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Tables of Affine Moment Invariants Generated by theGraph Method

Tom as Suk and Jan FlusserInstitute of Information Theory and Automation

Academy of Sciences of the Czech RepublicPod vod arenskou vez 4, 182 08 Praha 8, Czech Republic

E-mail: suk@utia.cas.cz, usser@utia.cas.cz

Abstract

These tables include all irreducible affine moment invariants up to the 10thorder. They are generated by a newly developed graph method. The techniqueof elimination of reducible invariants is described. There are also mentioned a fewapproaches to the selection of an independent set of the invariants. Finally, anexample of a few dependencies among irreducible invariants is presented.

Keywords: affine invariants, moments, graphs, independent features.

Acknowledgment

This work has been supported by the grant No. 201/03/0675 of the GrantAgency of the Czech Republic.

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The goal of these tables is to compute and publish affine moment invariants of higherorders and weights. Reiss in [20] presents 17 selected invariants of the 6th order inmaximum. These tables present all irreducible 362 invariants up to weight 10. They areconnected with our previous paper [13]. They also deal with the problem of selection of the independent invariants. They are organized by the following way: Section 2 deals with

derivation of the AMIs from algebraic invariants, Section 3 discusses various methods,how to nd the number of invariants, Section 4 describes the graph method used forderivation of the AMIs, Section 5 concerns with looking for dependencies between theinvariants and Section 6 describes a method, how to verify independency of the choseninvariants. It is followed by the tables of all irreducible AMIs up to the 10th order andsome dependencies between them.

2 The Fundamental Theorem

First, a few basic terms. The affine transformation can be expressed as

u = a0 + a1x + a2yv = b0 + b1x + b2y.

(1)

The general two-dimensional ( p+ q )-th order moments of an image f (x, y) are denedas

m pq =

x pyqf (x, y)dxdy p, q = 0 , 1, 2, . . . (2)

The fundamental theorem deals with the relation of the AMIs and algebraic invariants.

It was published in [1] and revisited independently by Reiss [12] and Flusser and Suk [14].It can be formulated as follows.

Theorem 1: If the binary form of order p has an algebraic invariant of weight w anddegree k

I (a p,0, , a 0,p) = wI (a p,0, , a0,p)( denotes the determinant of the respective affine transform) then the moments of order p have the same invariant but with the additional factor |J |k :

I ( p0, , 0 p) =w

|J |k

I ( p0, , 0 p),where |J | is the absolute value of the Jacobian. Actually = J = a1b2 a2b1.

This theorem makes possible to use results of the theory of algebraic invariants forcomputation of the AMIs. The algebraic invariants and the AMIs differ by the nor-malization to the scaling only. The AMIs can be derived directly, without the theoryof the algebraic invariants, by the following way. The affine transformation (1) can bedecomposed into six one-parameter transformations.

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Horizontal translation:u = x + v = y. (3)

Vertical translation:u = xv = y + . (4)

Scalingu = xv = y. (5)

Stretching:u = xv = 1 y.

(6)

Horizontal skew:u = x + tyv = y. (7)

Vertical skew: u = xv = y + sx. (8)

Any function F of moments is invariant under these six transformations if and only if itis invariant under the general affine transformation (1). Each of these transformationsimplies one constraint on the invariants. For completeness, the affine transform composedfrom these six elementary transforms cannot have negative Jacobian. To enable it, wewould have to insert possible mirror reection to the decomposition

u = x

v = zy,(9)

where z can be either 1 or -1. We will explain consequences of not doing it later.Invariancy to translation can be provided by using central moments instead of general

ones (2), any function of them is invariant under the translations (3) and (4). Centralmoments are dened as

pq =

(x xt ) p(y yt )qf (x, y)dxdy p, q = 0 , 1, 2, . . . , (10)

where xt = m10/m 00 and yt = m01/m 00 are the coordinates of the centroid.The constraint of the scaling implies the correct normalization of the moments. The

moments after scaling pq =

p+ q+2 pq, (11)

particularly00 =

200. (12)

Therefore the function F of quotients pq/( p+ q+2) / 200 is invariant under scaling.

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The moments after stretching (6) become

pq = pq pq. (13)

That is why the products of moments, where the sum of p-th indices equals the sum of q -thindices, are invariant under the stretching. The algebraic invariants are functions of binary

form coefficients and have similar form as the AMIs. They have the same coefficientsand the only difference is the normalization to scaling. The moment invariants use theexponent ( p + q + 2) / 2 while the algebraic invariants use ( p + q )/ 2. The weight w of theinvariant equals the sum of the p-th indices and the sum of q -th indices of one term of the corresponding moment invariant. It is important characterization of the invariant.Other attributes of the invariant are an order, that is the highest order of a momentin the invariant, and a degree. If we suppose the invariant in form of a polynomial of the moments, then the degree is the number of moments in one term of the polynomial.The more precise characteristics of the invariant is a structure, that is the orders of all

moments in one term.Now, something about the mirror reection. The invariants with odd weight change

their signs, when the affine transform has negative Jacobian. We can either use absolutevalues of them or suppose the affine transform has always positive Jacobian and twoimages differing by the mirror reection are two different images, how is suitable in manypractical tasks.

Two differential equations can be derived from the skew transformations (7) and (8):

p q p p1,q+1

F pq

= 0 (14)

and

p qq p+1 ,q1

F pq

= 0 . (15)

In the theory of algebraic invariants, they are called Cayley - Aronhold differential equa-tions. We can use these equations for computation of coefficients of terms of the invariantsas in [13], but another method for generation of the AMIs was used in this report, seeSection 4, where the algorithms are described.

3 The Number of the InvariantsThe number of independent invariants is very important for their using in pattern recog-nition. It can be computed by a few ways. The basic method is rule of thumb: Thenumber n of independent invariants equals

n = m p, (16)where m is the number of independent measurements of some object, in our case it isthe number of moments, and p is the number of independent constraints, which must be

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satised (see e.g. [10]). Mostly it equals the number of parameters of the transformation.This equation is called rule of thumb, because often it is not easy to nd, which mea-surements and constraints are independent and which not. Each dependency between themeasurements decreases the number of invariants by one, and each dependency betweenthe constraints increases the number of invariants by one. In our case, the moments are

independent. The zero-order moment is used to satisfy the scaling constraint and the tworst-order moments are used to satisfy two translation constraints. If we use moments of the second order only, then the differential equations (14) and (15) become:

211F

20+ 02

F 11

= 0 (17)

and211

F 02

+ 20F

11= 0 . (18)

If the derivative with respect to the 11 is excluded from these equations, we acquire

02F

02= 20

F 20

. (19)

It means, if there are terms of form

kr 120r 211

r 302, (20)

thenF

20= kr 1r 1 120

r 211

r 302 (21)

and F 02

= kr 3r 120r 211

r 3 102 . (22)

From (19) there iskr 1r 120

r 211

r 302 = kr 3

r 120

r 211

r 302 (23)

or r 1 = r 3. But then the sum of the p-th indices 2r 1 + r 2 equals the sum of the q -th indices2r 3 + r2. It means that the stretching constraint is satised automatically by satisfyingthe two skew constraints and we have 6 - 5 = 1 invariant from the second order moments.If we use moments of some higher order, then the stretching constraint must be satised

separately and we must subtract 6 from the number of moments. E.g. in case of momentsup to the third order there are 10 - 6 = 4 independent invariants.

The rule of thumb yields the whole number of independent invariants, but it doesnot say anything about their weights, orders, degrees and structures. If we need to nd thenumber of invariants with a specic structure, we can use the Cayley-Sylvester theorem.Its precise statement and proof can be found [17] or in [19]. It implies that if we denoteA(k,r,w ) the number of partitions of the number w to the sum of r integers from 0 to k,

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r,sr +3 s= w

N (2, r, 3, 2s; w) =r,s

r +3 s= w

2 + r2

3 + 2s3 w

=

=r,sr +3 s= w

(1 x2+ r )(1 x1+ r )(1 x3+2 s )(1 x2+2 s )(1 x1+2 s )(1

x)(1

x2)(1

x2)(1

x3)

w

=

=r,s

r +3 s= w

(1 x2+ r x1+ r + x3+2 r )(1 x3+2 s x2+2 s x1+2 s + x5+4 s + x4+4 s + x3+4 s x6+6 s )(1 x)(1 x2)2(1 x3) w

=

=r,s

r +3 s= w

1 x3+2 s x2+2 s x1+2 s + x5+4 s + x4+4 s + x3+4 s x6+6 sx2+ r + x5+2 s+ r + x4+2 s+ r + x3+2 s+ r

x7+4 s+ r x6+4 s+ r x5+4 s+ r + x8+6 s+ rx1+ r + x4+2 s+ r + x3+2 s+ r + x2+2 s+ r

x6+4 s+ r x5+4 s+ r x4+4 s+ r + x7+6 s+ r+ x3+2 r

x6+2 s+2 r

x5+2 s+2 r

x4+2 s+2 r

+ x8+4 s+2 r + x7+4 s+2 r + x6+4 s+2 r x9+6 s+2 r(1 x)(1 x2)2(1 x3) w

=

(31)

The members with the indices, which are always greater than w can be omitted.

=r,s

r +3 s= w

1 x3+2 s x2+2 s x1+2 s + x5+4 s + x4+4 s + x3+4 s x6+6 sx2+ r + x5+2 s+ r + x4+2 s+ r + x3+2 s+ rx1+ r + x4+2 s+ r + x3+2 s+ r + x2+2 s+ r+ x3+2 r x6+2 s+2 r x5+2 s+2 r x4+2 s+2 r

(1

x)(1

x2)2(1

x3) w

=

=[w/ 3]

s=0

1(1 x)(1 x2)2(1 x3) w

[w/ 3]

s=0

x + x2 + x3

(1 x)(1 x2)2(1 x3) w2s+

[w/ 3]

s=0

x3 + x4 + x5

(1 x)(1 x2)2(1 x3) w4s [w/ 3]

s=0

x6

(1 x)(1 x2)2(1 x3) w6s

[w/ 3]

s=0

x + x2

(1 x)(1 x2)2(1 x3) 3s+

[w/ 3]

s=0

x2 + 2 x3 + 2 x4 + x5

(1 x)(1 x2)2(1 x3) s+

[w/ 3]

s=0

x3

(1 x)(1 x2)2(1 x3) 6sw

[w/ 3]

s=0

x4 + x5 + x6

(1 x)(1 x2)2(1 x3) 4sw.

(32)

Indices can be moved: P (x)|w = P (x)x p|w+ p and it holds w 2r = 3 s r = 6 s w andw 2s 2r = s r = 4 s w. Each term is transformed separately in the following text.

[w/ 3]

s=0

1(1 x)(1 x2)2(1 x3) w

=1

(1 x)(1 x2)2(1 x3) w ([w/ 3] + 1) = (33)

The derivatives can be used for transformation of this formula. If we dene P (x) =n an xn , then P (x)|w = aw and ddx (xP (x))|w = ddx ( n an xn +1 ))|w = ( n (n +1) an xn ))|w =

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(w+1) aw = P (x)|w(w+1). The weight w can be expressed by integer division and moduloas w = 3 p + z , where p = [w/ 3].

=1

(1 x)(1 x2)2(1 x3) 3 p+ z ( p + 1) =

=(1

x3)(1

x6)2

(1 x)(1 x2)2(1 x3)2(1 x6)2 3 p+ z ( p + 1) ==

(1 + x + x2)(1 + x2 + x4)2

(1 x3)2(1 x6)2 3 p+ z ( p + 1) =

=1 + x + 3 x2 + 2 x3 + 5 x4 + 3 x5 + 5 x6 + 2 x7 + 3 x8 + x9 + x10

(1 x3)2(1 x6)2 3 p+ z ( p + 1) =

=

1 + 2x + 5 x2 + x3

(1 x)2(1 x2)2 p ( p + 1); z = 0

1 + 5x + 2 x2 + x3

(1 x)2(1 x2)2 p ( p + 1); z = 1

3 + 3x + 3 x2

(1 x)2(1 x2)2 p ( p + 1); z = 2 .

