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Consider a Boolean function $f(w,x,y,z)$ such that $$\begin{array}{lll} f(w,0,0,z) & = & 1 \\ f(1,x,1,z) & =& x+z \\  f(w,1,y,z) & = & wz +y \end{array}$$The number of literals in the minimal sum-of-products expression of $f$ is _________
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We can see $3$ squares, so number of literals $=3\times2 = 6.$

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To Understand the question :

Over boolean variables $w,x,y,z,$ we have been given the definition of a Boolean function $f.$

Over $4 $ boolean variables, we have $2^4 = 16$ different combinations (rows in truth table of $f$), and the value of function $f$ is given using three equations.

From equation $1,$ $i.e. f(w,0,0,z) = 1$ , We can find the value of  function $f$ to be $1$ for the combinations/rows in which $x=y=0.$

So,  $f(0,0,0,0) = 1 ;$ $f(0,0,0,1) = 1 ;$ $f(1,0,0,0) = 1 ;$ $f(1,0,0,1) = 1 ;$

From equation $2,$ $i.e. f(1,x,1,,z) = x+z$ , We can find the value of  function $f$ to be $x+z$ for the combinations/rows in which $w=y=1.$

So,  $f(1,0,1,0) = 0+0 = 0 ;$ $f(1,0,1,1) = 0+1 = 1;$ $f(1,1,1,0) = 1+0 = 1;$ $f(1,1,1,1) = 1+1 = 1;$

From equation $3,$ $i.e. f(w,1,y,z) = wz+y$ , We can find the value of  function $f$ to be $wz+y$ for the combinations/rows in which $x=1.$

So,  $f(0,1,1,0) = 0.0+1 = 1 ;$ and so on.

NOTE that from the given definition of function $f,$ we cannot find the value of function $f$ for the two combinations $0011,0010.$ Hence, we consider $(0011) , (0010) $ as “Don’t Cares Combinations” for which the value of function $f$ is “Don’t Care”.

Once understood the question, make K-map, and do minimization.

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let’s convert these equations into miniterms 

Equation 1 | x’y’

Equation 2 | wy(x+z)

Equation 3 | x(wz+y)

Adding all three together we have x’y’+wxy+wyz+wxz+xy. 2nd term is redundant because of the last term.

by removing and reordering we have xy+x’y’+wz(x+y)

 Now, if both x and y are false function equates true, if both false again true, only case remaining is either one of them is true in that case output is dependent on wz

so the final miniterms are xy+x’y’+wz – In total we have 6 literals

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