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Automatic Test Pattern Generation for Digital Circuits, Study notes of Spanish Language

University of California - Santa Barbara Spanish Language

The problem of automatic test pattern generation (atpg) for digital circuits, including the d-algorithm, podem, and boolean satisfiability approaches for combinational circuits, and time-frame expansion, nine-valued test generation, and potential detection for sequential circuits. Potential problems and solutions for the d-algorithm are also discussed.

Typology: Study notes

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Outline – Automatic Test Pattern

Generation

• Test generation systems

• Test generation for combinational ckts

–D-Algorithm

– PODEM

– Boolean Satisfiability approach

– Test compaction

• Test generation for sequential ckts

– Time-frame expansion & Extended D-Algorithm

– Nine-valued test generation

– Potential detection

– Issues of sequential ATPG

The ATPG Problem

• Problem definition: Given a logical fault model, and

a circuit, determine a small set of test vectors

that detect all faults in the circuit.

• Problem complexity: Under the stuck-at fault

model, the problem is NP-complete even for

combinational circuits

– However, commercial test generators that

efficiently generate tests for >10M-gate ckts

are in use today.

Test Generation Systems

Circuit

Description

Test

Patterns

Undetected

Faults

Redundant

Faults

Aborted

Faults

Backtrack

Distribution

Fault

List

Compacter

ATPG

With fault

simulator

Ack: Bushnell and Agrawal, Essential of Electronic Testing, 2000

Basics: Path Sensitization Method

Atomic operations:

1. Activation, fault excitation: Specify inputs so as to

generate the appropriate value at fault site for fault

excitation (I.e. set S to 1 for S-stuck-at-0 fault)

2. Error propagation: specify additional signal values to

propagate the fault effect from the fault site to the

outputs/observation points

3. Line justificatio n: Specify input values so as to produce

the signal values specified in (1) or (2)

4. Value implicatio n: unique determination of values at

other signals due to value assignments made in (1), (2), or (3)

(1) Fault activation : A = B = C = 1

(2) Have a choice of error propagating: through G5 or G6

(a) Propagating through G5 requires G2= 1 ⇒A=D=0 ⇒contradiction

(b) Propagating through G6 requires G4= 1 ⇒C=1, E=0

⇒Test ABCDE=(111x0)

G

1

G

5

G

2

G

6

G

3

A

C

B

D

E

a

f

2

f

1

A Simple Example

X

s.a.1

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Outline – Automatic Test Pattern

Generation

Test generation systems
Test generation for combinational ckts
- D-Algorithm
- PODEM
- Boolean Satisfiability approach
- Test compaction
Test generation for sequential ckts
- Time-frame expansion & Extended D-Algorithm
- Nine-valued test generation
- Potential detection
- Issues of sequential ATPG

The ATPG Problem

Problem definition: Given a logical fault model, and a circuit, determine a small set of test vectors that detect all faults in the circuit.
Problem complexity: Under the stuck-at fault model, the problem is NP-complete even for combinational circuits - However, commercial test generators that efficiently generate tests for >10M-gate ckts are in use today.

Test Generation Systems

Circuit Description

Test Patterns

Undetected Faults

Redundant Faults

Aborted Faults

Backtrack Distribution

Fault Compacter List

ATPG

With fault simulator

Ack: Bushnell and Agrawal, Essential of Electronic Testing, 2000

Basics: Path Sensitization Method

Atomic operations:

Activation, fault excitation: Specify inputs so as to generate the appropriate value at fault site for fault excitation (I.e. set S to 1 for S-stuck-at-0 fault)
Error propagation: specify additional signal values to propagate the fault effect from the fault site to the outputs/observation points
Line justification: Specify input values so as to produce the signal values specified in (1) or (2)
Value implication: unique determination of values at other signals due to value assignments made in (1), (2), or (3)

(1) Fault activation : A = B = C = 1 (2) Have a choice of error propagating: through G 5 or G 6 (a) Propagating through G 5 requires G 2 = 1 ⇒ A=D=0 ⇒ contradiction (b) Propagating through G 6 requires G 4 = 1 ⇒ C=1, E= ⇒ Test ABCDE=(111x0)

