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CS 829

Polynomial systems: geometry and algorithms

Lecture 3: Euclid, resultant and 2 × 2 systems

Eric Schost

[email protected]

1

Summary

In this lecture, we start actual computations (as opposed to Lectures 1 and 2,

where we discussed properties of univariate representations, but no any actual

way to compute them).

We discuss systems of 2 equations in 2 unknowns.

• These systems can be dealt with using resultants.

• Resultants can be computed using extensions of the Euclidean algorithm.

• The cost of resolution is O(d2M(d) log(d)), where M represents the cost of

multiplying univariate polynomials, and d is the total degree of the input.

2

An overview of Euclid’s algorithm

3

Euclid’s algorithm

Euclid’s algorithm is used to compute GCD’s in either Z of K[Y ].

By running it for random examples in K[X ][Y ], we observe:

• that the remainders are in K(X)[Y ];

• there are a lot of repetitions as denominators / common factors in

numerators;

• once all the junk has been cleaned, we get most of the information that we

need to solve the system.

The resultant (and subresultants) are what is left once all the junk has been

cleaned.

4


Euclid’s algorithm is used to compute GCD’s in either Z of K[Y ].

By running it for random examples in K[X ][Y ], we observe:

• that the remainders are in K(X)[Y ];

• there are a lot of repetitions as denominators / common factors in

numerators;

• once all the junk has been cleaned, we get most of the information that we

need to solve the system.

The resultant (and subresultants) are what is left once all the junk has been

cleaned.

In K[Y ], the complexity for degree-d inputs is O(d2). This can be made faster,

but the extension to K[X ][Y ] will add a new layer of difficulties.

5

Intersection of plane curves

6

Common factors of univariate polynomials

Let K be a field and let F, G be in K[Y ], with

deg(F ) = m, deg(G) = n.

F and G have a common factor ⇐⇒F

G=

B

A, deg(B) < m, deg(A) < n

7


Let K be a field and let F, G be in K[X ], with



G=

B


⇐⇒ AF − BG = 0

8


Let K be a field and let F, G be in K[X ], with



G=

B


⇐⇒ AF − BG = 0

In other words, gcd(F, G) 6= 1 if and only if the map

φ : K[Y ]<n × K[Y ]<m → K[Y ]<m+n

(A, B) 7→ AF + BG

has a non-trivial kernel, where

K[Y ]<ℓ = {a0 + · · · + aℓ−1Yℓ−1} ≃ Kℓ.

9

The Sylvester matrix

The Sylvester matrix represents the previous application. With

F = fmY m + · · · + f0, G = gnY n + · · · + g0,

their Sylvester matrix is

Syl(F, G) =

n︷︸︸︷

fm

... fm

... fm

f0. . .

...

f0

...

f0

m︷︸︸︷

gn

... gn

... gn

g0. . .

...

g0

...

g0

10

Resultant

Definition The resultant res(F, G) of F, G ∈ K[Y ] is the determinant of their

Sylvester matrix.

11

Resultant


Sylvester matrix.

Proposition res(F, G) = 0 ⇐⇒ gcd(F, G) 6= 1.

12

Resultant


Sylvester matrix.


We extend the definition to polynomials F, G with coefficients in a ring R.

(warning: GCD’s do not really make sense over a general ring).

13

Resultant


Sylvester matrix.




Useful proposition Let ϕ be a morphism from a ring R to a ring S:

ϕ(1) = 1, ϕ(a + b) = ϕ(a) + ϕ(b), ϕ(ab) = ϕ(a)ϕ(b).

14

Resultant


Sylvester matrix.






• If ϕ(fm)ϕ(gn) 6= 0, ϕ(res(F, G)) = res(ϕ(F ), ϕ(G)).

15

Resultant


Sylvester matrix.






• If ϕ(fm)ϕ(gn) 6= 0 then ϕ(res(F, G)) = res(ϕ(F ), ϕ(G)).

• If ϕ(fm) 6= 0 then ϕ(res(F, G)) = ϕ(fm)deg(G)−deg(ϕ(G))res(ϕ(F ), ϕ(G)).

