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# p1.4 - 2 n but in this case we have replaced all AND gates...

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Daniel Mihai Problem 1.4 Binary comparator for 3 1-bit numbers

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a 1 b 1 c 1 . . . . a n b n c n a n . . . . . a 2 a 1 a n . . . . . a 2 a 1 x 1 . . . . x n Binary comparator for 3 n-bit numbers n-ary AND
a b The depth of the circuit is 7+2*log 2 n because a a n-ary AND made from AND gates has depth log

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Unformatted text preview: 2 n, but in this case, we have replaced all AND gates with 2 NAND gates. For the second part, we first create a Binary comparator for 2 1-bit numbers a 1 b 1 . . a n b n c 1 b 1 . . c n b n a 1 c 1 . . a n c n The circuit has now depth 11+2*log 2 n....
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p1.4 - 2 n but in this case we have replaced all AND gates...

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