 ( y2  y3 )
1
2
3
4
Note:
We are after a minimum subset of cubes. Thus we want:
the largest prime of g (least literals).
Consider g : it is monotone decreasing in y, e.g.
g ( y1 , y2 , y3 , y4 )  y1  y4  y2 y3
11
Example
Create a Boolean matrix for g:
f  bc  a c  a b  b c
g ( y1 , y2 , y3 , y4 )  y1  y4  y2 y3
• Recall a minimal column cover of B is a prime of
g  (g)
• We want a minimum cover of B
Example 3: {1,2,4}  y1 y2 y4 (cubes 1,2,4)
  bc,a c,b c

12
Deriving g(y)
Modify tautology algorithm:
F = cover of =(f,d,r)
D = cover of d
Pick a cube ci  F.
(Note: ci  F  Fci  1)
Do the following for each
cube ci  F :
 A  FCi 
B  D 
   Ci 
13
Deriving g(y)
1. All leaves must be tautologies
2. G’ means how can we make it
not the tautology. We must
exactly delete all rows of all
2’s that are not part of D
3. Each row came from some row
of A/B
4. Each row of A is associated with some cube of F.
5. Each cube of B associated with some cube of D ( we
don’t need to know which, and we can’t delete its rows).
6. Rows that must be deleted are written as a cube, e.g.
y1y2y7  delete rows 1,3,7 of F.
14
Deriving g(y)
Example 4: Suppose unate leaf is in
subspace x1x’2x3 :
Thus we write down: y10 y18 (actually, yi
must be one of y10 , y18). Thus, F is not a
cover if we leave out cubes c10 , c18.
Note: If row of all
2’s in don’t cares,
then there is no
way not to have
tautology at that
leaf.
Unate leaf
Row of all 2’s
in don’t cares
15
Deriving g(y)
g ( y)  gi ( y)  g j ( y)  
gi ( y)  y10 y18 
gi ( y)
ci
F 
D 
 
g j ( y)
cj
x1
x2
x3
16
Summary
Convert g(y) into a Boolean matrix B (note that g(y) is
unate). Then find a minimum cover of B. For example,
if y1y3y18 is a minimum cover, then the set of cubes:
{c1 , c3 , c18 }
is a minimum sub cover of { ci | i=1,…,k}. (Recall that a
minimal cover of B is a prime of g(y), and g(y) gives all
possible sub-covers of F).
Note: We are just doing tautology in constructing g(y) ,
so unate reduction works:
A B 
T C  ,


Other efficiencies ?
A 1
17
Back to Q-M
All primes
B=
Minterms
of f
1011010
  
  
  
  
  
