2.2 Candidate Solutions
2.2.1 Candidate Solutions and Factorization
So far we have analyzed the properties of parametric root descriptions and candidate solutions. We did so for arbitrary quantifier-free Tarski formulas. Here, in contrast, we restrict ourselves to atomic formulas of the form f % 0, where f ∈ Z[u][x] and the relation % is one of {=, 6=, <, ≤, ≥, >}. We investigate the special case when we can factorize f into c · f1· · · fk, where c ∈ Z[u] and
fi∈ Z[u][x] for all i ∈ {1, . . . , k}.
Since f % 0 translates to a set of sign conditions involving fi, it is quite
natural to expect that it is sufficient to consider any sets of candidate solutions for atomic formulas fi% 0, i ∈ {1, . . . , k}, to derive a set of candidate solutions
for f % 0. We make things precise and show how to derive a set of candidate solutions for f % 0 from any sets of candidate solutions for atomic formulas involving the polynomials fi. We begin our exposition with the following key
Proposition 21. Let f ∈ Z[u][x] with deg f > 0. Let f = c · f1· · · fk be
a factorization of f such that c ∈ Z[u], fi ∈ Z[u][x], and deg fi > 0 for all
i ∈ K = {1, . . . , k}. Denote by cςi, where ς is one of {=, 6=, <, ≤, ≥, >}, a set of candidate solutions for the atomic formula fiς 0. Then the following hold:
(i) If % ∈ {=, 6=}, then the set c = [
i∈K
c%i is a set of candidate solutions for the atomic formula f % 0.
(ii) If % ∈ {<, >}, then the set c = [
i∈K
(c<i ∪ c>
i ) is a set of candidate solutions
for the atomic formula f % 0. (iii) If % ∈ {≤, ≥}, then the set c = [
i∈K
(c≤i ∪ c≥i ) is a set of candidate solutions
for the atomic formula f % 0.
Proof. Let a ∈ Rm and consider a boundary point β ∈ R of the satisfying set Φ(f % 0, a). In all three parts of the proposition we have to prove that β is properly covered by c.
To begin with, notice that the fact that β is a boundary point of the set Φ(f % 0, a) ensures that f hai is not identically zero, i.e., chai ∈ R \ {0} and
fihai ∈ R[x] \ {0} for every i ∈ K. Recall that Proposition 9 guarantees that
each boundary point of the set Φ(f % 0, a) is a root of f hai, so in particular
f hai(β) = 0. Since f = c · f1· · · fk and chai ∈ R \ {0}, this implies that
fihai(β) = 0 for at least one i ∈ K.
Now we are ready to prove the three parts of the proposition:
(i) We show (i) for the case when % is “=.” The proof for the case when % is “6=” is similar. Since we consider the atomic formula f = 0 and f hai is not identically zero, β has to be an isolated point such that fihai(β) = 0
for at least one i ∈ K. Recall that fihai ∈ R[x] \ {0}, so β is a boundary
point of the set Φ(fi = 0, a) as well. On the other hand, we assume
that c=i is a set of candidate solutions for fi = 0, so there exists either
(f, S, IP) ∈ c=i ⊆ c such that (f, S) covers β, or there exist (f1, S1, WLB),
(f2, S2, WUB) ∈ c=i ⊆ c such that both (f1, S1) and (f2, S2) cover β. In
either case we deduce that β is properly covered by c. This finishes the proof of (i).
(ii) We show (ii) for the case when % is “<.” The proof for the case when % is “>” is similar. Let I = {i ∈ K | fihai(β) = 0}. Since f = c · f1· · · fk and
chai ∈ R \ {0}, fihai ∈ R[x] \ {0} for every i ∈ K, we deduce that I 6= ∅.
In the following we denote by ε a positive infinitesimal. The definition of
I together with Lemma 18 ensure that the following two equations hold:
sgn f hai(β − ε) = sgn chai(β − ε) ·Y i∈K sgn fihai(β − ε) = sgn chai ·Y i∈I sgn fihai(β − ε) · Y i∈K\I sgn fihai(β)
and sgn f hai(β + ε) = sgn chai(β + ε) ·Y i∈K sgn fihai(β + ε) = sgn chai ·Y i∈I sgn fihai(β + ε) · Y i∈K\I sgn fihai(β).