G 1

G (^2) G 5

G 6

G 3

A CB

D

E

a f 2

f 1

A Simple Example

s.a.1^ X

E s-a-1 ⇒ E =
C = D = 1 to propagate through G 1.
To propagate G 4 , need G 2 = G 3 = 1 Attempt to line justify G 2 = G 3 = 1
- G 3 = 1possible if A = F = 1 or B = H = 1
  - If A = F = 1 ⇒ inconsistency since C = 1 so G 2 = 0. ⇒ Therefore, G 3 = 1 ⇒ B = H = 1
- G 2 = 1 need A = 0 or F = 0 ⇒ Tests are ABCDEH, BCDEFH

G 1

G 2 G^4

G 3 A

C

B

DE

H F

Line Justification

Completeness of ATPG Algorithms

A test generation algorithm is deemedcomplete iff it will find a test for a fault if exists or prove that there exists no test, given sufficient time. - Complete algorithms can identify untestable faults
Major complete algorithms for comb. ckts
- D-algorithm (Roth, 1966)
- PODEM (Goel, 1981)
- FAN (Fujiwara, 1983)
- Socrates (Schulz, 1988)
- Boolean-SAT-based ATPG (Larrabee, 1992)

Single Path Sensitization Is NOT

Complete

G 3

G 2

G 4 G 5

G (^1) f

C

E

A

B d s.a.0^ x^ d G^6

d s-a-0 : ⇒ A = B = 1
Propagate along G 3 , G 6 requires C =1, G 2 = G 4 = G 5 = 1.
- In order for G 4 = 1 either E = 0 or G 1 = 0 ⇒ inconsistency ⇒ E = 0, B = 1 ⇒ G 5 = 0 ⇒ inconsistency
Propagate along G 4 , G^6 ⇒^ E = 1 & G^2 = G^3 = G^5 = 1
- G 2 = 1 ⇒ C = 1, ABC = 111 ⇒ G 3 = 0 inconsistency ⇒ No test (^) © K.T. Tim Cheng, 05_comb_tg, v1.0 10

G 3

G 2

G 4

G 5

G 1

C

E

A B

(^1) d G 6 1

1

0

1

1/01/

0/

s-a-0 1/

But Boolean Difference Method

Finds Test ABCD=(1111)!!

Multiple Path Sensitization

G 3

G 2

G 4

G 5

G 1

C

E

A B

(^1) d G 6 1

1

0

1

1/ 1/

0/

s-a-0 1/

Two paths G 3 , G 2 and G 4 , G 6 are sensitized, i.e. error is

propagated alongboth paths.

The D-Algebra

Need to be able to deal with multiple “errors” at the inputs to a gate.
D represents a signal which has value 1 in normal (fault-free, good) ckt, and value 0 in faulty ckt. (I.e. D ≡ 1/0); Similarly, D ≡ 0/ D D D D D D

D D D D D D

D 0 D D 1 D

Behaves like a Boolean variable.

Flowchart for D-algorithm

start Initialize test cube (tc) Select a primitive D-cube of that as C D-intersect C with previous test cube tc and perform implication

Is there a D or D on any PO?

Select a gate from D-frontier and a propagation D-cube of the selected gate as C

Backtrack to the last point a choice exists

Line Justification

inconsistent consistent yes impossible Test hasbeen none exists generated

no done

Line Justification

Begin

Intersect C with previous test cube tc

Is there any line in tc which are not justified

Select an unjustified line and a primitive cube C to justify the line

Backtrack to the last point a choice exists

consistent inconsistent

yes

Test has generatedbeen

none exists

no

Line justification impossible

D

1 2 3 4 5 6 7 8 9 10 11 12 11 1 1 1

1 1 0 1 1 1 D

0 1 0

X

D 0

0 0 0 D 1

1

primitive D-cube (1) Select D-frontier G 5 (3)

Line justification (7)Implication

1 1 D D D D

D 0 1 1 1

1 D D

1 D 1 1 D 1 1 1 0 D 1 D 1 1 D 1 1 1 1 1 1 D 1 1 D 1 1 1 00 D 01 D 1 1 D 1 1 1 00 D 01 D 1 1 D 1 0 1 1 1 0 D 1 D 1 D D 1 1 1 10 D 1 DD 1 1 0 0 1 1 1 10 D 01 D D 1 D

implication (2)

implication (4)

implication (10)

