Consider a finite state machine (FSM) with one input \(X\) and one output…

Final answer: 5 states

Explanation:

1. Recognize the machine type: outputs depend on current state and input, so this is a Mealy machine. Equivalent states must have identical outputs for each input and their next states must also be equivalent.

2. Initial partition by output pairs (output for X=0, X=1):

   States with outputs (0,0): A, B, C, E. States with outputs (1,0): D, H. State with outputs (1,1): F. State with outputs (0,1): G.

3. Compute next-state transitions (from the table): A: X=0→F, X=1→B; B: X=0→D, X=1→C; C: X=0→F, X=1→E; D: X=0→G, X=1→A; E: X=0→D, X=1→C; F: X=0→F, X=1→B; G: X=0→G, X=1→H; H: X=0→G, X=1→A.

4. Refine the partition using these transitions (map next states to the current groups). The group {A,B,C,E} splits because A and C transition to F then to states in the (0,0) group differently than B and E. After refinement the groups become: {A, C}, {B, E}, {D, H}, {F}, {G}.

5. Check for further splits: within each refined group all states have identical output behavior and next-state mappings (relative to the refined groups), so no further splitting is needed.

6. Therefore the minimal number of states required to realize this Mealy FSM is 5.