All sign expressions in both of these equations are nonzero, because both sgn f hai(β −ε) and sgn f hai(β +ε) are nonzero. Isolating in both equa- tions the common term sgn(chai), and putting these together we obtain that sgn f hai(β − ε) Q i∈Isgn fihai(β − ε) = sgn f hai(β + ε) Q i∈Isgn fihai(β + ε) . (2.4)
Since we consider the atomic formula f < 0 and f hai is not identically zero, β has to be a strict lower bound, a strict upper bound, or an exception point of the set Φ(f < 0, a). We distinguish these three cases:
1. β is a strict lower bound: In this case we have sgn(f hai(β − ε)) = 1 and sgn(f hai(β +ε)) = −1: Plugging these values into Equation (2.4) yields −Y i∈I sgn fihai(β − ε) = Y i∈I sgn fihai(β + ε).
Thus, sgn(fihai(β − ε)) 6= sgn(fihai(β + ε)) for at least one i ∈ I.
Now there are two cases to consider:
If sgn(fihai(β + ε)) = −1, then β is a strict lower bound of the set
Φ(fi < 0, a). Since c<i is a set of candidate solutions for fi < 0,
c<i ⊆ c properly covers β.
If sgn(fihai(β + ε)) = 1, then β is a strict lower bound of the set
Φ(fi > 0, a). Since c>i is a set of candidate solutions for fi > 0,
c>i ⊆ c properly covers β.
We have proven that β is properly covered by c in both cases, so the proof of (ii) for the case when β is a strict lower bound is finished. 2. β is a strict upper bound: Now we have sgn(f hai(β − ε)) = −1 and
sgn(f hai(β + ε)) = 1. Similarly as in the previous case, plugging these values into Equation (2.4) we obtain that there exists at least one i ∈ I such that sgn(fihai(β − ε)) 6= sgn(fihai(β + ε)). Using
similar arguments as in the previous case, we show that β is properly covered by c. This finishes the proof of (ii) for the case when β is a strict upper bound.
3. β is an exception point: It holds that sgn(f hai(β − ε)) = −1 and sgn(f hai(β + ε)) = −1: Plugging these values into Equation (2.4) we obtain Y i∈I sgn fihai(β − ε) = Y i∈I sgn fihai(β + ε).
This implies that
(a) there exists at least one i ∈ I such that
(b) there exist i, j ∈ I, where i 6= j ∈ I such that sgn fihai(β − ε) = sgn fjhai(β + ε) = 1,
sgn fihai(β + ε) = sgn fjhai(β − ε) = −1.
In case (a), β is either an exception point of the set Φ(fi< 0, a), or
β is an exception point of the set Φ(fi> 0, a). Therefore, c = c<i ∪ c > i
properly covers β.
In case (b), β is a strict lower bound of the set Φ(fi < 0, a) and
a strict upper bound of the set Φ(fj < 0, a). Therefore, there ex-
ists (f, S, EP) ∈ c<i ∪ c>
j such that (f, S) covers β, or there exist
(f1, S1, SLB) and (f2, S2, SUB) from c<i ∪ c >
j such that both (f1, S1)
and (f2, S2) cover β. This means that β is properly covered by
c = c<i ∪ c> j.
(iii) The proof is similar to the proof of part (ii), so we omit it.
Let us at this point state a few well-known equivalences that hold in the theory of the real closed fields:
Proposition 22. The following equivalences hold in R for any f , g ∈ Z[u][x]
and any positive integer m:
f · g = 0 ←→ f = 0 ∨ g = 0, f · g 6= 0 ←→ f 6= 0 ∧ g 6= 0, f · g > 0 ←→ f > 0 ∧ g > 0 ∨ f < 0 ∧ g < 0, f · g < 0 ←→ f < 0 ∧ g > 0 ∨ f > 0 ∧ g < 0, g2m= 0 ←→ g = 0, g2m6= 0 ←→ g 6= 0, g2m< 0 ←→ false, g2m≤ 0 ←→ g = 0, g2m≥ 0 ←→ true, g2m> 0 ←→ g 6= 0, g2m+1% 0 ←→ g % 0, where % ∈ {=, 6=, <, ≤, ≥, >}.