16

Resultant


Sylvester matrix.








Examples

• R = Z and ϕ(K) = K mod p,

17

Resultant


Sylvester matrix.








Examples

• R = Z and ϕ(K) = K mod p,

• R = K[X ] and ϕ(P (X)) = P (x) = P mod (X − x).

18

Application to the intersection of curves

Let F, G be in K[X, Y ]:

F =m∑

i=0

fi(X)Y i G =n∑

i=0

gi(X)Yi.

19



F =m∑

i=0

fi(X)Y i G =n∑

i=0

gi(X)Yi.

If they have no common factor in K[X, Y ], then in K′[Y ], with K′ = K(X), we

have gcd(F, G) = 1.

Hence, R = resY (F, G) ∈ K[X ] is not zero.

20



F =m∑

i=0

fi(X)Y i G =n∑

i=0

gi(X)Yi.


have gcd(F, G) = 1.


Proposition. Let x be in K. Then R(x) = 0 if and only if

• fm(x) = gn(x) = 0

• or there exists y such that F (x, y) = G(x, y) = 0.

21



F =m∑

i=0

fi(X)Y i G =n∑

i=0

gi(X)Yi.


have gcd(F, G) = 1.


Proposition. Let x be in K. Then R(x) = 0 if and only if

• fm(x) = gn(x) = 0

• or there exists y such that F (x, y) = G(x, y) = 0.

Proof. If fm(x) = gn(x) = 0, R(x) = 0.

Suppose now that e.g. fm(x) 6= 0.

Then R(x) = fm(x)k res(F (x, Y ), G(x, Y )), so R(x) = 0 if and only if

F (x, Y ) and G(x, Y ) have a common factor.

22

A degenerate example

Consider

F = X2Y + X + 1, G = XY − 1

whose resultant with respect to Y is R = −X(2X + 1).

The root x = 0 of R does not correspond to any root of the system

F (X, Y ) = G(X, Y ) = 0 (or so it seems. . . )

23

Curves in generic position

Let F 6= 0 be in K[X, Y ]:

F = F0(X) + · · · + Fm(X)Y m.

The curve C = V (F ) is in Noether position (for the order Y > X) if Fm is a

constant.

NO YES

When not in Noether position, values of x where Fm(x) = 0 correspond to points

going to infinity above x (more on this later).

24



F = F0(X) + · · · + Fm(X)Y m.


constant.

Proposition. For a generic change of variable X = X ′ + αY , the curve C ′ is in

Noether position (but the degree in Y may change!)

25



F = F0(X) + · · · + Fm(X)Y m.


constant.



Proof. Let d be the total degree of F and let H be the homogeneous part of degree

d of F . Write H =∑

hiXiY d−i.

26



F = F0(X) + · · · + Fm(X)Y m.


constant.





hiXiY d−i.

Then

H(X ′ + αY, Y ) =∑

hi(X′ + αY )iY d−i = Y d

∑

hiαi + · · ·

27



F = F0(X) + · · · + Fm(X)Y m.


constant.





hiXiY d−i.

Then

H(X ′ + αY, Y ) =∑

hi(X′ + αY )iY d−i = Y d

∑

hiαi + · · ·

So it suffices that α is not a root of∑

hiAi.

28



F = F0(X) + · · · + Fm(X)Y m.


constant.



Proposition. When this is the case, the roots of the resultant of F ′, G′ are exactly

the X ′-coordinates of the solutions.

29

Finiteness of the solution set

Proposition. Let F, G be in K[X, Y ] with a common factor. Then V (F, G) is

finite.

30



finite.

Proof. Suppose that the curves C = V (F ) and C ′ = V (G) are both in Noether

position.

31



finite.


position.

Let R ∈ K[X ] be the resultant of F and G with respect to Y , so that R 6= 0.

Each solution of F (x, y) = G(x, y) = 0 satisfies R(x) = 0.

32



finite.


position.



• R has a finite number of roots.

33



finite.


position.



• R has a finite number of roots.

• For any root x of R, there is a finite number of y such that F (x, y) = 0.

34

Curves in generic position, continued

Proposition. Let F, G be in K[X, Y ] with a common factor, of total degrees at

most d.