Select D-frontier G 8 (6)

Line justification (8)Implication

backtrack to (6)inconsistent & Select D-frontierG 6 (9)

implication (8)

Test is found : 1 1 1 1

implication (5) 1 D

G (^5) G 8

G 4

G 6

G 7

G 2

4

2 6

1 3

8

11

9 s-a-

12 10

G (^15)

G (^37)

A D-Algorithm

Example

Potential Problems with D-Algorithm

Since the assignment of values is allowed to internal lines, more than one choice is available at such internal line/gate and backtracking could occur at each gate
Could result in inefficiency for large ckts and some special classes of ckts.
Example: An ECAT (error-correction-&-translation) ckt AB CE FG

s.a.0H J K N

L

Q

P (^) R

M

The PODEM Algorithm (Goel 1981)

Only allows assignment of values to primary inputs
The values assigned to primary inputs are then

propagated toward internal lines by theimplication.

AB CE

GF

H s.a.

J K N L Q

P

R

M

Example: First, a binary value is assigned to an unassigned PI to provide a fault effect at fault site: A = 1 Determine the implications of assigned PIs (only forward implication): A= 1 cause no implication Next, assign B = 1 A = B = 1 ⇒ imply H = D

Start

A = 1 B = 1

G = 1

G = 0

F = 0

E = 0

B = 0

Test has beenSuccessfully generated

A = 0

C = 1 E = 1

F = 1

AB CE FG

H s.a.

J K N L Q

P

R

M

PODEM Decision Tree for the Example

PODEM

Essentially a process of finding a PI & a binary value for initial assignment.
Continue assigning PI values, checking to see if the error is being propagated to outputs (after each PI assignment, perform forward implication)
If at any stage, either the fault cannot be excited or the error cannot be propagated further,

backtrack to the most recent PI assignment and

change it.

Flowchart of PODEM

Start Assign a binary valueto an unassigned PI

Determine implication of all PIs Is there a D or D on anyP0? Test possible with additionalassigned PIs? untried combinationIs there an of values on assignedPIs?

of values on assigned PIsSet untried combination

Test is found

No test exists

maybe

yes

no

no no

yes

Steps in PODEM

(1) Determine aninitial objective. If the fault

effect has not appeared at fault site, the initial objective is directed toward providing the fault effect on the faulty line. (2) Given the initial objective, a PI & a logic value are chosen that have a good likelihood of meeting the object.

Done using thebacktrace procedure.

Flowchart ofBacktrace

begin (^) Is objective line fed by a PI? Current objective value V and type of gate driving objective line? Next obj line isthe input of G which is at xand is the hardest to control

Next obj line isthe input of G which is at xand is the easiest to control Is G a NAND/NORgate?

Next obj value is the same asthe current objective value

Next obj value is the complementof the current objective value

Found PI initial assignmentis the current objective value

A

Exit

no (fed bygate G) OR/NAND & V=1AND/NOR & V=

yes

OR/NAND & V=0AND/NOR & V=

A

no yes

A PODEM Example

G (^5) G 8

G 4

G 6

G 7

G (^2) s-a-1 Z 1

G 1

G 3

X 2 X 3 X 4

X 1

Initial objective: (0, G 2 )
Backtrace to PI : X 2 = 1
Initial objective: (0, G 2 )
Backtrace: X 3 = 1
Implication: G 2 = D
D-frontier is { G 5 , G 6 }
Attempt to propagate through G 5
- Require X 1 = 1
Implication: G 1 = 0, G 4 = 1, G 5 = D
Attempt to propagate D through G 8
- Initial objective: ( 1, G 6 )
- Backtrace to set X 4 = 0
- Implication: G 3 = 1, G 7 = 0 & G 8 = 1 ⇒ failed in propagating error
- Backtrack to most recent PI assignment ⇒ reassign X 4 = 1
- Implication: G 3 = 0, G 6 = D, G 8 = D ⇒ Test is generated

Decision tree

A PODEM Example – Cont’d

X 2 = 1 X 3 = 1 X 1 = 1 X 4 = 0 test is found conflict &backtrack

X 4 = 1

G 8

G 5

G 4

G 6

G 7

G (^2) s-a-1 Z 1

G 1

G 3

X 2 X 3 X 4

X 1

(c) Construct the formula of the Boolean Difference: XOR of (a) & (b) and the output of XOR should be 1

