Cylindrical Algebraic Decomposition - an Introduction

Cylindrical Algebraic Decomposition - an Introduction

Mats Jirstrand

Department of Electrical Engineering Linkoping University

S-581 83 Linkoping Sweden

email:

matsj@isy.liu.se

1995-10-18

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Cylindrical Algebraic Decomposition - an Introduction

Mats Jirstrand

Department of Electrical Engineering Linkoping University, S-581 83 Linkoping, Sweden

email:

matsj@isy.liu.se

1995-10-18

Abstract. In this report we give an introduction to a constructive way of treating systems of polynomial equations and inequalities. We present a method called cylindrical algebraic decomposition (CAD) discovered 1973 by Collins. The method constructs a decomposition of^Rn such that a given set of polynomials have constant sign on each component. All concepts needed to understand the algorithm is presented, e.g., polynomial remainder sequences, subresultants, principal subresultant coe cients, Sturm chains and algebraic number representations.

Keywords: cad, inequalities, real polynomial systems, semi-algebraic sets, real algebra

1 Introduction

The aim of this report is to describe a constructive method in real algebraic geometry to obtain a so called Cylindrical Algebraic Decomposition, (CAD) of

^Rn

. A CAD is a decomposition of the n -dimensional real space into regions over which a given set of polynomials have constant sign. Given such a decomposition it is easy to give a solution of a system of inequalities and equations dened by the polynomials, so called real polynomial systems.

The CAD-algorithm was discovered by Collins in 1973 6] as a sub- algorithm in his work on an eective method for quantier elimination (QE) in real closed elds. The rst algorithm for the solution of this problem was given by Tarski 1948 18] but was completely impractical. The sub- algorithms of CAD has been improved over the years see e.g., 16, 14, 11]

and it is now possible to run nontrivial examples on a computer with realistic runtimes. For an extensive bibliography on QE and CAD see 1].

The report is organized as follows: section 2 presents an introductory

example of the use of CAD. In Section 3 and 4 we present the machinery

needed to understand the CAD algorithm. Section 5 is a presentation of the algorithm. In Section 6 we consider some examples and nally in Section 7 we give a summary of the report.

2 An Introductory Example

To get a feeling of what kind of problems CAD may solve we present an introductory example.

Example 2.1 Suppose we are given the following real polynomial system:

x

₂₁⁺

x

₂₂^-

1 < 0

x

₃₁^-

x

₂₂ ⁼

0 ⁽¹⁾

For system (1) the set of solutions is very easy to nd. It consists of the part of the cuspidal cubic inside the unit circle, see Figure 1.

x1

x2

Figure 1: The zero sets of the polynomials in system (1) and some projec- tions onto the x

1

-axis (gray dots).

In this case it was easy to solve the system by inspection but is there a systematic approach which can be used for more complicated problems?

The answer is yes and we will now give an outline of a procedure for nding the solution of system (1). Let f

1 ⁼

x

₂₁ ⁺

x

₂₂^-

1 and f

2 ⁼

x

₃₁ ^-

x

₂₂

. The procedure can be divided into three steps:

(i) Find the projections onto the x

1

-axis of all points of the zero sets of

f

1

and f

2

corresponding to vertical tangents, singularities and inter-

sections (gray dots). The projections of these points of system (1)

is

-

1 0  1

where  0:7549 is the real zero of x

³⁺

x

^2-

1 . The projected points induce a decomposition of the x

1

-axis into 9 components ( 4 points and

5 open intervals).

(ii) Evaluate f

1

and f

2

over a \sample point" of each component (dot or interval). This gives 2

^

9 polynomials in x

2

.

(iii) Evaluate the signs of the obtained polynomials, i.e., evaluate the signs of f

1

and f

2

on a vertical line over each \sample point".

Notice that a vertical line over each component intersects the zero sets of f

1

and f

2

a constant number of times, since the points of the x

1

-axis over which the zero sets change \character" is exactly the points picked out by the projection. Enumerating the components of the decomposition from left to right Table 1 summarizes the number of intersections over each component.

