Arithmetic Test Pattern Generation: A Bit Level Formulation of the

Optimization Problem

S. Manich, L. García and J. Figueras

Central idea• Embed a set of Test Vectors into a single adder test

pattern generator.

• Minimize length of test pattern sequence

Central idea• Embed a set of Test Vectors into a single adder test

pattern generator.

• Minimize length of test pattern sequence

Seed

Incr

emen

t

Clock i = 0,1,···,k

Load seed at preset

Testpatterns

Embedding TV in -positive numbers

• Apply Euclid’s algorithm to select GCD

Embedding TV in -positive numbers

Change 43 1

Embedding TV in -positive numbers

Exploiting GF(26), addition overflow

Exploiting GF(26), addition overflow

• Computation of GCD (Euclid’s algorithm)

does not properly minimize test sequence

length in GF(2n).

Minimizing length in GF(2n)

• How to compute the length?

0···0

TV

I I Length TVLength < 2n

Solving the length equation• Iterating up to a maximum of

n steps

?

IAt preset 0···0

+

n bits

n bi

ts

n bits

Clock

Shift 1 at clock

Shi

ft 1

at c

lock

TV

Len

gth

I Length TVLength < 2n

Example

I

+

n bits

n bi

ts

n bits

TV

Length101001

0010010

00

00

00

037

36

0 0 0 0 0 0

Example

I

+

n bits

n bi

ts

n bits

TV

Length101001

0010010

00

00

00

037

36

0 0 0 0 0 0

Example

I

+

n bits

n bi

ts

n bits

TV

Length101001

0010010

00

00

01

037

36

0 0 0 0 0 0

Example

I

+

n bits

n bi

ts

n bits

TV

Length101001

0010010

01

00

00

437

36

0 1 0 1 0 0

Example

I

+

n bits

n bi

ts

n bits

TV

Length101001

0010010

01

00

01

437

36

0 1 0 1 0 0

Example

I

+

n bits

n bi

ts

n bits

TV

Length101001

0010010

01

01

0 0

2037

36

1 0 0 1 0 0

Example

I

n bits

n bi

ts

n bits

TV

Length101001

0010010

01

01

0

2037

36

1 0 0 1 0 0

Let several TV = {V0, ···, VM-1}

• Assume V0 is the seed.

• Find increment I such that the length to generate all

test vectors is minimized.

I L1 (V1V0)

I L2 (V2V0)

I LM-1 (VM-1V0)···I?

max Length of test sequence

The optimization of the equations give I such that Length is minimized

Extract the loop out of the equations

• In order to isolate Li and make the optimization possible.

I

+

n bitsn

bits

n bits

TV

LengthTV

= Y

TV= 1

0 0 0 1

n bi

ts

n bits

n bits

TV

Length

Extract the loop out of the equations

• In order to isolate Li and make the optimization possible.

Length TV Y

Y I

1

I Y 1

Rewritting equations

Y (V1V0) L1

Y (V2V0) L2

Y (VM-1V0) LM-1

···Y?

max Length of test sequenceThe optimization of the equations give Y such that Length is minimized

Circuital implementation

(V1V0) (VM1V0)

Y

(mod 2n)

L1 LM1

MAXLength of test sequence

(mod 2n)

Optimizing binary search

0 1 0 1 ··· 1

Before evaluating, load 0

YES

NO

After evaluation,write 0

After evaluation, write 1

After ENDminimal test sequenceLength and optimal Y

(V1V0) (VM1V0)

Y

(mod 2n)

L1 LM1

MAXLength of test sequence

(mod 2n)

Optimal Igiven seed V0

Example: TV = {0,59,60,61,62}

1 1 1 1 1 1

1

59

(mod 26)

59

MAX

62

62

(mod 26)61

61

(mod 26)60

60

(mod 26)

YES

62 = 1 1 1 1 1 0

= 63

Example: TV = {0,59,60,61,62}

1

59

(mod 26)

59

62

62

(mod 26)61

61

(mod 26)60

60

(mod 26)

0 1 1 1 1 1

62 = 1 1 1 1 1 0

NO

MAX

= 31

Example: TV = {0,59,60,61,62}

1

59

(mod 26)

59

62

62

(mod 26)61

61

(mod 26)60

60

(mod 26)

0 1 1 1 1 1

62 = 1 1 1 1 1 0

NO

MAX

= 31

Example: TV = {0,59,60,61,62}

59

59

(mod 26)

25

62

10

(mod 26)61

15

(mod 26)60

20

(mod 26)

0 1 1 1 1 1

25 = 0 1 1 0 0 1

MAX

= 31

YES

Example: TV = {0,59,60,61,62}

59

59

(mod 26)

25

62

10

(mod 26)61

15

(mod 26)60

20

(mod 26)

25 = 0 1 1 0 0 1

MAX

= 15NO

0 0 1 1 1 1

Example: TV = {0,59,60,61,62}

59

59

(mod 26)

25

62

10

(mod 26)61

15

(mod 26)60

20

(mod 26)

25 = 0 1 1 0 0 1

MAX

= 15NO

0 0 1 1 1 1

Example: TV = {0,59,60,61,62}

61

59

(mod 26)

15

62

6

(mod 26)61

9

(mod 26)60

12

(mod 26)

15 = 0 0 1 1 1 1

MAX

= 150 0 1 1 1 1

YES

Example: TV = {0,59,60,61,62}

61

59

(mod 26)

15

62

6

(mod 26)61

9

(mod 26)60

12

(mod 26)