From (a): (X+D)x(X+E)x(X+D+E)x(D+A)x(D+B)x(D+A+B)x(C+E)x(C+E) From (b): (X’+D’)x(X’+E)x(X’+D’+E)xD’ From (c): (V 1 +X)x(V 1 +X’)x(V 1 +X+X’)x(V 2 +X)x(V 2 +X’)x(V 2 +X+X’)x(BD+V 1 )x(BD+V 2 )x (BD+V 1 +V 2 )xBD (note: BD=X⊕X’=(X+X’)x(X+X’)=V 1 xV 2 )

Step 1: Extracting the formula – Cont’d

X X’

BD=(X+X’)x(X+X’)=V 1 xV 2 =

Step 2: Satisfying the Formula

- Boolean Satisfiability

Given a suitable representation for a Boolean functionf(X):
- Find an assignmentX* such thatf(X*) = 1
- Or prove that such an assignment does not exist (i.e.f(X) = 0 for all possible assignments)
In the “classical” SAT problem,f(X) is represented in product-of- sums (POS) or conjunctive normal form (CNF)
Many decision (yes/no) problems can be formulated either directly or indirectly in terms of Boolean Satisfiability

Conjunctive Normal Form (CNF)

Clause

Positive Literal

Negative Literal

ϕ = ( a + c ) ( b + c ) (¬a + ¬b + ¬c )

Source: João Marques-Silva and Karem A. Sakallah, “Boolean Satisfiability in EDA”, CAV2000 tutorial. © K.T. Tim Cheng, 05_comb_tg, v1.0 40

Literal & Clause Classification

ϕ = (a + ¬b)(¬a + b + ¬c )(a + c + d )(¬a + ¬b + ¬c )

a assignedb assigned 0 c and d unassigned 1

violated satisfied unresolved satisfied

Source: João Marques-Silva and Karem A. Sakallah, “Boolean Satisfiability in EDA”, CAV2000 tutorial.

Basic Backtracking Search

(a + b + c) (a + b + ¬c) (¬a + b + ¬c) (a + c + d) (¬a + c + d) (¬a + c + ¬d) (¬b + ¬c + ¬d) (¬b + ¬c + d)

(a + b + c)^ a (a + b + ¬c) (¬a + b + ¬c) (a + c + d) (¬a + c + d) (¬a + c + ¬d) (¬b + ¬c + ¬d) (¬b + ¬c + d)

(a + b + c) (a + b + ¬c) (¬a + b + ¬c) (a + c + d) (¬a + c + d) (¬a + c + ¬d) (¬b + ¬c + ¬d) (¬b + ¬c + d)

(¬b + ¬c + ¬d)

(a + b + c) (a + b + ¬c) (¬a + b + ¬c) (a + c + d) (¬a + c + d) (¬a + c + ¬d)

(¬b + ¬c + d)

(¬b + ¬c + ¬d)

(a + b + c) (a + b + ¬c) (¬a + b + ¬c) (a + c + d) (¬a + c + d) (¬a + c + ¬d)

(¬b + ¬c + d)

(¬b + ¬c + ¬d)

(a + b + c) (a + b + ¬c) (¬a + b + ¬c) (a + c + d) (¬a + c + d) (¬a + c + ¬d)

(¬b + ¬c + d)

(¬b + ¬c + ¬d)

(a + b + c) (a + b + ¬c) (¬a + b + ¬c) (a + c + d) (¬a + c + d) (¬a + c + ¬d)

(¬b + ¬c + d)

(¬b + ¬c + ¬d)

(a + b + c) (a + b + ¬c) (¬a + b + ¬c) (a + c + d) (¬a + c + d) (¬a + c + ¬d)

(¬b + ¬c + d)

(¬b + ¬c + ¬d)

(a + b + c) (a + b + ¬c) (¬a + b + ¬c) (a + c + d) (¬a + c + d) (¬a + c + ¬d)

(¬b + ¬c + d)

(¬b + ¬c + ¬d)