Component 1 2 3 4 5 6 7 8 9

# of intersections 0 1 2 3 4 2 4 3 2

Table 1: Number of intersections with

⁽

x

1

x

2^)2R2 j

f

1⁼

0 and f

2 ⁼

0

^

. To illustrate step

⁽

ii

⁾

and

⁽

iii

⁾

we just consider \sample points" of com- ponent 2 and 5 , 

2 ^{= -}

1 (which is the only choice for this particular com- ponent) and 

5 ⁼ 12

. The evaluation of f

1

and f

2

is summarized in Table 2.

x

1

f

1⁽

x

1

x

2⁾

f

2⁽

x

1

x

2⁾

# real roots of f

1

 f

2

"

sign

⁽

f

1⁾

sign

⁽

f

2⁾

#

 x

2 ^:-1!+1



2

x

₂₂ ^-

1

^-

x

₂₂

2 0

^{ +}- -^{0 +}-





5

x

₂₂^-_{34 18} ^-

x

₂₂

2 2

^{ +}- -^{0 -}- ^-0 ^-+ ^-0 ^-- -^{0 +}-



Table 2: Evaluation of f

1

and f

2

over 

2

and 

5

.

Since f

1

< 0 and f

2 ⁼

0 for a solution of system (1) Table 2 shows that

there are two sets of solutions of (1) over component 5, i.e., over the interval

0 < x

1

<  which consists of parts of the zero set of f

2

. We also see that there are no solutions over component 2.

The whole solution of system (1) is given by

x

₃₁^-

x

₂₂ ⁼

0

0

^

x

1

<  ⁽²⁾

where  is the real zero of x

³⁺

x

^2-

1 .

In Figure 2

⁽

sign

⁽

f

1⁽

x

1

x

2⁾⁾

 sign

⁽

f

2⁽

x

1

x

2⁾⁾

along the lines x

1 ^=-

1 and

x

1 ⁼ 12

is displayed.

x₁ x2

2 ₅

(+^-)

(-⁺⁾

(+^-)

(-^-)

(0^-)

(-0⁾

(-0⁾

(-^-)

(0^-)

(+^-)

(+^-)

(0^-)

Figure 2: A part of the construction of the solution.

In this example it was easy to nd the projections and evaluate the

polynomials over dierent sample points. However, the general projection

operation is more intricate and for the evaluation step one usually need to

do calculations with algebraic numbers. In the rest of this report we will

treat these problems in detail.

3 A Semi-Algebraic Dictionary

In this section we will introduce the reader to some of the basic concepts of the CAD-machinery and their geometrical interpretation. The denitions essentially follows 2]. A very detailed description of CAD is given in 17]

and some briefer presentations may be found in 9, 10].

Since one of the aims of this report is to understand algorithmic ways of treating systems of equations and inequalities, important objects to study is so called semi-algebraic sets.

Denition 3.1 A set is semi-algebraic if it can be constructed by nitely many applications of union, intersection and complementation operations on sets of the form

f

x

^2Rnj

f

⁽

x

^)

0

^g



where f

^2R

x

1

::: x

n^]

.

An interesting property of semi-algebraic sets is that they are closed under projection, i.e., the projection of a semi-algebraic set to some lower dimensional space is again a semi-algebraic set. This is not true for algebraic sets, i.e., sets dened by a system of polynomial equations (consider the unit circle). We also observe that the set of points which satises a system of equalities and inequalities is a semi-algebraic set.

Example 3.1 An example of a semi-algebraic set

^{S 2R2}

,

S =



x

^{2R2 j }

x

₂₁⁺

x

₂₂^-

1

^

0

^{^}

x

₂₁^-

x

2⁼

0

^{ _}



(

x

1^-

1

⁾²⁺⁽

x

2^-

1

^)2-

1

^

0

^{^ (}

x

1^-

2

⁾²⁺⁽

x

2^-

2

^)2-

1

^

0

^

:

The projection of the set onto the x

1

-axis

Sx^{1 =}



x

^{2R2 j}

x

₂₁⁺

x

₄₁^-

1

^

0

^_

1

^

x

1 ^

2

^



is again a semi-algebraic set, see Figure 3.

We now introduce some terminology for describing how algebraic curves

and surfaces partition

^Rn

.

x

2

x

1

Figure 3: A semi-algebraic set (black region) and its projection onto the

x

1

-axis (gray region). The dashed sets are the zero sets of the dening polynomials.

Denition 3.2



A region, R is a connected subset of

^Rn

.



The set Z

⁽

R

^{) =}

R

^{R = f(}

x

^)j



²

R x

^2Rg

is called a cylinder over R .



Let ff

1

f

2

be continuous, real-valued functions on R . A f -section of Z

⁽

R

⁾

is the set

f(

f

⁽



^{)) j}



²

R

^g

and a

⁽

f

1

f

2⁾

-sector of Z

⁽

R

⁾

is the set

f(



^)j



²

R f

1⁽



⁾

< < f

2⁽



^)g

:

Notice that an algebraic equation implicitly denes a set of real-valued,

piecewise continuous functions. The real roots of an polynomial equation in

one variable are piecewise continuous functions of its coecients.

