Now consider constructing a cylindrical algebraic decomposition for
(note the decomposition is actually semi-algebraic). Figure
6.35 shows an example for
. First consider
how to iteratively project the polynomials down to
to ensure
that when the decomposition of
is constructed, the
sign-invariant property is maintained. The resulting decomposition
corresponds to a singular complex.
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There are two cases that cause cell boundaries to be formed, as
shown in Figure 6.33. Let
denote the
original set of polynomials in
that are used to
define the semi-algebraic set (or Tarski sentence) in
. Form a
single polynomial
. Let
, which is also a polynomial. Let
, which is the greatest common divisor of
and
. The set of zeros of
is the set of all points that are
zeros of both
and
. Being a zero of
means
that the surface given by
does not vary locally when perturbing
. These are places where a cell boundary needs to be formed
because the surface may fold over itself in the
direction, which
is not permitted for a cylindrical decomposition. Another place where
a cell boundary needs to be formed is at the intersection of two or
more polynomials in
. The projection technique from
to
generates a set,
, of polynomials in
that satisfies these requirements. The
polynomials
have the property that at least one
contains a zero point below every point in
for which
and
, or polynomials in
intersect. The projection method that constructs
involves computing principle subresultant coefficients, which
are covered in [77,853]. Resultants, of which the
subresultants are an extension, are covered in [250].
The polynomials in
are then projected to
to
obtain
. This process continues until
is
obtained, which is a set of polynomials in
. A
one-dimensional decomposition is formed, as defined earlier. From
, a single polynomial is formed by taking the product, and
is partitioned into 0-cells and
-cells. We next describe
the process of lifting a decomposition over
up to
.
This technique is applied iteratively until
is reached.
Assume inductively that a cylindrical algebraic decomposition has been
computed for a set of polynomials
in
. The decomposition consists of
-cells for
which
. Let
. For each one of the
-cells
, a
cylinder over
is defined
as the
-dimensional set
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(6.23) |
![]() ![]() |
(6.24) |
![]() ![]() |
(6.25) |
![]() ![]() |
(6.26) |
![]() ![]() |
(6.27) |
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Note that the cells do not necessarily project onto a rectangular set,
as in the case of a higher dimensional vertical decomposition. For
example, a generic -cell
for a decomposition of
is
described as the open set of
such that
The resulting decomposition is sign invariant, which allows the
decision and quantifier-elimination problems to be solved in finite
time. To solve a decision problem, the polynomials in
are
evaluated at every sample point to determine whether one of them
satisfies the Tarski sentence. To solve the quantifier-elimination
problem, note that any semi-algebraic sets that can be constructed
from
can be defined as a union of some cells in the
decomposition. For the given Tarski sentence,
is formed
from all polynomials that are mentioned in the sentence, and the cell
decomposition is performed. Once obtained, the sign information is
used to determine which cells need to be included in the union. The
resulting union of cells is designed to include only the points in
at which the Tarski sentence is TRUE.
Steven M LaValle 2020-08-14