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16 must be the answer. Both of the (0,0,0) & (1,0,1) will be mapped to either 0 or 1. Both of the (0,0,1) & (1,0,0) will be mapped to either 0 or 1. Both of the (0,1,0) & (1,1,1) will be mapped to either 0 or 1. Both of the (0,1,1) & (1,1,0) will be mapped to either 0 or 1. 4 pairs having 2 choices each, thus 4^2 = 16.

0

Since f is a 3 input boolean function, its domain is all possible 0, 1 combinations that can be applied to x, y, z simultaneously.so input must be in the form of an ordered triplet(x, y, z). So total 8(=2^3) such triplets are possible. Output of a Boolean function must be either 0 or 1. Now we can not map all these input triplets independently to 0 or 1 since we have a constraint that triplets (x, y, z) & (x', y, z') must produce same output, that is they must be mapped to the same element. Since we have 4 such triplet pairs( of the form {(x, y, z), (x', y, z')} & since each of them can independently produce either 0 or 1, so 2x2x2x2 = 16.

0 votes

f(x',y,z') = f(x,y,z) means y remains same and the for x z values the function does`nt have any effect .

so in kmap for 2 3 6 7 y is 1 . and the remaining 0 1 4 5 we put dontcare . so in the function 2 3 6 7 will be there but all these dont care have 2 choices either its present or not present . 2^{4} = 16 possibilities . ( i may be wrong :D )

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