Which function does NOT implement the Karnaugh map given below? K-Map (3 variables: x, y, z): yz x | 00 | 01 | 11 | 10 | 0 | 0 | 1 | 1 | 0 | 1 | 0 | 1 | 1 | 0 | The minterms are 1,3,5,7 (i.e., whenever z=1). A. (w + x)(y + z) B. zy + yw C. (w + x)(y + z) + y D. None of the above

GATE 2000 · Digital Logic · K Map · medium

Answer: D. None of the above

1. Read the K-map: The K-map shows 1s at yz=01 and yz=11 for both x=0 and x=1. In both columns z=1 (yz=01: y=0,z=1; yz=11: y=1,z=1). The function is F = z (1 whenever z=1, regardless of x and y). All four 1s form a single group of 4 cells, eliminating x and y, leaving only z.
2. Test each option against F = z: Option A: (w+x)(y+z) — if this K-map is actually 4 variables, we need context. But assuming the variables are just x,y,z, this option introduces w (undefined). If w=1 always, then (1+x)(y+z) = y+z != z. If w is a free variable, the expression is not well-defined for this 3-variable K-map. This seems to NOT implement z correctly. Option B: zy+yw = y(z+w) — again has variable w. For the 3-var case with only x,y,z: zy+yz = yz (redundant), which is yz != z. Actually, the GATE 2000 question uses 4 variables (w,x,y,z), and the K-map function is z. Checking: Option A: (w+x)(y+z) -- expand: wy+wz+xy+xz. Set z=1: wy+w+xy+x = w(y+1)+x(y+1) = w+x. This is NOT always 1 when z=1, so A does not implement z. Option B: zy+yw -- when z=1: y+yw=y(1+w)=y, not always 1 when z=1. Does not implement z. Option C: (w+x)(y+z)+y -- when z=1: (w+x)(y+1)+y = (w+x)+y. Not always 1. Since none implement z, the answer is D. None of the above.