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Guest

Sat Nov 30, 2019 1:51 am

When ATPG errors

suppose an ATPG errors with slight probability p

p->0

now suppose it is used to calculate untestability of a fault.

Let T be 1 if fault is testable, let T be 0 if fault is untestable.

Now, suppose we use an errorneous ATPG be T OR A, where A = and(x1,x2......x_n)

where n->infinity.

the average <T OR A> for an untestable fault if n->infinity = <T> = 0 in RTG

T = T OR A for exactly for every case except 1 case, for an untestable fault.

Now, T OR A can be solved by deterministic ATPG, T OR A = 1.

<T OR A> = <T> = 0 , is untestable by RTG ATPG. <T OR A> = 0 , therefore has 0 solutions.

Proof

<T OR A> = 1/2^n

no. of solutions = <T OR A> * 2^n = ( 1/2^n )*2^n

= lim episoln1,2->0 n->infinity (1/2^n -episoln1+episoln2)*2^n

as n->infinity select episoln1=1/2^n

= (0 +episoln2)*2^n

Select episoln2=0, such that episoln2*2^n =0

= 0

no. of solutions = 0;

T OR A has 1 solution by deterministic ATPG.

Therefore

solutions= 0 = 1

Suppose T is the output T + 0 = T + solutions = T + 1, if T is 0 = 1, if T- 0 = = T - solutions = T - 1 if T is 1 = 0

untestable is testable, testable is untestable!

Such effects may be heuristically observable.

ATPG will remain unsolved

Suppose a cripple found a solution and

says it is solved, since a cripple found it , it is not solved.

Guest

Sat Nov 30, 2019 1:53 am

On Friday, November 29, 2019 at 6:51:08 PM UTC-5, const...@gmail.com wrote:

When ATPG errors

suppose an ATPG errors with slight probability p

p->0

now suppose it is used to calculate untestability of a fault.

Let T be 1 if fault is testable, let T be 0 if fault is untestable.

Now, suppose we use an errorneous ATPG be T OR A, where A = and(x1,x2......x_n)

where n->infinity.

the average <T OR A> for an untestable fault if n->infinity = <T> = 0 in RTG

T = T OR A for exactly for every case except 1 case, for an untestable fault.

Now, T OR A can be solved by deterministic ATPG, T OR A = 1.

T OR A> = <T> = 0 , is untestable by RTG ATPG. <T OR A> = 0 , therefore has 0 solutions.

Proof

T OR A> = 1/2^n

no. of solutions = <T OR A> * 2^n = ( 1/2^n )*2^n

= lim episoln1,2->0 n->infinity (1/2^n -episoln1+episoln2)*2^n

as n->infinity select episoln1=1/2^n

= (0 +episoln2)*2^n

Select episoln2=0, such that episoln2*2^n =0

= 0

no. of solutions = 0;

T OR A has 1 solution by deterministic ATPG.

Therefore

solutions= 0 = 1

Suppose T is the output T + 0 = T + solutions = T + 1, if T is 0 = 1, if T- 0 = = T - solutions = T - 1 if T is 1 = 0

untestable is testable, testable is untestable!

Such effects may be heuristically observable.

ATPG will remain unsolved

Suppose a cripple found a solution and

says it is solved, since a cripple found it , it is not solved.

suppose an ATPG errors with slight probability p

p->0

now suppose it is used to calculate untestability of a fault.

Let T be 1 if fault is testable, let T be 0 if fault is untestable.

Now, suppose we use an errorneous ATPG be T OR A, where A = and(x1,x2......x_n)

where n->infinity.

the average <T OR A> for an untestable fault if n->infinity = <T> = 0 in RTG

T = T OR A for exactly for every case except 1 case, for an untestable fault.

Now, T OR A can be solved by deterministic ATPG, T OR A = 1.

T OR A> = <T> = 0 , is untestable by RTG ATPG. <T OR A> = 0 , therefore has 0 solutions.

Proof

T OR A> = 1/2^n

no. of solutions = <T OR A> * 2^n = ( 1/2^n )*2^n

= lim episoln1,2->0 n->infinity (1/2^n -episoln1+episoln2)*2^n

as n->infinity select episoln1=1/2^n

= (0 +episoln2)*2^n

Select episoln2=0, such that episoln2*2^n =0

= 0

no. of solutions = 0;

T OR A has 1 solution by deterministic ATPG.

Therefore

solutions= 0 = 1

Suppose T is the output T + 0 = T + solutions = T + 1, if T is 0 = 1, if T- 0 = = T - solutions = T - 1 if T is 1 = 0

untestable is testable, testable is untestable!

Such effects may be heuristically observable.

ATPG will remain unsolved

Suppose a cripple found a solution and

says it is solved, since a cripple found it , it is not solved.

-suresh

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