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

To design the counter, we first list the desired state transition sequence. The outputs of the three D flip-flops represent the state variables $Q_2$, $Q_1$, and $Q_0$, where the bits are given in the order $Q_2{Q}_1{Q}_0$.

For a D flip-flop, the next-state equation is given by:
$Q^{\text{next}}=D$
Therefore, the input $D_1$ to the second flip-flop must be equal to the next-state value of $Q_1$ (i.e., $Q_1^{\text{next}}$).

Let's map each current state ($Q_2{Q}_1{Q}_0$) to its corresponding next state ($Q_2^{\text{next}}{Q}_1^{\text{next}}{Q}_0^{\text{next}}$) based on the given periodic sequence:
1. Current State: 000 ⇒ Next State: 001 (so $D_1=Q_1^{\text{next}}=0$)
2. Current State: 001 ⇒ Next State: 011 (so $D_1=Q_1^{\text{next}}=1$)
3. Current State: 011 ⇒ Next State: 010 (so $D_1=Q_1^{\text{next}}=1$)
4. Current State: 010 ⇒ Next State: 110 (so $D_1=Q_1^{\text{next}}=1$)
5. Current State: 110 ⇒ Next State: 111 (so $D_1=Q_1^{\text{next}}=1$)
6. Current State: 111 ⇒ Next State: 101 (so $D_1=Q_1^{\text{next}}=0$)
7. Current State: 101 ⇒ Next State: 100 (so $D_1=Q_1^{\text{next}}=0$)
8. Current State: 100 ⇒ Next State: 000 (so $D_1=Q_1^{\text{next}}=0$)

We compile these values into a truth table to find the logic function for $D_1$ as a function of the current states ($Q_2,Q_1,Q_0$):

From the table, the minterms for which $D_1=1$ correspond to the following current states:
- $Q_2{Q}_1{Q}_0=001$ (minterm 1)
- $Q_2{Q}_1{Q}_0=011$ (minterm 3)
- $Q_2{Q}_1{Q}_0=010$ (minterm 2)
- $Q_2{Q}_1{Q}_0=110$ (minterm 6)

We can simplify this logic expression using a Karnaugh map (K-map):

Let's identify the prime implicants to group the 1s:
1. **First Group**: Grouping the two adjacent 1s in the first row ($Q_2=0$) at columns $Q_1{Q}_0=01$ and $Q_1{Q}_0=11$. This eliminates $Q_1$ and yields:
${\bar{Q}}_2{Q}_0$
2. **Second Group**: Grouping the adjacent 1s at column $Q_1{Q}_0=10$ across rows $Q_2=0$ and $Q_2=1$. This eliminates $Q_2$ and yields:
$Q_1{\bar{Q}}_0$

Combining these two groups gives the minimized sum-of-products expression for the combinational logic input $D_1$:
$D_1={\bar{Q}}_2{Q}_0+Q_1{\bar{Q}}_0$