(34)

If z = 0:1 + 2x + 5 x2 + x3

(1 x)2(1 x2)2 p ( p + 1) =

ddx

x + 2 x2 + 5 x3 + x4

(1 x)2(1 x2)2 p=

=

(1 + 4x + 15 x2 + 4 x3)(1 x)2(1 x2)2(x + 2 x2 + 5 x3 + x4)(2(1 x)(1 x2)2 + (1 x)22(1 x2)(2x))

(1

x)4(1

x2)4

p

=

=(1 + 4x + 15 x2 + 4 x3)(1 x2) + 2( x + 2 x2 + 5 x3 + x4)(1 + 3 x)

(1 x)2(1 x2)3 p=

=1 + 6x + 24 x2 + 22 x3 + 17 x4 + 2 x5

(1 x)2(1 x2)3 p=

1 + 6x3 + 24x6 + 22 x9 + 17 x12 + 2 x15

(1 x3)2(1 x6)3 3 p.

(35)If z = 1:

1 + 5x + 2 x2 + x3

(1 x)2(1 x2)2 p ( p + 1) =

ddx

x + 5 x2 + 2 x3 + x4

(1 x)2(1 x2)2 p=

=

(1 + 10x + 6 x2 + 4 x3)(1 x)2(1 x2)2(x + 5 x2 + 2 x3 + x4)(2(1 x)(1 x2)2 + (1 x)22(1 x2)(2x))(1 x)4(1 x2)4 p

=

=(1 + 10x + 6 x2 + 4 x3)(1 x2) + 2( x + 5 x2 + 2 x3 + x4)(1 + 3 x)

(1 x)2(1 x2)3 p=

=1 + 12x + 21 x2 + 28x3 + 8 x4 + 2 x5

(1 x)2(1 x2)3 p=

x + 12 x4 + 21 x7 + 28 x10 + 8 x13 + 2 x16

(1 x3)2(1 x6)3 3 p+1.

(36)

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Therefore the entire second term of (32) equals

[w/ 3]

s=0

x + x2 + x3

(1 x)(1 x2)2(1 x3) w2s=

x + x2 + x3

(1 x)(1 x2)

3(1 x

3) w

x5 + x7 + x8 + x9 + x10 + x11 + x12 + x13 + x15

(1 x3)(1 x

6)

3(1 x

9) w

.

(42)

The third term of (32)

[w/ 3]

s=0

x3 + x4 + x5

(1 x)(1 x2)2(1 x3) w4s=

[w/ 3]

s=0

(x3 + x4 + x5)x4s

(1 x)(1 x2)2(1 x3) w=

=(x3 + x4 + x5)(1 x4([w/ 3]+1) )

(1 x)(1 x2)2(1 x3)(1 x4) w=

x3 + x4 + x5

(1 x)(1 x2)2(1 x3)(1 x4) w,

(43)

4([w/ 3]+1) is always greater than w, that is why the second term can be omitted. Another

term of (32) is[w/ 3]

s=0

x6

(1 x)(1 x2)2(1 x3) w6s=

[w/ 3]

s=0

x6+6 s

(1 x)(1 x2)2(1 x3) w=

=x6(1 x6([w/ 3]+1) )

(1 x)(1 x2)2(1 x3)(1 x6) w=

x6

(1 x)(1 x2)2(1 x3)(1 x6) w,

(44)

6([w/ 3]+1) is always greater than w, that is why the second term can be omitted. Anotherterm of (32) is

[w/ 3]

s=0

x + x2(1 x)(1 x2)2(1 x3) 3s

=[w/ 3]

s=0

(x + x2)x3s(1 x)(1 x2)2(1 x3) 0

=

=(x + x2)(x3([w/ 3]+1) 1)

(1 x)(1 x2)2(1 x3)(x3 1) 0=

(x + x2)(x3[w/ 3] x3)(1 x)(1 x2)2(1 x3)2 0

=

=x + x2

(1 x)(1 x2)2(1 x3)2 3[w/ 3]=

(x + x2)(1 + x + x2)(1 + x2 + x4)2

(1 x3)3(1 x6)2 3[w/ 3]=

=x + 2 x2 + 4 x3 + 5 x4 + 7 x5 + 8 x6 + 8 x7 + 7 x8 + 5 x9 + 4 x10 + 2 x11 + x12

(1 x3)3(1 x6)2 3[w/ 3]=

= 4x3

+ 8 x6

+ 5 x9

+ x12

(1 x3)3(1 x6)2 3[w/ 3]= (4x

3+ 8 x

6+ 5 x

9+ x

12)(1 + x + x

2)

(1 x3)3(1 x6)2 w=

=(4x3 + 8 x6 + 5 x9 + x12)(1 x3)

(1 x)(1 x3)3(1 x6)2 w=

4x3 + 8 x6 + 5 x9 + x12

(1 x)(1 x3)2(1 x6)2 w.

(45)

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Another term

[w/ 3]

s=0

x2 + 2 x3 + 2 x4 + x5

(1 x)(1 x2)2(1 x3) s=

[w/ 3]

s=0

(x2 + 2 x3 + 2 x4 + x5)xs

(1 x)(1 x2)2(1 x3) 0=

=(x2 + 2 x3 + 2 x4 + x5)(x([w/ 3]+1) 1)

(1 x)(1 x2

)2

(1 x3

)(x1

1) 0=

(x2 + 2 x3 + 2 x4 + x5)(x[w/ 3] x)(1 x)

2

(1 x2

)2

(1 x3

) 0=

=x2 + 2 x3 + 2 x4 + x5

(1 x)2(1 x2)2(1 x3) [w/ 3]=

x6 + 2 x9 + 2 x12 + x15

(1 x3)2(1 x6)2(1 x9) 3[w/ 3]=

=(x6 + 2 x9 + 2 x12 + x15)(1 + x + x2)

(1 x3)2(1 x6)2(1 x9) w=

x6 + 2 x9 + 2 x12 + x15

(1 x)(1 x3)(1 x6)2(1 x9) w.

(46)The 7th term of (32)

[w/ 3]

s=0

x3

(1 x)(1 x2

)2

(1 x3

) 6sw=

[w/ 3]

s=0

x3+ w6s

(1 x)(1 x2

)2

(1 x3

) 0=

=x3+ w(x6([w/ 3]+1) 1)

(1 x)(1 x2)2(1 x3)(x6 1) 0=

x3+ w(x6[w/ 3] x6)(1 x)(1 x2)2(1 x3)(1 x6) 0

=

=x3+2 w6[w/ 3]

(1 x)(1 x2)2(1 x3)(1 x6) w=

x3+2 z

(1 x)(1 x2)2(1 x3)(1 x6) 3 p+ z=

=x3+ z(1 + x + x2)(1 + x2 + x4)2

(1 x3)2(1 x6)3 3 p=

x3+ z(1 + x + 3 x2 + 2 x3 + 5 x4 + 3 x5 + 5 x6 + 2 x7 + 3 x8 + x9 + x10)(1

x3)2(1

x6)3

3 p

=

=

z = 0 :x + 2 x2 + 5 x3 + x4

(1 x)2(1 x2)3 p=

x3 + 2 x6 + 5 x9 + x12

(1 x3)2(1 x6)3 3 pz = 1 :

3x2 + 3 x3 + 3 x4

(1 x)2(1 x2)3 p=

3x7 + 3 x10 + 3 x13

(1 x3)2(1 x6)3 3 p+1z = 2 :

x2 + 5 x3 + 2 x4 + x5

(1 x)2(1 x2)3 p=

x8 + 5 x11 + 2 x14 + x17

(1 x3)2(1 x6)3 3 p+2

=

=x3 + 2 x6 + 3 x7 + x8 + 5 x9 + 3 x10 + 5 x11 + x12 + 3 x13 + 2 x14 + x17

(1 x3)2(1 x6)3 w

(47)

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and last term

[w/ 3]

s=0

x4 + x5 + x6

(1 x)(1 x2)2(1 x3) 4sw[w/ 3]

s=0

(x4 + x5 + x6)xw4s

(1 x)(1 x2)2(1 x3) 0=

=(x4 + x5 + x6)xw(x4([w/ 3]+1) 1)(1 x)(1 x

2

)2

(1 x3

)(x4

1) 0=

(x4 + x5 + x6)xw(x4[w/ 3] x4)(1 x)(1 x

2

)2

(1 x3

)(1 x4

) 0=

=(x4 + x5 + x6)x3 p+ z4 p

(1 x)(1 x2)2(1 x3)(1 x4) w=

(x4 + x5 + x6)xz

(1 x)(1 x2)2(1 x3)(1 x4) p=

=

z = 0 :x12 + x15 + x18

(1 x3)(1 x6)2(1 x9)(1 x12) 3 pz = 1 :

x16 + x19 + x22

(1 x3)(1 x6)2(1 x9)(1 x12) 3 p+1z = 2 :

x20 + x23 + x26

(1

x3)(1

x6)2(1

x9)(1

x12)

3 p+2

=

=x12 + x15 + x16 + x18 + x19 + x20 + x22 + x23 + x26

(1 x3)(1 x6)2(1 x9)(1 x12) w.

(48)

Now all terms can be added up

r,sr +3 s= w

N (2, r, 3, 2s; w) =

=

1 + x + 3 x2 + 6 x3 + 12 x4 + 12 x5 + 24 x6 + 21 x7 + 30x8+22 x9 + 28 x10 + 18 x11 + 17 x12 + 8 x13 + 9 x14 + 2 x15 + 2 x16

(1 x3)2(1 x6)3 w

x + x2 + x3

(1 x)(1 x2)3(1 x3) w+

x5 + x7 + x8 + x9 + x10 + x11 + x12 + x13 + x15

(1 x3)(1 x6)3(1 x9) w+

x3 + x4 + x5

(1 x)(1 x2)2(1 x3)(1 x4) w x6

(1 x)(1 x2)2(1 x3)(1 x6) w

4x3 + 8 x6 + 5 x9 + x12

(1 x)(1 x3)2(1 x6)2 w+

x6 + 2 x9 + 2 x12 + x15

(1 x)(1 x3)(1 x6)2(1 x9) w+

x3 + 2 x6 + 3 x7 + x8 + 5 x9 + 3 x10 + 5 x11 + x12 + 3 x13 + 2 x14 + x17

(1 x3)2(1 x6)3 w

x12 + x15 + x16 + x18 + x19 + x20 + x22 + x23 + x26

(1 x3)(1 x6)2(1 x9)(1 x12) w=

(49)

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=

1 + x + 3 x2 + 7 x3 + 12 x4 + 12 x5 + 26 x6 + 24x7 + 31 x8 + 27 x9+31 x10 + 23 x11 + 18x12 + 11 x13 + 11 x14 + 2 x15 + 2 x16 + x17

(1 x3)2(1 x6)3 w

x + x2 + x3

(1 x)(1 x2)3(1 x3) w

+

(x5 + x7 + x8 + x9 + x10 + x11 + x12 + x13 + x15)(1 x12)(x12 + x15 + x16 + x18 + x19 + x20 + x22 + x23 + x26)(1 x6)(1 x3)(1 x6)3(1 x9)(1 x12) w

+(x3 + x4 + x5)(1 x6) x6(1 x4)

(1 x)(1 x2)2(1 x3)(1 x4)(1 x6) w

(4x3 + 8 x6 + 5 x9 + x12)(1 x9) (x6 + 2 x9 + 2 x12 + x15)(1 x3)(1 x)(1 x3)2(1 x6)2(1 x9) w