B  g ( y)  y1 y3 y4 y6  
We want a maximum prime of g(y).
Note: A row of B says if we leave out primes
{p1 , p3 , p4 , p6 }
we cease to have a cover.
So basically, the only difference between Q-M and IRREDUNDANT
is that for the latter, we just constructed a g(y) where we did
not consider all primes, but only those in some cover:
F = {c1 , c3 ,…, ck }
18
EXPAND
F  EXPAND(F,R)
Problem: Take a cube c and make it prime by removing
literals.
a). Greedy way: (uses D and not R)
 remove literal l i from c (results in c  c )
 test if c  f  d (same as (f  d)c  1)
 repeat, removing valid literals in order found.
19
EXPAND
F  EXPAND(F,R)
b). Better way: (uses R and not D)
Want to see all possible ways to remove maximal
subset of literals.
Idea: Create a function g(y) such that g(y)=1 iff
literals:
{li | yi  0}
can be removed.
or {li | yi  1} is a subset of literals
such that if kept in c, will still make
c~  f  d , i.e. c~  r  
20
Main Idea of EXPAND
Outline:
1. Expand one cube, ci , at a time
2. Build “blocking” matrix B = B ci
3. See which other cubes cj can be feasibly
covered using B.
4. Choose expansion (literals to be removed) to
cover most other cj .
Note:
 g (y ) is monotone increasing
 B  g (y ) is easily built if we have R , a cover of r .
 We do not need all of R . (reduced offset)
21
Reduced offset
on
off
Don’t care
r  a b  ab  ac
Make r unate by adding (1,1,1) to offset. Then
Rblack  a  b  g ( y ). This is simpler and easier to
deal with. (See paper A. Malik et. al.)
22
Blocking Matrix B (for cube C)
Given R = {ri }, a cover of r. [  = (f,d,r) ]
l j  C and l j  ri
Bij  1  
l j  C and l j  ri
What does row i of B say?
It says that if literals
j
| Bij  1 are removed from c,
then c  ri  , i.e. Bij  1 is one reason why c is
orthogonal to offset cube ri .
Thus B  g ( y )  y1 y3 y10   gives all ways that
literals of c can be removed to get c~  f  d
c~  r   
23
EXPAND example
c  abd
ri  abde
y1 y2 y3  a, b, d
y1  1  keep a
Suppose g(y)=1,
If y1 = 1, we keep literal
y2  1  keep b
a in cube c.
y3  1  keep c
Bi   101  y1 y3    gi(y)
Bi means do not keep literals 1 and 3 of c. (implies that
~ is not an implicant). If literals 1, 3 are
subsequent c
~  b . But c~  r   :
removed we get c  c
i
b  a bde  a bde  
so obviously b is not an implicant.
24
EXPAND continued
Thus all minimal column covers (  g(y) ) of B are the
minimal subsets of literals of c that must be kept to
ensure that c  f  d ( i.e. c  ri   )
Thus each minimal column cover is a prime p that covers
c, i.e. p  c.
p1
p2
c
p3
pi  c
25
End of lecture 4
26
Expanding ci
F = { ci },  = (f,d,r) f  F  f+d
Q: Why do we want to expand ci ?
A: To cover some other cj ‘s.
ci
p
cj
Q: Can we cover cj ?
A: If and only if (SCC = “smallest cube containing” also called
“supercube” )
equivalent to:
SCC ci  c j   f  d
SCC ci  c j   ri  
equivalent to:
literals ”conflicting” between ci, cj can be removed and still have
an implicant
27
Expanding ci
Can check SCC(ci, cj) with blocking matrix:
ci = 12012
cj = 12120
implies that literals 3 and 4 must be removed for
to cover cj . Check if column 3, 4 of B can be
removed without causing a row of all 0’s.
ci
p
c~i
cj
28
Covering function
The objective of EXPAND is to expand ci to cover as
many cubes cj as possible. The blocking function
g(y)=1 whenever the subset of literals {li | yi  1}
~  f  d. (Note that c~ 
yields a cube c
 l j).
( y j 1)
We now build the covering function,h , such that:
h(y) = 1
~  c covers another cube c  F.
whenever the cube c
j
i
Note: h(y) is easy to build.
Thus a minterm m  g(y) h(y) is such that it gives
c~  f  d ( g(m) =1 ) and covers at least one cube
(h(m) =1). In fact every cube c~m  cl is covered. We
seek m which results in the most cubes covered. 29
Covering function
Thus define h(y) by a set of cubes where dk = kth cube is:
d k   if SCC[ci  ck ]  f  d else



j
dk  



yj
2  1

if ckj  cij i.e. 0  1
1  0

2 otherwise
(Thus each dk gives minimal expansion to cover ck )
Thus dk   if cube ck can be feasibly covered by expanding cube ci.
dk says which literals we have leave out to minimally cover ck .
Thus h(y) is defined by h(y) = d1 + d2 +… + d|F|-1 (one for each cube
of F, except ci. It is monotone decreasing).
30
Covering function
We want m  g(y)h(y) contained in a maximum number of
dk ‘s.
In the Espresso, we build a Boolean covering matrix C
(note that h(y) is unate negative) representing h(y) and
solve this problem with greedy heuristics.
Note:
  
010110
B  g ( y)


~
but C  h ( y)  h( y) C  101011
100101


  
  
110110


B  101010
101001


  
31
Covering function
  
010110


C  101011
100101


  
  
