Multilevel Logic SynthesisMultilevel Logic Synthesis-- Algebraic Methods-- Algebraic Methods

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Why Algebraic Methods?Why Algebraic Methods?

What is our goal?What is our goal?

minimization, maximally factored form.minimization, maximally factored form. Algebraic methods provide Algebraic methods provide fastfast algorithms. algorithms.

Algebraic methods treat logic function like a Algebraic methods treat logic function like a polynomialpolynomial

Fast algorithms for polynomial manipulation exist. Fast algorithms for polynomial manipulation exist.

Algebraic methods may lose the optimality, but Algebraic methods may lose the optimality, but the the resultresults are quite s are quite goodgood in general. in general.

Algebraic methods can be used to iterate and Algebraic methods can be used to iterate and interleave with interleave with Boolean operationsBoolean operations..

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What is Division?What is Division? 7 = 2 x 3 + 17 = 2 x 3 + 1 Divisor: what to divide with?Divisor: what to divide with? Quotient and remainder: how to divide?Quotient and remainder: how to divide? F = G F = G · H + R· H + R

Why Division? Why Division? Division is central in many operations: Division is central in many operations: factoring and decomposition factoring and decomposition resubstitution resubstitution extractionextraction

Division: Division: OutlineOutline

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ProductProduct

The The productproduct of two algebraic expressions F = X of two algebraic expressions F = X11+X+X22++…+X…+Xmm and G = Y and G = Y11+Y+Y22+…+Y+…+Ynn is the SOP expression is the SOP expression obtained by simplifying the result of (Xobtained by simplifying the result of (X11+X+X22+…+X+…+Xmm))(Y(Y11+Y+Y22+…+Y+…+Ynn) using xx’=0 and xx=x.) using xx’=0 and xx=x. Example: (a+b)(a+c) = a + ab + ac + bcExample: (a+b)(a+c) = a + ab + ac + bc

The product FG is The product FG is algebraicalgebraic if F and G are if F and G are orthogonal, otherwise, it is called a orthogonal, otherwise, it is called a Boolean productBoolean product.. Example: (a+b)(c+d) = ac + ad + bc + bdExample: (a+b)(c+d) = ac + ad + bc + bd (algebraic)(algebraic)

(a+b)(a+c) = aa+ab+ac+bc(a+b)(a+c) = aa+ab+ac+bc (Boolean)(Boolean)

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Algebraic and Boolean ExpressionsAlgebraic and Boolean Expressions

ff is an is an algebraic expressionalgebraic expression if if f f is a set of cubes is a set of cubes (SOP), such that no single cube contains another (SOP), such that no single cube contains another (minimal with respect to single cube (minimal with respect to single cube containment). Otherwise, containment). Otherwise, ff is called a is called a Boolean Boolean expressionexpression.. Example: a+bc is algebraic, a+ab is Boolean.Example: a+bc is algebraic, a+ab is Boolean.

The The supportsupport of an expression of an expression f, supp(f),f, supp(f), is the set is the set of variables that of variables that ff explicitly depends on. Two explicitly depends on. Two expressions f and g are said to be expressions f and g are said to be orthogonal orthogonal if if supp(f)supp(f)supp(g)=supp(g)=, denoted by , denoted by ffg.g. Example: a+b Example: a+b c+d c+d

Recall:Recall:

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DivisionDivision

DivisionDivision is an operation that takes two SOP is an operation that takes two SOP expressions F and P, and generates SOP expressions F and P, and generates SOP expressions Q and R, s.t. F = PQ +R.expressions Q and R, s.t. F = PQ +R. The division is The division is algebraicalgebraic is PQ is an algebraic product; is PQ is an algebraic product;

otherwise it is a otherwise it is a Boolean divisionBoolean division.. Q is called the Q is called the quotientquotient, R is called the , R is called the remainderremainder.. If R=0, P is a If R=0, P is a factorfactor, otherwise it is called a , otherwise it is called a divisordivisor..

