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A polynomial time algorithm for the equivalence problem of weighted automata over a field

April 06, 2019 | by Nathanaël Fijalkow

We present an algebraic algorithm solving the equivalence problem for weighted automata over a field (in particular probabilistic automata) in polynomial time.

Edits

29/03/2020: Clarified the model to avoid a terminology clash

We consider weighted automata over an arbitrary semiring. The key assmuption in this post is that the semiring is a field. For simplicity we will use the reals \(\R\), but everything readily extends to any field. The algorithm does not extend to semirings which are not fields, for instance it is well known that for the semiring \((\N,\min,+)\), the equivalence problem is undecidable.

A formal series (here over the reals) is a function \(\Sigma^* \to \R\). For recognising formal series we use weighted automata:

\(\A = (Q, \alpha \in \R^Q, (\Delta(a) \in \R^{Q \times Q})_{a \in A}, \eta \in \R^Q)\).

Such an automaton recognises the function

\[\A(a_1 \cdots a_n) = \alpha \cdot \underbrace{\Delta(a_1) \cdots \Delta(a_n)}_{\Delta(a_1 \cdots a_n)} \cdot \eta\]

Probabilistic automata are a special case of weighted automata.

Theorem: (Schützenberger ‘61) The following problem can be solved in polynomial time: given two weighted automata $\A$ and $\B$ over a field, determine whether $\A = \B$, meaning for all words $w \in \Sigma^*$, we have $\A(w) = \B(w)$.

As evidence that this is an important theorem, it has been rediscovered a number of times. The techniques described here are close to the original ones, which where about minimising weighted automata.

We simplify the problem by constructing a weighted automaton recognising the formal series $\A - \B$: the automaton has size the sum of the two automata. The question reduces to whether $\A = 0$, where $\A$ is given by a weighted automaton.

We let $\Delta(a)(s)$ denote the vector in $\R^Q$ whose entry for $s’$ is $\Delta(a)(s,s’)$.

We define the following sequence of sets in $\R^Q$: \(\begin{array}{lll} u \in X_0 & \Longleftrightarrow & \alpha \cdot u = 0 \\ u \in X_{k+1} & \Longleftrightarrow & u \in X_k \text{ and for all } a \in \Sigma,\ u \cdot \Delta(a) \in X_k. \end{array}\)

Proposition:

  • $X_0 \supseteq X_1 \supseteq X_2 \supseteq \cdots$,
  • $X_k$ is a vector space in $\R^Q$,
  • if $X_{k+1} = X_k$, then $X_{k+2} = X_{k+1}$,
  • $X = X_{Q - 1} = X_{Q} = \cdots$,
  • for all $k$, we have the equivalence between for all $w \in \Sigma^{\le k}, \A(w) = 0$ and $\eta \in X_k$.

Before going into the proof, we explain how this implies an algorithm. The algorithm iteratively compute bases, starting from $X_0$, then $X_1$, and so on until it stabilises, which happens in at most $Q$ steps. We let $X$ denote the limit. The final step is to check whether $\eta$ belongs to $X$.

The core of the algorithm relies on polynomially calls (specifically, at most $Q$) to the following task: given a matrix $M$ in $\R^{Q \times Q}$ and a basis for a vector space $V$, compute a basis for the vector space $\set{u \mid u \cdot M \in V}$, which can be performed in polynomial time.

Indeed, let $M$ be the matrix with $\Sigma + 1$ blocks each of dimension $Q \times Q$: the first one is the identity, and for each $a \in A$ a block equal to $\Delta(a)$. By construction $X_{k+1} = \set{u : u \cdot M \in X_k^{1 + \Sigma}}$.

We proceed to the proof of the proposition. The first two items are easy to see, so we focus on the remaining three.

  • This is a symbolic argument, better understood using the remark that $X_{k+1} = \set{u \mid u \cdot M \in X_k^p}$, for some matrix $M$ and some $p$ in $\N$. Assume that $X_{k+1} = X_k$. We already know that $X_{k+1} \supseteq X_{k+2}$, so we prove the converse implication. Let $u$ in $X_{k+1}$, since $X_{k+1} = X_k$ we have that $u$ is in $X_k$, so by definition $u \cdot M$ is in $X_{k+1}^p$, which means that $u$ is in $X_{k+2}$.
  • This is a dimension argument: the sequence $(\text{dim}(X_k))_{k \in \N}$ is non-increasing, takes values in $[0,Q-1]$, and stabilises as soon as it hits twice the same value. Note that we do need the property above saying that once the same value is hit twice then the sequence stabilises. There is a special place in hell for people who forget to state (and prove) that property.
  • We first prove by induction on $k$ that $X_k$ is the set of $u \in \R^Q$ such that for all $w \in \Sigma^{\le k}$, we have $\alpha \cdot \Delta(w) \cdot u = 0$. Both the base and the inductive cases are easy. This readily implies that for all $w \in \Sigma^{\le k}, \A(w) = 0$ if and only if $\eta \in X_k$.