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

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This is the second of a two-part series on quantifier elimination over the reals. In the first part, we introduced the Tarski-Seidenberg quantifier elimination theorem. The goal of this post is to give a direct and completely elementary proof of this result.

Before embarking on the proof journey, there’s something we can immediately note. It’s enough to prove quantifier elimination for formulas of the following type:

$latex \displaystyle \phi(y_1,\dots,y_m) = \exists x \bigwedge_{i=1}^n \left( p_i(x,y_1,y_2,\dots, y_m)~\Sigma_i~ 0\right)&fg=000000$

where each $latex {\Sigma_i \in \{<,=,>\}}&fg=000000$ and each $latex {p_i \in {\mathbb R}[x][y_1, \dots, y_m]}&fg=000000$ is a polynomial over the reals. Why? Well, if one quantifier can be eliminated, then we can inductively eliminate all of them. And we can always convert the single quantifier to an $latex {\exists}&fg=000000$ by negation if necessary and then distribute the $latex {\exists}&fg=000000$ over disjunctions to arrive at a boolean combination of formulas of the type above.

I will try to describe now how we could “rediscover” a quantifier elimination procedure due to Albert Muchnik. My description is based on a very nice exposition by Michaux and Ozturk but with some shortcuts and a more staged revealing of the full details. The specific algorithm for quantifier elimination we describe will be much less efficient than the current state-of-the-art described in the previous post. The focus here is on clarity of presentation instead of performance.

We will complete the proof in three stages.

1. $latex {m = 0}&fg=000000$ and linear polynomials

First, let’s restrict ourselves to the simplest case: $latex {m = 0}&fg=000000$ and all the polynomials $latex {p_i}&fg=000000$ are of degree one. In fact, let’s even start with a specific example:

$latex \displaystyle \phi = \exists x ((x+1 > 0) \wedge (-2x+3>0) \wedge (x > 0))&fg=000000$

We will partition the real line into a constant number of intervals so that none of the functions changes sign inside any of the intervals. If we do this, then to decide $latex {\phi}&fg=000000$, we merely have to examine the finitely many intervals to see if on any of them, the three functions ($latex {x+1, -2x+3}&fg=000000$, and $latex {x}&fg=000000$) are all positive.

Each of the linear functions has one root and is always negative on one side of the root and always positive on the other. We will keep ourselves from using the exact value of the root, even though finding it here is trivial, because this will help us in the general case. Whether the function is negative to the left or right side of the root depends on the sign of the leading coefficient. So, the first function $latex {x+1}&fg=000000$ has a root at $latex {\gamma_1}&fg=000000$, is always negative to the left of $latex {\gamma_1}&fg=000000$ and always positive to the left.

The second function, $latex {-2x+3}&fg=000000$, has another root $latex {\gamma_2}&fg=000000$, is always negative to its right, and always positive to its left. How do we know if $latex {\gamma_2}&fg=000000$ is to the left or right of $latex {\gamma_1}&fg=000000$? Well, $latex {-2x+3 = -2(x+1) + 5}&fg=000000$, and since $latex {\gamma_1}&fg=000000$ is a root of $latex {x+1}&fg=000000$, it follows that $latex {-2\gamma_1 + 3 = 5 > 0}&fg=000000$. So, $latex {\gamma_1}&fg=000000$ is to the left of $latex {\gamma_2}&fg=000000$. This results in the following sign configuration diagram (for each pair of signs, the first indicates the sign of $latex {x+1}&fg=000000$, the second the sign of $latex {-2x+3}&fg=000000$):

Finally, the third function $latex {x}&fg=000000$ has a root $latex {\gamma_3}&fg=000000$, is always negative to the left, and always positive to the right. Also, $latex {x = (x+1)-1}&fg=000000$ and $latex {x=-\frac{1}{2}(-2x+3) + 1.5}&fg=000000$, and so $latex {x}&fg=000000$ is negative at $latex {\gamma_1}&fg=000000$ and positive at $latex {\gamma_2}&fg=000000$. This means the following sign configuration diagram:

Since there exists an interval $latex {(\gamma_3,\gamma_2)}&fg=000000$ on which the sign configuration is the desired $latex {(+,+,+)}&fg=000000$, it follows that $latex {\phi}&fg=000000$ evaluates to true. The algorithm we described via this example will obviously work for any $latex {\phi}&fg=000000$ with $latex {m=0}&fg=000000$ and linear inequalities.

