ATPG stuff, tell me what you think

[firewireblue]

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 mentally retarded found a solution and
says it is solved, since a mentally retarded found it , it is not solved.

[firewireblue]

Maple finite precision arithmetic errors of the form exists


evalf[4000](limit((evalf[100](limit(1 - 1/2^n, n = 2000)) - 1)*2^k, k = 2000)=0

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 mentally retarded found a solution and
says it is solved, since a mentally retarded found it , it is not solved.

- - - Updated - - -

also buffer overflow errors in calculation of 2^n
take program

#include <stdio.h>

int main()
{
unsigned long long x1 = 0;
x1 = ~x1+1;
printf("%lu", x1*x1);

return 0;
}

[firewireblue]

register overflow errors in the calculation of 2^n such as
#include <stdio.h>

int main()
{
unsigned long long x1 = ((unsigned long long )1) << 32;
- hide quoted text -

printf("%lu", x1*x1);

return 0;
}


-suresh

[ThisIsNotSam]

what I think is that you are incapable of comprehending how a forum works. you are talking to yourself, mate. and it is not pretty from what I can tell.