Denition 3.3 Let X

^Rn

. A decomposition of X is a nite collection of disjoint regions (components) whose union is X :

X

^{= k}

i⁼1

X

i

 X

i^\

X

j⁼

 i

⁶⁼

j

Denition 3.4 A stack over R is a decomposition which consists of

f

i

-sections and

⁽

f

i

f

i⁺1⁾

-sectors, where f

0

< ::: < f

k⁺1

for all x

²

R and

f

0 ^=-1

 f

k⁺1 ⁼⁺¹

.

Observe the strict inequalities in Denition 3.4. Geometrically this means that the graphs of dierent functions may not intersect each other over R .

x

1

x

2

x

3

f

1-section

f

2-section

(

f

1

f

2⁾-sector

R

-region

Z

⁽

R

⁾-cylinder

Figure 4: A geometrical interpretation of the denitions of region, cylinder,

sections, sectors and stack.

Denition 3.5 A decomposition

^D

of

^Rn

is cylindrical if

n

⁼

1

^D

is a partition of

^R1

into a nite set of numbers, and the nite and innite open intervals bounded by these numbers.

n > 1

^{D0 =}

F

1

:::



F

m

is a cylindrical decomposition of

^Rn-1

and over each F

i

there is a stack which is a subset of

^D

.

From the denition of a cylindrical decomposition it is clear that any cylindrical decomposition of

^Rn

induces a cylindrical decomposition of

^Rn-1

etc. down to

^R1

.

Denition 3.6 Let X

^Rn

and f

²

k

^

x

1

::: x

n^]

. Then f is invariant on

X if one of the following conditions holds:

(

i

^{) 8}

x

²

X

^:

f

⁽

x

⁾

> 0

(

ii

^{) 8}

x

²

X

^:

f

⁽

x

⁾⁼

0

(

iii

^{) 8}

x

²

X

^:

f

⁽

x

⁾

< 0

The set

^{F = f}

f

1

::: f

r^{g2 R}

x

1

::: x

n^]

of polynomials is invariant on

X if each f

i

is invariant on X . We also say that X is

^F

-invariant if

^F

is invariant on X .

Example 3.2 Let

^{F =f}

x

₂₁⁺

x

₂₂^-

1

⁽

x

1^-

1

^)2-

x

₂₂^-

1

^g

. Then the set

^I

is

^F

-invariant, see Figure 5.

Denition 3.7 A decomposition is algebraic if each of its components is a semi-algebraic set.

Example 3.3 Let f

⁽

x

⁾⁼⁽

x

1^-

2

⁾²⁺⁽

x

2^-

2

^)2-

1 . Then

f

x

^j

f

⁽

x

⁾

> 0

^{g  f}

x

^j

f

⁽

x

⁾⁼

0

^{g  f}

x

^j

f

⁽

x

⁾

< 0

^g

is an algebraic decomposition of

^R2

.

I

x

1

x

2

Figure 5: The zero set of

^F

in Example 3.2 (dashed lines) and the

^F

- invariant set

^I

.

Notice that when the decomposition is dened by a set of polynomials it is algebraic since all boundaries are zero sets of the dening polynomials.

Denition 3.8 A Cylindrical Algebraic Decomposition (CAD) of

^Rn

is a decomposition which is both cylindrical and algebraic. The components of a CAD is called cells.

Example 3.4 Let

^{F =f(}

x

1^-

2

⁾²⁺⁽

x

2^-

2

^)2-

1

⁽

x

1^-

3

⁾²⁺⁽

x

2^-

2

^)2-

1

^g

. The CAD consists of the distinct black \dots", \arcs" and \patches of white space" in Figure 6. The induced CAD of

^R1

consists of the gray dots on the

x

1

-axis and the intervals between.

x1

x2

Figure 6: A CAD of

^R2

and the induced CAD of

^R1

.

One of the components may be characterized semi-algebraically as

f

x

^{2R2 j (}

x

1^-

2

⁾²⁺⁽

x

2^-

2

^)2-

1 > 0

^{^}

(

x

1^-

3

^)2-(

x

2^-

2

^)2-

1 < 0

^{^}

x

1

< 3

^{^}

x

2

< 2

^g

:

Which one?

