This post has 0 ideas. At least, no ideas that are younger than 50 years or so. I'm talking about the two most fundamental models of probabilistic computation and the two most fundamental results about them, according to my subjective judgement.

Every computer scientist knows finite automata: State $1$ is the start state, and state $3$ is the only accepting state. Some words are accepted (i.e., have an accepting run), for instance the word $aa$. Other words are rejected (i.e., have no accepting run), for instance the word $ba$. Most questions about finite automata are decidable. For instance, the emptiness problem, which asks whether the automaton accepts at least one word. This problem is decidable in polynomial time: view the automaton as a graph, and search the graph for a path from the initial state to an accepting state. One can also decide whether two given automata are equivalent, i.e., do they accept the same words? This problem is computationally hard though: PSPACE-complete (never mind if you don't know what that means).

Let's slap on probabilities:

This is a

While probabilistic automata are reactive,

Now for each state, the probabilities of all outgoing transitions sum up to $1$. There are no input words and no accepting states. Instead, in any state, the labelled Markov chain autonomously picks an outgoing transition and emits the letter that the transition is labelled with. For instance, in state $1$, it emits $a$ with probability $\frac14$, and $b$ with probability $\frac12 + \frac14$. This generalizes to longer words in the natural way. For instance, starting in state $1$, the word $bb$ is emitted with probability $\frac12 \cdot \frac12 + \frac12 \cdot \frac14 + \frac14 \cdot \frac13 + \frac14 \cdot \frac16$. In this way, for each number $n$, the labelled Markov chain defines a probability distribution over words of length $n$.

As a rule of thumb, equality questions about probabilistic models are decidable, and inequality questions are undecidable. More about that in later posts, such as this one.

Every computer scientist knows finite automata: State $1$ is the start state, and state $3$ is the only accepting state. Some words are accepted (i.e., have an accepting run), for instance the word $aa$. Other words are rejected (i.e., have no accepting run), for instance the word $ba$. Most questions about finite automata are decidable. For instance, the emptiness problem, which asks whether the automaton accepts at least one word. This problem is decidable in polynomial time: view the automaton as a graph, and search the graph for a path from the initial state to an accepting state. One can also decide whether two given automata are equivalent, i.e., do they accept the same words? This problem is computationally hard though: PSPACE-complete (never mind if you don't know what that means).

Let's slap on probabilities:

This is a

*probabilistic automaton*. Previously each state and each letter "activated" transitions, one of which was nondeterministically taken. A probabilistic automaton, in addition, determines a probability distribution over the activated transitions. There is one such probability distribution for each state and each letter. For instance, if in state $1$ the letter $b$ is input then the probabilistic automaton moves to state $3$ with probability $\frac13$, and remains in state $1$ with probability $\frac23$. Probabilistic automata are*reactive*: someone inputs letters, and the probabilistic automaton reacts to the input by taking transitions probabilistically. As a consequence, each input word is accepted with some probability. For instance, consider the input word $bb$. The probabilistic automaton takes the path $1 \xrightarrow{b} 1 \xrightarrow{b} 3$ with probability $\frac23 \cdot \frac13 = \frac29$. Similarly, it takes the path $1 \xrightarrow{b} 3 \xrightarrow{b} 3$ with probability $\frac13 \cdot \frac34 = \frac14$. These two paths are accepting, because state $3$ is accepting. The other two $bb$-labelled paths, $1 \xrightarrow{b} 1 \xrightarrow{b} 1$ and $1 \xrightarrow{b} 3 \xrightarrow{b} 2$, are rejecting, because states $1$ and $2$ are rejecting. It follows that the probabilistic automaton accepts $bb$ with probability $\frac29 + \frac14 = \frac{17}{36}$. Whereas a nonprobabilistic automaton either accepts or rejects the input word, a probabilistic automaton accepts the input with some probability.
Given a probabilistic automaton, the

The following theorem appears in the 1971 textbook **emptiness problem**asks if there exists a word that is accepted with probability $\ge \frac12$.*Introduction to Probabilistic Automata*by Azaria Paz:
The emptiness problem is undecidable.

The emptiness problem remains undecidable when the nonstrict inequality $\ge$ is replaced by strict inequality $>$, and also when the threshold $\frac12$ is replaced by another number strictly between $0$ and $1$.
With threshold $0$ or $1$, the emptiness problem becomes decidable: apply what I wrote above about nonprobabilistic automata.
While probabilistic automata are reactive,

*labelled Markov chains*are*generative:*Now for each state, the probabilities of all outgoing transitions sum up to $1$. There are no input words and no accepting states. Instead, in any state, the labelled Markov chain autonomously picks an outgoing transition and emits the letter that the transition is labelled with. For instance, in state $1$, it emits $a$ with probability $\frac14$, and $b$ with probability $\frac12 + \frac14$. This generalizes to longer words in the natural way. For instance, starting in state $1$, the word $bb$ is emitted with probability $\frac12 \cdot \frac12 + \frac12 \cdot \frac14 + \frac14 \cdot \frac13 + \frac14 \cdot \frac16$. In this way, for each number $n$, the labelled Markov chain defines a probability distribution over words of length $n$.

Given two labelled Markov chains, the

The following theorem goes back to a 1961 paper by Marcel-Paul Schützenberger and was subsequently rediscovered many times:
**equivalence problem**asks if for each word $w$ the chains emit $w$ with the same probability.
The equivalence problem is decidable in polynomial time.

The algorithm is based on linear algebra.
Loosely speaking, it iteratively discovers equivalent states and, more generally, equivalent linear combinations of states.
In this way one can also check two probabilistic automata for equivalence, where equivalence means that for each word $w$ the automata accept $w$ with the same probability.
More generally, the equivalence-checking algorithm works for *weighted*automata, which are similar to labelled Markov chains and probabilistic automata except that the numbers do not have to be probabilities. Equivalence checking is used to check randomized protocols for anonymity and obliviousness, properties that are hard to formalize otherwise.As a rule of thumb, equality questions about probabilistic models are decidable, and inequality questions are undecidable. More about that in later posts, such as this one.

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