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Short Killing Words

Given a finite set $\mathcal{M}$ of $n \times n$ matrices over the integers, can you express the zero matrix as a product of matrices in $\mathcal{M}$? This is known as the mortality problem. Michael Paterson showed in 1970 that it is undecidable, even for $n=3$. Later it was shown that mortality remains undecidable for $n=3$ and $|\mathcal{M}| = 7$, and for $n=21$ and $|\mathcal{M}| = 2$. Decidability of mortality for $n=2$ is open.

If all matrices in $\mathcal{M}$ are nonnegative, the mortality problem becomes decidable, because then it matters only whether matrix entries are $0$ or not. Specifically, view the given matrices as transition matrices of a nondeterministic finite automaton where all states are initial and accepting, and check whether there is a word that is not accepted. This problem is decidable and PSPACE-complete. One can construct cases where the shortest word that is not accepted by the automaton has exponential length.

In this post we focus on the nonnegative ca…
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Calibrating Assessments

Last year I wrote a post on how Oxford selects undergraduate students for computer science. Oxford undergraduate admissions are politically charged: the British public suspects that Oxbridge is biased towards privileged applicants.

This post is unpolitical though. I'm advocating a better way of solving a typical problem in academia: you have many candidates (think: papers, grant proposals, or Oxford applicants) and few assessors (think: members of program committees, assessment panels, or admission tutors), and the goal of the assessors is to select the best candidates. Since there are many candidates, each assessor can assess only some candidates. From a candidate's point of view: a paper may get 3 reviews, a grant proposal perhaps 5, and a shortlisted Oxford applicant may get 3 interviews. Each such review comes with a score that reflects the assessor's estimate of the candidate's quality. The problem is that different assessors will score the same candidate differe…

Equivalence Checking

There is a compact data structure for words: automata. Finite automata can “store” many words, even infinitely many—in the sense that the language of a finite automaton can be infinite. This powerful feature makes automata ubiquitous in computer science. Lecture notes by Javier Esparza emphasize this view of automata as data structures.

The most fundamental algorithmic problems on automata are:
operations, such as computing set union or complement of the languages of automata; tests, such as checking their languages for equality. In this post I focus on the latter: equivalence checking. In general, equivalence checking for automata is PSPACE-complete, which is excusable considering how compact and flexible data structures they are. One can check two deterministic finite automata for equivalence in polynomial time, by searching their product.

Less known is the fact that one can efficiently check much wider classes of automata for equivalence, including unambiguous automata, probab…

Oxford Admissions

An initial draft of this post was about how to make rankings more meaningful, a piece of mathsy computer science, as you expect it from this blog. My motivation is personal: I'm involved in selecting CS undergraduate students. Some people might be curious how Oxford admissions work: it's certainly a favourite subject for UK politicians and news media. Therefore I decided to postpone my original plan and to write this post about Oxford admissions. I hope I'm not playing with too much fire and the blog will not suddenly become famous for the wrong reasons!

The tenor in newspapers and political campaigns is that the Oxbridge admissions process is “opaque” and unfair. Here I am going to offer transparency and some comments on fairness. The views expressed here are my own, and may not necessarily coincide with the university's views.

Oxford's goal is to admit the strongest applicants. One can disagree with this basic premise, but we try to admit those applicants who w…

The Price of Memory

In an earlier post I defined the price of memory, $u$, and asked:
Is $u < 1$? Worth €100.
Determine $u$. Worth another €100, even if $u=1$. I'm happy to announce that there is now a solution:
The price of memory, $u$, is 1. I don't want to repeat the precise definition of $u$, but here is a short recap: Suppose you have an acyclic finite MDP and a strategy $\sigma$ that guarantees to visit a red transition with probability 1. This strategy $\sigma$ might use memory, i.e., might base its behaviour in a state on the history of the run, as indicated in yellow:
Any such strategy $\sigma$ induces a randomized strategy $\overline\sigma$ that acts like $\sigma$ in average, but is memoryless. In each state, when averaged over all runs, the strategies $\sigma$ and $\overline\sigma$ take the same actions with the same probabilities:
One can show that $\sigma$ and $\overline\sigma$ visit red transitions equally often in average; in particular, $\overline\sigma$ visits at least…

Probabilistic Models of Computation

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-co…

Tell me the price of memory and I give you €100

Markov Decision Processes (MDPs) are Markov chains plus nondeterminism: some states are random, the others are controlled (nondeterministic). In the pictures, the random states are round, and the controlled states are squares:
The random states (except the brown sink state) come with a probability distribution over the successor states. In the controlled states, however, a controller chooses a successor state. What does the controller want to achieve? That depends. In this blog post, the objective is very simple: take a red transition. The only special thing about red transitions is that the controller wants to take them. We consider only MDPs with the following properties:
There are finitely many states and transitions.The MDP is acyclic, that is, it has no cycles.There are a unique start state from which any run starts (in the pictures: blue, at the top) and a unique sink state where any run ends (in the pictures: brown, at the bottom). No matter what the controller does, the sink…

Infinitely many states and no memory

Markov Decision Processes (MDPs) are Markov chains plus nondeterminism: some states are random, the others are controlled (nondeterministic). In the pictures, the random states are round, and the controlled states are squares:
The random states come with a distribution over successor states, but in the controlled states a controller chooses a successor state (or a probability distribution over the successor states). For instance, the controller could stay on the leftmost column forever, by always choosing to go one state down. Or the controller could go right at some point; in the random state a successor is picked randomly, either the initial state or the state on the right, according to the blue probabilities.

What objective should the controller aim at? In this post, the objective will be the following: visit green states infinitely often, and red states only finitely often. Here is the previous MDP with colours:
Since the controller wants to visit (infinitely many) green state…

Model Checking Markov Chains Against Unambiguous Automata: The Quantitative Case

In my first post I promised to give a polynomial-time procedure to compute the probability that an infinite word (chosen uniformly at random) is accepted by a given unambiguous Büchi automaton. For example, let's take this Büchi automaton again:
Let's call the probability $x_1, x_2, x_3$, depending on the start state: \[ \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} := \begin{pmatrix} \text{Pr(state $1$ accepts a random word)}\\ \text{Pr(state $2$ accepts a random word)}\\ \text{Pr(state $3$ accepts a random word)} \end{pmatrix} \] I showed in the first post that $x_1, x_2, x_3$ are nonzero in this example. In order to compute the $x_i$ we set up a linear system of equations and solve it. We have \[ x_1 \ = \ \frac12 \cdot x_{1,a} + \frac12 \cdot x_{1,b}\,, \] where $x_{1,a}$ and $x_{1,b}$ denote the (conditional) probabilities that state $1$ accepts a random word, under the condition that the word starts with $a$ and $b$, respectively.
$x_{1,b} = 0$ because state $1$ has…

Unambiguous Automata: From PSPACE to P

Computer science is about leveraging mathematical structure, for instance, in order to accelerate algorithms. More often than we'd like we can't accelerate much: For instance, checking two NFAs for language inclusion or equivalence is PSPACE-complete. Even the problem whether an NFA accepts all words is PSPACE-complete. For DFAs, those problems are easy. This means that for NFAs we can't do much better than determinizing (using the subset construction and incurring an exponential blowup) and solving the resulting DFA problem.

In this post I'll discuss an automata problem that seems to call for determinization and is, in fact, PSPACE-complete for NFAs. But there is an efficient algorithm for unambiguous automata.

Here is an example of an unambiguous automaton:
The notion of accepting states is not used in this post. By unambiguousness I just mean that the automaton has no diamonds, that is, for any states $q, r$, any word $w$ labels at most one path from $q$ to $r$. (…