15 = 0 0 1 1 1 1

MAX

= 70 0 0 1 1 1NO

Example: TV = {0,59,60,61,62}

61

59

(mod 26)

15

62

6

(mod 26)61

9

(mod 26)60

12

(mod 26)

15 = 0 0 1 1 1 1

MAX

= 70 0 0 1 1 1NO

Example: TV = {0,59,60,61,62}

63

59

(mod 26)

5

62

2

(mod 26)61

3

(mod 26)60

4

(mod 26)

5 = 0 0 0 1 0 1

MAX

= 70 0 0 1 1 1

YES

Example: TV = {0,59,60,61,62}

63

59

(mod 26)

5

62

2

(mod 26)61

3

(mod 26)60

4

(mod 26)

5 = 0 0 0 1 0 1

MAX

= 30 0 0 0 1 1NO

Example: TV = {0,59,60,61,62}

63

59

(mod 26)

5

62

2

(mod 26)61

3

(mod 26)60

4

(mod 26)

5 = 0 0 0 1 0 1

MAX

= 30 0 0 0 1 1NO

Example: TV = {0,59,60,61,62}

63

59

(mod 26)

5

62

2

(mod 26)61

3

(mod 26)60

4

(mod 26)

5 = 0 0 0 1 0 1

MAX

= 70 0 0 1 1 1NO

Example: TV = {0,59,60,61,62}

63

59

(mod 26)

5

62

2

(mod 26)61

3

(mod 26)60

4

(mod 26)

5 = 0 0 0 1 0 1

MAX

= 50 0 0 1 0 1

YES

Example: TV = {0,59,60,61,62}

63

59

(mod 26)

5

62

2

(mod 26)61

3

(mod 26)60

4

(mod 26)

5 = 0 0 0 1 0 1

MAX

= 50 0 0 1 0 1

YES

Example: TV = {0,59,60,61,62}

63

59

(mod 26)

5

62

2

(mod 26)61

3

(mod 26)60

4

(mod 26)

5 = 0 0 0 1 0 1

MAX

= 40 0 0 1 0 0NO

Example: TV = {0,59,60,61,62}

63

59

(mod 26)

5

62

2

(mod 26)61

3

(mod 26)60

4

(mod 26)

5 = 0 0 0 1 0 1

MAX

= 40 0 0 1 0 0NO

Example: TV = {0,59,60,61,62}

63

59

(mod 26)

5

62

2

(mod 26)61

3

(mod 26)60

4

(mod 26)

5 = 0 0 0 1 0 1

MAX

= 50 0 0 1 0 1NO

Example: TV = {0,59,60,61,62}

63

59

(mod 26)

5

62

2

(mod 26)61

3

(mod 26)60

4

(mod 26)

5 = 0 0 0 1 0 1

MAX

Minimum test sequence length

Example: TV = {0,59,60,61,62}

63 = Y

59

(mod 26)

5

62

2

(mod 26)61

3

(mod 26)60

4

(mod 26)

5 = 0 0 0 1 0 1

MAX

Minimum test sequence length

I = 63

I Y 1

Example: TV = {0,59,60,61,62}

63 = Y

59

(mod 26)

5

62

2

(mod 26)61

3

(mod 26)60

4

(mod 26)

5 = 0 0 0 1 0 1

MAX

Minimum test sequence length

I = 63 Optimal increment

Example: TV = {0,59,60,61,62}

How to evaluate max(Length)i = 0

(mod 26)

(mod 26)

(mod 26)

(mod 26)

MAX

OR 0

0 0 0 0 1 1

+ ++0

+0

+0

+0

How to evaluate max(Length)i = 0

• Place Stuck-1 at output.

• Detect fault using ATPG methodologies:– Approximate methods.– SAT techniques optimal solution.

Cost of the methodology

• SAT problem is -complete.

• For our problem: cost indicator probability of finding

a max-term at random, number of bits (nstable) and

number of test vectors (Mdown).

M-1n-1

(M-1)n

1 1P[MT]= 2 +

22

1E-10

1E-09

1E-08

1E-07

1E-06

1E-05

0,0001

0,001

0,01

0,1

1

0 5 10 15 20 25 30 35

Graphically

P[MT] n (number of bits)

M (number of test vectors to embed)

Critical boundary

Critical boundary

• MTmax 1

n M -1 n M -1s820 23 24 c3540 50 51

s1196 31 32 s641 54 55c6288 32 33 c880 60 61s1238 32 33 s838 66 67c1908 33 34 s1423 91 92s420 34 35 c5315 178 179c432 36 37 c7552 207 208c499 41 42 s5378 214 215

c1355 41 42 c2670 233 234s953 45 46 s9234 247 248

1E-10

1E-09

1E-08

1E-07

1E-06

1E-05

0,0001

0,001

0,01

0,1

1

0 5 10 15 20 25 30 35

Cost: step i

MSBs (0...0

)

step i

MSBs (1...1)

1E-10

1E-09

1E-08

1E-07

1E-06

1E-05

0,0001

0,001

0,01

0,1

1

0 5 10 15 20 25 30 35

Restriction: (0...01...10...01...10...0) step i

Conclusions

• Minimization of test sequence length for embedding

of test vectors in a single arithmetic additive

generator is defined.

• The cost of the methodology is estimated.

• The methodology allows embedding of partially

defined test vectors.

• Theoretic background of methodology gives

strategies to improve existing heuristic methods

based on multiple-arithmetic generators (re-loading).