2. $latex {m=0}&fg=000000$ and general polynomials

Now, let’s see what changes if some of the polynomials are of higher degree with $latex {m}&fg=000000$ still equal to $latex {0}&fg=000000$:

$latex \displaystyle \psi = \exists \left( \bigwedge_{i=1}^n p_i(x) \Sigma_i 0\right)&fg=000000$

where each $latex {\Sigma_i \in \{<,>,=\}}&fg=000000$. Again, we’ll try to find a sign configuration diagram and check at the end if any of the intervals is described by the wanted sign configuration.

If we try to implement this approach, a fundamental issue crops up. Suppose $latex {p_i}&fg=000000$ is a degree $latex {d}&fg=000000$ polynomial and we know that $latex {p_i(\gamma_1) > 0}&fg=000000$ and $latex {p_i(\gamma_2)>0}&fg=000000$. In the case $latex {p_i}&fg=000000$ was linear, we could safely conclude that $latex {p_i(x)}&fg=000000$ is positive everywhere in the interval $latex {(\gamma_1,\gamma_2)}&fg=000000$. Not so fast here! It could very well be for a higher degree polynomial $latex {p_i}&fg=000000$ that it becomes negative somewhere in between $latex {\gamma_1}&fg=000000$ and $latex {\gamma_2}&fg=000000$. But note that this would mean $latex {p_i’ = \frac{\partial}{\partial x} p_i}&fg=000000$ must change sign somewhere at $latex {\gamma_3 \in (\gamma_1, \gamma_2)}&fg=000000$ by Rolle’s theorem. New idea then! Let’s also partition according to the sign changes of $latex {p_i’}&fg=000000$, a degree $latex {(d-1)}&fg=000000$ polynomial. Suppose we have done so successfully by an inductive argument on the degree, and $latex {\gamma_3}&fg=000000$ is the only sign change occurring in $latex {(\gamma_1, \gamma_2)}&fg=000000$ for $latex {p_i’}&fg=000000$.

We now want to also know the sign of $latex {p_i(\gamma_3)}&fg=000000$. Imitating what we did for $latex {d=1}&fg=000000$, note that if $latex {q_i = (p_i \text{ mod } p_i’)}&fg=000000$, then $latex {p_i(\gamma_3) = q_i(\gamma_3)}&fg=000000$ since $latex {\gamma_3}&fg=000000$ is a root of $latex {p_i’}&fg=000000$. The polynomial $latex {q_i}&fg=000000$ is degree less than $latex {d-1}&fg=000000$, so let’s assume by induction that we know the sign pattern of $latex {q_i}&fg=000000$ as well. Thus, we have at hand the sign of $latex {p_i(\gamma_3)}&fg=000000$.

There are now two situations that can happen. First is if $latex {p_i(\gamma_3) > 0}&fg=000000$ or $latex {p_i(\gamma_3) = 0}&fg=000000$.

This is because $latex {p_i}&fg=000000$ reaches the minimum in $latex {(\gamma_1, \gamma_2)}&fg=000000$ at $latex {\gamma_3}&fg=000000$, and since $latex {p_i(\gamma_3) \geq 0}&fg=000000$, $latex {p_i}&fg=000000$ is not negative anywhere in $latex {(\gamma_1, \gamma_2)}&fg=000000$.