4 Basic Tools

In this section we will present some theoretical tools which are important for the understanding of the dierent steps of the CAD-algorithm. These are polynomial remainder sequences, subresultants, principal subresultant coecients, Sturm chains and representations of algebraic numbers. All concepts are treated in 17]. These are also the main tools for other con- structive methods in real algebra and real algebraic geometry, see 4, 3, 12].

4.1 Common Factors of Multivariate Polynomials

We start with the univariate case and then generalize the ideas to multivari- ate polynomials. Given two polynomials fg

²

k

^

x

^]

. How many common zeros do f and g have? How many distinct zeros do f have? These two questions may be answered knowing the greatest common divisor (gcd) of two polynomials. In the rest of this section k denotes a eld of characteristic zero.

Lemma 4.1 The number of common zeros of f g

²

k

^

x

^]

is deg

⁽

gcd

⁽

fg

⁾⁾

Proof. The lemma follows from the denition of gcd and the fact that the number of zeros (counting multiplicity) of a univariate polynomial is equal to its degree.

Lemma 4.2 The number of distinct zeros of f

²

k

^

x

^]

is deg

⁽

f

^)-

deg

⁽

gcd

⁽

ff

⁰⁾⁾

:

Proof. Let f

²

k

^

x

^]

have l distinct zeros 

1

:::

l

. Then

f

⁽

x

⁾⁼

a

⁽

x

^-



1⁾e¹(

x

^-



l⁾e^l

:

Dierentiating f

⁽

x

⁾

it is easy to see that

deg

⁽

gcd

⁽

ff

⁰⁾⁾⁼

deg

⁽

f

^)-

l

and the lemma follows.

How do one calculate the gcd of two univariate polynomials? The an- swer is provided by the Euclidean algorithm for polynomials. By repeated polynomial division we end up with the gcd of the two original polynomials.

The process may be described as

f

1 ⁼

q

1

f

2⁺

f

3

 deg

⁽

f

3⁾

< deg

⁽

f

2⁾

f

2 ⁼

q

2

f

3⁺

f

4

 deg

⁽

f

4⁾

< deg

⁽

f

3⁾

f

p^-2 ⁼

... q

p^-2

f

p^-1⁺

f

p

 deg

⁽

f

p⁾

< deg

⁽

f

p^-1⁾

f

p^-1 ⁼

q

p^-1

f

p⁺

0

where f

p ⁼

gcd

⁽

f

1

f

2⁾

, see 13] or 8] for a detailed treatment. The al- gorithm terminates since the degree decreases in each step. The sequence

(

f

1

f

2

::: f

p⁾

is called a polynomial remainder sequence (PRS).

Example 4.1 Let f

1⁼⁽

1

⁺

x

⁾⁽

3

⁺

x

⁾⁽

1

⁺

x

⁺

x

²⁾

and f

2 ⁼⁽

1

⁺

x

⁾⁽

2

⁺

x

⁾²

. The PRS of f

1

f

2

in expanded form is

f

1⁽

x

^{) =}

9

⁺

24x

⁺

31x

²⁺

23x

³⁺

8x

⁴⁺

x

⁵

f

2⁽

x

^{) =}

4

⁺

8x

⁺

5x

²⁺

x

³

f

3⁽

x

^{) =}

9

⁺

12x

⁺

3x

²

f

4⁽

x

^{) =}

1

⁺

x

where f

4⁽

x

⁾⁼

1

⁺

x

⁼

gcd

⁽

f

1

f

2⁾

.

We now extend the above ideas to multivariate polynomials. Let fg

²

k

^

x

1

::: x

n^-1^]

x

n^]

, i.e., f and g is considered as polynomials in the variable

x

n

with polynomial coecients in the variables x

1

::: x

n^-1

. Hence the

coecients of f and g is no longer xed numbers but depends on over which

point 

^{= (}



1

::: 

n^-1^{) 2 R}n^-1

they are evaluated. This implies that

both the number of zeros and their location varies for dierent 

^2Rn-1

.

Example 4.2 Let

f

⁼⁽

x

₂₁⁺

x

₂₂^-

1

⁾

x

₂₃⁺⁽

x

2^-

1

⁾

x

3⁺

x

₂₂

:

Consider f as a polynomial in x

3

. Then deg

_x³⁽

f

⁾⁼

2 everywhere except on the cylinder x

₂₁ ⁺

x

₂₂^-

1

⁼

0 . On the cylinder deg

x³⁽

f

^{) =}

1 except on the line

⁽

x

1

x

2⁾⁼⁽

01

⁾

where deg

_x³⁽

f

⁾⁼

0 .