Example:Example: F = ad + ae + bcd + jF = ad + ae + bcd + j For P = a, Q = d+e, R = bcd+jFor P = a, Q = d+e, R = bcd+j For P = a+bc, Q=d, R=ae+jFor P = a+bc, Q=d, R=ae+j Both a and a+bc are divisors, not factors.Both a and a+bc are divisors, not factors.

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Weak DivisionWeak Division

For division F = PQ+R, it is For division F = PQ+R, it is weakweak if if PQ is an algebraic product;PQ is an algebraic product; R has as few cubes as possible;R has as few cubes as possible; PQ+R and F are the same expression (same cubes).PQ+R and F are the same expression (same cubes).

{Q,R} is unique given F and P{Q,R} is unique given F and P, denote Q=F/P., denote Q=F/P. Example:Example:

For F=For F=adad+abc++abc+bcdbcd, P=a+bc: Q=d, R=abc, P=a+bc: Q=d, R=abc For F=For F=ac+adac+ad++bc+bdbc+bd+e, P=a+b: Q=c+d,R=e+e, P=a+b: Q=c+d,R=e For F=ad+aef+ab+b’cd+b’cef, P=a+b’cFor F=ad+aef+ab+b’cd+b’cef, P=a+b’c

• If P=a: Q=d+ef+bIf P=a: Q=d+ef+b• If P=b’c, Q=d+efIf P=b’c, Q=d+ef

Q=d+ef,R=abQ=d+ef,R=ab

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Kernel and Co-Kernel: Kernel and Co-Kernel: MotivationMotivation

We know how to divide a given expression F by We know how to divide a given expression F by another expression G. (division)another expression G. (division)

How to find G?How to find G? Too many Boolean divisors: a+bc’Too many Boolean divisors: a+bc’ Restrict to algebraic divisors, e.g. cubesRestrict to algebraic divisors, e.g. cubes

Problem: Problem:

Given a set of functions, find common weak Given a set of functions, find common weak (algebraic) divisors.(algebraic) divisors.

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Kernels: Kernels: DefinitionsDefinitions

An expression is An expression is cube-freecube-free if no cube divides the if no cube divides the expression evenly (i.e. no remainder, or there is no expression evenly (i.e. no remainder, or there is no literal that is common to all the cubes).literal that is common to all the cubes). Example: ab+c is cube-freeExample: ab+c is cube-free

ab+ac is not (a is a common factor)ab+ac is not (a is a common factor)

abc is not (must have two or more abc is not (must have two or more cubes)cubes)

The The primary divisorsprimary divisors of an expression F are the set of an expression F are the set of expressions of expressions D(F) = {F/c | c is a cube}.D(F) = {F/c | c is a cube}. Example: F = (a+b+c)(d+e)f + bfg + hExample: F = (a+b+c)(d+e)f + bfg + h

df+ef (=F/a) is a primary divisordf+ef (=F/a) is a primary divisor

d+e+g (=F/bf) is a primary divisord+e+g (=F/bf) is a primary divisor

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Kernels: Kernels: Definitions (cont’d)Definitions (cont’d)

The The kernelskernels of an expression F are of an expression F are

K(F) = {G | G K(F) = {G | G D(F) and G is cube-free}. D(F) and G is cube-free}. Example:Example: F = (a+b+c)(d+e)f + bfg + h F = (a+b+c)(d+e)f + bfg + h

df+ef (=F/a) df+ef (=F/a) D(F), but not a kernel. D(F), but not a kernel.

d+e+g (=F/bf) d+e+g (=F/bf) D(F) and is cube-free. D(F) and is cube-free.

A cube c used to obtain the kernel K=F/c is A cube c used to obtain the kernel K=F/c is called a called a co-kernelco-kernel of K. of K. C(F) = {c | F/c C(F) = {c | F/c K(F)}. K(F)}.

