30 An Introduction to Logic Circuit Testing Figure 2.13: Multiple backtracks along and H−E−C H−G−F−C. done via both and the at can be set so that the value at is justified. In PODEM, H−E−C H−G−F−C, C H a logic value assigned to a primary input in order to achieve one objective may in turn result in the failure of satisfying another objective, thereby forcing a backtrack. We illustrate the application of the FAN algorithm by deriving a test for the fault s-a-0 in Figure Z 2.14. First, the value is assigned to the line and the value 1 to each of the inputs and The initial D Z M N. objectives are to set and to 1. By the multiple backtrack, and are assigned 1 (note that instead of M N G I G and could be assigned logic 1). Again, by the multiple backtrack, we have the final objectives I, L A=l, B=l and The assignment makes and the assignment makes E=1, F=l. A=1, B=l J=1, M=1, E=1, F=1 I=1, N=1. Thus, the assignments constitute a test for the fault s-a-0. It is easy to see that if the first A=B=E=F=1 Z multiple backtracks stopped at and the second multiple backtrack at the test for the fault would be L H, C=D=1.
30 An Introduction to Logic Circuit Testing Figure 2.13: Multiple backtracks along H−E−C and H−G−F−C. done via both H−E−C and H−G−F−C, the at C can be set so that the value at H is justified. In PODEM, a logic value assigned to a primary input in order to achieve one objective may in turn result in the failure of satisfying another objective, thereby forcing a backtrack. We illustrate the application of the FAN algorithm by deriving a test for the fault Z s-a-0 in Figure 2.14. First, the value D is assigned to the line Z and the value 1 to each of the inputs M and N. The initial objectives are to set M and N to 1. By the multiple backtrack, G and I are assigned 1 (note that instead of G and I, L could be assigned logic 1). Again, by the multiple backtrack, we have the final objectives A=l, B=l and E=1, F=l. The assignment A=1, B=l makes J=1, M=1, and the assignment E=1, F=1 makes I=1, N=1. Thus, the assignments A=B=E=F=1 constitute a test for the fault Z s-a-0. It is easy to see that if the first multiple backtracks stopped at L and the second multiple backtrack at H, the test for the fault would be C=D=1.