With these equivalences at hand, we sketch an alternative proof of Propo- sition 21: Consider an atomic formula of the form c · f1· · · fk = 0. The first
equivalence of Proposition 22 ensures that this atomic formula is equivalent to
c = 0 ∨ f1 = 0 ∨ · · · ∨ fk = 0. Now we repeatedly apply Proposition 15 to
deduce that the union of candidate solutions for formulas fi= 0 yields a set of
candidate solutions for c = 0 ∨ f1= 0 ∨ · · · ∨ fk= 0. Proposition 14 then ensures
that this set is also a set of candidate solutions for the formula c · f1· · · fk= 0.
This proves part (i) of Proposition 21 for the case when % is “=.”
Consider now an atomic formula of the form c · f1· · · fk < 0. Applying
repeatedly the third and the fourth equivalence of Proposition 22 we end up with a formula of length O(2k) containing only atomic formulas of the form c><0
and fi><0. Repeatedly applying Proposition 15 then ensures that the union of
formula c · f1· · · fk < 0. This proves part (ii) of Proposition 21 for the case
when % is “<.”
Observe that we do not need to write down the equivalent formula of length
O(2k) explicitly to construct a set of candidate solutions for c · f1· · · fk < 0.
In fact, the formula c · f1· · · fk < 0 hides the Boolean complexity behind the
algebraic complexity exhibited by a polynomial of higher degree. This way of “hiding” the Boolean complexity behind the algebraic complexity has an inter- esting consequence: A set of candidate solutions for c · f1· · · fk < 0 obtained by
Proposition 21, i.e., by considering the factors of f , possibly contains redundant candidate solutions. The actual reason for this will become clear in Chapter 3 in the context of conjunctive associativity, where we will analyze the Boolean structure of such a formula. Here we illustrate this redundancy on a simple example:
Example 23. Consider the atomic formula f1·f2< 0, where f1= x2−47 ∈ R[x]
and f2= −x + 5 ∈ R[x]. Sets of candidate solutions for atomic formulas f1><0
and f2><0 are listed in Table 2.2. Note that these sets are minimal, i.e., no
candidate solution can be removed from them.
Proposition 21 guarantees that the union of the sets listed in Table 2.2 is a set of candidate solutions for atomic formula f1· f2 < 0. This is indeed
the case, because the satisfying set Φ(f1· f2 < 0) is ]−
√
47, 5[ ∪ ]√47, ∞[, and (f1, ((1, 0, −1, 0, 1), 1), SLB) covers −
√
47, (f1, ((1, 0, −1, 0, 1), 2), SLB) cov-
ers √47, and (f2, ((1, 0, −1), 1), SUB) covers 5. Therefore, all boundary points
are properly covered. At the same time, we see that the candidate solution (f2, ((1, 0, −1), 1), SLB) is redundant, because it covers the boundary point 5,
which is not a strict lower bound. Similarly, (f1, ((1, 0, −1, 0, 1), 1), SUB) cov-
ers −√47, and (f1, ((1, 0, −1, 0, 1), 2)}, SUB) covers
√
47, but neither −√47 nor √
47 is a strict upper bound of the satisfying set Φ(f1· f2 < 0). It would be
actually sufficient to take
(f1, ((1, 0, −1, 0, 1), 1), SLB),
(f1, ((1, 0, −1, 0, 1), 2), SLB),
(f2, ((1, 0, −1), 1), SUB)
to obtain a set of candidate solutions for f1· f2< 0.
Our example also shows that Proposition 21 cannot be strengthened in the sense that we need to include both sets of candidate solutions generated by “<” as well as by “>:” In our example, taking the union of the sets of candidate solutions only for f1< 0 and f2< 0 does not yield a set of candidate solutions
for f1· f2< 0, because the strict lower bound
√
47 is not properly covered, and the strict upper bound 5 is not properly covered. Similarly, taking the union of the sets of candidate solutions for f1> 0 and f2 > 0 does not yield a set of
candidate solutions for f1· f2< 0, because the strict lower bound −
√
47 is not
properly covered. 3