Introduction

Hey DFA devotees! Time to slim things down: the Minimization Problem asks how to transform any given DFA into an equivalent one with the fewest states possible. Good news: not only is this decidable, but we have an efficient, classic algorithm to do it.

Formal Setup

Given a DFA

$$ M = (Q, Σ, δ, q_0, F), $$

the goal is to produce a minimal DFA

$$ M_{min} = (Q_{min}, Σ, δ_{min}, q_{0,min}, F_{min}) $$

such that:

  1. $L(M_{min}) = L(M)$.
  2. For any other DFA $M’$ equivalent to $M$, $|Q_{min}| \le |Q’|$.

Example DFA

# Original DFA D:
# Q={A,B,C,D}, Σ={0,1}, q0=A, F={C,D}
# Transitions:
# A--0-->B, A--1-->C
# B--0-->A, B--1-->D
# C--0-->D, C--1-->A
# D--0-->C, D--1-->B

This DFA accepts all strings whose number of 1’s is odd or whose number of 0’s is odd. It’s known this can be done with just 2 states tracking parity pair, so we’ll minimize it.

Algorithm (Hopcroft’s Partition Refinement)

  1. Initialize a partition $P = {F, Q\setminus F}$.

  2. Worklist initially contains $F$ and $Q\setminus F$.

  3. While the worklist isn’t empty:

    1. Remove a set $A$ from the worklist.

    2. For each input symbol $a \in Σ$:

      • Let $X = {q \in Q \mid δ(q,a) \in A}.$

      • For each $Y$ in the current partition $P$ that intersects $X$ nontrivially but isn’t contained in $X$:

        1. Replace $Y$ in $P$ with the two sets $Y \cap X$ and $Y \setminus X$.
        2. If $Y$ was in the worklist, replace it by those two sets; otherwise add the smaller of the two to the worklist.

  4. Construct the minimized DFA by collapsing each block in $P$ into a single state.

This runs in $O(|Σ|\cdot |Q| \log |Q|)$ time, which is optimal for general DFAs.

Example Walkthrough

  1. Start with partitions: ${C,D}$ (accepting) and ${A,B}$ (non-accepting).
  2. Process symbol $0$ or $1$, refining until partition stabilizes.
  3. End up with two blocks: one for states {A,B} and one for {C,D}.
  4. Build a 2-state DFA that distinguishes even/odd counts.

Why It Matters

  • State Reduction: Smaller DFAs mean faster matching in regex engines and less memory.
  • Canonical Form: A minimal DFA is unique (up to renaming states), great for checking equivalence.

Wrap-Up

DFA minimization gives us a canonical, smallest automaton for any regular language. That’s it for DFAs! Next up: the Post Correspondence Problem, an undecidable gem in the world of strings.