Question

Consider a Boolean function F with the following minterm expression:
𝐹(𝑃, 𝑄, 𝑅, 𝑆) = ∑𝑚 (1, 2, 3, 4, 5, 7, 10, 12, 13, 14)
Which of the following options is/are the minimal sum-of-products expression(s)
of F ?

• A. 𝑃̅𝑆+ 𝑄𝑅̅ + 𝑃̅𝑄̅𝑅+ 𝑄̅𝑅𝑆̅
• B. 𝑃̅𝑆+ 𝑄𝑅̅ + 𝑃̅𝑄̅𝑅+ 𝑃𝑅𝑆̅
• C. 𝑃̅𝑆+ 𝑄𝑅̅ + 𝑃𝑄𝑆̅ + 𝑃𝑅𝑆̅
• D. 𝑃̅𝑆+ 𝑄𝑅̅ + 𝑃𝑄𝑆̅ + 𝑄̅𝑅𝑆̅

Hint:

First pull out the two essential prime implicants \(P'S\) (covers minterms 1,3,5,7) and \(QR'\) (covers 4,5,12,13) -- these appear in every correct option. The remaining minterms 2, 10, 14 cannot form a quad, so they must be covered by two overlapping pairs; there are two equally valid ways to pair them up, which is why two options are simultaneously correct.

Answer 1

The Correct Option is B, D

Step 1: Set up the K-map. \(F(P,Q,R,S)=\sum m(1,2,3,4,5,7,10,12,13,14)\). Take rows as \(PQ = 00,01,11,10\) and columns as \(RS = 00,01,11,10\). Placing the 1's for the given minterms gives:
Row \(PQ=00\): 0,1,1,1 (cells 0,1,3,2)
Row \(PQ=01\): 1,1,1,0 (cells 4,5,7,6)
Row \(PQ=11\): 1,1,0,1 (cells 12,13,15,14)
Row \(PQ=10\): 0,0,0,1 (cells 8,9,11,10)

Step 2: Find the essential prime implicants. The four cells \(m1,m3,m5,m7\) (all have \(P=0, S=1\)) form a quad that reduces to \(P'S\). Minterms \(m1\) and \(m7\) are covered only by this quad, so \(P'S\) is essential.
The four cells \(m4,m5,m12,m13\) (all have \(Q=1,R=0\)) form a quad that reduces to \(QR'\). Minterm \(m4\) and \(m13\) are covered only by this quad, so \(QR'\) is essential. Together \(P'S+QR'\) is common to every valid answer.

Step 3: Cover the leftover minterms 2, 10, 14. After removing the cells already covered by \(P'S\) and \(QR'\), minterms 2 (0010), 10 (1010) and 14 (1110) still need covering, and each can only be grouped in pairs (no quad is possible for them). Two different valid pairings exist:
Pairing A: \(P'Q'R\) (pairs \(m2\) with \(m3\)) together with \(PRS'\) (pairs \(m10\) with \(m14\)) covers all of 2, 10, 14.
Pairing B: \(Q'RS'\) (pairs \(m2\) with \(m10\)) together with \(PQS'\) (pairs \(m12\) with \(m14\)) also covers all of 2, 10, 14.
Both pairings use valid prime implicants (each checked to not expand into a larger group), so both give equally minimal 4-term SOP expressions.

Step 4: Match against the options.
Option 1: \(P'S+QR'+P'Q'R+Q'RS'\) -- covers \(m2\) twice but leaves \(m14\) uncovered. Invalid.
Option 2: \(P'S+QR'+P'Q'R+PRS'\) -- this is Pairing A above, covers 2, 10, 14 correctly. Valid minimal SOP.
Option 3: \(P'S+QR'+PQS'+PRS'\) -- covers \(m14\) twice but leaves \(m2\) uncovered. Invalid.
Option 4: \(P'S+QR'+PQS'+Q'RS'\) -- this is Pairing B above, covers 2, 10, 14 correctly. Valid minimal SOP.

Step 5: Conclusion. Both Option 2 and Option 4 are correct minimal sum-of-products expressions for F, since the leftover minterms 2, 10 and 14 can be legitimately grouped in two different ways while keeping the same essential terms \(P'S\) and \(QR'\).

\[\boxed{\text{Correct options: (2) and (4)}}\]