Lattice-Based Computation of Boolean Functions

Mustafa Altun and Marc Riedel

University of Minnesota

Switch-based Boolean computation

Shannon’s work: A Symbolic Analysis of Relay and Switching Circuits(1938)

x1

x2

x3

x1 x2 x3

x1

x2

x1 x2

Parallel: x1 + x2Series: x1 . x2

1D and 2D switches

2D switch

ON OFF

1D switch

A lattice of 2D switches

3 × 3 2D switching network and its lattice form

BOTTOM

LE

FT

RIG

HT

TOP

LE

FT

RIG

HT

BOTTOM

TOP

Boolean functionality and paths Switches are controlled by Boolean literals. fL evaluates to 1 iff there exists a top-to-bottom path.

gL evaluates to 1 iff there exists a left-to-right path.

x9

x1 x4

x2 x5

x7

x8

x3 x6

LE

FT

TOP

RIG

HT

BOTTOM

f L

gL

0

1 0

1 1

1

0

0 1

LE

FT

TOP

RIG

HT

BOTTOM

f L

gL

fL = 1gL = 0

Logic synthesis problemHow can we implement a given target Boolean function fT with a lattice of 2D switches?

Example: fT = x1x2x3+x1x4

x1x2x3 + x1x4 + x1x2 + x1x2x3x4 x1x2x3 + x1x4 + x1x2x4 + x1x2x3x4fL1 = fL2 =fL2 = x1x2x3 + x1x4fL1 = x1x2 + x1x4

x2 x1

x1 x4

x3 x1

BOTTOM

RIG

HTL

EF

T

TOP

x2 x4

x1 x1

x3 x4

BOTTOM

RIG

HTL

EF

T

TOP

Logic synthesis problem

x1

x1 x1

x2 x4

x5

x5

x3 x4

LE

FT

TOP

RIG

HT

BOTTOM

Example: fT = x1x2x3+x1x4+x1x5

9 TOP-TO-BOTTOM PATHS!

Our synthesis method Example: fT = x1x2x3+x1x4+x1x5

x1 x2 x3

x1 x4

x1 x5

x1

x2 x4 x5

x3 x4 x5 {x5}

{x1} {x1}

{x2} {x4}

{x1}

{x5}

{x3} {x4}

x1 x2 x3

x1 x4

x1 x5

x1

x2 x4 x5

x3 x4 x5

fTD = (x1+x2+x3)(x1+x4)(x1+x5)

fTD = x1 + x2x4x5 + x3x4x5

x5

x1 x1

x2 x4

x1

x5

x3 x4

x1 x2 x3

x1 x4

x1 x5

x1

x2 x4 x5

x3 x4 x5

Obtain the dual of fT.

Assign each product of fT to a

column. Assign each product of fT

D to

a row. Compute an intersection set

for each site. Arbitrarily select a literal from

an intersection set and assign it to the corresponding site.

x1 x1

x1 x1

x2

x3

x3 x4 x2

x1 x2 x3

x2

x4

x5

x3

x2 x3

x2 x4 x4 x5 x5

x1x4

x2 x3x4

x2 x4x5

x3x5

x1 x2 x5

x1 x3 x4

x2 x3 x4

x2 x4 x5

{x2, x3, x4}

Our synthesis methodExample: fT = x1x2x3+x1x4+x2x3x4+x2x4x5+x3x5

fTD= x1x2x5+x1x3x4+x2x3x4+x2x4x5

x1 x1

x1 x1

x2

x3

x3 x4 x2

x1 x2 x3

x2

x4

x5

x3

x2 x3

x2 x4 x4 x5 x5

x1x4

x2 x3x4

x2 x4x5

x3x5

x1 x2 x5

x1 x3 x4

x2 x3 x4

x2 x4 x5

Math behind the method – Theorem 1Example: fT = x1x2x3+x1x4+x2x3x4+x2x4x5+x3x5

Theorem 1 allows us to only consider column-paths. We do not need to enumerate all paths!

x1 x1

x1 x1

x2

x3

x3 x4 x2

x1 x2 x3

x2

x4

x5

x3

x2 x3

x2 x4 x4 x5 x5

x1x4

x2 x3x4

x2 x4x5

x3x5

x1 x2 x5

x1 x3 x4

x2 x3 x4

x2 x4 x5

Math behind the method – Theorem 2Example: fT = x1x2x3+x1x4+x2x3x4+x2x4x5+x3x5

Theorem 2 explains the relation between intersection sets and column-paths. Each column is for each product!

Our method’s performance

Area of the lattice: m×n

The time complexity:O(m2n2)

n and m are the number of products of the target function fT and its dual fT

D, respectively.

Future work We are investigating our method’s

applicability to different technologies. We are studying the applicability of the

Theorems to the famous problem of testing whether two given monotone Boolean functions are mutually dual.

Thank you!