For a generic choice of X ′ = X − αY , X ′ is a separating element for

V = V (F, G).

35



most d.


V = V (F, G).

Proof. Let {(xi, yi)}i≤N be the finite set of common solutions.

36



most d.


V = V (F, G).


Through the change of variables X ′ = X − αY , the solution set becomes

{xi − αyi, yi}i≤N .

37



most d.


V = V (F, G).


Through the change of variables X ′ = X − αY , the solution set becomes

{xi − αyi, yi}i≤N .

Let

∆(A) =∏

i 6=j

(xi − x′i) − A(yi − y′).

X ′ is separating if and only if ∆(α) 6= 0.

38

Cardinality of the intersection


most d. Then V (F, G) has cardinality at most d2.

39




Proof. By a generic change of variables, we can suppose that X is a separating

element for V = V (F, G) and that the curves are in Noether position.

40






This does not change the total degree, or the number of solutions!

41







Then, then number of solutions of the system is bounded by the degree of

R = resY (F, G).

42








R = resY (F, G).

The resultant R is the determinant of a matrix of size 2d, with entries of

degree at most d. So deg(R) ≤ 2d2.

43








R = resY (F, G).

The resultant R is the determinant of a matrix of size 2d, with entries of

degree at most d. So deg(R) ≤ 2d2.

Exercise. Work out the bound d2 by looking more carefully at the matrix.

44

Some properties of the resultant

45

Properties of the resultant

Theorem. Let f, u1, . . . , um and g, v1, . . . , vn be new variables and

F = f(X − u1) · · · (X − um), G = g(X − v1) · · · (X − vn).

Then

R := res(F, G) = fngm∏

i

∏

j

(ui − vj).

Proof. Let Fi,j = subs(ui = vj , F). Then

res(Fi,j , G) = 0 but also res(Fi,j , G) = subs(ui = vj , R).

Hence, R is divisible by all (ui − vj).

Working a little bit more (counting degrees and checking the leading terms)

finishes the proof.

46


Theorem. Let F, G be in K[X ] with factorization in K:

F = f(X − u1) · · · (X − um), G = g(X − v1) · · · (X − vn).

Then


i

∏

j

(ui − vj).

Proof. Specialization properties.

Corollary res(F, GH) = res(F, G)res(F, H).

Proof. True for polynomials with coefficients in a field (cf. above). What about

more general cases? It is enough to prove it for polynomials with generic

coefficients

F = f0 + · · · + fmXm, G = g0 + · · · + gnXn, H = h0 + · · · + hℓXℓ.

Because then we can use specialization properties.

What we want to prove is that a big polynomial ∆ in Z[fi, gj , hk] is identically

zero. Since ∆ vanishes for all possible evaluations in e.g. Q, it is zero.47



F = f(X − u1) · · · (X − um), G = g(X − v1) · · · (X − vn).

Then


i

∏

j

(ui − vj).



Corollary For F, G with coefficients in a field,

R := res(F, G) = fn∏

F (ui)=0

G(ui).

48



F = f(X − u1) · · · (X − um), G = g(X − v1) · · · (X − vn).

Then


i

∏

j

(ui − vj).




R := res(F, G) = fn∏

F (ui)=0

G(ui).


res(F, G) = (−1)deg(F ) deg(G)LeadCoeff(G)deg(F )−deg(R)res(G, R),

for R such that F = QG + R.49

Computing resultants

50


Input: F, G in K[Y ].

• F1 := F ;

• F2 := G;

• i := 2;

• while Fi <> 0

• Fi+1 := Fi−1 mod Fi;

• i++;

• return Fi−1;

51



• F1 := F ;

• F2 := G;

• i := 2;

• while Fi <> 0


• i++;

• return Fi−1;

Complexity.

• Euclidean division in degrees n ≥ m has cost O((n − m)m);

• With di = deg Fi, Euclid’s algorithm has cost (up to big-Oh)∑

i≥2

(di−1 − di)di ≤ d1

∑

i≥

(di−1 − di) ≤ d21.