For some 

^2Rn-1

there might be common factors of f and g but not for others, i.e., the gcd

⁽

fg

⁾

and hence its degree depends on  . To calculate a gcd of two polynomials we use the Euclidean algorithm with pseudo division in each step since the polynomial coecients no longer belongs to a eld.

Here is a brief exposition of pseudo division.

Lemma 4.3 Let fg

²

k

^

x

1

::: x

n^-1^]

x

n^]

,

f

⁼

f

p

x

pn⁺

:::f

0

g

⁼

g

m

x

_mn⁺

:::g

0

i.e., f

i

g

j

are polynomials in k

^

x

1

::: x

n^-1^]

and m

^

p . Then

g

_sm

f

⁼

qg

⁺

r

where qr

²

k

^

x

1

::: x

n^-1^]

x

n^]

are unique, s

^

0 and deg

_xⁿ⁽

r

⁾

< m .

Proof. A proof can be found in e.g., 8].

The remainder, r of the pseudo division is called the pseudo remainder of f

1

and f

2

and is denoted prem

⁽

f

1

f

2⁾

.

If we allow denominators pseudo division can be seen as



Ordinary polynomial division for polynomials in x

n

with coecients from the rational function eld k

⁽

x

1

::: x

n^-1⁾

, followed by



Clearing denominators. The only term which needs to be inverted

in the division is the leading coecient, g

m

of g . Hence we get the

equation g

_sm

f

⁼

qg

⁺

r .

Denition 4.1 Let R be a unique factorization domain. Two polynomials in R

^

x

^]

are similar, denoted

f

⁽

x

^)

g

⁽

x

⁾

if there exist ab

²

R such that af

⁽

x

⁾⁼

bg

⁽

x

⁾

.

We can now generalize the concept of polynomial remainder sequences.

Denition 4.2 Let R

⁼

k

^

x

1

::: x

n^-1^]

and f

1

f

2 ²

R

^

x

n^]

with deg

_xⁿ⁽

f

1^)

deg

_xⁿ⁽

f

2⁾

. The sequence f

1

f

2

::: f

k

is a polynomial remainder sequence (PRS) for f

1

and f

2

if:

(i) For all i

⁼

3::: k

f

i^

prem

⁽

f

i^-2

f

i^-1⁾

(ii) The sequence terminates with

prem

⁽

f

k^-1

f

k⁾⁼

0

Since we only claim similarity there are innitely many PRS to every pair f

1

f

2²

k

^

x

1

::: x

n^-1^]

x

n^]

of polynomials.

Immediately one thinks of the following two sequences:



Euclidean Polynomial Remainder Sequence (EPRS):

f

i⁼

prem

⁽

f

i^-2

f

i^-1⁾⁶⁼

0

prem

⁽

f

k^-1

f

k⁾⁼

0



Primitive Polynomial Remainder Sequence (PPRS):

f

i ⁼

pp

⁽

prem

⁽

f

i^-2

f

i^-1⁾⁾⁶⁼

0

prem

⁽

f

k^-1

f

k⁾⁼

0

where pp stands for primitive part, i.e. the remaining part of the polynomial when we have removed all common factors of the coecients.

We observe that a PRS is unique up to similarity since pseudo division is unique. Furthermore,

gcd

⁽

f

1

f

2^)

:::

^

gcd

⁽

f

k^-1

f

k^)

f

k



i.e., PRS essentially computes the gcd of two polynomials up to similarity.

From a computational point of view both EPRS and PPRS suers from complexity problems. EPRS has an exponential coecient growth and the PPRS algorithm involves calculations of gcd of the coecients, which in our case again is polynomials. However, this computational complexity is not inherent in the problem. The Subresultant Polynomial Remainder Sequence (SPRS) oers a tradeo between coecient growth and the cost of gcd cal- culations of the coecients. We will now take a closer look at subresultants and SPRS.

4.2 Subresultants

In this subsection we will present a way of calculating the SPRS, see e.g.,

13, 15, 17].

To do this we observe that the process of pseudo division may be or- ganized as row operations on a matrix containing the coecients of the polynomials.