110110


B  101010
101001


  
Want set of columns such that if eliminated from B and C
results in no empty rows of B and a maximum of empty
rows in C.
Note:
~does not
a 1 in C can be interpreted as a reason why c
cover cj .
32
Endgame
What do we do if h(y)  0 ?
Some things to try:
• generate largest prime covering ci
• cover most care points of another cube ck
• coordinate two or more cube expansions, i.e. try to cover
another cube by a combination of several other cube expansions
So far we have not come up with a really satisfactory
endgame. This could be important in many hard
problems, since it is often the case that h(y)  0 .
33
REDUCE
Problem: Given a cover F and c  F, find the smallest
cube c  c such that
F\{ c } + { c }
is still a cover. c is called the maximally reduced cube
of c.
REDUCE is
order dependent
on
off
Don’t care
BAD
GOOD
34
REDUCE Example
Example 5:
F  ac  bc  b c  a c
Two orders:
BAD
GOOD
   abc  bc  abc  ac
 
2. REDUCE  F  bc, bc , ac, ac    abc  ac  abc  ac
1. REDUCE F  ac, bc, bc , ac
REDUCE is order dependent !
35
REDUCE Algorithm
REDUCE  F , D

{
F  ORDER( F)
for (1  j  F ) do {
c j  MAX_REDUCE( c, F, D)
F  (F  {c j }) \ {c j }
}
return F
}
36
REDUCE
Main Idea: Make a prime not a prime but still maintain
cover:
{c1 ,…, ci,…, ck}  {c1 ,…,ci,ci+1 ,…,ck }
i 1
k
But
f  c j  ci   c j  f  d
j 0
j  i 1
• Get out of a local minimum (prime and
irredundant is local minimum)
• Then have nonprimes, so can expand
again in different directions. (Since
EXPAND is “smart”, it may know best
direction)
37
REDUCE
F = {c1 ,c2, …,ck}
F (i) = (F + D) \ { ci }
= {c1 ,c2,…,ci-1,ci+1,…, ck}
• Reduced cube:
c i = smallest cube containing (ci F(i) )
• Note that ci F(i) is the set of points uniquely covered
by ci (and not by any other cj or D).
• Thus, ci is the smallest cube containing the minterms of
ci which are not in F(i).
38
REDUCE
C1
C1
C2
on
C2
off
Don’t care
 F (i ) 
 SCC  c F (i ) 
 c SCC  F (i ) 
c i  SCC ci
i
i
ci
ci
 ci SCCC  Fci (i ) 
• SCC == “supercube”
• SCCC = “smallest cube containing complement”
39
Efficient Algorithm for SCCC
Unate Recursive Paradigm:
• select most binate variable
• cofactor
• until unate leaf
x1
x1
x2
x3
x2
x3
unate
What is SCCC (unate cover) ?
• Note that for a cube c with at least 2 literals,
SCCC(c) is the universe:
12222
cube 
cube  01222
20222
So SCCC(cube) = 22222
40
SCCC
SCCC (U) = 
Claim: if unate cover has row of
– all 2’s except one 0, then complement is in x i , i.e. i = 1
– all 2’s except one 1, complement is in x i, i.e. i = 0
otherwise in both subspaces, i.e. i = 2
Finally
SCCC  c1  c2 
 ck   SCC  c1c2
 SCC  c1 
ck 
SCC  ck 
Implies that only need to look at 1-literal cubes.
41
SCCC Example
Example1:
f  a  bc  d  f  a (b  c )d  a d
Note: 0101 and 0001 are both inf . So SCCC could not have literal
b or b in it.
Example2:
2
0
U (unate )  2
2

2
2
1
1
2
2
1
2
2
2
2
1
0
2
2
0

Note that columns 1 and 5 are essential: they must be in every
minimal cover. So U = x1x5(...). Hence SCCC(U) = x1x5
42
SCCC Example
Proof. (sketch):
U  x1  x5  x2 ( x3  x4 )
U  x1 x5 ( x2  x3 x4 )
1 0 2 2 1
U (unate)  1 2 0 0 1  12221
  
