Moore machine vs Mealy machine in Verilog - Performance Comparison
We want to see how the time to run Moore and Mealy machines changes as inputs grow.
How does the design affect the number of steps the machine takes?
Analyze the time complexity of the following Verilog Moore and Mealy machine snippets.
// Moore machine example
always @(posedge clk) begin
state <= next_state;
output_signal <= state_output;
end
// Mealy machine example
always @(posedge clk) begin
state <= next_state;
output_signal <= output_logic(input_signal, state);
end
These snippets show how outputs depend on state only (Moore) or state and input (Mealy).
Both machines update state and output every clock cycle.
- Primary operation: State update and output assignment on each clock edge.
- How many times: Once per clock cycle, repeated for each input event.
Each input triggers one state and output update, so operations grow linearly with input size.
| Input Size (n) | Approx. Operations |
|---|---|
| 10 | 10 state and output updates |
| 100 | 100 state and output updates |
| 1000 | 1000 state and output updates |
Pattern observation: Operations increase directly with input size, no extra loops inside.
Time Complexity: O(n)
This means the time grows in a straight line as inputs increase, one step per input.
[X] Wrong: "Mealy machines are always faster because outputs change immediately."
[OK] Correct: Both machines update outputs every clock cycle, so speed depends on clock, not output logic style.
Understanding how Moore and Mealy machines handle inputs helps you explain timing and design choices clearly in real projects.
"What if the Mealy machine had multiple inputs changing asynchronously? How would that affect time complexity?"