A kernel is of A kernel is of level 0level 0 if it has no kernels except if it has no kernels except itself. A kernel of itself. A kernel of level nlevel n has at least one kernel has at least one kernel of level n-1, but no kernel (except itself) of level n of level n-1, but no kernel (except itself) of level n or greater. or greater.

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Kernels: Kernels: ExampleExample

F = adf + aef + bdf + bef + cdf + cef + bfg + hF = adf + aef + bdf + bef + cdf + cef + bfg + h

= (a+b+c)(d+e)f + bfg + h= (a+b+c)(d+e)f + bfg + hcubecube Prim. Div.Prim. Div. KernelKernel Co-kernelCo-kernel levellevel

aa df+efdf+ef NONO NONO ----

bb df+ef+fgdf+ef+fg NONO NONO ----

bfbf d+e+gd+e+g YESYES YESYES 00

cfcf d+ed+e YESYES YESYES 00

dfdf a+b+ca+b+c YESYES YESYES 00

fgfg bb NONO NONO ----

ff (a+b+c)(d+e)+bg(a+b+c)(d+e)+bg YESYES YESYES 11

11 FF YESYES YESYES 22

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Kernel Computation: Kernel Computation: Cube IntersectionCube Intersection

Given F = XGiven F = X11+X+X22+…+X+…+Xnn, compute K(F) and C(F), compute K(F) and C(F) C*(F) = {c | c= XC*(F) = {c | c= XiiXXjj for some i and j} for some i and j}

(This contains all the level-0 co-kernels.)(This contains all the level-0 co-kernels.) weak_div(F,c) = F/c is a kernel with c as the weak_div(F,c) = F/c is a kernel with c as the

corresponding co-kernel.corresponding co-kernel.Example: F = abcd + abce + efgExample: F = abcd + abce + efg

C*(F) = {abc, e}C*(F) = {abc, e} F/abc = d+e,F/abc = d+e, F/e = abc+fgF/e = abc+fg

Problem: gives only level-0 co-kernels, to get for Problem: gives only level-0 co-kernels, to get for example level-1 co-kernels, example level-1 co-kernels,

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Kernel Computation: Kernel Computation: ExampleExample

Find all the level-0 co-kernels forFind all the level-0 co-kernels for

abc+abd’+bcd’+ade+cde+a’ceabc+abd’+bcd’+ade+cde+a’ce

--ceceeecc--cca’cea’ce

--dedecc--cccdecde

----aaaaadeade

--bd’bd’bcbcbcd’bcd’

--abababd’abd’

--abcabc

a’cea’cecdecdeadeadebcd’bcd’abd’abd’abcabc

00a’+da’+dcece

11ad+cd+a’cad+cd+a’cee

00a+ca+cdede

00a+ca+cbd’bd’

11ab+bd’+deab+bd’+de+a’e+a’e

cc

11bc+bd’+debc+bd’+deaa

00a+d’a+d’bcbc

00c+d’c+d’abab

levlevelel

primary primary divisordivisor

cubecube

How about level-1 kernels?How about level-1 kernels?

Repeat Above RecursivelyRepeat Above Recursively

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Fundamental TheoremFundamental Theorem

Theorem: If two expressions F and G have the property Theorem: If two expressions F and G have the property that that kkFFK(F), kK(F), kGGK(G) K(G) | k | kGG k kF F ||1 (k1 (kG G and kand kFF have at have at

most one term in common), most one term in common),

then F and G have no common nontrivial algebraic then F and G have no common nontrivial algebraic divisors (i.e. with at least two terms) else it does.divisors (i.e. with at least two terms) else it does.

Importance: If we “kernel” all functions and there are no Importance: If we “kernel” all functions and there are no nontrivial intersections, then the only common algebraic nontrivial intersections, then the only common algebraic divisors left are single cube divisors. This is used for divisors left are single cube divisors. This is used for quickly finding common subfunctions. quickly finding common subfunctions.

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Factoring and Factoring AlgorithmFactoring and Factoring Algorithm

Factoring is the process of deriving a factored Factoring is the process of deriving a factored form for a given logic function in SOP form. form for a given logic function in SOP form.

