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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 singleline 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 singleline conflicts a = a = 1 and c = c = 1 . There exists a conflicting assignment that is not covered by any singleline 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
 elbarki

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