=

(50)

=

(1 + x + 3 x2 + 7 x3 + 12 x4 + 12 x5 + 26x6 + 24 x7 + 31 x8 + 27 x9 + 31 x10

+23 x11

+ 18 x12

+ 11 x13

+ 11 x14

+ 2 x15

+ 2 x16

+ x17

)(1 x)(1 x2

)3

(1 x4)(1 x9)(1 x12)(x + x2 + x3)(1 x3)(1 x4)(1 x6)3(1 x9)(1 x12)+( x5 + x7 + x8 + x9 + x10 + x11 + x13 x16 x17 2x19 2x20

x22 2x23 x27 + x28 + x29 + x32)(1 x)(1 x2)3(1 x3)(1 x4)+( x3 + x4 + x5 x6 x9 x11)(1 x2)(1 x3)(1 x6)2(1 x9)(1 x12)(4x3 + 7 x6 + 4 x9 3x12 7x15 4x18 x21)(1 x2)3(1 x4)(1 x6)(1 x12)

(1 x)(1 x2)3(1 x3)2(1 x4)(1 x6)3(1 x9)(1 x12) w=

=

(1 + x + 3 x2 + 7 x3 + 12 x4 + 12 x5 + 26 x6 + 24 x7 + 31 x8 + 27 x9 + 31 x10

+23 x11 + 18 x12 + 11 x13 + 11 x14 + 2 x15 + 2 x16 + x17)(1 x2)2(1 x4)(1 x9)(1 x12)(x + x2 + x3)(1 + x + x2)(1 + x2)(1 x6)3(1 x9)(1 x12)+( x5 + x7 + x8 + x9 + x10 + x11 + x13 x16 x17 2x19 2x20x22 2x23 x27 + x28 + x29 + x32)(1 x2)2(1 x3)(1 x4)+( x3 + x4 + x5 x6 x9 x11)(1 + x + x2)(1 x6)2(1 x9)(1 x12)

(4x3 + 7 x6 + 4 x9 3x12 7x15 4x18 x21)(1 + x)(1 x2)(1 x4)(1 x6)(1 x12)(1 x2)2(1 x3)2(1 x4)(1 x6)3(1 x9)(1 x12) w

=

(51)

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=

1 + x + x2 + 5 x3 + 6 x4 2x5 + 4 x6 + 2 x7 16x8 9x9 11x10 15x1110x12 6x13 + 12 x14 + x15 + 3 x16 + 34 x17 + 5 x18 + 5 x19 + 34 x20 + 6 x21+3 x22 + 4 x23 3x24 7x25 29x26 7x27 6x28 36x29 2x30 18x32+5 x33 + 9 x34 + 11 x35 + 12 x36 + 12 x37 + 16 x38 + x39 3x40 + 4 x41 7x42

7x43 2x45 x46

(x + 2 x2 + 4 x3 + 4 x4 + 4 x5 + 2 x6

2x7

6x8

12x9

13x10

14x11

10x12 5x13 + 6 x15 + 10 x16 + 14 x17 + 16 x18 + 16 x19 + 16 x20 + 14 x21+9 x22 + 4 x23 4x24 9x25 14x26 16x27 16x28 16x29 14x3010x31 6x32 + 5 x34 + 10 x35 + 14x36 + 13 x37 + 12 x38 + 6 x39 + 2 x40

2x41 4x42 4x43 4x44 2x45 x46)+ x5 x7 x9 x12 + x13 + x14 + x15 2x17 + 2 x18 + x19 2x20 + 2 x212x23 2x27 + 3 x29 2x30 x31 + 2 x32 x33 + 2 x35 + x36 x37 x38+ x39 2x41 + x43+ x3 + 2 x4 + 3 x5 + x6 x8 3x9 5x10 8x11 4x12 3x13 x14 + x15+2 x16 + 5 x17 + 5 x18 + 7 x19 + 8 x20 + 5 x21 + 5 x22 + 4 x23 2x25 6x265x27 6x28 7x29 5x30 6x31 5x32 x33 + x34 + 4 x35 + 4 x36+5 x37 + 7 x38 + 3 x39 + 2 x40

x41

x42

x43

2x44

x45

x46

(4x3 + 4 x4 4x5 + 3 x6 + 3 x7 11x8 3x9 3x10 7x11 3x12 3x13+10 x14 5x15 5x16 + 25 x17 3x18 3x19 + 23x20 x21 x22 + 2 x2320x26 + 4 x27 + 4 x28 26x29 + 3 x30 + 3 x31 13x32 + 5 x33 + 5 x34 + 5 x35+3 x36 + 3 x37 + 10 x38 3x39 3x40 + 5 x41 3x42 3x43 + x44 x45 x46)

(1 x2)2(1 x3)2(1 x4)(1 x6)3(1 x9)(1 x12) w=

(52)

=

1 x2 2x3 + 2 x5 + 2 x9 2x11 2x12 + 2 x14 + 2 x15 2x17 x18 + x20+ x24 x26 2x27 + 2 x29 + 2 x30 2x32 2x33 + 2 x35 + 2 x39 2x41 x42+ x44(1

x2)2(1

x3)2(1

x4)(1

x6)3(1

x9)(1

x12) w

=

=1 2x3 + 2 x9 2x12 + 2 x15 x18 + x24 2x27 + 2 x30 2x33 + 2 x39 x42

(1 x2)(1 x3)2(1 x4)(1 x6)3(1 x9)(1 x12) w=

=1 x3 x6 + x9 x12 + x15 + x24 x27 + x30 x33 x36 + x39

(1 x2)(1 x3)(1 x4)(1 x6)3(1 x9)(1 x12) w=

=1 x6 x12 + x24 + x30 x36

(1 x2)(1 x4)(1 x6)3(1 x9)(1 x12) w=

=1 x12 x18 + x30

(1

x2)(1

x4)(1

x6)2(1

x9)(1

x12) w

=

=1 x18

(1 x2)(1 x4)(1 x6)2(1 x9) w=

=1 + x9

(1 x2)(1 x4)(1 x6)2 w.

(53)The result in form 1 / (1 xn )|w can be understood as the sum of a geometric series

1 + xn + x2n + x3n + |w . It means an invariant with a weight n and all its powers.

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Similarly, the result in form 1 / ((1 xn 1 )(1 xn 2 ))|w means two independent invariantswith weights n1 and n2, all their powers and products. The polynomials in the numeratorexpress invariants, which are linearly independent, but some of their powers or productsare linearly dependent. Such invariants together with the independent ones are calledirreducible. Generally, the reducible invariant can be expressed as a polynomial of other

invariants, the irreducible one cannot.In our case there is one invariant of weight 2, one invariant of weight 4 and two

invariants of weight 6. There is also one invariant of weight 9, but its second poweris dependent. Therefore its sign is independent, but its absolute value is algebraicallydependent on the other four invariants.

There is another method, how to acquire expressions as the last one in (53). We cancompute the beginning of the sequence in (30) for adequate amount of w and then searcha combination of weights, which causes in polynomial multiplication that last membersof the sequence are zero and all members are nonnegative. The following expressions

were acquired by this method. The invariants from the moments of the 2nd, 3rd and 4thorders:

N (2, r, 3, 2s, 4, t ; w) =r,s,t

r +3 s+2 t= w=

=2 + r

23 + 2s

34 + t

4 wr,s,tr +3 s+2 t= w=

=

1 + x4 + 4 x6 + 2 x7 + 5 x8 + 9 x9 + 9 x10 + 11 x11 + 18 x12 + 20 x13 + 21x14 + 34 x15+30 x16 + 36 x17 + 44 x18 + 45 x19 + 46 x20 + 60x21 + 54 x22 + 57 x23 + 66 x24+59 x25 + 59 x26 + 66 x27 + 57 x28 + 54 x29 + 60x30 + 46 x31 + 45 x32 + 44 x33

+36 x34 + 30 x35 + 34 x36 + 21 x37 + 20 x38 + 18 x39 + 11 x40 + 9 x41 + 9 x42 + 5 x43+2 x44 + 4 x45 + x47 + x51

(1 x2)(1 x4)2(1 x6)2(1 x8)(1 x10)(1 x12)(1 x18) w.

(54)The numbers of invariants from the moments of the 2nd, 3rd, 4th and 5th orders were

acquired by similar method:

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N (2, r, 3, s, 4, t, 5, u; w) =r,s,t,u

2r +3 s+4 t+5 u =2 w=

=2 + r

23 + s

34 + t

45 + u

5 wr,s,t,u2r +3 s+4 t+5 u=2 w=

=

1 + x4 + x5 + 7 x6 + 6 x7 + 19x8 + 28 x9 + 51 x10 + 71 x11 + 128x12 + 180x13+284 x14 + 414x15 + 609x16 + 854x17 + 1241x18 + 1673x19 + 2322x20 + 3119x21

+4158x22 + 5439x23 + 7122x24 + 9095x25 + 11606x26 + 14601x27 + 18195x28+22428x29 + 27529x30 + 33282x31 + 40054x32 + 47774x33 + 56473x34

+66257x35 + 77232x36 + 89180x37 + 102379x38 + 116695x39 + 131987x40+148324x41 + 165669x42 + 183664x43 + 202381x44 + 221699x45 + 241124x46+260807x47 + 280390x48 + 299499x49 + 318072x50 + 335954x51 + 352493x52+367915x53 + 381767x54 + 393829x55 + 403917x56 + 412146x57 + 417835x58+421456x59 + 422688x60 + 421456x61 + 417835x62 + 412146x63 + 403917x64+393829x65 + 381767x66 + 367915x67 + 352493x68 + 335954x69 + 318072x70

+299499x71 + 280390x72 + 260807x73 + 241124x74 + 221699x75 + 202381x76+183664x77 + 165669x78 + 148324x79 + 131987x80 + 116695x81 + 102379x82

+89180x83 + 77232x84 + 66257x85 + 56473x86 + 47774x87 + 40054x88+33282x89 + 27529x90 + 22428x91 + 18195x92 + 14601x93 + 11606x94 + 9095x95

+7122x96 + 5439x97 + 4158x98 + 3119x99 + 2322x100 + 1673x101 + 1241x102+854 x103 + 609x104 + 414x105 + 284x106 + 180x107 + 128x108 + 71 x109 + 51x110

+28 x111 + 19 x112 + 6 x113 + 7 x114 + x115 + x116 + x120

(1 x2)(1 x4)2(1 x6)2(1 x7)(1 x8)(1 x10)(1 x11)(1 x12)(1 x15)2(1 x16)(1 x18)(1 x20)w

.