A factoring algorithm is algebraic if it guarantees A factoring algorithm is algebraic if it guarantees an an algebraic factored formalgebraic factored form from an from an algebraic algebraic SOP expressionSOP expression; otherwise it is Boolean.; otherwise it is Boolean.

F = ab+ac+bc F = a(b+c)+bcfactoringfactoring

SOP formSOP form factored formfactored form

goal: minimize goal: minimize the number of the number of

literalsliterals

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Generic Factoring AlgorithmGeneric Factoring Algorithm

Factor(F) Factor(F)

{ if (F has no factor) { if (F has no factor)

return F;return F;

D = D = DivisorDivisor(F);(F);

(Q,R) = (Q,R) = DivideDivide(F,D);(F,D);

return return

Factor(Q)Factor(D) Factor(Q)Factor(D)

+ Factor(R);+ Factor(R);

}}

Example:Example:

F = abc+abd+ae+af+gF = abc+abd+ae+af+g

D = c+dD = c+d

F = DQ+RF = DQ+R

Q = abQ = ab

R = ae+af+gR = ae+af+g

= a(e+f) + g= a(e+f) + g

((Factor(R)Factor(R)))

O = ab(O = ab(c+dc+d) + ) + a(e+f)+ga(e+f)+g

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Maximal FactorizationsMaximal Factorizations

A factored form is A factored form is maximally factoredmaximally factored if if For every sum of products, there are no two For every sum of products, there are no two

syntactically equivalent factors in the products;syntactically equivalent factors in the products; For every product of sums, there are no two For every product of sums, there are no two

syntactically equivalent factors in the sums. syntactically equivalent factors in the sums.

Example:Example: ab+ac is not maximally factored,ab+ac is not maximally factored,

a(b+c) is.a(b+c) is. (a+b)(a+c) is not maximally factored,(a+b)(a+c) is not maximally factored,

a+bc is.a+bc is.

bb

++

··ccaaaa

··

Recall:Recall:

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Problem with Algorithm FactorProblem with Algorithm Factor

O = O = aabb(c+d) + (c+d) + aa(e+f)+g(e+f)+g is not optimal since it is is not optimal since it is not maximally factored. not maximally factored. a(b(c+d)+e+f)+ga(b(c+d)+e+f)+g

The problem occurs whenThe problem occurs when Quotient Q is a single cube, andQuotient Q is a single cube, and Some of the literals of Q appear in the remainder RSome of the literals of Q appear in the remainder R

Solution to the problem:Solution to the problem: If the quotient Q is not a single cube, done; elseIf the quotient Q is not a single cube, done; else Pick a literal l in Q that occurs most frequently in the cubes of F;Pick a literal l in Q that occurs most frequently in the cubes of F; Divide F by l to obtain a new divisor DDivide F by l to obtain a new divisor D11;;

F = l DF = l D11 + R + R11, where literal l does not appear in R, where literal l does not appear in R11..

Lemma: partial factored form Lemma: partial factored form l Dl D11 + R + R11 is maximally factored if Q is a is maximally factored if Q is a single cubesingle cube

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Another ExampleAnother Example

FF ==ace+adeace+ade++bce+bdebce+bde+cf+df+cf+df

DD =a+b=a+b

QQ =ce+de=ce+de

RR =cf+df=cf+df

PP =QD+R=QD+R

=(ce+de)(a+b)+(cf+df)=(ce+de)(a+b)+(cf+df)

OO =e(c+d)(a+b)+(c+d)f=e(c+d)(a+b)+(c+d)f

Problem: Problem:

O=e(c+d)(a+b)+(c+d)fO=e(c+d)(a+b)+(c+d)f

is not maximally factored is not maximally factored because it can be further because it can be further factored to (c+d)(e(a+b)+f).factored to (c+d)(e(a+b)+f).

The problem occurs when The problem occurs when Q and R have a common Q and R have a common factor. (c+d in this case).factor. (c+d in this case).

