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 for an untestable fault if n->infinity = = 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.
= = 0 , is untestable by RTG ATPG. = 0 , therefore has 0 solutions.
Proof
= 1/2^n
no. of solutions = * 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.