52

Euclid’s algorithm for resultant in K[Y ]


• F1 := F ;

• F2 := G;

• i := 2;

• R1 := 1;

• while deg(Fi) > 0


• Ri := (−1)didi−1LeadCoeff(Fi)di−1−diRi−1;

• i++;

• if Fi <> 0 return Ri−1 LeadCoeff(Fi)deg(Fi−1).

• else return 0;

53


Correctness

As long as deg(Fi) > 0, res(F, G) = Ri res(Fi, Fi+1) (by induction).

When deg(Fi) ≤ 0:

• If Fi = 0, res(Fi−1, Fi) = 0

• Else, res(Fi−1, Fi) is LeadCoeff(Fi)deg(Fi−1).

54


Correctness

As long as deg(Fi) > 0, res(F, G) = Ri res(Fi, Fi+1) (by induction).

When deg(Fi) ≤ 0:

• If Fi = 0, res(Fi−1, Fi) = 0

• Else, res(Fi−1, Fi) is LeadCoeff(Fi)deg(Fi−1).

Complexity

Still O(d2).

55

Resultants of bivariate polynomials

Euclid’s algorithm in K(Y )[X ] creates huge intermediate polynomials.

With inputs of total degree d:

• It takes some work (introducing subresultants) to prove that all intermediate

degrees are in O(d4).

• However, the output we want is a single univariate polynomial of degree d2.

56

Resultants of bivariate polynomials

Euclid’s algorithm in K(Y )[X ] creates huge intermediate polynomials.

With inputs of total degree d:

• It takes some work (introducing subresultants) to prove that all intermediate

degrees are in O(d4).

• However, the output we want is a single univariate polynomial of degree d2.

Two workarounds:

• Understand, predict and pre-clean the common factors and denominators;

• Use a modular algorithm.

57

Modular algorithm, plain version

Input: F, G in K[X, Y ] of total degrees ≤ d.

• Compute res(F (c, Y ), G(c, Y )) for d2 + 1 values of c (which do not cancel a

leading term);

• Interpolate the result.

58

Modular algorithm, plain version

Input: F, G in K[X, Y ] of total degrees ≤ d.

• Compute res(F (c, Y ), G(c, Y )) for d2 + 1 values of c (which do not cancel a

leading term);

• Interpolate the result.

Complexity.

• O(d2 × d2) + O(Costinterpolation(d2)) ∈ O(d4) (proof upcoming).

59

Interpolating polynomials

Given a0, . . . , aℓ pairwise distinct sample points and values v0, . . . , vℓ, the unique

polynomial F such that F (ai) = vi is

ℓ∑

i=0

vi

∏

j 6=i(X − aj)∏

j 6=i(ai − aj)

60

Interpolating polynomials

Given a0, . . . , aℓ pairwise distinct sample points and values v0, . . . , vℓ, the unique

polynomial F such that F (ai) = vi is

ℓ∑

i=0

vi

∏

j 6=i(X − aj)∏

j 6=i(ai − aj)

To compute it:

• Compute M =∏

i(X − ai),

• Deduce all∏

j 6=i(X − aj),

• Deduce all∏

j 6=i(ai − aj),

• Combine everybody,

for a cost of O(ℓ2).

61

Switching to fast algorithms

62

Speeding things up

Let M(d) denote the cost of polynomial multiplication in degree d:

• M(d) ∈ O(d2) for a naive algorithm

• M(d) ∈ O(d log d) using Fast Fourier Transform (if the field has roots of 1)

• M(d) ∈ O(d log d log log d) using Fast Fourier Transform in general.

Technically, we ask M(d + d′) ≥ M(d) + M(d′).

63

Speeding things up

Let M(d) denote the cost of polynomial multiplication in degree d:

• M(d) ∈ O(d2) for a naive algorithm

• M(d) ∈ O(d log d) using Fast Fourier Transform (if the field has roots of 1)

• M(d) ∈ O(d log d log log d) using Fast Fourier Transform in general.

Technically, we ask M(d + d′) ≥ M(d) + M(d′).

