ENGG3190 Logic Synthesis Computational Boolean Algebra Winter 2014 S. Areibi School of Engineering University of Guelph

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Outline

• Computational Boolean Algebra “Basics” • Shannon Cofactors and Properties • The Boolean Difference Operator • Quantification Operators • Recursive Tautology • URP (Unate Recursive Paradigm)

Computational Boolean Algebra

  

Computational Boolean Algebra are algorithmic Circuits.

and computational strategies for manipulating Boolean Functions and It is a set of

  

Data structures , Techniques and theorems , Algorithms , It will help us to simplify and optimize circuits, repair circuits, prove that they perform certain tasks.

y + z’ + yz’

Co-factors of F()

xz’

Definition

The cofactor C xj of a cube C • • • with respect to a literal x j C if x C\{x  j j and x j } if x if x j j is do not appear in C appears positively in C, i.e. x j  C appears negatively in C, i.e. x j  C

Example

If C = x 1 x 4 ‘ x 6 C x2 = C (x 2 does not appear in C ) C x1 = x 4 C x4 =  ‘x 6 (x (x 1 4 appears positively in C) appears negatively in C) 8

F xi = Positive Co-factor F xi’ = Negative Co-factor

Not a proof but a case study Let xi = 1 F(x1, x2, …, xi=1, ….) = 1.F(xi=1) + 0.F(xi=0) = F(xi=1) Which is the same!!

The same can be obtained by letting xi=0

Shannon (Boole) Cofactors

The cofactor f xi f xi (x 1 , x 2 , …, x n ) = f (x 1 , …, x i-1 , 1, x i+1 ,…, x n ) (x 1 f a , x of f by a literal a=x 2 , …, x n ) = f (x 1 i , …, x or a=x i-1 , 0, x i is i+1 ,…, x n ) The computation of the cofactor is a fundamental operation in Boolean reasoning!!!!

Example:

f = abc + abc

F no longer depends on a

f a = bc c a b 11 c a b

Shannon Cofactor

f



abc



a b c



a b c



a b c

on off Don’t care

a f a



bc



b c

12

Shannon Expansion Definitions:

f : B

n 

B

f



x f i x i

Theorem:

F is a cover of f. Then F  x i   is also a cover of f.

We say that f (F) is expanded about x i . x i is called the splitting variable .

x i

x i 13

Example

F F



aF

a



aF

a



)



bc c b Cube

bc

a got split into two cubes c a b 14

Generalized Cofactor

• The generalized cofactor f C of f by a cube C is f with the fixed values indicated by the literals of C, e.g. if C=x i x j , then x i =1, and x j =0.

• if C= x 1 x 4 x 6 f C is just the function f restricted to the subspace where x 1 =x 6 =1 and x 4 =0.

• As a function,  f C does not depend on x 1 ,x 4 or x 6 anymore (However, we still consider f C as a function of all be independent of x

1 ,x 4 and x 6

).

n

variables, it just happens to 15

Assume we have a mux with two Inputs (a,b).

Let

variable x

control the selection Line.

If x = 0, we send a If x = 1, we send b F = a.x’ + b.x

Negative Co-factor x’ . F(x=0) + x . F(x=1) This is Shannon Expansion!! Positive Co-factor

F(x=1) = y . F(x=1,y=1) + y’ . F(x=1,y=0) F(x=0) = y . F(x=0,y=1) + y’ . F(x=0,y=0) x.y.F(x=1,y=1) + x.y’.F(x=1,y=0) + x’.y.F(x=0,y=1) + x’.y’.F(x=0,y=0)

Notice we have four sub-Boolean Functions when cofactoring around two vars

Shannon Cofactors: Example

   

F = a.b + a.c + a.b’.c’ + a’

  

F a F a ’ = b + c + b’.c’ (a’ is gone since a=1) = 1 (first three terms dissapear) F = a.Fa + a’Fa’

F

a

= b + c + b’.c’ Decompose F further around b bF

 

ab F ab F ab’

+ b’F

ab ’ = 1 + c = c + c’

(function of c only)

Recursive?

