In the Boolean Algebra S

In the Boolean Algebra S - ’[d.3.3 =[a.2[d.3 = 1...

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In the Boolean Algebra S={0,1,2,3} and f:S 2 →S, we want to find BFs having the same values at these 4 points: f(2,0) = 3 ; f(2,3) = 0 ; f(1,3) = 3 ; f(3,3) = 1 Using expansion-1, the general expression is: f(x 1 ,x 2 ) = [a.x 1 .x 2 ]+[b.x 1 .x 2 ]+[c.x 1 .x 2 ]+[d.x 1 .x 2 ]. Using the 4 points: f(2,0)= [a.2 .0 ]+[b.2 .0 ]+[c.2.0 ]+[d.2.0] = [a.3]+[c.2] = 3 ………………. (1) f(2,3)= [a.2 .3 ]+[b.2 .3]+[c.2.3 ]+[d.2.3] = [b.3]+[c.2] = 0 ………………. (2) f(1,3)= [a.1 .3 ]+[b.1 .3]+[c.1.3 ]+[d.1.3] = [c.2]+[d.3] = 3 ………………. (3) f(3,3)= [a.3 .3 ]+[b.3 .3]+[c.3.3
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Unformatted text preview: ’ ]+[d.3.3] = [a.2]+[d.3] = 1 ………………. (4) From equation (2): [b.3]=0 & [c.2]=0 ise; b = {0,2}, c = {0,3}, d = {3,1} and a = {3,1} but a=3 is not satisfied in 4th equation so a=1 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....
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In the Boolean Algebra S - ’[d.3.3 =[a.2[d.3 = 1...

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