Weikang Qian. Outline Intersection Pattern and the Problem Motivation Solution 2.

Weikang Qian

OutlineIntersection Pattern and the ProblemMotivationSolution

2

Intersection Pattern of CubesSum-of-product Boolean function: A set of

cubesExample:

Notation V( f ): number of minterms covered by f

Intersection pattern: numbers of minterms covered by the intersections of all subsets of cubes

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Example of Intersection Pattern

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3 cubes on 4 variables

Intersection pattern depends on the total number of variables!

Reverse ProblemGiven the number of variables and an

intersection pattern, synthesize cubes to satisfy that pattern

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Example: Synthesize 3 cubes on 4 variables so that

Sol.: One solution is

Reverse problem (cont.)Sometimes, these is no solution!

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Example: Synthesize 2 cubes on 4 variables so that

Sol.: No solution! Why?

00 01 11 10

00

01

11

10

x0x1

x2x3

1 1

1 1

1 1

1 1

c0

No way to put the rectangle of 4 ones so the intersection has 1 minterms!


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Why Intersection Pattern?Probabilistic computation

Digital circuit operating on random bit streamsTransform probabilities into probabilities

ApplicationIn test: Weighted random pattern generationHardware implementation of probabilistic

algorithms8

AND

xy z

P(x = 1) = 0.4

P(z = 1) = 0.2P(y = 1) = 0.5

0,1,0,1,0,0,1,1,0,0

1,0,1,1,0,0,1,0,0,1

0,0,0,1,0,0,1,0,0,0

A Fundamental Synthesis Problem

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Independent Combinational

Logic

0.5

0.5

0.5

0.5

q

?

Output Probability: AnalysisIf the number of inputs is n, then q = m/2n

Each input combination has probability 1/2n

If the Boolean function has m minterms, then q = m/2n

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5 mintermsq = 5/8

Output Probability: SynthesisImplement m/2n

Choose m mintermsBut, which m minterms to choose?

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Example: q = 7/16

00 01 11 10

00

01

11

10

x0x1

x2x3

1 1

1 1

1 1 1

00 01 11 10

00

01

11

10

x0x1

x2x3

1 1

1 1 1

1 1

Two Level Logic Synthesis ProblemGiven number of inputs n and m, find a SOP

Boolean function with minimum cubes to cover m minterms.

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Optimization Flow

λ = Lower bound on the number of cubes to cover m

minterms

Exist λ cubes to cover m interms?

endY

λ = λ + 1

N Related

Exist λ cubes to cover m miterms?Step 1: Construct an intersection pattern so

that it covers m minterms by Inclusion-Exclusion Principle:

Step 2: Synthesize cubes to satisfy the intersection pattern (Focus of this work)

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PreliminariesLiterals: If cube c has k literals, then V(c) = 2n − k

Empty cube: c = 0, V(c) = 0For a cube c,

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Cube-Variable Matrix Dλ rows, each corresponding to a cuben columns, each corresponding to a variable

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Example:

If a row vector has k *’s, V(c) = 2k

Synthesis a cube-variable matrix to satisfy theintersection pattern

V(c) = 2

PropertiesNegation

Column negation and column permutation of matrix maintain the intersection pattern.

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exchangecolumn 1 & 2

negatecolumn 1

Same pattern:

Key Ideas of SolutionDetermine what each column should beEach column has 3λ candidate choices:

Only need to consider a subset of candidate choicesFor example, by column negation invariability,

choice equivalent to choice Column permutation invariability: position does

not matter; only number of occurrence of each candidate choice in cube-variable matrix matters

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A Special CaseA case where all-cube intersection is non-

emptyCandidate choices: those columns that do not

contain both a 0 and a 1

Further reduction on candidate choices: only consider columns without 0’s

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XIntersection Empty!}

By column negationinvariability:

Replaced by

Special Case: Illustration with λ = 3Candidate choices:

Number of occurrences in cube-variable matrix matters

Set equations between unknowns and knowns

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y0 y1 y2 y3 y4 y5 y6 y7unknowns:

knowns: n, V(c0), V(c1), V(c0c1), V(c2), V(c0c2), V(c1c2), V(c0c1c2)

c0

c1

c2

Set up Equations

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c0

c1

c2

Equation on n

Equation on V(c0)• If a row vector of c has k *’s, V(c) = 2k

• Relates to total number of columns with first entry *

y0 + y1 + y2 + y3 + y4 + y5 + y6 + y7 = n

Each column belongs to one of the candidate choices

y1 + y3 + y5 + y7 = k1 = log2V(c0)

y2 + y3 + y6 + y7 = log2V(c1)Equation on V(c1)

y4 + y5 + y6 + y7 = log2V(c2)Equation on V(c2)

Set up Equations (cont.)

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c0

c1

c2

Equation on V(c0c1)

• If a row vector of c has k *’s, V(c) = 2k

• Relates to total number of columns with 1st and 2nd entry both *

c0

c1

c2

c0c1

contribute to * in c0c1

y3 + y7 = k3 = log2V(c0c1)

Equation on V(c0c2)

y5 + y7 = log2V(c0c2)

Equation on V(c1c2)

y6 + y7 = log2V(c1c2)

Set up Equations (cont.)

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c0

c1

c2

Equation on V(c0c1c2)• Relates to total number of columns with 1st, 2nd, and 3rd entry all *

y7 = log2V(c0c1c2)

System of Linear Equations

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y0 + y1 + y2 + y3 + y4 + y5 + y6 + y7 = n = k0

y1 + y3 + y5 + y7 = log2V(c0) = k1

y2 + y3 + y6 + y7 = log2V(c1) = k2

y4 + y5 + y6 + y7 = log2V(c2) = k4

y3 + y7 = log2V(c0c1) = k3

y5 + y7 = log2V(c0c2) = k5

y6 + y7 = log2V(c1c2) = k6

y7 = log2V(c0c1c2) = k7

Matrix form

Solution

Solution of the Special CaseSolutionIf all entries of are non-negative, then we

find one set of cubes satisfying the intersection pattern

Otherwise, there is no set of cubes satisfying the intersection pattern

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Example of Special Case

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Synthesize 3 cubes on 5 variables so that V(c0) = V(c1) = 16, V(c0c1) = V(c2) = 8, V(c0c2) = V(c1c2) = 4, V(c0c1c2) = 2

Sol.:

y0 y1 y2 y3 y4 y5 y6 y7

Cube-variable matrix

General CasesCases where all-cube intersection could be

emptySame ideas

Reduce candidate choicesTake numbers of occurrences as unknowns Set up equations between unknowns and

knownsA system of linear equalities and inequalities

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ConclusionProblem: Synthesize a set of cubes to satisfy

a given intersection patternImportant step in the two-level logic

synthesis for probabilistic computationSolution: Determine the number of

occurrences of each candidate choiceFuture work: integrate into the optimization

problem

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Thank You!

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Weikang Qian. Outline Intersection Pattern and the Problem Motivation Solution 2.

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