Binary Recursion Tree

The recursive Shannon expansion corresponds to a binary recursion tree

f



xf x



x f x



xy



y f x y

) 

xy



y f x y

)

x

Example:

1 x 1 0 1 0

 (v)

1 x 2 0 1 y 0 1 x 3 0

f xy

v

Path  (v) to node v corresponds to cube c  (v) Example: c  (v) = x 1 x 2 ‘x 3

f x y

1 y 0

f xy f x y

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Computational Boolean Algebra

  

Use of Shannon Cofactors Properties of Cofactors Boolean Difference

F G H

(F x )’

We are measuring the sensitivity of f to small changes in x

F

x

xor F

x’ XOR is the ODD Function

Like Additions

f = x’, df/dx = f x xor f x’ f x =0, f x’ =1 df/dx = 0 xor 1 = 1 what make this function == 1?

We must make y=0, Any change on x will change f f = x+y, f x =1, f x’ =y df/dx = 1 xor y = y’, f = x.y, f x What makes this function == 1?

We must make y=1, if you change x



f will change = y and f df/dx = y xor 0 = y f = x xor y, f x = y’, f x’ df/dx = y xor y’ = 1, =y, x’ =0,

APPLICATIONS?

When does a change in C in make a change to C out ?

((C out ) C in =1) = ab + (a + b).1 = ab + (a+b) = a(b+1) + b = a + b ((C out ) Cin =0) = ab dC out /dC in = C out (c in =1)

xor

= (a + b)

xor

C out (cin=0) = ab = (a+b)’ ab + (a+b) (ab)’ = a xor b

What does this mean?

When a != b any change c in will cause change in C out

Computational Boolean Algebra

 

New Quantification Operators

 

Existential Quantification Universal Quantification A Realistic Application



Application to Logic Network Repair

ALSO CALLED CONSENSUS OF F wrt Xi

ALSO CALLED the SMOOTHING OF F wrt Xi

Universal Quantification What makes this a 1?

These other inputs make F=1 FOR ALL values of x Existential Quantification What makes this a 1?

There exits a value of ‘x’ that makes F=1 for this input Pattern of other variables.

Quantify the ‘x’ then quantify the ‘y’

What if we quantify the A Operands?

Independent of A1 A0!!

0

X + D 0 XD 0

X + C + 0 + XD + 0 = X + D

F(a,b) = ab + b’ == a + b’ Note: This is a simple and Trivial Example!!

What we want?

Values of constants d0,d1,d2,d3 that make Z == 1 For all possible values of original inputs a, b!!

Universally quantify the variables ‘a’ and ‘b’ out of the Z

function, this function will then depend only on d0,d1,d2,d3 == 1 (a, b) will disappear by cofactors

MUX Eqn

We will first use the above expression of the MUX written and then substitute:



a.b for s 1 and



b’ for s 0

G = d 0 (ab)’b’’ + d 1 (ab)’b’ +d 2 = d 0 (ab)’b + d 1 (ab)’b’ + d 2 (ab) b’’ + d ab b + 0 3 = d 0 (a’ + b’)b + d = d 0 a’b + d 1 b’ + d 2 1 (a’ + b’)b’ + d ab 2 ab (ab)b’

So now Z = G xnor F == GF + G’F’ What I want is to go calculate ** (For all a b Z) = Z ab . Z a’b . Z ab’ . Z a’b’

If we Calculate Z then find Co-factors, this will be complicated!!

Is there an easier way!

When a=0,b=0 then Z a’b’ = d 1 xnor 1 = d 1 When a=0,b=1 then Z a’b = d 0 xnor 0 = d 0 ’ When a=1,b=0 then Z ab’ = d 1 xnor 1 = d 1 When a=1,b=1 then Z ab = d 2 xnor 1 = d 2

The thing that I need to calculate is the universal quantification (FOR ALL ab Z) (d0,d1,d2,d3) = AND of all cofactors=

d0’d1d2

!!

What did we get??

How do we make this a ‘1’??

To make this a 1 (i.e. repair it) we need to make d 0 =0, d 1 =1,d 2 =1, d 3 =x

0 1 1 1 0 1 1 x 0 1 1 0

Computational Boolean Algebra

  

Positional Cube Notation (PCN) Tautology Checking

 

Recursive Tautology Concrete data structure URP (Unate Recursive Paradigm)

Tautology

F



acd



bcd



abd

 

ad



bcd



abd



abc acd



cd



ac

Given a cover F, we want to know if F = 1? NOT EASY !!!

