Reset Word Length 2F31Ae

1. **Problem Statement:** We want to find the minimum length of a reset word for a deterministic finite automaton (DFA) defined on $n$-bit arrays with transformations $f1l$, $f1r$, $f2$, and $f3$. 2. **Definitions and Transformations:** - $f1l$ and $f1r$ are shift moves: shifting the array left or right by one index with wrap-around. - $f2$ flips bits at indices 0 and 2. - $f3$ flips bits at indices 0, 1, and 2. - For $n > 3$, the new bits at indices $\\geq 3$ are unaffected by $f2$ and $f3$ but are shifted by $f1l$ and $f1r$. 3. **Reset Word:** A reset word is a sequence of transformations that, when applied from any state except the all-ones state ($111...1$), passes through the all-ones state at least once. 4. **Known Observations for $n=3$ to $5$:** - The number of flip moves ($f2$ and $f3$) in the shortest reset word is exactly $2^n - 1$. - The number of shift moves ($f1l$ and $f1r$) is at least $(2^n // 3) - 1$ (a lower bound). - The state space is partitioned into $2^{n}/4$ cosets. 5. **Reasoning:** - Flip moves change bits within cosets. - Shift moves move between cosets. - To cover all states and ensure the reset word passes through the all-ones state, the reset word must include enough flip moves to cover all states within cosets and enough shift moves to move between cosets. 6. **Formula for Minimum Length:** - Total flip moves: $2^n - 1$ - Total shift moves: at least $(2^n // 3) - 1$ Therefore, the minimum length $L_{min}$ of a reset word satisfies: $$ L_{min} \geq (2^n - 1) + \left(\frac{2^n}{3} - 1\right) = \frac{4 \cdot 2^n}{3} - 2 $$ 7. **Summary:** The minimum length of a reset word for given $n$ is at least approximately $\frac{4}{3} \cdot 2^n - 2$ moves, combining flip and shift moves. This matches the observed patterns for small $n$ and aligns with the structure of cosets and transformations. **Final answer:** $$ \boxed{L_{min} \geq \frac{4}{3} \cdot 2^n - 2} $$