(55)

These expressions are not too useful, because they do not say anything about thestructure of the invariants and the weights of the invariants in the denominator can bechanged. The expressions for homogeneous invariants, i.e. invariants from the moments of the same order, are more interesting. They can be calculated by the simpler way, becausethe summation is omitted:

N (2, r ; w) = 2 + r2 w=

11 x2 w

w = r (56)

N (3, 2s; w) =3 + 2s

3 w =1

1 x6 w w = 3 s (57)N (4, r ; w) = 4 + r4 w

=1

(1 x4)(1 x6) ww = 2 r (58)

N (5, 2s; w) = 5 + 2s5 w=

1 + x45

(1 x10)(1 x20)(1 x30) ww = 5 s (59)

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N (6, r ; w) = 6 + r6 w=

1 + x45

(1 x6)(1 x12)(1 x18)(1 x30) ww = 3 r

(60)

N (7, 2s; w) = 7 + 2s7 w=

=

1 + 2x28 + 4 x42 + 4 x49 + 5 x56 + 9 x63 + 6 x70 + 9 x77 + 8 x84 + 9 x91+6 x98 + 9 x105 + 5 x112 + 4 x119 + 4 x126 + 2 x140 + x168

(1 x14)(1 x28)(1 x42)2(1 x70) ww = 7 s

(61)

N (8, r ; w) = 8 + r

8 w=

1 + x32 + x36 + x40 + x72

(1 x8)(1 x

12)(1 x

16)(1 x

20)(1 x

24)(1 x

28) ww = 4 r

(62)

N (9, 2s; w) = 9 + 2s9 w=

=

1 + x18 + 5 x36 + 4 x45 + 17 x54 + 20x63 + 47 x72 + 61 x81 + 97 x90 + 120x99+165 x108 + 189x117 + 223x126 + 241x135 + 254x144 + 254x153 + 241x162

+223 x171 + 189x180 + 165x189 + 120x198 + 97 x207 + 61x216 + 47 x225

+20 x234

+ 17 x243

+ 4 x252

+ 5 x261

+ x279

+ x297

(1 x18)(1 x36)(1 x45)(1 x54)2(1 x63)(1 x72) ww = 9 s

(63)

N (10, r ; w) = 10 + r10 w=

=

1 + 2x30 + 4 x40 + 4 x45 + 7 x50 + 8 x55 + 15 x60 + 15 x65 + 20 x70 + 27 x75+29 x80 + 35 x85 + 40 x90 + 44 x95 + 47 x100 + 55 x105 + 52 x110 + 57 x115

+56 x120 + 57 x125 + 52 x130 + 55 x135 + 47 x140 + 44 x145 + 40 x150 + 35x155

+29 x160

+ 27 x165

+ 20 x170

+ 15 x175

+ 15 x180

+ 8 x185

+ 7 x190

+ 4 x195

+4 x200 + 2 x210 + x240

(1 x10)(1 x20)(1 x30)2(1 x40)(1 x45)(1 x50)(1 x70) ww = 5 r

(64)

It corresponds with results published in [15]. Table 1 presents weights of the indepen-dent homogeneous invariants falling under the limit 10.

The way of expressing the number of invariants in (53), (54) and (55) does not expressthe structure of the invariants. We can use another way. E.g. the number of invariants

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Table 1: The weights 10 of the independent homogeneous invariantsorder 2 3 4 5 6 7 8 9 10

weight 2 6 4,6 10 6 - 8 - 10

of the 2nd and 3rd orders (53) can be expressed as

1 + x3y6

(1 x2)(1 y6)(1 xy3)(1 x3y3) r, 3s=

1 + x3y4

(1 x2)(1 y4)(1 xy2)(1 x3y2) r, s,

(65)where |r,s denotes the coefficient at the term, where the exponent of the x equals r

and the exponent of the y equals s. We use the convention that x corresponds to the 2ndorder moments, y to the 3rd order, z to the 4th order and v to the 5th order.

The rst expression uses weights, i.e. it expresses the number of invariants, where

the contribution of the 2nd order moments to the weight is r and the contribution of the3rd order moments to the weight is 3 s. The second expression uses the numbers of themoments in one term, i.e. it expresses the number of invariants, where the number of the2nd order moments in one term is r and the number of the 3rd order moments in oneterm is s. Since the weight of s 3rd order moments is 3s/ 2, it holds s = 2 s.

The structure of the invariant means the numbers of moments of the individual ordersin each term of the invariant. It can be expressed as a list of the orders or as a list of thenumbers of the orders from 2. This latter way is used in this paper, i.e. the structure is avector showing the number of moments of the second, third, fourth, etc. orders involved

in one term, e.g. the structure 0,1,0,3 means that each term of the invariant is the productof one 3rd order moment and three 5th order moments. There are no 2nd and 4th ordermoments.

The following expressions were obtained by the similar method as the previous morecomplicated ones, i.e. the array of the numbers of linearly independent invariants wascomputed by means of Gauss polynomials and then the expressions in the denominatorwere tipped and their contributions were subtracted from the array. The homogeneousmembers were used as the expressions in the denominator rst, other were tipped fromperiodicity of the difference until no innite residue remains. The number of invariants

of the 2nd and 4th orders

1 + x3z 6

(1 x2)(1 z 4)(1 z 6)(1 x2z 2)(1 x2z 4) r,, 2t=

=1 + x3z 3

(1 x2)(1 z 2)(1 z 3)(1 x2z )(1 x2z 2) r,, t,

(66)

where t = 2 t. It means there is one invariants of the second order with weight 2, twoinvariants of the fourth order with weights 4 and 6, two simultaneous invariants second

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and fourth order with weights 4 and 6 and one dependent simultaneous invariant withweight 9.

The number of invariants of the 3rd and 4th orders

1 + y6z 4 + 2 y6z 6 + 2 y6z 8 + y6z 10 + y9z 4 + 3 y9z 6 + 2 y9z 8 + y9z 10

y12z 8 2y12z 10 3y12z 12 y12z 14 y15z 8 2y15z 10 2y15z 12 y15z 14x21y18(1 y6)(1 z 4)(1 z 6)(1 y6z 2)(1 y6z 4)(1 y3z 6)(1 y6z 6) ,3s, 2t

=

=

1 + y4z 2 + 2 y4z 3 + 2 y4z 4 + y4z 5 + y6z 2 + 3 y6z 3 + 2 y6z 4 + y6z 5

y8z 4 2y8z 5 3y8z 6 y8z 7 y10z 4 2y10z 5 2y10z 6 y10z 7 x14y9(1 y4)(1 z 2)(1 z 3)(1 y4z )(1 y4z 2)(1 y2z 3)(1 y4z 3) ,s, t

.

(67)We were not successful in removing negative coefficients in the numerator. It means

that not all invariants in the denominator are independent. These results correspond with

that published in [16].There is used number notation in the next expressions, not weight one. The number

of invariants of the 2nd, 3rd and 4th orders

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1 + x3y4 + 2 xy2z + 2 x2y2z + x3y2z + x4y2z + 2 xy4z + x2y4z + x3y4z x5y4z x3y6z x4y6z x3y8z + 3 xy2z 2 + 3 x2y2z 2 + x3y2z 2 + x4y2z 2 + y4z 2 + 4 xy4z 2+3 x2y4z 2 + 2 x3y4z 2 x5y4z 2 + y6z 2 2x3y6z 2 3x4y6z 2 xy6z 2 x2y6z 2

3x3y6z 2 2x4y6z 2 + x3z 3 + 2 xy2z 3 + 3 x2y2z 3 x4y2z 3 x6y2z 3 + 2 y4z 3+5 xy

4

z 3

+ 4 x2

y4

z 3

+ x3

y4

z 3

x5

y4

z 3

+ x7

y4

z 3

+ 3 y6

z 3

x2

y6

z 3

2x3

y6

z 3

4x4y6z 3 + x6y6z 3 3xy8z 3 3x2y8z 3 6x3y8z 3 4x4y8z 3 x2y10z 3 x5y10z 3+ x3y12z 3 + xy2z 4 + x2y2z 4 x3y2z 4 x4y2z 4 + 2 y4z 4 + 3 xy4z 4 + 2 x2y4z 4x3y4z 4 2x4y4z 4 x5y4z 4 + 2 y6z 4 2x2y6z 4 2x3y6z 4 3x4y6z 4 + 2 x6y6z 4y8z 4 3xy8z 4 5x2y8z 4 5x3y8z 4 4x4y8z 4 + 2 x6y8z 4 y10z 4 2x2y10z 4

x3y10z 4 x4y10z 4 2x5y10z 4 + xy12z 4 + x2y12z 4 + 2 x3y12z 4 + x4y12z 4 x3y2z 5x4y2z 5 + y4z 5 + xy4z 5 x2y4z 5 3x3y4z 5 3x4y4z 5 2x5y4z 5y6z 5 3x2y6z 52x3y6z 5 2x4y6z 5 x5y6z 5 + 2 x6y6z 5 2y8z 5 3xy8z 5 5x2y8z 5 4x3y8z 5x4y8z 5 + x5y8z 5 + 3 x6y8z 5 + x7y8z 5 2y10z 5 xy10z 5 3x2y10z 5 2x3y10z 5

x5y10z 5 + x6y10z 5 + x7y10z 5 + xy12z 5 + 2 x2y12z 5 + 3 x3y12z 5 + 2 x4y12z 5+ x5y12z 5 + x2y14z 5 + x3y14z 5 + x4y14z 5 + x5y14z 5

x3y2z 6

2x2y4z 6

3x3y4z 6

2x4y4z 6 2x5y4z 6 2x2y6z 6 2x3y6z 6 2x4y6z 6 2x5y6z 6 3x2y8z 62xy8z 6 2x2y8z 6 2x3y8z 6 + x4y8z 6 + 2 x5y8z 6 + 2 x6y8z 6 + 2 x7y8z 6 2y10z 62xy10z 6 2x2y10z 6 x3y10z 6 + 2 x4y10z 6 + 2 x5y10z 6 + 2 x6y10z 6 + 3 x7y10z 6+2 x2y12z 6 + 2 x3y12z 6 + 2 x4y12z 6 + 2 x5y12z 6 + 2 x2y14z 6 + 2 x3y14z 6 + 3 x4y14z 6

+2 x5y14z 6 + x4y16z 6 x2y4z 7 x3y4z 7 x4y4z 7 x5y4z 7 x2y6z 7 2x3y6z 73x4y6z 7 2x5y6z 7 x6y6z 7 y8z 7 xy8z 7 + x2y8z 7 + 2 x4y8z 7 + 3 x5y8z 7+ x6y8z 7 + 2 x7y8z 7 y10z 7 3xy10z 7 x2y10z 7 + x3y10z 7 + 4 x4y10z 7 + 5 x5y10z 7+3 x6y10z 7 + 2 x7y10z 7 2xy12z 7 + x2y12z 7 + 2 x3y12z 7 + 2 x4y12z 7 + 3 x5y12z 7x7y12z 7 + 2 x2y14z 7 + 3 x3y14z 7 + 3 x4y14z 7 + x5y14z 7 x6y14z 7 x7y14z 7+ x3y16z 7 + x4y16z 7 x3y6z 8 2x4y6z 8 x5y6z 8 x6y6z 8 + 2 x2y8z 8 + x3y8z 8+ x4y8z 8 + 2 x5y8z 8 + x7y8z 8

2xy10z 8 + 4 x2y10z 8 + 5 x3y10z 8 + 5 x4y10z 8

+3 x6y10z 8 + x7y10z 8 2xy12z 8 + 3 x3y12z 8 + 2 x4y12z 8 + 2 x5y12z 8 2x7y12z 8+ x2y14z 8 + 2 x3y14z 8 + x4y14z 8 2x5y14z 8 3x6y14z 8 2x7y14z 8 + x3y16z 8+ x4y16z 8 x5y16z 8 x6y16z 8 x4y6z 9 + x2y8z 9 + x5y8z 9 + 4 x3y10z 9 + 6 x4y10z 9+3 x5y10z 9 + 3 x6y10z 9 xy12z 9 + 4 x3y12z 9 + 2 x4y12z 9 + x5y12z 9 3x7y12z 9y14z 9 + x2y14z 9 x4y14z 9 4x5y14z 9 5x6y14z 9 2x7y14z 9 + xy16z 9 + x3y16z 93x5y16z 9 2x6y16z 9 x4y18z 9 + 2 x3y10z 10 + 3 x4y10z 10 + x5y10z 10 + x6y10z 10+3 x3y12z 10 + 2 x4y12z 10 x7y12z 10 + x2y14z 10 2x4y14z 10 3x5y14z 104x6y14z 10 x7y14z 10 x3y16z 10 x4y16z 10 3x5y16z 10 3x6y16z 10 + x4y10z 11+ x3y12z 11 + x4y12z 11 + x2y14z 11 x4y14z 11 x5y14z 11 2x6y14z 11 x3y16z 11

x4y16z 11 2x5y16z 11 2x6y16z 11 x4y14z 12 x7y18z 12(1 x2)(1 y4)(1 z 2)(1 z 3)(1 xy2)(1 x3y2)(1 x2z )(1 x2z 2)(1 y4z )(1 y4z 2)(1 y2z 3)(1 y4z 3)

r, s, t .