Example 4.3 Pseudo division applied to f

⁼

x

³⁺

2x

²⁺

3x

⁺

1 and

g

⁼

2x

²⁺

x

⁺

2 gives

2

²

f

⁼⁽

2x

⁺

3

⁾

g

⁺

5x

^-

2 :

One way of organizing the calculations is

M

⁼

2

6

4

1 2 3 1 2 1 2 0 0 2 1 2

3

7

5

 2

6

4

2 1 2 0 0 2 1 2 0 0 5

^-

2

3

7

5

=

M

^-

where M

^-

is obtained from M by row operations. Observe the correspon- dence between the pseudo remainder coecients and the last row of M

^-

. In fact multiplying the rst row of M with 2

²

, M

^-

can be obtained by just subtracting integer multiples of the other rows from the rst row. Finally, permuting the rows gives the triangular form. According to the triangular structure of M

^-

the pseudo remainder coecients may be found, modulo a common factor, by determinant calculations on a matrix consisting of the

rst two columns of M

^-

augmented by either column three or four.

Since M and M

^-

only diers by row operations their minors are strongly

related. Hence the coecients of a polynomial similar to the pseudo re-

mainder may be calculated as minors of the original matrix M . In fact,

by calculating minors of matrices whose elements are the coecients of two polynomials f and g we can construct a whole PRS of f and g . To see this we rst need a couple of denitions. Throughout the rest of this subsection

R is a unique factorization domain (UFD).

Denition 4.3 Let f

i ^{= P}nj⁼ⁱ0

f

ij

x

^{j 2}

R

^

x

^]

 i

⁼

1::: k . The matrix associated with f

1

::: f

k

is

mat

⁽

f

1

::: f

k⁾⁼

h

f

il^-j

i



where l

⁼

1

⁺

max

1^i^k⁽

n

i⁾

.

Denition 4.4 Let M

²

R

^kl

l

^

k . The determinant polynomial of M

is detpol

⁽

M

⁾⁼

det

⁽

M

^(k))

x

^l-k+

:::

⁺

det

⁽

M

^(l))



where M

^(j)=h

M

^1

::: M

^k^-1

M

^j

i

:

In Example 4.3 we have M

⁼

mat

⁽

fxgg

⁾

, detpol

⁽

M

^{) =}

5x

^-

2

⁼

prem

⁽

fg

⁾

and detpol

⁽

M

^-)=

2

²⁽

5x

^-

2

⁾

.

Denition 4.5 Let fg

²

R

^

x

^]

and deg

⁽

f

⁾⁼

m deg

⁽

g

⁾⁼

n m

^

n .



The k

^th

subresultant of f and g is

subres

_k⁽

fg

⁾⁼

detpol

⁽

M

k⁾



where M

k ⁼

mat

⁽

x

^n-k-1

f::: fx

^m-k-1

g::: g

⁾

:



The subresultant chain of f and g is

⁽

S

j⁾nj⁼⁺01

, where

S

n⁺1 ⁼

f

S

n ⁼

g

S

n^-1 ⁼

subres

n^-1⁽

fg

⁾



S

0 ⁼

... subres

0⁽

fg

⁾

:

Observe that M

0

is the Sylvester matrix of f and g and subres

0 ⁼

detpol

⁽

M

0^{) =}

det

⁽

M

0⁾

is the resultant of f and g . Furthermore, the M

k

matrices is obtained by deleting rows and columns of M

0

.

The importance of the denition of subresultants and subresultant chains lies in the fact that all PRS of f and g is embedded in the subresultant chain of f and g up to similarity.

We illustrate the denitions with a numerical example.

Example 4.4 Let f

⁼

x

^5-

2x

⁴⁺

3x

^3-

4x

²⁺

5x

^-

6 and

g

⁼

3x

³⁺

5x

²⁺

7x

⁺

9 . Then

M

0 ⁼

2

6

6

6

6

6

6

6

6

6

6

6

6

4

1

^-

2 3

^-

4 5

^-

6 0 0 0 1

^-

2 3

^-

4 5

^-

6 0 0 0 1

^-

2 3

^-

4 5

^-

6 3 5 7 9 0 0 0 0 0 3 5 7 9 0 0 0 0 0 3 5 7 9 0 0 0 0 0 3 5 7 9 0 0 0 0 0 3 5 7 9

3

7

7

7

7

7

7

7

7

7

7

7

7

5

M

1 ⁼

2

6

6

6

6

6

6

6

4

1

^-

2 3

^-

4 5

^-

6 0 0 1

^-

2 3

^-

4 5

^-

6 3 5 7 9 0 0 0 0 3 5 7 9 0 0 0 0 3 5 7 9 0 0 0 0 3 5 7 9

3

7

7

7

7

7

7

7

5

 M

2⁼

2

6

6

6

4

1

^-

2 3

^-

4 5

^-

6 3 5 7 9 0 0 0 3 5 7 9 0 0 0 3 5 7 9

3

7

7

7

5

:

Observe how M

1

and M

2

is obtained by deleting rows and columns of M

0

. The M

⁽_k^l)

matrices are formed by the left part of the partitioned matrices and one column from the right part. Now, the subresultants is the determinant polynomials of M

0

M

1

and M

2

, i.e.,

S

4 ⁼

x

^5-

2x

⁴⁺

3x

^3-

4x

²⁺

5x

^-

6 S

3 ⁼

3x

³⁺

5x

²⁺

7x

⁺

9

S

2 ⁼

det

⁽

M

⁽₂⁴⁾⁾

x

²⁺

det

⁽

M

⁽₂⁵⁾⁾

x

⁺

det

⁽

M

⁽₂⁶⁾⁾⁼

263x

^2-

5x

⁺

711 S

1 ⁼

det

⁽

M

⁽₁⁶⁾⁾

x

⁺

det

⁽

M

⁽₁^7))=-

2598x

^-

11967

S

0 ⁼

det

⁽

M

0⁾⁼

616149

⁼

3

^2

223

^

307

Compare the subresultant chain with the EPRS and PPRS for f and g :

f

1 ⁼

x

^5-

2x

⁴⁺

3x

^3-

4x

²⁺

5x

^-

6 f ^~

1 ⁼

x

^5-

2x

⁴⁺

3x

^3-

4x

²⁺

5x

^-

6 f

2 ⁼

3x

³⁺

5x

²⁺

7x

⁺

9 f ^~

2 ⁼

3x

³⁺

5x

²⁺

7x

⁺

9

f

3 ^=-

263x

²⁺

5x

^-

711 f ^~

3 ^=-

263x

²⁺

5x

^-

711 f

4 ^=-

70146x

^-

323109 f ^~

4 ^=-

866x

^-

3989 f

5 ^=-

3

^8

223

^

263

^2

307 f ^~

5 ^=-

1

where f

4 ⁼

81 f ^~

4 ⁼

3S

1

. Notice the tremendous coecient growth in the EPRS compared with the modest growth in the subresultant chain. In this example there is a one to one correspondence between the PRS and the subresultant chain. This is due to the fact that the degree drops by one for each pseudo division. If the degree drops with more than one some of the subresultants becomes similar.

What is the connection between a polynomial in a PRS and the deter- minant polynomial of M

k

? By row operations on M

k

a number of pseudo divisions may be carried out consecutively. The process is described by the following example which is a generalization of Example 4.3.

Example 4.5 Let f

⁼

a

5

x

⁵⁺

a

4

x

⁴⁺

a

3

x

³⁺

a

2

x

²⁺

a

1

x

⁺

a

0

and

g

⁼

b

3

x

³⁺

b

2

x

²⁺

b

1

x

⁺

b

0

. The corresponding M

1

matrix becomes

M

1 ⁼

2

6

6

6

6

6

6

6

4

a

5

a

4

a

3

a

2

a

1

a

0

a

5

a

4

a

3

a

2

a

1

a

0

b

3

b

2

b

1

b

0

b

3

b

2

b

1

b

0

b

3

b

2

b

1

b

0

b

3

b

2

b

1

b

0

3

7

7

7

7

7

7

7

5 (1⁾

 2

6

6

6

6

6

6

6

4

b

3

b

2

b

1

b

0

b

3

b

2

b

1

b

0

b

3

b

2

b

1

b

0

b

3

b

2

b

1

b

0

a ~

2

a ~

1

a ~

0

a ~

2

a ~

1

a ~

0

3

7

7

7

7

7

7

7

5 (2⁾



2

6

6

6

6

6

6

6

4

b

3

b

2

b

1

b

0

b

3

b

2

b

1

b

0

b

3

b

2

b

1

b

0

a ~

2

a ~

1

a ~

0

a ~

2

a ~

1

a ~

0

~ b

1

b ~

0

3

7

7

7

7

7

7

7

5

=

M

^-₁

Row operations in detail:

(1) Multiply the rst row with b

₃₃

and subtract multiplicities of row 34

and 5 to eliminate a

5

a

4

and a

3

. A similar treatment of the second row

followed by interchanging rows gives the second matrix. The resulting matrix elements ~ a

i

are the coecients of prem

⁽

fg

⁾

.

(2) The same process as in (1) repeated on row 45 and 6. The matrix elements ~ b

i

are the coecients of prem

⁽

g prem

⁽

fg

⁾⁾

.