Using the fact that Euclidean division can be made in time O(M(d)), both parts

can be made faster:

• Euclid’s algorithm: divide-and-conquer and half-GCD techniques,

O(d2 × M(d) log(d))

• Interpolation using subproduct trees techniques.

O(M(d2) log(d))

64

FFT in a nutshell

Suppose you want to evaluate F (X) ∈ C[X ] at all N -roots of 1

1, exp2iπ

N, exp

4iπ

N, . . . , exp

2(N − 1)iπ

N,

with deg(F ) < N .

65

FFT in a nutshell


1, exp2iπ

N, exp

4iπ

N, . . . , exp

2(N − 1)iπ

N,

with deg(F ) < N .

Write F = Feven(X2) + XFodd(X2). Then

F (exp2ikπ

N) = Feven(exp

2ikπ

N ′) +

2ikπ

NFodd(exp

2ikπ

N ′),

with N ′ = N/2.

66

FFT in a nutshell


1, exp2iπ

N, exp

4iπ

N, . . . , exp

2(N − 1)iπ

N,

with deg(F ) < N .


F (exp2ikπ

N) = Feven(exp

2ikπ

N ′) +

2ikπ

NFodd(exp

2ikπ

N ′),

with N ′ = N/2.

So it suffices to

• Feven at all N ′-roots of 1;

• Fodd at all N ′-roots of 1;

• combine the values.

67

FFT in a nutshell


1, exp2iπ

N, exp

4iπ

N, . . . , exp

2(N − 1)iπ

N,

with deg(F ) < N .


F (exp2ikπ

N) = Feven(exp

2ikπ

N ′) +

2ikπ

NFodd(exp

2ikπ

N ′),

with N ′ = N/2.

So it suffices to

• Feven at all N ′-roots of 1;

• Fodd at all N ′-roots of 1;

• combine the values.

The complexity satisfies T (N) ≤ 2T (N/2) + CN so T (N) ∈ O(N log N).

68

FFT in a nutshell

Proposition The inverse FFT can be performed for the same cost as the direct

FFT.

Corollary One can multiply F (X), G(X) ∈ C[X ], both of them having degree

< N , in O(N log N) operations

• Evaluate F and G at 2N -th roots of 1

• Multiply the values

• Do inverse-FFT to interpolate the product FG.

Extension to any field having “roots of unity”.

69

Towards a fast Euclidean algorithm

We want to make Euclid’s algorithm faster, reducing its complexity to

O(M(d) log(d)).

Warning! Computing all intermediate remainders is too slow (the total number of

coefficients is ≃ d2).

Workaround: Work with the quotients Qi such that

Fi−1 = QiFi + Fi+1

Main idea: knowing Qi makes it possible to do a single step

Fi

Fi+1

=

0 1

1 −Qi

Fi−1

Fi

70



O(M(d) log(d)).






Fi

Fi+1

=

0 1

1 −Qi

Fi−1

Fi

One can do several steps at once:

Fj−1

Fj

=

0 1

1 −Qj

· · ·

0 1

1 −Qi

Fi−1

Fi

71



O(M(d) log(d)).






Fi

Fi+1

=

0 1

1 −Qi

Fi−1

Fi

One can do several steps at once:

Fj−1

Fj

= M[i → j]

Fi−1

Fi

72

Half GCD: specifications and applications

Let F, G ∈ K[Y ] with d = deg F > deg G, and let

F1 = F, F2 = G, . . . , FN = 0

be the remainders met during Euclid’s algorithm.

73



F1 = F, F2 = G, . . . , FN = 0

be the remainders met during Euclid’s algorithm. There exists a unique ℓ such

that

deg(Fℓ−1) ≥ d/2 > deg(Fℓ).

74



F1 = F, F2 = G, . . . , FN = 0


that


The half-GCD algorithm compute the matrix M[2 → ℓ], so that

Fℓ−1

Fℓ

= M[2 → ℓ]

F1

F2

.

75



F1 = F, F2 = G, . . . , FN = 0


that



Fℓ−1

Fℓ

= M[2 → ℓ]

F1

F2

.

• If Fℓ = 0, Fℓ−1 is the GCD,

76



F1 = F, F2 = G, . . . , FN = 0


that



Fℓ−1

Fℓ

= M[2 → ℓ]

F1

F2

.