(a + b + c) (a + b + ¬c) (¬a + b + ¬c) (a + c + d) (¬a + c + d) (¬a + c + ¬d)

(¬b + ¬c + d)

(¬b + ¬c + ¬d)

(a + b + c) (a + b + ¬c) (¬a + b + ¬c) (a + c + d) (¬a + c + d) (¬a + c + ¬d)

(¬b + ¬c + d)

(¬b + ¬c + ¬d)

(a + b + c) (a + b + ¬c) (¬a + b + ¬c) (a + c + d) (¬a + c + d) (¬a + c + ¬d)

(¬b + ¬c + d)

b

c

d d

b

c

d d

c

(¬b + ¬c + ¬d) d

(a + b + c) (a + b + ¬c) (¬a + b + ¬c) (a + c + d) (¬a + c + d) (¬a + c + ¬d)

(¬b + ¬c + d)

(¬b + ¬c + ¬d)

(a + b + c) (a + b + ¬c) (¬a + b + ¬c) (a + c + d) (¬a + c + d) (¬a + c + ¬d)

(¬b + ¬c + d)

(¬b + ¬c + ¬d)

(a + b + c) (a + b + ¬c) (¬a + b + ¬c) (a + c + d) (¬a + c + d) (¬a + c + ¬d)

(¬b + ¬c + d)

(¬b + ¬c + ¬d)

(a + b + c) (a + b + ¬c) (¬a + b + ¬c) (a + c + d) (¬a + c + d) (¬a + c + ¬d)

(¬b + ¬c + d)

(¬b + ¬c + ¬d)

(a + b + c) (a + b + ¬c) (¬a + b + ¬c) (a + c + d) (¬a + c + d) (¬a + c + ¬d)

(¬b + ¬c + d)

(¬b + ¬c + ¬d)

(a + b + c) (a + b + ¬c) (¬a + b + ¬c) (a + c + d) (¬a + c + d) (¬a + c + ¬d)

(¬b + ¬c + d)

(¬b + ¬c + ¬d)

(a + b + c) (a + b + ¬c) (¬a + b + ¬c) (a + c + d) (¬a + c + d) (¬a + c + ¬d)

(¬b + ¬c + d)

(¬b + ¬c + ¬d)

(a + b + c) (a + b + ¬c) (¬a + b + ¬c) (a + c + d) (¬a + c + d) (¬a + c + ¬d)

(¬b + ¬c + d)

(¬b + ¬c + ¬d)

(a + b + c) (a + b + ¬c) (¬a + b + ¬c) (a + c + d) (¬a + c + d) (¬a + c + ¬d)

(¬b + ¬c + d)

(¬b + ¬c + ¬d)

(a + b + c) (a + b + ¬c) (¬a + b + ¬c) (a + c + d) (¬a + c + d) (¬a + c + ¬d)

(¬b + ¬c + d)

(¬b + ¬c + ¬d)

(a + b + c) (a + b + ¬c) (¬a + b + ¬c) (a + c + d) (¬a + c + d) (¬a + c + ¬d)

(¬b + ¬c + d)

(¬b + ¬c + ¬d)

(a + b + c) (a + b + ¬c) (¬a + b + ¬c) (a + c + d) (¬a + c + d) (¬a + c + ¬d)

(¬b + ¬c + d)

(¬b + ¬c + ¬d)

(a + b + c) (a + b + ¬c) (¬a + b + ¬c) (a + c + d) (¬a + c + d) (¬a + c + ¬d)

(¬b + ¬c + d)

(¬b + ¬c + ¬d)

(a + b + c) (a + b + ¬c) (¬a + b + ¬c) (a + c + d) (¬a + c + d) (¬a + c + ¬d)

(¬b + ¬c + d)

(¬b + ¬c + ¬d)

(a + b + c) (a + b + ¬c) (¬a + b + ¬c) (a + c + d) (¬a + c + d) (¬a + c + ¬d)

(¬b + ¬c + d)

(¬b + ¬c + ¬d)

(a + b + c) (a + b + ¬c) (¬a + b + ¬c) (a + c + d) (¬a + c + d) (¬a + c + ¬d)

(¬b + ¬c + d)

(¬b + ¬c + ¬d)

(a + b + c) (a + b + ¬c) (¬a + b + ¬c) (a + c + d) (¬a + c + d) (¬a + c + ¬d)

