1

Equivalence Checking Problems

Graph equivalence connectivity analysis (little or no functional analysis)

Combinational functional equivalence pin boundary around each combinational block is identical

in the two designs and the correspondence is known

Latch correspondence exists but is unknown tool needs to find a correspondence

Sequential equivalence with initial states requires traversal of implicit FSM

Sequential equivalence with no initial state notion of equivalence is not uniform

2

Combinational Verification Algorithms

BDDs with decomposition points (Berman/Trevillyan ‘86)

ATPG (Automatic Test Pattern Generation) with decomposition points (Brand ‘93)

Recursive Learning (Kunz ‘93) etc...

3

Binary Decision Diagrams (Akers’78, Bryant’86)

Ordered decision tree for f = ab + cd

0 0 0 1 0 0 0 1 0 0 0 1 1 1 1 1

d d d d d d d d

c c c c

0 1

0 1 0 1

0 1 0 1 0 1 0 1

b b

a

4

Subgraph reduction

Reduced Ordered Binary Decision Diagram (BDD) form:

0 1

d

c

01

0 1

0 1

b

a

0

1

Key idea: combine equivalent subcases

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BDD properties Same order of variables on all paths Reduced w.r.t. isomorphism Variable ordering crucial for BDD size Advantages:

canonical trivial equivalence check

Disadvantages: may be exponential in circuit size

(especially multipliers)

ac

b

f

f = ac + abc

ab

f

a

c

b

01

00

01

1

1

c

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Verification by monolithic BDDs

Single monolithic BDD for each output pair BDD size may depend on variable ordering; some

designs may have no efficient ordering

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Verification by decomposed BDDs

How to generate good decomposition point pairs? False negatives

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The false negative problem

Efficient false negative resolution is the key to equivalence checking on large combinational cones

Say f1 = f2

If g1 is not equal to g2, then (f1 || g1) may or may not be equal to (f2 || g2)

g2

g1f1

f2

f1 || g1

f2 || g2

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False negative example

Is [f = (d1 = d2)] equal to 1?

f = d1 d2 + d1 d2 = b c + a g + b g + a b g

Substitute value of g into f:

f = b c + a (a b) + b (a b) + a b (a b) = 1

ab

c

d1

ba

c

d2

g = a b

d1 = bcg + (b + c) g d2 = b + cg + a c

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False negative resolution

Is f(x1, x2, ..., xm, y1, y2, ... yk) = 1 ?

where yi = gi(x1, x2, ..., xm, yi+1, ... yk)

BDDs for f(x1, x2, ..., xm, y1, y2, ... yk)

for each i, [yi = gi(x1, …, xm, yi+1, …, yk)]

If f = 1 (false negative), decomposition points are equal, create a new decomposition point

If f = 0 (real negative), we have a witness to prove that it is a real negative

11

Decomposition point selection

Tradeoff: more decomposition points cause more false negative

resolutions fewer decomposition points cause larger sub-problems

Important to select more meaningful decomposition point candidates: Separate primary input variables Belong to the corresponding pairs of primary input cones

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Candidates for decomposition points

From design methodology: pairs of nets with same or related names

Through simulation: First set by simulation of random/user-supplied vectors Later simulations of witnesses of counterexamples for

false negative resolution

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Refining decomposition pairs by simulation

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Refining decomposition pairs by simulation

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Refining decomposition pairs by simulation

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Refining decomposition pairs by simulation

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Refining decomposition pairs by simulation

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Refining decomposition pairs by simulation

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Refining decomposition pairs by simulation

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Refining decomposition pairs by simulation

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Refining decomposition pairs by simulation

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qg

ris

xh

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Candidates for decomposition points

From design methodology: pairs of nets with same or related names

Through simulation: First set by simulation of random/user-supplied vectors Later simulations of witnesses of counterexamples for

false negative resolution

23

Candidates for decomposition points

Real negatives can refine many other equivalence classes:

initial randomsimulation

real negativewitnesssimulation

a

c

d

b A

D

Be

f

P

Q

e

P

Q

f

c

1

e

P

Q

f

c

1

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False negative resolution example

h = pq p = p

= ab(c+d) = ab

Counterexample: a=1 b=1 c=0 d=0 e=0

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1

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11

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pf x

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===?

