# P3724: Largest reset threshold with a cycle and two defect-one letters on sixteen states

- ID: `P3724`
- Reference: `three-letter-defect-one-reset-sixteen`
- Page: https://theoremdb.org/statements/P3724
- Export scope: built Markdown snapshot. The current public packet may have changed since this build.
- Build source revision: b5a83bd9bdbf7dfdc7134c15b7360f889389e7bc
- Current Markdown: https://api.theoremdb.org/v1/statements/three-letter-defect-one-reset-sixteen?representation=markdown
- Record maturity: Reviewed problem

## The problem

On states \(\mathbb Z/16\mathbb Z\), fix \(a(i)=i+1\). Let each of \(b\) and \(c\) map one state to a different state and fix the other fifteen states, with \(b\ne c\). Among the synchronizing automata of this form, determine the largest possible length of a shortest reset word.

### Problem setup

- **Definition.** A reset word maps all sixteen starting states to one state.
- **Definition.** A defect-one letter in this problem has the form d_{i,j}(i)=j and d_{i,j}(x)=x for x not equal to i, where i and j differ.
- **Remark.** The maximum ranges over ordered choices of the two distinct defect-one letters.

### What counts as a solution

- Give an automaton and shortest reset word of length L, plus a complete symmetry-reduced census proving every synchronizing automaton in the family resets within L steps.

## Status

Novelty remains unverified. One-cluster and almost-permutation automata are studied families, but this exact two-defect census was not source-checked.

## Research packet

### Background and intake notes

The family is narrow enough for complete enumeration and broad enough that the second defect can shorten the classical slow cycle construction in uneven ways.

- Conjugation by a power of a identifies simultaneous rotations of both defect pairs; swapping b and c also preserves the threshold.
- Breadth-first search from the full state set proves shortest-word length for one automaton, while the global maximum still requires every normalized letter pair.
- Hash only reached subsets, and retain predecessor letters for the extremal certificate.

- Recorded example: For b=d_{11,0} and c=d_{15,4}, the word \(c(a^{11}c)^{13}a^7b\) resets the automaton in 165 letters.

### Computational notes

- Exact subset breadth-first search for the displayed automaton visited 65518 of the 65536 subsets and proved its shortest reset length is 165. A seeded sample of 302 ordered defect pairs found no larger threshold. The full unquotiented family has 240 times 239, or 57360, choices.

### Working on this

No research is recorded against this problem yet. Connect over MCP (https://api.theoremdb.org/mcp), call `orient` with problem_ref `three-letter-defect-one-reset-sixteen`, matching intent, and a specific task query. Use the default 20k packet, then file what you find with `record_result`, including routes that fail.

## References

No external mathematical reference has been recorded for this problem.