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Unformatted text preview: 204 VLSI Test Principles and Architectures line assignments b = b = 1 and c = c = 1 as b = b = 1 c = and c = c = 1 a = 1 b = 1 , respectively. Note that there exist other sets of impossible value combinations not covered by any of these three single-line conflicts. Not all remaining conflicting combinations are nontrivial. For example, consider the conflicting scenario a = c = 1 . This is a trivial value conflict because a = → c = and c = 1 → a = 1 . There- fore, a = c = 1 is already covered by the single-line conflicts a = a = 1 and c = c = 1 . There exists a conflicting assignment that is not covered by any single-line conflicts: a = 1 b = 1 c = 0 . In order to compute the corresponding impossible value assignment set, it is necessary to compute the following implications: impl a = impl b = 0 , and impl c = 1 . By traversing the implication edges in the graph, the impossible value assignment set a = b = c = c = 1 a = 1 b = 1 is obtained. This set has not been covered in any of the previous impossible valueobtained....
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- Spring '08