• If Fℓ = 0, Fℓ−1 is the GCD,

• Else, compute Fℓ+1 (to be sure that all degrees are < d/2), and continue with

Fℓ, Fℓ+1.

77

Extension to resultant computation

Suppose that the half-GCD algorithm outputs as well:

• the leading coefficients of F1, . . . , Fℓ and

• their degrees.

78




• their degrees.

Then collecting the leading coefficients and degree of Fℓ+1, we get a recursive

algorithm for computing all leading coeffcients and their degrees.

79




• their degrees.



This leads to the resultant (cf. formula in the plain version).

80




• their degrees.




Complexity

Costresultant(d) ≤ CostHalf−gcd(d)+O(M(d))+CostEuclidean division(d)+Costresultant(d/2)

81




• their degrees.




Complexity

Costresultant(d) ≤ CostHalf−gcd(d) + CM(d) + Costresultant(d/2)

82




• their degrees.




Complexity

Costresultant(d) ≤ CostHalf−gcd(d) + CM(d) + Costresultant(d/2)

so

Costresultant(d) ∈ O(CostHalf−gcd(d) + M(d))

83

The idea of half-GCD

Intuition 1. In the first steps of the Euclidean algorithm, the quotients do not

depend on the lower degree terms.

X1000 + X999 + · · · = (X + 2)(X999 − X998 + · · · ) + R

84




X1000 + X999 + · · · = (X + 2)(X999 − X998 + · · · ) + R

Intuition 2. The degrees of the remainders decrease by one at each step and the

quotients have degree 1.

85




X1000 + X999 + · · · = (X + 2)(X999 − X998 + · · · ) + R



Hence, a transition matrix of degree ℓ yields remainders of degree ≃ d − ℓ.

86




X1000 + X999 + · · · = (X + 2)(X999 − X998 + · · · ) + R




Intuition 3. The half-GCD matrix of F1, F2 has entries of degrees ≃ d/2.

87




X1000 + X999 + · · · = (X + 2)(X999 − X998 + · · · ) + R




Intuition 3. The half-GCD matrix of F1, F2 has entries of degrees ≃ d/2.

Intuition 4. Let m ≃ d/2 and Gi = Fi div Xm.

Then the half-GCD matrix of G1, G2 has entries of degrees ≃ m/2. So

applying it to F1, F2 gives remainders of degrees d − m/2 ≃ 3d/4.

88

The half-GCD (sketch)

Input F1, F2 with d := deg(F1) > deg(F2) and m := ⌈d/2⌉.

• M := halfGCD(F1 div Xm, F2 div Xm);

• [F ′0, F

′1]

t := M[F1, F2]t;

• Q := F ′0 quo F ′

1;

• F ′2 := F ′

0 − QF ′1;

• N = halfGCD(F ′1 div Xm, F ′

2 div Xm);

• return N[

0 11 −Q

]M;

89




• [F ′0, F

′1]

t := M[F1, F2]t;

• Q := F ′0 quo F ′

1;

• F ′2 := F ′

0 − QF ′1;


2 div Xm);

• return N[

0 11 −Q

]M;

Remark. The actual algorithm is slightly more subtle (early exits, more precise

choice in the degrees)

Complexity.

CosthalfGCD(d) ≤ 2CosthalfGCD(d/2) + CM(d)

so CosthalfGCD(d) ∈ O(M(d) log(d)).

90




• [F ′0, F

′1]

t := M[F1, F2]t;

• Q := F ′0 quo F ′

1;

• F ′2 := F ′

0 − QF ′1;


2 div Xm);

• return N[

0 11 −Q

]M;

Remark. The actual algorithm is slightly more subtle (early exits, more precise

choice in the degrees)

Complexity.

CosthalfGCD(d) ≤ 2CosthalfGCD(d/2) + CM(d)

so CosthalfGCD(d) ∈ O(M(d) log(d)) and Costresultant(d) ∈ O(M(d) log(d)).

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an overview of euclid’s algorithm - western universityeschost/publications/cs829-lecture3.pdf ·...

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