(68)The negative coefficients in the numerator again mean not all invariants in the de-

nominator are independent. The denominator also suggests there is no other independentinvariant from moments of all 3 of 2nd, 3rd and 4th orders, but all independent invariantsare either homogeneous or from moments of two orders only.

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The number of invariants from moments of 2nd and 5th orders

1 + x2v4 + x3v4 + x4v4 + x5v4 + x7v4 + xv6 + x2v6 + x3v6 + x4v6 + x6v6 x8v6+ xv8 + x2v8 + x3v8 + x5v8 x7v8 + xv10 + x2v10 + x4v10 x6v10 + xv12 + x3v12x5v12 x7v12 + x2v14 x4v14 x6v14 x7v14 + xv16 x3v16 x5v16 x6v16x7v16 + v18 x2v18 x4v18 x5v18 x6v18 x7v18 xv20 x3v20 x4v20

x5

v20

x6

v20

x8

v24

(1 x2)(1 v4)(1 v8)(1 v12)(1 xv2)(1 x3v2)(1 x5v2) r,,, u.

(69)The numerator can also be expressed by a table (Table 2).

Table 2: The numerator - 2nd and 5th order

x0 x1 x2 x3 x4 x5 x6 x7 x8

v0 1 0 0 0 0 0 0 0 0v2 0 0 0 0 0 0 0 0 0

v4

0 0 1 1 1 1 0 1 0v6 0 1 1 1 1 0 1 0 -1v8 0 1 1 1 0 1 0 -1 0v10 0 1 1 0 1 0 -1 0 0v12 0 1 0 1 0 -1 0 -1 0v14 0 0 1 0 -1 0 -1 -1 0v16 0 1 0 -1 0 -1 -1 -1 0v18 1 0 -1 0 -1 -1 -1 -1 0v20 0 -1 0 -1 -1 -1 -1 0 0v22 0 0 0 0 0 0 0 0 0v24 0 0 0 0 0 0 0 0 -1

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The number of invariants from moments of 3rd and 5th orders

n35(y, v)(1 y4)(1 v4)(1 v8)(1 v12)(1 y3v)(1 y5v)(1 y2v2)(1 yv3)(1 y5v3) ,s,, u

.

(70)The numerator n35(y, v) can be seen in Table 3.

Table 3: The numerator - 3rd and 5th order

y0 y1 y2 y3 y4 y5 y6 y7 y8 y9 y10 y11 y12 y13 y14 y15 y16

v0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0v1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0v2 0 0 1 0 1 0 1 0 0 0 0 0 0 0 0 0 0v3 0 0 0 3 0 2 0 1 0 0 0 0 0 0 0 0 0v4 0 0 1 0 4 0 2 0 -1 0 -1 0 -1 0 0 0 0v5 0 1 0 3 0 5 0 1 0 -2 0 -2 0 0 0 0 0

v6

0 0 3 0 4 0 3 0 -2 0 -4 0 -1 0 0 0 0v7 0 1 0 4 0 3 0 0 0 -4 0 -3 0 0 0 -1 0v8 0 0 2 0 4 0 2 0 -2 0 -3 0 -2 0 -1 0 0v9 0 2 0 3 0 4 0 0 0 -3 0 -3 0 -3 0 0 0v10 0 0 2 0 3 0 1 0 -2 0 -4 0 -3 0 -1 0 1v11 0 1 0 2 0 2 0 -1 0 -4 0 -5 0 -1 0 0 0v12 0 0 2 0 2 0 0 0 -3 0 -5 0 -2 0 1 0 0v13 0 1 0 1 0 0 0 -3 0 -5 0 -1 0 0 0 1 0v14 0 0 0 0 1 0 -2 0 -4 0 -2 0 1 0 0 0 0v15 0 1 0 0 0 -1 0 -5 0 -3 0 0 0 1 0 1 0v16 0 0 1 0 -2 0 -5 0 -3 0 0 0 2 0 2 0 0v17 0 0 0 -1 0 -5 0 -4 0 -1 0 2 0 2 0 1 0v18 1 0 -1 0 -3 0 -4 0 -2 0 1 0 3 0 2 0 0v19 0 0 0 -3 0 -3 0 -3 0 0 0 4 0 3 0 2 0v20 0 0 -1 0 -2 0 -3 0 -2 0 2 0 4 0 2 0 0v21 0 -1 0 0 0 -3 0 -4 0 0 0 3 0 4 0 1 0v22 0 0 0 0 -1 0 -4 0 -2 0 3 0 4 0 3 0 0v23 0 0 0 0 0 -2 0 -2 0 1 0 5 0 3 0 1 0v24 0 0 0 0 -1 0 -1 0 -1 0 2 0 4 0 1 0 0v25 0 0 0 0 0 0 0 0 0 1 0 2 0 3 0 0 0v26 0 0 0 0 0 0 0 0 0 0 1 0 1 0 1 0 0v27 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0v28 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1

The number of invariants from moments of 4th and 5th orders

n45(z, v)(1 z 2)(1 z 3)(1 v4)(1 v8)(1 v12)(1 z 3v2)(1 z 5v2)(1 zv4)(1 z 2v4)(1 z 3v4)(1 z 5v4)

,, t, u. (71)

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The numerator n45(z, v) can be seen in Table 4.We cannot see directly how many irreducible invariants exists from expressions with

more complicated numerator. The following algorithm was developed for their retrieval.

1. Two arrays are needed, one contains the number of linearly independent invariants,it can be computed by means of Gauss polynomials. The second array containscounted invariants, it is lled by zeros, only rst element equals 1 (the element withall zero indices). The size is theoretically innite, we choose some size leading to anacceptable computing time in practice, but all structures we want to search must beincluded in it. The size of both arrays must be the same. The number of dimensionsof the arrays equals the number of orders of the moments. The indices equal thenumber of moments of the corresponding order in one term of the invariant.

2. Find another irreducible invariant from the difference of both arrays. Find thenonzero element with the maximum number of zero indices. If there is more than

one such element, choose that with minimum weight. If there is no other nonzeroelement, than stop.

3. Count powers and products of the new invariant, i.e. copy the array of countedinvariants, shift it so the rst element is at the newly found invariant and add.Then shift the original copy twice and add, shift three times and add etc. Forgetelements outside the chosen size of the array. Stop shifts and additions, when allelements are outside. Compare the sum with the array of linearly independentinvariants. If it is greater, then the new invariant is dependent. Substitute the

element of the sum with the number of linearly independent invariants in this case.4. Go to the point 2.

The results of this algorithm up to the 5th order and weight 10 are in Table 5. Theresults with the highest order 6, 7, 8, 9 and 10 are in Tables 6, 7, 8, 9 and 10, respectively.The structures of the invariants are represented by the numbers of moments of individ-ual orders in one term of the invariant. The numbers and structures of the invariantscorrespond to the results of the next section.

The only ve-tuple of orders is 2,3,4,5,6 with weight (2+3+4+5+6)/2=10, all other

ve-tuples have higher weight and similarly structures of more orders. Therefore we cansearch four-tuples only for orders higher than 6. Nevertheless, the invariants with thehighest 7th order were searched completely. The results with the highest orders 8, 9 and10 were computed for four-tuples of orders only because of saving time.

These results are interesting, nevertheless, the possibilities of the Cayley-Sylvestertheorem for the computation of the structure of all irreducible invariants are limited. Letus imagine such a situation, we have an array of the numbers of all linearly independentinvariants and we subtract the numbers of the products of the found irreducible invariants

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Table 5: The numbers of the irreducible invariants up to the 5th order and up to theweight 10. They are presented in the form weight : (structure ) number of invariants .

orders w : (s) n2 2 : (2, 0, 0, 0) 13 6 : (0, 4, 0, 0) 12,3 4 : (1, 2, 0, 0) 1 6 : (3, 2, 0, 0) 1 9 : (3, 4, 0, 0) 14 4 : (0, 0, 2, 0) 1 6 : (0, 0, 3, 0) 12,4 4 : (2, 0, 1, 0) 1 4 : (2, 0, 2, 0) 1 9 : (3, 0, 3, 0) 13,4 8 : (0, 4, 1, 0) 1 9 : (0, 2, 3, 0) 1 10 : (0, 4, 2, 0) 22,3,4 6 : (1, 2, 1, 0) 2 7 : (2, 2, 1, 0) 2 8 : (1, 2, 2, 0) 3 8 : (3, 2, 1, 0) 19 : (1, 4, 1, 0) 2 9 : (2, 2, 2, 0) 3 9 : (4, 2, 1, 0) 1 10 : (1, 2, 3, 0) 210 : (2, 4, 1, 0) 1 10 : (3, 2, 2, 0) 15 10 : (0, 0, 0, 4) 12,5 6 : (1, 0, 0, 2) 1 8 : (3, 0, 0, 2) 1 10 : (5, 0, 0, 2) 13,5 7 : (0, 3, 0, 1)

1 8 : (0, 2, 0, 2)

2 9 : (0, 1, 0, 3)

1 10 : (0, 5, 0, 1)

1

2,3,5 5 : (1, 1, 0, 1) 1 6 : (2, 1, 0, 1) 1 7 : (3, 1, 0, 1) 1 8 : (1, 3, 0, 1) 28 : (4, 1, 0, 1) 1 9 : (1, 2, 0, 2) 2 9 : (2, 3, 0, 1) 2 10 : (1, 1, 0, 3) 210 : (2, 2, 0, 2) 3 10 : (3, 3, 0, 1) 22,4,5 8 : (1, 0, 1, 2) 2 9 : (2, 0, 1, 2) 2 10 : (1, 0, 2, 2) 4 10 : (3, 0, 1, 2) 23,4,5 6 : (0, 1, 1, 1) 1 8 : (0, 1, 2, 1) 2 9 : (0, 3, 1, 1) 3 10 : (0, 1, 3, 1) 210 : (0, 2, 1, 2) 52,3,4, 7 : (1, 1, 1, 1) 3 8 : (2, 1, 1, 1) 4 9 : (1, 1, 2, 1) 5 9 : (3, 1, 1, 1) 35 10 : (1, 3, 1, 1) 7 10 : (2, 1, 2, 1) 6 10 : (4, 1, 1, 1) 2

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Table 8: The numbers of the irreducible invariants with the highest 8th order andthe weight up to 10. They are presented in the form weight : (structure ) number of invariants .

orders w : (s) n8 8 : (0, 0, 0, 0, 0, 0, 2) 12,8 8 : (4, 0, 0, 0, 0, 0, 1)

1 10 : (2, 0, 0, 0, 0, 0, 2)

1

3,8 10 : (0, 4, 0, 0, 0, 0, 1) 14,8 8 : (0, 0, 2, 0, 0, 0, 1) 1 10 : (0, 1, 0, 0, 0, 0, 2) 1 10 : (0, 0, 3, 0, 0, 0, 1) 16,8 10 : (0, 0, 0, 0, 2, 0, 1) 12,3,8 8 : (1, 2, 0, 0, 0, 0, 1) 1 9 : (2, 2, 0, 0, 0, 0, 1) 1 10 : (3, 2, 0, 0, 0, 0, 1) 22,4,8 8 : (2, 0, 1, 0, 0, 0, 1) 1 9 : (1, 0, 2, 0, 0, 0, 1) 1 9 : (3, 0, 1, 0, 0, 0, 1) 110 : (2, 0, 2, 0, 0, 0, 1) 2 10 : (4, 0, 1, 0, 0, 0, 1) 13,4,8 9 : (0, 2, 1, 0, 0, 0, 1) 12,5,8 10 : (1, 0, 0, 2, 0, 0, 1) 23,5,8 8 : (0, 1, 0, 1, 0, 0, 1) 12,6,8 8 : (1, 0, 0, 0, 1, 0, 1)