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Solution to the ProblemSolution to the Problem

Make Q cube-free to obtain QMake Q cube-free to obtain Q11;;

Divide F by QDivide F by Q11 to obtain a new divisor D to obtain a new divisor D11;;

If DIf D11 is cube-free, we get a partial factored form is cube-free, we get a partial factored form

F=QF=Q11DD11+R+R11 and recursively factor Q and recursively factor Q11,D,D11, and R, and R11;;

If DIf D11 is not cube-free, let D is not cube-free, let D11=cD=cD22 and D and D33=Q=Q11DD22, we , we

have the partial factoring F=cDhave the partial factoring F=cD33+R+R11 and can and can

recursivelyrecursively factor Dfactor D33 and R and R11..

Lemma: if Q is not a single cube, F is maximally Lemma: if Q is not a single cube, F is maximally factored at this level.factored at this level.

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Application: Application: DecompositionDecomposition

Given an expression F and one of its divisors D, Given an expression F and one of its divisors D, the the decompositiondecomposition associates a new variable, G, associates a new variable, G, with the divisor and reduce F to QG+R, where with the divisor and reduce F to QG+R, where Q=F/G and R is the remainder in the division. Q=F/G and R is the remainder in the division.

Decomposition vs. factorizationDecomposition vs. factorization Divisor is added as a new variable.Divisor is added as a new variable. The new variable may fan out elsewhere in both The new variable may fan out elsewhere in both

regular or complement forms.regular or complement forms.

Why decomposition: Why decomposition: reduce the size of the expression reduce the size of the expression (e.g. to typical library cells).(e.g. to typical library cells).

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Example: Example: DecompositionDecomposition

F=ace+bce+de+gF = Ge+g

G=ac+bc+ddecompositiondecomposition

SOP formSOP form factored formfactored form

goal: minimize goal: minimize the size of the the size of the

expressionexpression

F = ab+ac+bc F = a(b+c)+bcfactoringfactoring

SOP formSOP form factored formfactored form

goal: minimize goal: minimize the number of the number of

literalsliterals

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Application: Application: Algebraic ResubstitutionAlgebraic Resubstitution

Goal: Goal: reduce number of literalsreduce number of literals

Idea: Idea: (Reuse)(Reuse) For multi-output For multi-output function, one output (or its complement) function, one output (or its complement) may be a useful divisor in another output.may be a useful divisor in another output.

Procedure: Procedure: substitution of Fsubstitution of Fjj into F into Fii

Divide FDivide Fii by F by Fjj: : FFii = QF = QFjj + R + R

Divide R by F’Divide R by F’jj: : R = QR = QCCF’F’jj+R+RCC

Result: Result: FFii = QF = QFjj + Q + QCCF’F’jj+R+RCC

ffiiffjj

ffjj

In practiceIn practice: this is tried for each output pair, therefore requires : this is tried for each output pair, therefore requires O(nO(n22) divisions when there are n outputs.) divisions when there are n outputs.

Filters:Filters: criteria that reduce the number of divisions. For example, criteria that reduce the number of divisions. For example, don’t try Fdon’t try Fii/F/Fjj if 1) F if 1) Fjj contains a literal not in F contains a literal not in F ii, or 2) F, or 2) Fjj has more has more

terms than Fterms than Fii, or 3) there is a literal occurs more often in F, or 3) there is a literal occurs more often in F jj than in F than in Fii..

