Consider a Boolean function f(w, x, y, z). Suppose exactly one input is…
2006
Consider a Boolean function f(w, x, y, z). Suppose exactly one input is allowed to change at a time. If f is true for two input vectors i1 = (w1, x1, y1, z1) and i2 = (w2, x2, y2, z2), and i1 and i2 differ in exactly one bit position, we want f to remain true during the transition without becoming false momentarily. Let f(w, x, y, z) = ∑(5, 7, 11, 12, 13, 15). Which of the following cube covers of f ensures this required property?
- A.
w'xz, wxy', xy'z, xyz,wyz
- B.
wxy,w'xz,wyz
- C.
wx(yz)', xz, wx'yz
- D.
wxy', wyz, wxz, w'xz, xy'z, xyz
Attempted by 44 students.
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Correct answer: D
For a static-1-hazard-free SOP/cube cover, every pair of 1-minterms that differ in exactly one bit must be covered by at least one common cube.
The true minterms are:
5 = 0101, 7 = 0111, 11 = 1011, 12 = 1100, 13 = 1101, 15 = 1111.
Adjacent 1-minterm pairs are:
(5,7), (5,13), (7,15), (11,15), (12,13), and (13,15).
Check Option D:
w'xz covers (5,7).
xy'z covers (5,13).
xyz covers (7,15).
wyz covers (11,15).
wxy' covers (12,13).
wxz covers (13,15).
Thus, every adjacent pair of true minterms is covered by a common cube, so the function remains true during any allowed single-bit transition.
Why Option A is not sufficient:
Option A covers (5,7), (5,13), (7,15), (11,15), and (12,13), but it does not cover the adjacent pair (13,15) with a common cube. The missing cube is wxz. Hence Option A can still have a static-1 hazard.
Therefore, the correct cube cover is Option D.