Unformatted text preview: d = {3,1} and (3) gives c = {0,3}. c = {0,3} and (2) gives b = {0,2} Hence, solution: a=1, b={0,2}, c= {0,3}, d= {1,3} The corresponding function numbers: (1001) 4 =65 (1003) 4 =67 (1031) 4 =77 (1033) 4 =79 (1201) 4 =97 (1203) 4 =99 (1231) 4 =109 (1233) 4 =111 which verifies ryavuz2002.exe. results: USING CHOICE 3 and f(2,0)=3 ; f(2,3)=0 ; f(1,3)=3 ; f(3,3)=1 ; the same functions are obtained. Problem (if your faculty number is odd): Use expansion1: f=ax 1 ′x 2 ′+b x 1 ′x 2 + cx 1 x 2 ′ + dx 1 x 2 obtain 4 equations and solve them. Problem (if your faculty number is even): Use expansion3: f=a ⊕ bx 1 ⊕ cx 2 ⊕ dx 1 x 2 obtain 4 equations and solve them...
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 Spring '04
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 Expression, Boolean Algebra S=

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