As pointed out earlier the only operations needed to get from M

k

to the last triangular matrix, M

^-_k

are row operations. It is then clear that the minors det

⁽

M

⁽_k^l))

and the corresponding minors of the triangular matrix only diers by a common factor.

Our original motivation for calculating PRS was to determine the gcd of two polynomials and especially its degree. Hence we are interested in the coecient of the highest power of a gcd.

Denition 4.6 Let fg

²

R

^

x

^]

and deg

⁽

f

⁾⁼

m deg

⁽

g

⁾⁼

n m

^

n . The

k

^th

principal subresultant coecient of f and g is psc

_k⁽

fg

^{) =}

det

⁽

M

⁽_k^k))

 0

^

k

^

n:

In other words psc

_k⁽

fg

⁾

is the coecient of x

^k

in subres

_k⁽

fg

⁾

. The following lemma, which is not hard to believe in knowing the connection be- tween a PRS and the corresponding subresultant chain, tells us that knowing the psc chain of two polynomials we know the degree of their gcd.

Lemma 4.4 Let f g

²

R

^

x

^]

where deg

⁽

f

^{) =}

m and deg

⁽

g

^{) =}

n . Then for all 0 < i

^

min

⁽

mn

⁾

f and g have a common factor of degree

⁼

i

()

psc

_j⁽

fg

⁾⁼

0j

⁼

0::: i

^-

1 and psc

_i⁽

fg

⁾⁶⁼

0 Proof. See Corollary 7.7.9. in 17].

In this subsection we have worked with polynomials in R

^

x

^]

where R is a

unique factorization domain (UFD). For simplicity the examples were done

for polynomials over the integers but any UFD will do e.g., multivariate

polynomials. In the following we use R

^=R

x

1

::: x

n^-1^]

, i.e., polynomials

in

^R

x

1

::: x

n^-1^]

x

n^]

.

4.3 Delineable Sets

The key idea in the so called projection phase of the CAD-algorithm is to

nd regions over which the given polynomials have a constant number of real roots, cf. Example 2.1. This is formalized by the concept of delineability.

Let f

i^2R

x

1

::: x

n^-1^]

x

n^]

be a real polynomial in n -variables:

f

i⁽

x

1

::: x

n^-1

x

n⁾⁼

f

^d_iⁱ⁽

x

1

::: x

n^-1⁾

x

^d_n^i++

f

_0i⁽

x

1

::: x

n^-1⁾

and 

⁼⁽



1

::: 

n^-1^)2Rn^-1

. Then we write f

i⁽

x

n⁾⁼

f

⁽



1

::: 

n^-1

x

n⁾

for the univariate polynomial obtained by substituting  for the rst n

^-

1

variables.

We also introduce the reductum, ^ f

^k_iⁱ

of a polynomial. Let

f ^

^k_iⁱ⁽

x

1

::: x

n^-1

x

n⁾⁼

f

^k_iⁱ⁽

x

1

::: x

n^-1⁾

x

^k_n^i++

f

_0i⁽

x

1

::: x

n^-1⁾



where 0

^

k

i ^

d

i

. Thus, if f

^d_iⁱ⁽



^{) =  =}

f

^k_iⁱ⁺¹⁽



^{) =}

0 and f

^k_iⁱ⁽



^{) 6=}

0 , then f

i⁽

x

n⁾⁼

f ^

^k_iⁱ⁽

x

n⁾⁼

f

^k_iⁱ⁽



⁾

x

^k_n^i++

f

_0i⁽



⁾

:

Denition 4.7 Let

^{F =f}

f

1

f

2

::: f

r^{g R}

x

1

::: x

n^-1^]

x

n^]

be a set of multivariable real polynomials and C

^Rn-1

. We say that

^F

is delineable on C , if it satises the following invariant properties:

(i) For every 1

^

i

^

r , the total number of complex roots of f

i

(counting multiplicity) remains invariant as  varies over C .

(ii) For every 1

^

i

^

r , the number of distinct complex roots of f

i

remains invariant as  varies over C .

(iii) For every 1

^

i < j

^

r , the total number of common complex roots of

f

i

and f

j

(counting multiplicity) remains invariant as  varies over

C .

Consider one of the polynomials f

i^2F

. Since it has real coecients its complex roots have to occur in conjugate pairs. Let  and 

⁰

be two points in C

^Rn-1

. The only way for a pair of complex conjugated roots of f

i

to become real when  varies to 

⁰

is to coalesce into a real double root, i.e.

the number of distinct roots of f

i

drops. Hence a transition from a nonreal

root to a real root is impossible over C . Similar arguments holds for the

reversed problem.