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False negative resolution example

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1

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00 0

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False negative resolution example

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1

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00 0

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False negative resolution - Solution 1

Is f(y1, y2, ... yk, x1, x2, ..., xm) = 1 ?

where yi = gi(yi+1, ... yk, x1, x2, ..., xm)

Compose each decomposition point into f yi [f(y1, ... yk, x1, x2, ..., xm) (yi = gi(yi+1, ... yk, x2, ..., xm))]

Order of compositions is important Stop if BDD = constant 1 or if there exists a path to 0-vertex

consisting of only primary inputs

Best when resulting BDD = constant 1 (false negative)

[Intuition: final BDD is small (size = 1), so there might exist an order of compositions where no intermediate BDD blows up]

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Order of compositions

Ordering heuristics: compose yi so that the new support is

minimized compose yi such that the BDD for (yi = gi) is

smallest if yi is in support of both yj and f, compose yj

y1, y2, ... yk: have to compose each point at most once

Stop if BDD = constant 1 or if there exists a path to 0-vertex consisting of only primary inputs

f

x1

y3

x2

y4

x5x3 x4

y2

y1

01

0 0

01

1

1x2

x5

y3

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False negative resolution - Solution 2

k+1 BDDs: f, (y1 = g1), …, (yk = gk)

Find an assignment to all variables (y1, ... yk, x1, ..., xm) such that the corresponding path in each BDD leads to 1-vertex.

If such an assignment exists, we have a witness for f = 0 (real negative); otherwise, f = 1 for all possible assignments (false negative)

Search for an assignment among the set of all assignments. Works best if there exists one (or many) assignments (real negative).

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Search strategies

Backtrack search: pick an unassigned variable x Set x=0, solve sub-problem If fail, set x=1 and solve sub-problem fail if any path in any BDD ends in 0-

vertex, success if all paths end in 1-vertex

Local search: pick a starting assignment to all variables make a local move (flip a variable) so that

cost function improves success if all paths end in 1-vertex

0

0

fail fail

fail01

1

1

1

x4

y1x7

y2

(x1 = 0, x2 = 1, y1 = 0) flip x2

y1x1

x2

y1y1

success

success

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Summary of algorithm

Perform random simulation to obtain candidates for decomposition points

Verify decomposition point candidates: if false negative, create a decomposition point if real negative, refine future decomposition point

candidates

Heuristics for: selecting a “good” subset of candidates false negative resolution: “good” order of compositions or

search for real negative using real negatives to refine future candidates, to detect

bugs early

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ATPG with decomposition points (Brand’93)

Stuck-at-1 fault redundancy => functional equivalence F

G

X

X

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Recursive Learning (Kunz’93)

Case analysis to learn implications between nets in the designs

a dbc

e

f

x1

gx2

x2 = 1

1

f = 1 g = 1

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Recursive Learning (Kunz’93)

Case analysis to learn implications between nets in the designs

a dbc

e

f

x1

gx2

x2 = 1

1

f = 1 g = 1

1

35

Recursive Learning (Kunz’93)

Case analysis to learn implications between nets in the designs

a dbc

e

f

x1

gx2

x2 = 1

1

f = 1 g = 1

1

a = 1 b = 1

36

Recursive Learning (Kunz’93)

Case analysis to learn implications between nets in the designs

a dbc

e

f

x1

gx2

x2 = 1

1

f = 1 g = 1

1

a = 1 b = 1

1

37

Recursive Learning (Kunz’93)

Case analysis to learn implications between nets in the designs

a dbc

e

f

x1

gx2

x2 = 1

1

f = 1 g = 1

1

a = 1 b = 1

10

1

x1 = 1

38

Recursive Learning (Kunz’93)

Case analysis to learn implications between nets in the designs

a dbc

e

f

x1

gx2

x2 = 1

1

f = 1 g = 1

1

a = 1 b = 1

1

x1 = 1

39

Recursive Learning (Kunz’93)

Case analysis to learn implications between nets in the designs

a dbc

e

f

x1

gx2

x2 = 1

1

f = 1 g = 1

1

a = 1 b = 1

1 1

x1 = 1

0

x1 = 1

40

Recursive Learning (Kunz’93)

Case analysis to learn implications between nets in the designs

a dbc

e

f

x1

gx2

x2 = 1

1

f = 1 g = 1

a = 1 b = 1

x1 = 1 x1 = 11

41

Recursive Learning (Kunz’93)

Case analysis to learn implications between nets in the designs

a dbc

e

f

x1

gx2

x2 = 1

1

f = 1 g = 1

a = 1 b = 10

1

x1 = 1

0

x1 = 11

x1 = 1