The marked columns contain both 0’s and 1’s. But every prime ofU
contains literals x1 , x5
minterms of U 
1
1
0
0
1
0
1
1
1
1
1
1
0
0
1
0
0
0
1
1
1
0
0
0
1
1
1
0
0
1
43
SCCC
Thus
γ  SCCC(unate)   if row of all 2's
 x j if column j has a row singleton with a 0 in it

γ j   x j if column j has a row singleton with a 1 in it
 2
otherwise

Thus unate leaf is easy !
44
Merging
We need to produce
SCCC  f   SCC  xi c1  xi c2   

xi  
xi  
l j i  
 l1l 2
lk
 c2  
 c1  
 (l j  c1 )  (l j  c2 )
If c1  c2  , then points in both xi andxj, so
( SCC(xic1 +xic2) )i = 2
If ljc1 , or ljc2 , then both xj andxj points exit, hence i =2.
Also if ljc1 and ljc2 , then i =2.
45
ESPRESSO()
{
(F,D,R)  DECODE()
F  EXPAND(F,R)
F  IRREDUNDAN T(F, D)
E  ESSENTIAL_ PRIMES(F, D)
F  F - E; D  D  E
do{
do{
F  REDUCE(F,D)
F  EXPAND(F,R)
F  IRREDUNDAN T(F, D)
}while fewer terms in F
//LASTGASP
G  REDUCE_GASP(F,D)
G  EXPAND(G,R)
F  IRREDUNDANT(F  G,D)
//LASTGASP
}while G  
F  F  E; D  D-E
LOWER_OUTPUT(F,D)
RAISE_INPUTS(F,R)
error  (Fon  F) or (F  Fold  D)
46
return(F,error)
ESPRESSO - Heuristic
Two-Level Minimization
}
LASTGASP
Reduce is order dependent:
Expand can’t do anything with the
reduction produced by
REDUCE 2.
Maximal Reduce:

c i  SCCC ci
M

F ( i )  ci
REDUCE
2
REDUCE
1
GOOD
BAD
SCCC  F (i )ci 
i
i.e. we reduce all cubes as if each were the first one.
Note:
{c1M ,c2M ,...} is not a cover.
47
LASTGASP
Now expand use EXPAND, but try to cover only cjM‘s. (Note here
we call EXPAND(G,R), where G = {c1M ,c2M ,…, ckM} ) If a covering
is possible, take the resulting prime:
f  d  pi  c i  c j
and add to F:
~
F  F  { pi }
Since F is a cover, so is F . Now make F irredundant (using
M
M
IRREDUNDANT).
What about “supergasp” ?
Main Idea: Generally, think of ways
to throw in a few more primes
and then use IRREDUNDANT. If
generate all primes, then just
Quine-McCluskey
48
Goldberg Variation
Property of LASTGASP:
Let Fi = maximal reduction of Fi  F. LASTGASP searches for
maximally reduced cubes Fi and Fk such that
Fik  supercube( Fi, Fk)
is an implicant of . It then produces a new prime (which is not in
F) which covers supercube( Fi, Fk).
However, this does not guarantee that the size of F decreases by
at least one on the next IRREDUNDANT.
EXAMPLE:
100
F
F  110 F
1
122
221
2
D  010
001
 011
F12  212  prime
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Goldberg Theorem
Theorem 1:
Let Aik = (f Fi Fk ) - { F\ {Fi,Fk} }
= ( Fi Fk ) - { D+F\ {Fi,Fk} }
(i.e. Aik is the set of ON-set minterms covered by Fi Fk and not
intersected with any other cube from F.)
Then both Fi and Fk can be replaced by a prime
Pik  supercube( Fi, Fk) if and only if:
Pik  Aik
Thus for a reduction by at least one in the size of the cover, we
need a prime with the property
Pik  supercube( Fi, Fk)  Aik
An algorithm based on this was proposed by E. Goldberg of the Byelorussion
Academy of Sciences, Institue of Engineering Cybernetics in Minsk, FSU
(now at Cadence Berkeley Labs). He reported it was less expensive than
SUPER_GASP and more effective than LASTGASP.
50