1 9 : (2, 0, 0, 0, 1, 0, 1)

1 10 : (3, 0, 0, 0, 1, 0, 1)

1

3,6,8 10 : (0, 2, 0, 0, 1, 0, 1) 24,6,8 9 : (0, 0, 1, 0, 1, 0, 1) 13,7,8 9 : (0, 1, 0, 0, 0, 1, 1) 15,7,8 10 : (0, 0, 0, 1, 0, 1, 1) 12,3,4,8 10 : (1, 2, 1, 0, 0, 0, 1) 42,3,5,8 9 : (1, 1, 0, 1, 0, 0, 1) 2 10 : (2, 1, 0, 1, 0, 0, 1) 33,4,5,8 10 : (0, 1, 1, 1, 0, 0, 1) 32,4,6,8 10 : (1, 0, 1, 0, 1, 0, 1) 32,3,7,8 10 : (1, 1, 0, 0, 0, 1, 1) 3

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Table 9: The numbers of the irreducible invariants with the highest 9th order andthe weight up to 10. They are presented in the form weight : (structure ) number of invariants .

orders w : (s) n2,9 10 : (1, 0, 0, 0, 0, 0, 0, 2) 13,9 9 : (0, 3, 0, 0, 0, 0, 0, 1) 12,3,9 9 : (3, 1, 0, 0, 0, 0, 0, 1) 1 10 : (1, 3, 0, 0, 0, 0, 0, 1) 110 : (4, 1, 0, 0, 0, 0, 0, 1) 13,4,9 10 : (0, 1, 2, 0, 0, 0, 0, 1) 12,5,9 9 : (2, 0, 0, 1, 0, 0, 0, 1) 1 10 : (3, 0, 0, 1, 0, 0, 0, 1) 13,5,9 10 : (0, 2, 0, 1, 0, 0, 0, 1) 14,5,9 9 : (0, 0, 1, 1, 0, 0, 0, 1) 13,6,9 9 : (0, 1, 0, 0, 1, 0, 0, 1) 15,6,9 10 : (0, 0, 0, 1, 1, 0, 0, 1)

12,7,9 9 : (1, 0, 0, 0, 0, 1, 0, 1) 1 10 : (2, 0, 0, 0, 0, 1, 0, 1) 14,7,9 10 : (0, 0, 1, 0, 0, 1, 0, 1) 13,8,9 10 : (0, 1, 0, 0, 0, 0, 1, 1) 12,3,4,9 9 : (1, 1, 1, 0, 0, 0, 0, 1) 1 10 : (2, 1, 1, 0, 0, 0, 0, 1) 22,4,5,9 10 : (1, 0, 1, 1, 0, 0, 0, 1) 22,3,6,9 10 : (1, 1, 0, 0, 1, 0, 0, 1) 2

Table 10: The numbers of the irreducible invariants with the highest 10th orderand the weight up to 10. They are presented in the form weight : (structure ) number of invariants .

orders w : (s) n10 10 : (0, 0, 0, 0, 0, 0, 0, 0, 2) 12,10 10 : (5, 0, 0, 0, 0, 0, 0, 0, 1) 15,10 10 : (0, 0, 0, 2, 0, 0, 0, 0, 1) 12,3,10 10 : (2, 2, 0, 0, 0, 0, 0, 0, 1) 12,4,10 10 : (3, 0, 1, 0, 0, 0, 0, 0, 1) 1 10 : (1, 0, 2, 0, 0, 0, 0, 0, 1) 13,4,10 10 : (0, 2, 1, 0, 0, 0, 0, 0, 1) 12,6,10 10 : (2, 0, 0, 0, 1, 0, 0, 0, 1) 14,6,10 10 : (0, 0, 1, 0, 1, 0, 0, 0, 1) 13,7,10 10 : (0, 1, 0, 0, 0, 1, 0, 0, 1) 12,8,10 10 : (1, 0, 0, 0, 0, 0, 1, 0, 1) 12,3,5,10 10 : (1, 1, 0, 1, 0, 0, 0, 0, 1) 1

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from it. Let us suppose we have zero at a moment at some place. It can mean that there isno other invariant with this structure, but it can also mean there is a dependency betweeninvariants and another irreducible invariant and we cannot decide, what possibility fromthese two has occurred. The dependency with the least weight has weight 12, it is thethird in our tables and really, while Reiss [20] found 65 irreducible invariants up to the

4th order by means of the Cayley-Sylvester theorem, the actual number of them is 66,the extra invariant has the same structure as the dependency. It means we must computeall invariants up to some limit to nd all irreducible ones between them and using theCayley-Sylvester theorem for the computation of the structure of all irreducible invariantsis unreliable for weights 12. On the other hand, we have weight limit 10 in these tables,so we can verify the numbers of the invariants by means of the Cayley-Sylvester theoremand it really agrees.

4 The Graph Method

We generate the invariants by a way that we call the graph method. Let us consider animage f and two arbitrary points ( x1, y1), (x2, y2) from its support. Let us denote thecross-product of these points as C 12:

C 12 = x1y2 x2y1.After an affine transform it holds C 12 = J C 12, it means C 12 is a relative affine

invariant. It has also geometric meaning as the area of the parallelogram, whose one vertexis the origin of the coordinate system (centroid of the image f ) and two other vertices arepoints ( x1, y1) and (x2, y2). The basic idea of the AMIs generating is the following. Weconsider various numbers of points and we integrate their cross-products (or some powersof their cross-products) on the support of f . These integrals can be expressed in terms of moments and, after eliminating the Jacobian by proper normalization, they yield affineinvariants.

More precisely, having N points ( N 2) we dene functional I depending on N andon non-negative integers nkj as

I (f ) =

N k,j =1 C

n kjkj

N i=1 f (xi , yi)dxidyi . (72)

Note that it is meaningful to consider only j > k , because C kj = C jk and C kk = 0.After an affine transform, I becomes

I = J w|J |N I,where w = k,j nkj is the weight of the invariant and N is the degree of the invariant.

If I is normalized by w+ N 00 we get a desirable affine invariant

I w+ N 00

=

I w+ N 00

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(if w is odd and J < 0 there is an additional factor 1).We illustrate this idea on two simple invariants. First, let N = 2 and n12 = 2. Then

I (f ) =

(x1y2 x2y1)2f (x1, y1)f (x2, y2)dx1dy1dx2dy2 = 2( m20m02 m211). (73)

Similarly, for N = 3 and n12 = 2 , n13 = 2 , n 23 = 0 we get

I (f ) = (x1y2 x2y1)2(x1y3 x3y1)2f (x1, y1)f (x2, y2)f (x3, y3)dx1dy1dx2dy2dx3dy3

= m220m04 4m20m11m13 + 2 m20m02m22 + 4 m211m224m11m02m31 + m202m40 . (74)

Each invariant generated by formula (72) can be represented by a planar connectedgraph, where each point ( xk , yk) corresponds to one node and each cross-product C kj

corresponds to one edge of the graph. If nkj > 1, the respective term C n kjkj corresponds

to nkj edges connecting k-th and j -th nodes. Thus, the number of nodes equals thedegree of the invariant and the total number of the graph edges equals the weight w of the invariant. From the graph one can also learn about the orders of the moments theinvariant is composed from and about its structure. The number of edges originatingfrom each node equals the order of the moments involved, the number of nodes equals theorder of the invariant.

Particularly, for the invariants (73), (74) given above the corresponding graphs areshown in Fig. 1 and the list of edges for (73) is

1 12 2 (75)

and for (74) it is

1 1 1 12 2 3 3. (76)

Now one can see that the problem of derivation of the AMIs up to the given weightw is equivalent to generating all connected graphs with at least two nodes and at most wedges. Let us denote this set of graphs as Gw .

All graphs with 10 edges or less were generated and corresponding invariants werecomputed. More precisely, if we need to generate all invariants of some weight w, it issatisfactory to generate all graphs from

1 1 1 ... 1 1 12 2 2 ... 2 2 2 (77)

to

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Figure 1: The graphs corresponding to the invariants (73) (top) and (74) (bottom)

1 1 3 4 ... w

2 w

1 w

1

2 3 4 5 ... w 1 w w, (78)where w is the number of edges. E.g. if w = 10, then it is from

1 1 1 1 1 1 1 1 1 12 2 2 2 2 2 2 2 2 2 (79)

to

1 1 3 4 5 6 7 8 9 92 3 4 5 6 7 8 9 10 10. (80)

The following algorithm was used to create the next graph:1. Find the rst element from the end of the second row, which can be increased.

2. If it exists, then increase it by one to v1. Fill the rest of the row by the greater of two values: v1 and a + 1, where a is the value in the rst row above the lled one.

3. If it does not exist, then nd the rst element from the end of the rst row, whichcan be increased.

4. If it exists, then increase it by one to v2. Fill the rest of the row by v2.

5. If it does not exist, then stop.

This algorithm does not guarantee no isomorc graphs are generated, but their numberis not too high. The isomorc graphs lead to identical invariants and they are eliminatedlater. In fact, the graphs form (77) and (78) are not ever non-zero, but they guaranteethat no relevant graph will be omitted. Actually, the rst graph generating a non-zeroinvariant in case of even w is (77) and the last one is

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1 1 3 3 ... w 3 w 3 w 1 w 12 2 4 4 ... w 2 w 2 w w(81)

and in case of odd w is the rst graph generating a non-zero invariant

1 1 1 ... 1 1 12 2 2 ... 2 3 3 (82)

and the last one

1 1 3 3 ... w 4 w 4 w 3 w 3 w 32 2 4 4 ... w 3 w 3 w 2 w 2 w 2,(83)

but if we use it as the limit, some graphs would be omitted.Unfortunately, most resulting graphs are useless because they generate invariants,

which are dependent. Including dependent invariants into the feature set which we wantto use for object description and recognition would be a serious mistake. Dependent

features do not increase the discrimination power of the system at all and may even leadto misclassications.

There might be various kinds of dependencies in the set of all AMIs (i.e. in the setGw of all graphs). Let us categorize them into several groups.

1. Identical invariants. Isomorphic graphs (and rarely some non-isomorphic graphs)generate identical invariants. Elimination is done by comparing the invariants term-wise.

2. Zero invariants. Some AMIs might equal identically zero regardless of the imagethey are calculated from. If there are one or more nodes with one adjacent edge only,then all terms of the invariants contain rst-order moment(s). When using centralmoments, all rst-order moments are zero by denition and, consequently, suchinvariants are zero too and must be eliminated. However, also some other graphsmay generate zero invariants and should be eliminated, for instance the graph inFig. 2 leads to

I (f ) = (x1y2 x2y1)3f (x1, y1)f (x2, y2)dx1dy1dx2dy2 =

= m30m03

3m21m21 + 3 m21m21

m30m03 = 0 .

3. Products. Some invariants might be equal to products of other invariants. Elimina-tion of these is done by incremental exhaustive search. All possible pairs, triples,quadruples, etc. of the admissible invariants (the sum of their individual weightsmust not exceed w) are multiplied and the independence of the result is checked.

4. Linear combinations. Some invariants might be equal to linear combinations of otherinvariants or to linear combinations of products of other invariants. All possible

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Figure 2: The graph leading to the zero invariant

combinations of the admissible invariants (all terms must have the same numbers of moments of the same order and the sum of the individual weights must not exceedw) are calculated and their independence is checked.

