Something you should know about: Quantifier Elimination (Part I)

by Arnab Bhattacharyya

About a month ago, Ankur Moitra dropped by my office. We started chatting about what each of us was up to. He told me a story about a machine learning problem that he was working on with Sanjeev Arora, Rong Ge, and Ravi Kannan. On its face, it was not even clear that the problem (non-negative rank) was decidable, let alone solvable in polynomial time. But on the other hand, they observed that previous work had already shown the existence of an algorithm using quantifier elimination. Ankur was a little taken aback by the claim, by the power of quantifier elimination. He knew of the theory somewhere in the back of his mind, in the same way that you probably know of Brownian motion or universal algebra (possible future topics in this “Something you should know about” series!), but he’d never had the occasion to really use it till then. On the train ride back home, he realized that quantifier elimination not only showed decidability of the problem but could also be helpful in devising a more efficient algorithm.

Quantifier elimination has a bit of a magical feel to it. After the conversation with Ankur, I spent some time revisiting the area, and this post is a consequence of that. I’ll mainly focus on the theory over the reals. It’s a remarkable result that you definitely should know about!

1. What is Quantifier Elimination?

A zeroth-order logic deals with declarative propositions that evaluate to either true or false. It is defined by a set of symbols, a set of logical operators, some inference rules and some axioms. A first-order logic adds functions, relations and quantifiers to the mix. Some examples of sentences in a first-order logic:

$latex \displaystyle \forall x~ \exists y~(y = x^2)&fg=000000$

$latex \displaystyle \forall y~ \exists x~(y = x^2)&fg=000000$

$latex \displaystyle \forall x,y~((x+y)^2 > 4xy \wedge x-y>0)&fg=000000$

The above are examples of sentences, meaning that they contain no free variables (i.e., variables that are not bounded by the quantifiers), whereas a formula $latex {\phi(x_1,\dots,x_n)}&fg=000000$ has $latex {x_1, \dots, x_n}&fg=000000$ as free variables. A quantifier-free formula is one in which no variable in the formula is quantified. Thus, note that a quantifier-free sentence is simply a proposition.

Definition 1 A first-order logic is said to admit quantifier elimination if for any formula $latex {\phi(x_1,\dots,x_n)}&fg=000000$, there exists a quantifier-free formula $latex {\psi(x_1,\dots,x_n)}&fg=000000$ which is logically equivalent to $latex {\phi(x_1,\dots,x_n)}&fg=000000$.

If the quantifier elimination process can be described algorithmically, then decidability of sentences in the logic reduces to decidability of quantifier-free sentences which is often a much easier question. (Note though that algorithmic quantifier elimination of formulas is a stronger condition than decidability of sentences.)

2. Quantifier Elimination over the Reals

The real numbers, being an infinite system, cannot be exactly axiomatized using first-order logic because of the Löwenheim-Skolem Theorem. But the axioms of ordered fields along with the intermediate value theorem yields a natural first-order logic, called the real closed field. The real closed field has the same first-order properties as the reals.

Tarski (1951) showed that the real closed field admits quantifier elimination. As a consequence, one has the following (Seidenberg was responsible for popularizing Tarski’s result):

Theorem 2 (Tarski-Seidenberg) Suppose a formula $latex {\phi(y_1,\dots,y_m)}&fg=000000$ over the real closed field is of the following form: \begin{equation*} Q_1x_1~Q_2x_2~\cdots Q_nx_n~(\rho(y_1,\dots,y_m,x_1,\dots,x_n)) \end{equation*} where $latex {Q_i \in \{\exists,\forall\}}&fg=000000$ and $latex {\rho}&fg=000000$ is a boolean combination of equalities and inequalities of the form:

$latex \displaystyle f_i(y_1,\dots,y_m,x_1,\dots,x_n) = 0&fg=000000$

$latex \displaystyle g_i(y_1,\dots,y_m,x_1,\dots,x_n) > 0&fg=000000$

where each $latex {f_i}&fg=000000$ and $latex {g_i}&fg=000000$ is a polynomial with coefficients in $latex {{\mathbb R}}&fg=000000$, mapping $latex {{\mathbb R}^{m+n}}&fg=000000$ to $latex {{\mathbb R}}&fg=000000$. Then, one can explicitly construct a logically equivalent formula $latex {\phi'(y_1, \dots, y_m)}&fg=000000$ of the same form but quantifier-free. Moreover, there is a proof of the equivalence that uses only the axioms of ordered fields and the intermediate value theorem for polynomials.

I should be more explicit about what “explicitly construct” means. Usually, it is assumed that the coefficients of the polynomials $latex {f_i}&fg=000000$ and $latex {g_i}&fg=000000$ are integers, so that there is a bound on the complexity of the quantifier elimination algorithm in terms of the size of the largest coefficient. (If the coefficients are integers, all the computations only involve integers.) But, even when the coefficients are real numbers, there is a bound on the “arithmetic complexity” of the algorithm, meaning the number of arithmetic operations $latex {+, -, \times, \div}&fg=000000$ performed with infinite precision.

A quick example. Suppose we are given an $latex {r}&fg=000000$-by-$latex {s}&fg=000000$ matrix $latex {M = (M_{i,j})}&fg=000000$, and we’d like to find out whether the rows of $latex {M}&fg=000000$ are linearly dependent, i.e. the following condition:

$latex \displaystyle \exists \lambda_1, \dots, \lambda_r \left( \neg \left(\bigwedge_{i=1}^r (\lambda_i = 0)\right) \wedge \bigwedge_{j=1}^s (\lambda_1 M_{1,j}+\lambda_2 M_{2,j} + \cdots \lambda_m M_{r,j} = 0) \right)&fg=000000$

Tarski-Seidenberg immediately asserts an algorithm to transform the above formula into one that does not involve the $latex {\lambda_i}&fg=000000$’s, only the entries of the matrix. Of course, for this particular example, we have an extremely efficient algorithm (Gaussian elimination) but quantifier elimination gives a much more generic explanation for the decidability of the problem.