The other situation is the case $latex {p_i(\gamma_3)<0}&fg=000000$. By the intermediate value theorem, $latex {p_i}&fg=000000$ has a root $latex {\gamma_4}&fg=000000$ in $latex {(\gamma_1,\gamma_3)}&fg=000000$ and a root $latex {\gamma_5}&fg=000000$ in $latex {(\gamma_3, \gamma_2)}&fg=000000$.

Note again that because $latex {\gamma_3}&fg=000000$ is the only root of $latex {p_i’}&fg=000000$ in the interval $latex {(\gamma_1, \gamma_2)}&fg=000000$, there cannot be more than one sign change of $latex {p_i}&fg=000000$ between $latex {\gamma_1}&fg=000000$ and $latex {\gamma_3}&fg=000000$ and between $latex {\gamma_3}&fg=000000$ and $latex {\gamma_2}&fg=000000$.

So, from these considerations, it’s clear that knowing the signs of the derivatives of the functions $latex {p_i}&fg=000000$ and of the remainders with respect to these derivatives is very helpful. But of course, obtaining these signs might recursively entail working with the second derivatives and so forth. Thus, let’s define the following hierarchy of collections of polynomials:

$latex \displaystyle \mathcal{P}_0 = \{p_1, \dots, p_n\}&fg=000000$

$latex \displaystyle \mathcal{P}_1 = \mathcal{P}_0 \cup \{p’ : p \in \mathcal{P}_0\} \cup \{(p \text{ mod } q) : p,q \in \mathcal{P}_0, \deg(p)\geq \deg(q)\}&fg=000000$

$latex \displaystyle \dots&fg=000000$

$latex \displaystyle \mathcal{P}_{i} = \mathcal{P}_{i-1} \cup \{p’ : p \in \mathcal{P}_{i-1}\} \cup \{(p \text{ mod } q) : p,q \in \mathcal{P}_{i-1}, \deg(p)\geq \deg(q)\}&fg=000000$

$latex \displaystyle \dots&fg=000000$

The degree of the new polynomials added decreases at each stage. So, there is some $latex {k}&fg=000000$ such that $latex {\mathcal{P}_k}&fg=000000$ doesn’t grow anymore. Let $latex {\mathcal{P}_k = \{t_1, t_2, \dots, t_N\}}&fg=000000$ with $latex {\deg(t_1) \leq \deg(t_2) \leq \cdots \leq \deg(t_N)}&fg=000000$. Note that $latex {N}&fg=000000$ is bounded as a function of $latex {n}&fg=000000$ (though the bound is admittedly pretty poor).

Now, we construct the sign configuration diagram for the polynomials $latex {\{t_1, t_2, \dots, t_N\}}&fg=000000$, starting from that of $latex {t_1}&fg=000000$ (a constant), then adding $latex {t_2}&fg=000000$, and so on. When adding $latex {t_i}&fg=000000$, the real line has already been partitioned according to the roots of $latex {t_1, \dots, t_{i-1}}&fg=000000$. As described earlier, $latex {\text{sign}(t_i)}&fg=000000$ at a root $latex {\gamma}&fg=000000$ of $latex {t_j}&fg=000000$ is the sign of $latex {(t_i \text{ mod } t_j)}&fg=000000$ at $latex {\gamma}&fg=000000$, which has already been computed inductively. Then, we can proceed according to the earlier discussion by partitioning into two any interval where there’s a change in the sign of $latex {t_i}&fg=000000$. The only other technicality is that the sign of $latex {t_i}&fg=000000$ at $latex {-\infty}&fg=000000$ and its sign at $latex {+\infty}&fg=000000$ are not determined by the other polynomials. But these are easy to find given the parity of the degree of $latex {t_i}&fg=000000$ and the sign of its leading coefficient.

