State count of a Johnson (twisted-ring) counter In a 6-bit Johnson counter, how many distinct states (bit patterns) are produced in the complete sequence?

Introduction / Context:A Johnson counter (twisted-ring) is a shift register with complemented feedback. Its key property is that an n-stage Johnson counter cycles through 2n unique states, making it efficient for sequence generation and timing signals.

Given Data / Assumptions:

• Number of stages n = 6.
• Inverted feedback from last stage to input.
• Clocking advances the pattern by one bit each cycle.

Concept / Approach:

Unlike a one-hot ring counter (n states), the Johnson counter gradually fills with 1s and then empties, creating two distinct sequences of length n each. Therefore, the total number of unique states is 2n.

Step-by-Step Solution:

Verification / Alternative check:

Enumerating states for a smaller n (e.g., n = 3) yields 6 unique patterns, reinforcing the 2n rule; scaling to n = 6 yields 12.

Why Other Options Are Wrong:

• 2 and 6: too few; confuse with ring or stage count.
• 24: would correspond to 4n or 2^(n+1), which is not a Johnson property.

Common Pitfalls:

• Mixing Johnson (2n states) with regular ring (n states) or binary counters (2^n states).

Final Answer:

12