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# sol6 - Chapter 6 6.1 w3 w2 w1 w0 w1 w2 y0 y1 y2 y3 y4 y5 y6...

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Chapter 6 6.1. w 0 En y 0 w 1 y 1 y 2 y 3 y 7 y 6 y 5 y 4 w 2 f 1 w 1 w 2 w 3 6.2. w 0 En y 0 w 1 y 1 y 2 y 3 y 7 y 6 y 5 y 4 w 2 f 1 w 1 w 2 w 3 6.3. 0 0 0 1 1 0 1 1 1 0 1 1 0 0 0 1 1 0 1 1 0 0 1 0 w 1 w 2 w 3 f 0 0 0 0 1 1 1 1 0 1 f w 1 w 2 w 3 + w 2 w 3 f w 3 w 1 w 2 6-1

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6.4. 0 0 0 1 1 0 1 1 1 0 0 0 0 0 0 1 1 0 1 1 1 0 1 1 w 1 w 2 w 3 f 0 0 0 0 1 1 1 1 0 1 f w 1 w 2 w 3 w 2 w 3 + f w 3 w 1 w 2 6.5. The function f can be expressed as f = w 1 w 2 w 3 + w 1 w 2 w 3 + w 1 w 2 w 3 + w 1 w 2 w 3 Expansion in terms of w 1 produces f = w 1 ( w 2 + w 3 ) + w 1 ( w 2 w 3 ) The corresponding circuit is f w 2 w 1 w 3 6.6. The function f can be expressed as f = w 1 w 2 w 3 + w 1 w 2 w 3 + w 1 w 2 w 3 + w 1 w 2 w 3 Expansion in terms of w 2 produces f = w 2 ( w 3 ) + w 2 ( w 1 ) The corresponding circuit is f w 2 w 3 w 1 6-2
6.7. Expansion in terms of w 2 gives f = w 2 (1 + w 1 w 3 + w 1 w 3 ) + w 2 ( w 1 w 3 + w 1 w 3 ) = w 1 w 2 w 3 + w 1 w 2 w 3 + w 2 + w 1 w 2 w 3 + w 1 w 2 w 3 Further expansion in terms of w 1 gives f = w 1 ( w 2 w 3 + w 2 w 3 + w 2 ) + w 1 ( w 2 w 3 + w 2 w 3 + w 2 ) = w 1 w 2 w 3 + w 1 w 2 w 3 + w 1 w 2 + w 1 w 2 w 3 + w 1 w 2 w 3 + w 1 w 2 Further expansion in terms of w 3 gives f = w 3 ( w 1 w 2 + w 1 w 2 + w 1 w 2 + w 1 w 2 ) + w 3 ( w 1 w 2 + w 1 w 2 + w 1 w 2 + w 1 w 2 ) = w 1 w 2 w 3 + w 1 w 2 w 3 + w 1 w 2 w 3 + w 1 w 2 w 3 + w 1 w 2 w 3 + w 1 w 2 w 3 6.8. Expansion in terms of w 1 gives f = w 1 w 2 + w 1 w 3 + w 1 w 2 Further expansion in terms of w 2 gives f = w 2 ( w 1 w 3 ) + w 2 ( w 1 + w 1 + w 1 w 3 ) = w 1 w 2 + w 1 w 2 w 3 + w 1 w 2 w 3 + w 1 w 2 Further expansion in terms of w 3 gives f = w 3 ( w 1 w 2 + w 1 w 2 + w 1 w 2 + w 1 w 2 ) + w 3 ( w 1 w 2 + w 1 w 2 ) = w 1 w 2 w 3 + w 1 w 2 w 3 + w 1 w 2 w 3 + w 1 w 2 w 3 + w 1 w 2 w 3 6.9. Proof of Shannon’s expansion theorem f ( x 1 , x 2 , ..., x n ) = x 1 · f (0 , x 2 , ..., x n ) + x 1 · f (1 , x 2 , ..., x n ) This theorem can be proved using perfect induction , by showing that the expression is true for every possible value of x 1 . Since x 1 is a boolean variable, we need to look at only two cases: x 1 = 0 and x 1 = 1 . Setting x 1 = 0 in the above expression, we have: f (0 , x 2 , ..., x n ) = 1 · f (0 , x 2 , ..., x n ) + 0 · f (1 , x 2 , ..., x n ) = f (0 , x 2 , ..., x n ) Setting x 1 = 1 , we have: f (1 , x 2 , ..., x n ) = 0 · f (0 , x 2 , ..., x n ) + 1 · f (1 , x 2 , ..., x n ) = f (1 , x 2 , ..., x n ) This proof can be performed for any arbitrary x i in the same manner.

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