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Example: Example: Algebraic ResubstitutionAlgebraic Resubstitution

F/G:F/G: F/ab’ = d+fgF/ab’ = d+fg F/c = d+fgF/c = d+fg

F = (d+fg)G + (a’c’e+bc’e+aef) = QG + RF = (d+fg)G + (a’c’e+bc’e+aef) = QG + R R/G’:R/G’:

G’ = (ab’+c)’ = (a’+b)c’ = a’c’ + bc’G’ = (ab’+c)’ = (a’+b)c’ = a’c’ + bc’ R/a’c’ = eR/a’c’ = e R/bc’ = eR/bc’ = e

R = eG’ + aefR = eG’ + aef

F = (d+fg)G + eG’ + aefF = (d+fg)G + eG’ + aef

9 literals9 literals

21 literals21 literals

saving of saving of 12 literals12 literals

(#23 on Page 448) Resubstitution of (#23 on Page 448) Resubstitution of G=ab’+cG=ab’+c into into F=ab’d+cd+ab’fg+cfg+a’c’e+bc’e+aefF=ab’d+cd+ab’fg+cfg+a’c’e+bc’e+aef

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Idea: replace a literal (and its complement) in an Idea: replace a literal (and its complement) in an expression by its algebraic SOP form.expression by its algebraic SOP form.

Goal: prepare for a different factorization or Goal: prepare for a different factorization or decomposition.decomposition.

Elimination value:Elimination value: FFii = Q = QiiG + QG + QiCiCG’ + RG’ + Rii

Let nLet nii be the number of times that G or G’ appears in be the number of times that G or G’ appears in

the factored form Fthe factored form Fii, L, LGG be the number of literals in the be the number of literals in the

factored form for G, then approximately, (nfactored form for G, then approximately, (n iiLLGG-n-nii))--

literals will be added after the elimination. This value is literals will be added after the elimination. This value is defined as the defined as the elimination value of Gelimination value of G..

In practice, don’t eliminate node with large In practice, don’t eliminate node with large e_valuee_value..

Application: Application: Node EliminationNode Elimination

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Example: Example: Node EliminationNode Elimination

F = e(af+c’G) + (d+fg)(c+G’)F = e(af+c’G) + (d+fg)(c+G’) G = a’+bG = a’+b

e_valuee_valueGG = = (n(niiLLGG-n-nii)) = 2x2-2 = 2= 2x2-2 = 2

elimination elimination of G (and G’)of G (and G’)

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Application: Application: ExtractionExtraction

Idea: Idea: identifies common sub-expressions and identifies common sub-expressions and

manipulates the Boolean network.manipulates the Boolean network. We can combine decomposition, substitution, We can combine decomposition, substitution,

and node elimination to provide an effective and node elimination to provide an effective extraction algorithm:extraction algorithm:Quick_ExtractionQuick_Extraction For each node in the network, apply decomposition;For each node in the network, apply decomposition; For all profitable pairs, apply algebraic substitution;For all profitable pairs, apply algebraic substitution; Eliminate all single literal functions;Eliminate all single literal functions; Eliminate all functions with small e_value;Eliminate all functions with small e_value;

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Example: Example: ExtractionExtraction

Decomposition:Decomposition:

OO11 = aY= aY11 + g + g

YY11 = bX= bX11 + e + + e +

ff

XX11 = c + d= c + d

OO22 = X= X22ZZ22

ZZ22 = eY= eY22 + f + f

YY22 = a + b= a + b

XX22 = c + d = c + d

(18 literals)(18 literals)

FF11 = abc+abd+ae+af+g= abc+abd+ae+af+gFF22= ace+ade+bce+bde+cf+df= ace+ade+bce+bde+cf+df

(11+16 = 27 literals)(11+16 = 27 literals)

Resubstitution:Resubstitution:

(X(X11 = X = X22))

OO11 = aY= aY11 + g + g

YY11 = bX= bX11 + e + e

+ f+ f

XX11 = c + d= c + d

OO22 = X= X11ZZ22

ZZ22 = eY= eY22 + f + f

YY22 = a + b= a + b

(16 literals)(16 literals)

Node Elimination:Node Elimination:

(e_value(e_valueY2Y2=-1)=-1)

OO11 = aY= aY11 + g + g

YY11 = bX= bX11 + e + e

+ f+ f

XX11 = c + d= c + d

OO22 = X= X11ZZ22

ZZ22 = e(a+b) + = e(a+b) +

ff

(15 literals)(15 literals)