F= 1311 3111 2132 2322 3321 1323 1332 3212 2231 2213 == 1?

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a’ bc ab Is this a minimal representation?

NO

a’ bc

[a b c ] a’ = [10 11 11 ] [a b c ] bc = [11 01 01 ]

Anding Two Cubes and getting 00 (void) means kill the cube

EX: f(a,b,c) = a’ + bc + ab  [10 11 11], [11 01 01], [01 01 11]

What if the function I am looking at is too complex? Break it into two pieces (cofactors) thereby decomposing it into simpler problems!!

Steps

abd bc’ a



1 c



1

 [11 01 11 01] (change 01 to 11)  [11 01 10 11] (nothing !) 

[01 01 11 01] (nothing!)



Cube Vanishes!!

is unate? Why? Because you see variable ‘a’ but not a’ you see c’ but not c … is not unate? Why? Because you see variable ‘x’ and x’ … but it is unate in variable y!!

Analogy to Continuous Functions?

x 1 x 2 A Boolean function is called positive unate in x if the variable x orginally 0 changes to 1 and therefore the value of f(…,x,…) will remain flat or increases from

0 to 1

a b’c ac a b’c bc

Why do we care about this result?

1 1 1 1 1 1 1 1

1

Tautology Not a Tautology

x x’

You can have many rules but It will be computationally Expensive to check all of them!!

x + x’ = 1

The Binate Select Heuristic

Example Unate and non-unate covers: a b c d G = ac+cd 1 - 1 - - 1 0

is unate

F = ac+cd+bcd a b c d 1 - 1 - - 0 1 - 1 1 0

is not unate

=> Choose c for splitting !

• • The

binate variables

of a cover are those with both 1’s and 0’s in the corresponding column.

In the unate recursive paradigm, the BINATE_SELECT heuristic chooses a (most) binate variable for splitting, which is thus eliminated from the sub-covers. 66

Heuristic !!

Rules of Termination Selection Rule ..

Recursion ..

Generic Tautology Check

Algorithm

CHECK_TAUTOLOGY

(List_of_Cubes C) { if(C ==  ) return FALSE; if(C == {-...-})return TRUE; // cube with all ‘-’ x i =

SELECT_VARIABLE

C 0 =

COFACTOR

(C,^X i ) (C) if(

CHECK_TAUTOLOGY

(C 0 ) == FALSE) { print x i = 0 return FALSE; } C 1 =

COFACTOR

(C,X i ) if(

CHECK_TAUTOLOGY

(C 1 ) == FALSE) { print x i = 1 return FALSE; } return TRUE; } 69

when we set a=1 b [11 01 11] c [11 11 01] b’c’[11 10 10] fourth disappears Should we Stop?

when we set a=0

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Tautology …

n n n n n n

Check if a function is always TRUE.

Plays an important role in all algorithms for logic optimization.

Recursive paradigm

• Expand about a variable.

• If all cofactors are TRUE then function is a tautology.

TAUTOLOGY

• The cover has a row of all 1s (Tautology cube).

• The cover depends on one variable only, and there is no column of 0s in that field.

NO TAUTOLOGY

• The cover has a column of 0s (A variable that never takes a certain value).

When a cover is the union of two subcovers that depend on disjoint subsets of variables, then check tautology in both subcovers.

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… Tautology

n n

Unate heuristics

• If cofactors are unate functions, additional criteria to determine tautology.

• Faster decision.

A cover is not tautology if it is unate row of all 1’s.

and there is not a

n

If a function is expanded in a unate variable , only one cofactor needs to be checked for tautology

• Positive unate in variable xi, f xi checked for tautology.

 f xi’ ; only f xi’ needs to be • Negative unate in variable xi, f xi checked for tautology.

 f xi’ ; only f xi needs to be 75

Tautology Example

n n

f = ab+ac+ab’c’ +a’ Select variable a.

• Cofactor w.r.to a’ • 11 11 11 => Tautology.

• Cofactor w.r.to a is: n

Select variable b.

• Cofactor w.r. to b’ is: n • • Depends on a single variable, no column of 0’s => Tautology.

Cofactor w.r. to b is: 11 11 11 => Tautology

Function is a TAUTOLOGY.

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