A counter is constructed with three D flip-flops. The input-output pairs are named (D₀, Q₀), (D₁, Q₁), and (D₂, Q₂), where the subscript 0 denotes the least significant bit. The output sequence is desired to be the Gray-code sequence 000, 001, 011, 010, 110, 111, 101, and 100, repeating periodically. Note that the bits are listed in the Q₂ Q₁ Q₀ format. The combinational logic expression for D₁ is

The correct option is Q̅₂ Q₀ + Q₁ Q̅₀.

Let's derive this step-by-step by constructing the state transition table for the counter.
A D flip-flop takes the next-state value directly from its input, which means for each flip-flop, the input $D_i$ is equal to the next state $Q_i^{\text{next}}$.

The counter follows a repeating periodic Gray-code sequence specified as:
000 → 001 → 011 → 010 → 110 → 111 → 101 → 100 → (back to 000)

We represent this sequence in terms of the current state $(Q_2,Q_1,Q_0)$ and the next state $(Q_2^{\text{next}},Q_1^{\text{next}},Q_0^{\text{next}})$:
Since $D_1=Q_1^{\text{next}}$, we focus on finding the value of $Q_1^{\text{next}}$ for each state:

• For current state 000, next state is 001, so $D_1=0$.
• For current state 001, next state is 011, so $D_1=1$.
• For current state 011, next state is 010, so $D_1=1$.
• For current state 010, next state is 110, so $D_1=1$.
• For current state 110, next state is 111, so $D_1=1$.
• For current state 111, next state is 101, so $D_1=0$.
• For current state 101, next state is 100, so $D_1=0$.
• For current state 100, next state is 000, so $D_1=0$.

Now, let's list the values of $D_1$ corresponding to the minterms of $(Q_2,Q_1,Q_0)$:
• $m_0$ (000): $D_1=0$
• $m_1$ (001): $D_1=1$
• $m_2$ (010): $D_1=1$
• $m_3$ (011): $D_1=1$
• $m_4$ (100): $D_1=0$
• $m_5$ (101): $D_1=0$
• $m_6$ (110): $D_1=1$
• $m_7$ (111): $D_1=0$

Thus, $D_1=\sum m(1,2,3,6)$.

Let's simplify this expression using a Karnaugh map (K-map) for $D_1$ with variables $Q_2$ (rows) and $Q_1{Q}_0$ (columns):

$\begin{array}{ccccc}{Q}_2\backslash Q_1{Q}_0& 00& 01& 11& 10\\ 0& 0& 1& 1& 1\\ 1& 0& 0& 0& 1\end{array}$

We can group the 1s to find the simplified SOP expression:
1. Group 1: A pair formed by $m_1$ (001) and $m_3$ (011). This group lies in the row where $Q_2=0$ and columns where $Q_0=1$. This gives the term Q̅₂ Q₀.
2. Group 2: A pair formed by $m_2$ (010) and $m_6$ (110). This group lies in columns where $Q_1=1$ and $Q_0=0$, spanning both rows of $Q_2$. This gives the term Q₁ Q̅₀.

Combining these terms, the logic expression for $D_1$ is:
$D_1={\overline{Q}}_2{Q}_0+Q_1{\overline{Q}}_0$