2.1. Semialgebraic sets

If $latex {m=0}&fg=000000$ in Theorem 2, then Tarski-Seidenberg produces a proposition with no variables that can then be evaluated directly, such as in the linear dependency example. This shows that the feasibility of semialgebraic setsis a decidable problem. A semialgebraic set $latex {S}&fg=000000$ is a finite union of sets of the form:

$latex \displaystyle \{x \in {\mathbb R}^n~|~f_i(x) = 0, g_j(x) > 0 \text{ for all }i=1,\dots, \ell_1, j= 1,\dots, \ell_2\}&fg=000000$

where $latex {f_1,\dots, f_{\ell_1}, g_1,\dots, g_{\ell_2}: {\mathbb R}^n \rightarrow {\mathbb R}}&fg=000000$ are $latex {n}&fg=000000$-variate polynomials over the reals. The feasibility problem for a semialgebraic set $latex {S}&fg=000000$ is deciding if $latex {S}&fg=000000$ is empty or not. An algorithm to decide feasibility directly follows from applying quantifier elimination.

Note that this is in stark contrast to the first-order theory over the integers for which Godel’s Incompleteness Theorem shows undecidability. Also, over the rationals, Robinson proved undecidability (in her Ph.D. thesis advised by Tarski) by showing how to express any first-order sentence over the integers as a first-order sentence over the rationals. This latter step now has several different proofs. Poonen has written a nice survey of (un)decidability of first-order theories over various domains.

2.2. Efficiency

What about the complexity of quantifier elimination over the reals? The algorithm proposed by Tarski does not even show complexity that is elementary recursive! The situation was much improved by Collins (1975) who gave a doubly-exponential algorithm using a technique called “cylindrical algebraic decomposition”. More precisely, the running time of the algorithm is $latex {poly(C,(\ell d)^{2^{O(n)}})}&fg=000000$, where $latex {C}&fg=000000$ measures the size of the largest coefficient in the polynomials, $latex {\ell}&fg=000000$ is the number of polynomials, $latex {d}&fg=000000$ is the maximum degree of the polynomials, and $latex {n}&fg=000000$ is the total number of variables.

A more detailed understanding of the complexity of quantifier elimination emerged from an important work of Ben-Or, Kozen and Reif (1986). Their main contribution was an ingenious polynomial time algorithm to test the consistency of univariate polynomial constraints: given a set of univariate polynomials $latex {{f_i}}&fg=000000$ and a system of constraints of the form $latex {f_i(x) \leq 0}&fg=000000$, $latex {f_i(x) = 0}&fg=000000$ or $latex {f_i(x) > 0}&fg=000000$, does the system have a solution $latex {x}&fg=000000$? Such an algorithm exists despite the fact that it’s not known how to efficiently find an $latex {x}&fg=000000$ that makes the signs of the polynomials attain a given configuration. Ben-Or, Kozen and Reif also claimed an extension of their method to multivariate polynomials, but this analysis was later found to be flawed.

Nevertheless, subsequent works have found clever ways to reduce to the univariate case. These more recent papers have shown that one can make the complexity of the quantification elimination algorithm be only singly exponential if the number of quantifier alternations (number of switches between $latex {\exists}&fg=000000$ and $latex {\forall}&fg=000000$) is bounded or if the model of computation is parallel. See Basu (1999) for the current record and for references to prior work.

As for lower bounds, Davenport and Heintz (1988) showed that doubly-exponential time is required for quantifier elimination, by explicitly constructing formulas for which the length of the quantifier-free expression blows up doubly-exponentially. Brown and Davenport (2007) showed that the doubly exponential dependence is necessary even when all the polynomials in the first order formula are linear and there is only one free variable. I do not know if a doubly exponential lower bound is known for the decision problem when there are no free variables.

Thanks to Ankur for helpful suggestions. Part II of this post will contain a relatively quick proof of Tarski’s theorem! Stay tuned.

Alfred Tarski (1951). A Decision Method for Elementary Algebra and Geometry Rand Corporation

• Max says:

I think there’s a typo in definition 1: the two formulas are both called phi

• says:

There were two different formulas in the definition, “phi” and “phi-prime” but perhaps the distinction was not rendering on your browser. I changed the phi-prime formula to “psi” to improve readability.

• The align* environment is not getting LaTeXed for me/

• says:

What browser and system are you using, please? It is rendering correctly for me.

• Kaveh says:

I have the same problem. Firefox on Windows.

• says:

We’ve removed the align environment. Hopefully it renders correctly now, for everyone. Thanks for the heads up!

• Diego says:

I asked J. Heintz and he thinks the best lower bound for sentences is single-exponential (from Fischer-Rabin paper), so possibly it is an open problem to find a tight bound.

• Arnab says:

Thanks! It’s an interesting problem, I think.

• Rod Carvalho says:

Just to point out a typo:

In the example in which one is interested in determining whether the rows of matrix M are linearly dependent, we should have $\lambda_r M_{r,j}$ instead of $\lambda_m M_{r,j}$.

• Mahesha999 says:

• Gsz says:

Something seems to have gone wrong since this post was first published because LaTeX equations don’t render (they get displayed as raw code).

The source of the page contains:

which might point to the source of the problem.

• Gsz says:

Oops, the second part of the previous comment should read:

The source of the page contains: < ! – – MathJax Latex Plugin installed: Disabled as no shortcodes on this page – – > etc.

• Comments have been closed for this post