(¬b + ¬c + d)

(¬b + ¬c + ¬d)

(a + b + c) (a + b + ¬c) (¬a + b + ¬c) (a + c + d) (¬a + c + d) (¬a + c + ¬d)

(¬b + ¬c + d)

(¬b + ¬c + ¬d)

(a + b + c) (a + b + ¬c) (¬a + b + ¬c) (a + c + d) (¬a + c + d) (¬a + c + ¬d)

(¬b + ¬c + d)

(¬b + ¬c + ¬d)

(a + b + c) (a + b + ¬c) (¬a + b + ¬c) (a + c + d) (¬a + c + d) (¬a + c + ¬d)

(¬b + ¬c + d)

(¬b + ¬c + ¬d)

(a + b + c) (a + b + ¬c) (¬a + b + ¬c) (a + c + d) (¬a + c + d) (¬a + c + ¬d)

(¬b + ¬c + d)

(¬b + ¬c + ¬d)

(a + b + c) (a + b + ¬c) (¬a + b + ¬c) (a + c + d) (¬a + c + d) (¬a + c + ¬d)

(¬b + ¬c + d)

Source: João Marques-Silva and Karem A. Sakallah, “Boolean Satisfiability in EDA”, CAV2000 tutorial. © K.T. Tim Cheng, 05_comb_tg, v1.0 42

An unresolved clause isunit if it has exactly one unassigned literal

ϕ = (a +c)(b +c)(¬a + ¬b + ¬c)

A unit clause has exactly one option for being satisfied

a b → ¬c

i.e.c must be set to 0.

Unit Clause Rule - Implications

Source: João Marques-Silva and Karem A. Sakallah, “Boolean Satisfiability in EDA”, CAV2000 tutorial.

Basic Search with Implications

(a + b + c) (a + b + ¬c) (¬a + b + ¬c) (a + c + d) (¬a + c + d) (¬a + c + ¬d) (¬b + ¬c + ¬d) (¬b + ¬c + d)

(a + b + c)^ a (a + b + ¬c) (¬a + b + ¬c) (a + c + d) (¬a + c + d) (¬a + c + ¬d) (¬b + ¬c + ¬d) (¬b + ¬c + d)

b

(a + b + c) (a + b + ¬c) (¬a + b + ¬c) (a + c + d) (¬a + c + d) (¬a + c + ¬d) (¬b + ¬c + ¬d) (¬b + ¬c + d)

c

(a + b + c) (a + b + ¬c) (¬a + b + ¬c) (a + c + d) (¬a + c + d) (¬a + c + ¬d) (¬b + ¬c + ¬d) (¬b + ¬c + d) d

b c

(a + b + c) (a + b + ¬c) (¬a + b + ¬c) (a + c + d) (¬a + c + d) (¬a + c + ¬d) (¬b + ¬c + ¬d) (¬b + ¬c + d) 8 8

(a + b + c) (a + b + ¬c) (¬a + b + ¬c) (a + c + d) (¬a + c + d) (¬a + c + ¬d) (¬b + ¬c + ¬d) (¬b + ¬c + d)

(a + b + c) (a + b + ¬c) (¬a + b + ¬c) (a + c + d) (¬a + c + d) (¬a + c + ¬d) (¬b + ¬c + ¬d) (¬b + ¬c + d) d

a c

(a + b + c) (a + b + ¬c) (¬a + b + ¬c) (a + c + d) (¬a + c + d) (¬a + c + ¬d) (¬b + ¬c + ¬d) (¬b + ¬c + d) 6 6

(a + b + c) (a + b + ¬c) (¬a + b + ¬c) (a + c + d) (¬a + c + d) (¬a + c + ¬d) (¬b + ¬c + ¬d) (¬b + ¬c + d)

(a + b + c) (a + b + ¬c) (¬a + b + ¬c) (a + c + d) (¬a + c + d) (¬a + c + ¬d) (¬b + ¬c + ¬d) (¬b + ¬c + d) c

a b

(a + b + c) (a + b + ¬c) (¬a + b + ¬c) (a + c + d) (¬a + c + d) (¬a + c + ¬d) (¬b + ¬c + ¬d) (¬b + ¬c + d) 5

d

(a + b + c) (a + b + ¬c) (¬a + b + ¬c) (a + c + d) (¬a + c + d) (¬a + c + ¬d) (¬b + ¬c + ¬d) (¬b + ¬c + d) 6