After eliminating all the above dependencies, we get a set of irreducible invariants.However, there may be higher-order polynomial dependencies among irreducible invariantsbut they are very time-consuming to identify, especially for high w.

The method of the elimination of the linearly dependent invariants in detail:

1. The input is the list of the generated invariants. The zero and identical invariantsare eliminated and products are marked.

2. All invariants with the same structure (the numbers of moments of the same orderin one term) are found. The linear combination of such invariants can be computed.

3. The matrix of all coefficients is constructed. It has coefficients of one invariant inone column. The coefficient corresponding the term, which is not included in theinvariant, is zero. There are the coefficients of one term in different invariants inone row of the matrix. The rank of the matrix of coefficients equals the number of the linearly independent invariants of the current structure.

4. The invariants of the current structure are divided into two groups: the productsand the others. Now, one invariant, which is not the product, is removed from thematrix and its rank is checked. If it decreased, then the invariant is returned to thematrix, else another invariant is removed. If no other non-product invariant can beremoved, then the products are removed. The removing nishes, when the matrix

is regular, i.e. the number of columns equals its rank.

5. If there are some invariants, which were not removed and are not products, thenthey are the irreducible ones. All other invariants are eliminated from the set.

Generating of all graphs by this method is a combinatorial task with exponentialcomplexity, but for the weight 10 it has still acceptable computing time. It was appliedfor each w from 2 to 10. The corresponding invariant was computed for each graph. Theminimum absolute value of the coefficients was found and the coefficients were divided by

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products until weight 15. The computing time exceeded 9 days and the original number1519 invariants had to be appended by 17151 products. This computation was more time-and memory-consuming than that of generation of original invariants. In spite of this,many important dependencies did not t to the limit and remained hidden.

The algorithm was similar to that for elimination of reducible invariants from previous

section with small modications. The limit of the weight of the product was changed(from 10 to 15) and a list of invariants, which are simultaneously products and linearlydependent, was created before their elimination. The found dependencies must be fartherprocessed. If some factor is also product, it must be substituted by its factors, if it islinearly dependent, it must be substituted by this dependency so we obtain dependencyamong irreducible invariants only. By this way, we obtain relatively complicated tree datastructure, which must be simplied. It is done by the special recursive algorithm. Thetree is searched from its root. If some product of sums is found, then the multiplicationis executed so one sum of products is obtained. If some factor is not mere invariant, but

still subtree, the same procedure is called recursively on it. The factors of the originalproduct are moved to the other side of the equation and identical terms are added so weobtain the dependency in form of sum of products of irreducible invariants.

It often happens that no term remains after last addition and we obtain dependencyin form 0=0. Some dependencies are identical and some differ only in multiplicativeconstant. All these dependencies are eliminated and only one is preserved from the set of identical dependencies. We obtained 122 dependencies after this elimination in our case.

To remove non-integer coefficients, the special procedure was proposed. The tolerancewas set to 0.001. Its inverse value is 1000. The coefficients of the dependencies are divided

by all integers from 1 to this limit 1000 and the divider with minimum deviation is usedas denominator. So, all coefficients of the dependency are converted to fractions. Then,least common multiple from all denominators is found and the dependency is multipliedby it. Integer coefficients are the result.

The algorithm for elimination of the dependent dependencies is similar to that forelimination of linearly dependent invariants. All products of dependencies and invariantswith weight limit 22 are computed (the limit 22 is the least, when all reducible dependen-cies are eliminated). Then the dependencies and products originated from products withthe same structure of orders of moments are found. The matrix of their coefficients is

created. The coefficients of one dependency are in one column and the coefficients of iden-tical terms are in one row. The columns are removed until their number equals the rankof the matrix. If the rank decreased, the removed column must be returned back. Thedependencies with the biggest number of terms are removed rst. Some dependencies areexpressed in form invariant multiplied by another dependency. These dependencies areremoved last from the matrix, but they are eliminated from the set after this computation.

As the result, we have found second-order dependencies. We decided not to computehigher-order dependencies, but to eliminate one rst-order dependency for each second-

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order dependency by a special heuristic algorithm: we nd the rst-order dependency withmaximum weight in each second-order dependency. If there is more such dependencies, wend that with maximum number of terms. This dependency is eliminated and also eachmultiple that contains it is eliminated from following search. 54 dependencies remainedafter this algorithm, 12 of them of 4th order, 36 of the 5th order and 6 of the 6th order.

To append this number of dependencies, other computations with limited order of moments were carried out to decrease the number of products. When the limit of theorder was 3 and the limit of the weight was 18, one new dependency was found. Whenthe limit of the order was 4 and the limit of the weight was 16, 194 new dependencieswere found after elimination of the identical ones. Another dependency was obtained,when the selection conditions were dened as to use invariants of 2nd and 4th ordersonly and the limit of the weight was 18. From these 196 dependencies 31 ones remainedafter elimination of the dependent ones. It would be 85 dependencies together, but if 12overlaying dependencies of 4th order are omitted, the result is 73 dependencies.

Now, the algorithm for elimination of the dependent dependencies in detail:

1. Compute the product of each pair invariant dependency and dependency de-pendency satisfying some conditions, e.g. the sum of weights is less than or equalto some limit, in our case 22.

2. All dependencies with the same structure (the sum of the structures of the invari-ants in one term) are found. The linear combination of such dependencies can becomputed. The structures with minimum weight are processed rst.

3. The matrix of all coefficients is constructed. It has coefficients of one dependencyin one column. The coefficient corresponding the term, which is not included in thedependency, is zero. There are the coefficients of one term in different dependenciesin one row of the matrix. The rank of the matrix of coefficients equals the numberof the linearly independent dependencies of the current structure.

4. The dependencies of the current structure are divided into two groups: the productsinvariant dependency and the others. The dependencies are also sorted by thenumber of terms. Now, the dependency, which is not the product, with the max-imum number of terms is removed from the matrix and its rank is checked. If itdecreased, then the dependency is returned to the matrix, else another dependencyis removed. If no other non-product dependency can be removed, then the productsare removed, that with the maximum number of terms rst. The removing nishes,when the matrix is regular, i.e. the number of columns equals its rank.

5. If there are some dependencies, which were not removed and are not products, thenthey are linearly independent. All other dependencies are eliminated from the set.

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6. If the removed dependencies are used as right side of the system of linear equations,solving it we can obtain coefficients of the second order dependencies.

7. The second order dependencies, whose each term contains the same dependency orinvariant, are expunged. They were processed before.

8. In each second order dependency, the rst order dependency with maximum weightis found. If there is more such dependencies, we nd that with maximum numberof terms. This dependency is eliminated and also each multiple that contains it iseliminated from the following search.

9. The search is stopped, when all multiples are processed.

During multiplication of the dependencies in this algorithm we work with labels of theinvariants only, not with original coefficients and moments.

Now we have subtracted the second-order dependencies from the rst-order ones andstill too much dependencies remained. More precisely, we have 31 dependencies of the 3rdand 4th orders for 32 invariants with 9 of them independent. Thereto the invariant I 14does not occur in the dependencies, therefore only 22 dependencies are useful. We decidednot to compute higher-order dependencies, but to eliminate the redundant dependenciesby the following heuristic algorithm in which each invariant supposed to be dependentchooses its dependency. The dependencies were sorted by the highest order, by the numberof orders, by the weight and by the number of terms. Then each invariant gets assignedthe rst dependency, which it is included in. When all such dependencies are occupied,some invariant gets assigned a substitutional dependency. The dependencies without anyassignation are then omitted.

The remainder 64 dependencies are published in the appendix only for the sake of interest. In detail, there is 1 dependency of the 3rd order, 21 dependencies of the 4thorder, 36 dependencies of the 5th order and 6 dependencies of the 6th order. The set isincomplete and cannot be used for selection of the independent invariants even the 4thorder.

6 Selection of Independent Invariants

There is another approach to the selection of independent invariants. We can select someof them and verify their independency by comparison with the normalized moments.Firstly, we should choose irreducible invariants only and respect the rule of thumb notonly for the whole number of chosen invariants, but also for each subset. Let us supposewe have no restriction on the order of the invariants. For instance, we have 10 momentsup to the 3rd order, i.e. 10-6=4 independent invariants. We begin with I 1, I 2, I 3 andI 4. Now we would like to insert invariants of 4th order. There is 5 moments of 4th orderplus 3 ones of zero and rst orders, i.e. 8-6=2 homogenous invariants of of 4th order at

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maximum ( I 6 and I 7). If we take simultaneous invariants of the 4th and 2nd orders, thereare 11 moments, i.e. 11-6=5 invariants. The invariants used must be subtracted fromthis number, if we use I 1, I 6 and I 7, then 2 new invariants remain ( I 8 and I 9). I 10 mustnecessarily be dependent and we need at least one simultaneous invariant of orders 3 and4 or 2, 3 and 4.

Generally, If we have invariants up to the order r 1 and need to insert invariantsof the order r , we need to insert r + 1 invariants. The r of them should be homogenousor simultaneous of orders r and 2, with maximally r 2 of them homogenous and onesimultaneous of order r and some order higher than 2. We choose the invariants withminimum weight. In Table 11, there are the maximum numbers of the independentinvariants that can be constructed from moments of specic orders. There are the wholenumbers there, e.g. the value 9 from orders 2,3,4 means the whole number of the invariantsincluding homogeneous ones, simultaneous invariants from two orders and simultaneousones from three orders.

Table 11: The maximum numbers of the independent invariants up to the 5th order bythe rule of thumb.

orders 2 3 2,3 4 2,4 3,4 2,3,4 5 2,5 3,5 2,3,5 2,4,5 3,4,5 2,3,4,5numbers 1 1 4 2 5 6 9 3 6 7 10 11 12 15

The proof of independency and completeness of the chosen set is difficult and needsanother mathematical tools than the previous sections. Its main idea: if we have another

complete and independent set of affine invariants and we can compute unambiguouslyvalues of our invariants from them and them from our invariants, then our set is completeand independent too. This alternative set can be set of normalized moments. We cannormalize geometric moments, but we obtain simpler equations, when we use complexmoments. The complex moment c pq of order ( p + q ) is dened as

c pq =

(x + iy) p(x iy)qf (x, y)dxdy , (85)

where i denotes imaginary unit. Each complex moment can be expressed in terms of geometric moments m pq as

c pq = p

k=0

q

j =0

pk

q j

(1)q j i p+ qk j mk+ j,p + qk j . (86)

If we use central moments pq instead of the m pq in (86), we obtain complex momentsnormalized to translation.

To express geometric moments in terms of complex moments, we can substitute newvariables for (x + iy) and (x iy) into (85) and the denition of the geometric moments

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and, consequently, derive an inverse form of (86):

m pq =1

2 p+ qiq p

k=0

q

j =0

pk

q j

(1)q j ck+ j,p + qk j . (87)

In polar coordinates, (85) becomes the form

c pq = 0 2

0r p+ q+1 ei( pq)f (r, )drd. (88)

It follows immediately from (88) that c pq = cqp (the asterisk denotes complex conju-gate). After the rotation by an angle the complex moment becomes

c pq = ei( pq) c pq . (89)

If we use suitable products of the complex moments, we obtain rotation invariants.The product

=n

i=1ck i pi qi (90)

is invariant to rotation, if n

i=1ki( pi q i) = 0 ,

where n 1, ki , pi , and q i (i = 1 , , n ) are non-negative integers. Some other detailsabout rotation moment invariants can be found in [23].

The affine transform is decomposed by another way than in Section 2 during thenormalization of the moments: into two translations, scaling, rotation, stretching and

second rotation; the stretching must be between two rotations.Horizontal translation:

u = x + v = y. (91)

Vertical translation:u = xv = y + . (92)

Scalingu = xv = y. (93)

First rotationu = x cos y sin v = x sin + y cos. (94)

Stretching:u = xv = 1 y.