3. $latex {m>0}&fg=000000$ and general polynomials

Finally, consider the most general case, $latex {m>0}&fg=000000$ and arbitrary bounded degree polynomials. Let’s interpret $latex {{\mathbb R}[x,y_1,y_2,\dots, y_m]}&fg=000000$ as polynomials in $latex {x}&fg=000000$ with coefficients that are polynomials in $latex {y_1, \dots, y_m}&fg=000000$, and try to push through what we did above. We can start by extending the definitions of the hierarchy of collections of polynomials. Suppose $latex {p(x,\mathbf{y}) = p_D(\mathbf{y}) x^D + p_{D-1}(\mathbf{y}) x^{D-1} + \cdots + p_0(\mathbf{y})}&fg=000000$ and $latex {q(x,\mathbf{y}) = q_E(\mathbf{y}) x^E + q_{E-1}(\mathbf{y}) x^{E-1} + \cdots + q_0(\mathbf{y})}&fg=000000$ with $latex {d \leq D}&fg=000000$ the largest integer such that $latex {p_d(\mathbf{y}) \neq 0}&fg=000000$, $latex {e \leq E}&fg=000000$ the largest integer such that $latex {q_e(\mathbf{y}) \neq 0}&fg=000000$, and $latex {d \geq e}&fg=000000$. One problem is that $latex {(p \text{ mod } q)}&fg=000000$, as a polynomial in $latex {x}&fg=000000$, may not have polynomials in $latex {y}&fg=000000$ as its coefficients. To skirt this issue, let:

$latex \displaystyle (p\text{ zmod } q) = q_e(\mathbf{y}) p(x,\mathbf{y}) – p_d(\mathbf{y})x^{d-e} q(x,\mathbf{y})&fg=000000$

Then, whenever $latex {q(x,\mathbf{y}) = 0}&fg=000000$, the sign of $latex {p(x,\mathbf{y})}&fg=000000$ equals the sign of $latex {(p\text{ zmod } q)}&fg=000000$ at $latex {(x,\mathbf{y})}&fg=000000$ divided by the sign of $latex {q_e(\mathbf{y})}&fg=000000$. Moreover, $latex {(p \text{ zmod } q)}&fg=000000$ is a degree $latex {(d-1)}&fg=000000$ polynomial. So, we will want to replace mod by zmod when constructing the collections $latex {\mathcal{P}_i}&fg=000000$. All this is fine except that we have no idea what $latex {d}&fg=000000$ and $latex {e}&fg=000000$ are and what specifically the sign of $latex {q_e(\mathbf{y})}&fg=000000$ is. Recall that $latex {\mathbf{y}}&fg=000000$ is an unknown parameter! So, what to do? Well, the most naive thing: for each polynomial $latex {p}&fg=000000$, let’s just guess whether each coefficient of $latex {p}&fg=000000$ is positive, negative, or zero.

Now, we can define the hierarchy of collections of polynomials $latex {\mathcal{P}_0, \mathcal{P}_1, \dots}&fg=000000$ in the same way as in the previous section, except that here, the polynomials are in $latex {{\mathbb R}[x,y_1, \dots, y_m]}&fg=000000$, we use zmod instead of mod, and each time we add a polynomial, we make a guess about the signs of its coefficients. The degree of the newly added polynomials decreases at each stage, so that there’s some finite bound $latex {k}&fg=000000$ such that $latex {\mathcal{P}_k}&fg=000000$ doesn’t grow anymore. We can now run the procedure described in the previous section to find the sign configuration diagram for $latex {\mathcal{P}_k}&fg=000000$. In the course of this procedure, whenever we need to know the sign of the leading coefficient of a polynomial, we use the value already guessed for it. Finally, we check whether any interval in the sign configuration diagram contains the desired sign pattern.

Consider the whole set of guesses we’ve made in the course of arriving at this sign configuration diagram. Notice that the total number of guesses is finite, bounded as a function of $latex {n}&fg=000000$. Now, let’s call the set of guesses good if it results in a sign configuration diagram that indeed contains an interval with the wanted sign pattern. Our final quantifier-free formula is a disjunction over all good sets of guesses that checks whether the given input values of $latex {y_1, \dots, y_m}&fg=000000$ is consistent with a good set of guesses.

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