(a + b + c) (a + b + ¬c) (¬a + b + ¬c) (a + c + d) (¬a + c + d) (¬a + c + ¬d) (¬b + ¬c + ¬d) (¬b + ¬c + d)

b

(a + b + c) (a + b + ¬c) (¬a + b + ¬c) (a + c + d) (¬a + c + d) (¬a + c + ¬d) (¬b + ¬c + ¬d) (¬b + ¬c + d)

c

(a + b + c) (a + b + ¬c) (¬a + b + ¬c) (a + c + d) (¬a + c + d) (¬a + c + ¬d) (¬b + ¬c + ¬d) (¬b + ¬c + d)

(a + b + c) (a + b + ¬c) (¬a + b + ¬c) (a + c + d) (¬a + c + d) (¬a + c + ¬d) (¬b + ¬c + ¬d) (¬b + ¬c + d) d

a c

(a + b + c) (a + b + ¬c) (¬a + b + ¬c) (a + c + d) (¬a + c + d) (¬a + c + ¬d) (¬b + ¬c + ¬d) (¬b + ¬c + d)

Source: João Marques-Silva and Karem A. Sakallah, “Boolean Satisfiability in EDA”, CAV2000 tutorial. © K.T. Tim Cheng, 05_comb_tg, v1.0 44

Public Released SAT Solvers by UCSB

C-SAT: Combinational Circuit-based SAT Solver
- F. Lu, Li-C. Wang, K.-T Cheng, and R. Huang, “A Circuit SAT solver with Signal Correlation guided learning,” DATE, March 2003.
- F. Lu, L.-C. Wang, K.-T. Cheng, J. Moondanos and Z. Hanna, "A Signal Correlation Guided ATPG Solver and Its Applications for Solving Difficult Industrial Cases," DAC, Jun. 2003.
Satori & Satori2: Sequential Circuit-based SAT Solver
- M. K. Iyer, G. Parthasarathy, and K.-T Cheng, “SATORI—A Fast Sequential SAT solver for Circuits,” ICCAD, Nov. 2003.
- F. Lu, G Parthasarathy, M. K. Iyer, L.-C. Wang, K.-T. Cheng, "A Efficient Sequential SAT Solver With Improved Search Stategies", UCSB Technical Report, Dec. 2003

Test Compaction for Comb. Tests

Fault simulate test patterns in reverse order of generation - ATPG patterns go first - Randomly-generated patterns go last (because they may have less coverage) - When coverage reaches 100%, drop remaining patterns (which are the useless random ones) - Could significantly shortens test sequence – reducing test application time

Automatic Test Pattern Generation for Digital Circuits, Study notes of Spanish Language

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Outline – Automatic Test Pattern

Generation

The ATPG Problem

Test Generation Systems

ATPG

Basics: Path Sensitization Method

A Simple Example

Line Justification

Completeness of ATPG Algorithms

Single Path Sensitization Is NOT

Complete

B d s.a.0^ x^ d G^6

But Boolean Difference Method

Finds Test ABCD=(1111)!!

Multiple Path Sensitization

propagated alongboth paths.

The D-Algebra

Flowchart for D-algorithm

Line Justification

A D-Algorithm

Example

Potential Problems with D-Algorithm

The PODEM Algorithm (Goel 1981)

propagated toward internal lines by theimplication.

PODEM Decision Tree for the Example

PODEM

backtrack to the most recent PI assignment and

Flowchart of PODEM

Steps in PODEM

(1) Determine aninitial objective. If the fault

Flowchart ofBacktrace

A

A

A PODEM Example

A PODEM Example – Cont’d

Step 1: Extracting the formula – Cont’d

Step 2: Satisfying the Formula

- Boolean Satisfiability

Conjunctive Normal Form (CNF)

Literal & Clause Classification

Basic Backtracking Search

ϕ = (a +c)(b +c)(¬a + ¬b + ¬c)

a b → ¬c

i.e.c must be set to 0.

Unit Clause Rule - Implications