(95)

Second rotation:u = x cos y sin v = x sin + y cos. (96)

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The moments are successively normalized to these elementary transforms. The nor-malization to the translation is done by translation the origin of the coordinate systemto the centroid of the image (using central moments in (86)). The normalization to thescaling is similar to that of AMIs, we use ratios

c pq/c

p + q +22

00 . (97)The complex moment c20 can be used for normalization to the rst rotation. The

straight line passing through the centroid on the angle equaling a half of the phase of c20is called principle axis. If an image is rotated by angle so the principal axis agrees withthe axis x, then the c20 is real and positive. It holds

sin = s 12 (1 c20 + c022 c20 c02 ), s = sign( c20 c02i )cos = 12 (1 + c20 + c022 c20 c02 )(98)

for the angle . Maybe, better known formula is

tan 2 =211

20 02=

(c20)(c20)

=c20 c02

i(c20 + c02), (99)

where (c20) and (c20) are real and imaginary parts of the c20 .We can normalize the image to the stretching by scaling by the coefficient

= 4 c11 c20c02c11 + c20c02 (100)horizontally and by the coefficient 1 / vertically. Now, the image is normalized to the

affine transform up to the second rotation. During this normalization, the c10, c01, c20 aswell as c02 become zero, the c00 becomes one and rotation invariants from (90) becomeaffine invariants. We can either continue with the normalization to the second rotationby the c21 or to use . It emerged that the latter possibility leads to simpler equations,because we can choose the most appropriate set of s.

There is a question why to use affine moment invariants I computed from geometricmoments and not these normalized rotational invariants . One reason could be betternumerical behavior in some situations, another one easier computation; if we unite allexpressions for normalization to one formula, it is more complicated than that for AMIs.

We can compute values of the affine moment invariants directly from the complexmoments by means of the following theorem.

Theorem 2: Let us denote the value of the invariant computed from the geometricmoments I ( pq) and I (c pq) the value obtained by substitution the complex moments c pqinstead of pq. Then there is relation between them

(2i)wI ( pq) = I (c pq) , (101)where w is the weight of the invariant.

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Proof: We use a special affine transform

x = x + iyy = x iy.

(102)

Its Jacobian J = 2i. According to (85) the value of the moment after the transform pq = (x + iy) p(x iy)q|J |f (x, y)dxdy = |J |c pq (103)

(the coordinates of the centroid remains unchanged). From Theorem 1 we have for thevalues of the invariants without normalization by 00

I ( pq) = J w|J |kI ( pq) (104)

and from (103)we haveI ( pq) = |J |kI (c pq) (105)

Since c00 = 00, it imply directly (101).2

From now c pq will denote the complex moments after the former normalization, forsimplicity, they will not have any special notation. The former normalization is a specialcase of an affine transform, so we can compute affine moment invariants by Theorem 2with constraints c20 = c02 = 0 and c00 = 1

(2i)2I 1 = c211(2i)6I 2 = c230c203 + 6 c30c21c12c03 4c30c312 4c321c03 + 3 c221c212(2i)4I 3 = c11c30c03 + c11c21c12(2i)6I 4 = 2 c311c30c03 + 6 c311c21c12.(106)

Solving these equations we obtainc11 = 2 I 1c21c12 = 1 I 1 (2I 3 I 4I 1 )c30c03 = 1 I 1 (6I 3 + I 4I 1 )

(c30c312) = 8 I 2 12I 23I 1 +

I 24I 31

.

(107)

The set c11 , c21c12, c30c03 and (c30c312) is used as the other set of invariants. This systemof rotation invariants is independent and complete except the sign of (c30c312). It re-lates with fact that the set I 1, I 2, I 3, I 4 cannot distinguish two objects differing by mirror

reection. The general affine transform include the mirror reection, therefore the setI 1, I 2, I 3, I 4 is complete from this point of view, obversely, we should use absolute valuesof the invariants with odd weights. If we insert I 5 there, we can compute

(c30c312) =4I 5

( I 1)3 , (108)but we can compute absolute value of (c30c312) also as

| (c30c312)| = c30c03(c21c12)3 2(c30c312) , (109)43

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or

32I 16c11

= 2 (c31c212) (c30c312c221c212 1 ) +2 (c31c

212) (

c30c312c221c212 1) c30c03c22 + c21c12c22 . (117)

If we substitute (c31c212) from (112), we obtain a quadratic equation

2(c31c212)( 2( c30 c312

c221 c212 1) + 2( c30 c312c221 c212 1)) (c31c212) ( c30 c312c221 c212 1)(32 I 16c11 + c30c03c22

c21c12c22) + 14 (32 I 16c11 + c30c03c22 c21c12c22)2 2(c30 c312c221 c

212 1)(c31c13c221c212) = 0 .

(118)It has two solutions

(c31c212) =

( c30 c312

c221 c212 1)(32 I 16c11 + c30c03c22 c21c12c22)

(c30 c312c221 c

212 1) (32

I 16c11 + c30c03c22 c21c12c22)2

+4 c31c13c221c212( 2(c30 c312c221 c

212 1) + 2(

c30 c312c221 c

212 1))

2(2(

c30 c312c221 c212 1) +

2(

c30 c312c221 c212 1))

. (119)

You can see that a real solution always exists, i.e. the set set I 1, I 2, I 3, I 4, I 6, I 7, I 8, I 9, I 16is independent, but also increasing complexity of the equations. If we use I 11 insteadof I 16 (e.g. it has moments of two orders only), the nal equation would be quartic. Itleads to effort for simplication of the equations. If we need not have general formulas,but only dependency test, the solution in some points might be satisfactory. We obtainsimple equations, if we choose the values of the affine invariants so the values of rotationinvariants would be 1.

It relates with the fact that the sum of the coefficients of an invariant is zero. Theinvariant can be written as a weighted sum of products of moments

I =t

j =1c j

s

=1 pj ,qj . (120)

The t is the number of terms and the s is the degree of the invariant. If we substitute itinto Cayley - Aronhold differential equation (14), we obtain

n

i=1 pi pi 1,qi +1

t

j =1c j e( pi , q i , j )

1 pi ,qi

s

=1 pj ,qj = 0 , (121)

where e( pi , q i , j ) is the number of occurrences of the pi ,qi in the j -th term. The order of the summation can be exchanged

t

j =1

n

i=1 pi pi 1,qi +1 c j e( pi , q i , j )

1 pi ,qi

s

=1 pj ,qj = 0 , (122)

But n

i=1 pi e( pi , q i , j ) (123)

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equals the weight w of the invariant. If we substitute the same value, let us say , for allmoments, we obtain

s wt

j =1c j = 0 (124)

or the sum of the coefficients of an invariant is zero. It is an easy test, if some polynomial

can be affine invariant, and it also means that if all moments have the same values, thenall invariants are zero, and vice versa. In our case, we must compensate the fact that11 = 0 because of the normalization by non-zero values of some invariants.

In our case, if we choose

I 1 =14

, I 2 = 0 , I 3 = 0 , I 4 = 18

,

then we obtainc11 = 1 , c21c12 = 1 , c30c03 = 1 , c30c312 = 1 .

Similarly, if I 6 = 0 , I 7 = 0 , I 8 =

14

, I 9 = 0 ,

thenc22 = 1 , c31c13 = 1 , c40c04 = 1 , c40c213 = 1 .

Now, if we use I 10, then the last equation becomes

(c40c213) = 64I 10 ,

but from the previous equations we have

(c40c213) = 0 ,

therefore we cannot choose I 10 freely, it is dependent. If we use I 22 instead of I 10, thenthe last equation becomes

(c31c212) = 1 16I 22 .We can choose I 22 freely, if I 22 = 0, then c31c212 = 1. The I 22 is independent. If we useI 16, then the last equation becomes I 16 = 0. It looks like the I 16 would be dependent, butif we change e.g. I 2 = 14 , then c30c312 = 1 and we obtain

(c31c212) = 8I 16 .The I 16 can be chosen freely, it is independent. If we look at the general formula (119), wefound that the values of the affine invariants must be chosen so the denominator wouldbe non-zero. This is the biggest problem with limited solution in one point. There aresingular points in the space of invariants and if we choose such a point, the dependencytest fails in spite of the invariants are independent.

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The I 11 is similar to I 16 in this sense. For I 2 = 0 we obtain last equation I 11 = 0 likeit would be dependent, but for I 2 = 14 it becomes

256I 11 = 8 2(c31c212),

I 11 can be chosen freely and it means it is independent. In reality, it must be from 0 to1

32 , but it is whole interval, not only nite number of values, therefore it is satisfactory.The following analysis of higher orders will not be so detailed, we will describe one

choice of invariants with one test of independency only (except 8th order). We willsuppose choice I 2 = 14 and I 16 = 18 , i.e. c30c312 = 1, c31c212 = 1 and from beforec11 = 1, c21c12 = 1, c30c03 = 1, c22 = 1, c31c13 = 1, c40c04 = 1 and c40c213 = 1.

In the 5th order, we choose I 33, I 34, I 35, I 36 , I 42, I 43. We obtain equations

(2i)10I 33 = c250c205 + 10 c50c05c41c14 4c50c05c32c23 16c50c32c214 16c241c23c059c241c214 + 12c50c223c14 + 12 c41c232c05 + 76 c41c14c32c23 48c41c323

48c332

c14

+ 32 c232

c223(2i)6I 34 = c11c50c05 + 3 c11c41c14 2c11c32c23(2i)8I 35 = 4c311c41c14 + 4 c311c32c23(2i)10I 36 = 2 c511c50c05 + 10 c511c41c14 + 20 c511 c32c23(2i)5I 42 = 2c11c03c41 + 2 c11c30c14 + 6 c11c12c32 6c11c21c23(2i)6I 43 = 2 c211c03c41 + 2 c211c30c14 2c211c12c32 2c211c21c23 . (125)

From I 34 = 0 , I 35 = 0 , I 36 = 132 we havec50c05 = 1 , c41c14 = 1 , c32c23 = 1

and from I 42 = 0 , I 43 = 0 we obtain two solutions

a) c41c03 = 1 , c32c12 = 1b) c41c03 = 1, c32c12 = 1

and in both casesc41c323 = 1 .

The last equation becomes

1024I 33 = 200 56 (c50c523) ,it means the set is independent, we use (c50c523), c50c05, c41c14, c32c23, (c41c323) and

(c32c12) as the other set of invariants with values c50c523 = 1 for I 33 = 964 and c50c05 = 1,c41c14 = 1, c32c23 = 1, c41c323 = 1, c32c12 = 1 from before for computation of higherorders.

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In the 6th order, we choose I 112 , I 113 , I 114 , I 115 , I 116 , I 117 , I 124 . We obtain equations

(2i)6I 112 = c60c06 6c51c15 + 15 c42c24 10c233(2i)6I 113 = 8c311c33(2i)8I 114 = 4 c211 c51c15 16c211c42c24 + 12 c211c233(2i)10I 115 = c11c60c06c33 + c11c60c24c15 + c11c51c42c06 + 8 c11c51c33c159c11c51c

224 9c11c

242c15 + 17 c11 c42c33c24 8c11c

333

(2i)10I 116 = 16c411c42c24 16c411c233(2i)6I 117 = c230c06 6c30c21c15 + 6 c30c12c24 2c30c03c33 + 9 c221c24 18c21c12c33+6 c21c03c42 + 9 c212c42 6c12c03c51 + c203c60(2i)7I 124 = c11c230c06 4c11c30c21c15 + 2 c11c30c12c24 + 3 c11c221c242c11c21c03c42 3c11c212c42 + 4 c11c12c03c51 c11c203c60 (126)

From I 113 = 18 , I 116 = 0 , I 114 = 0 and I 112 = 0 we have

c33 = 1 , c42c24 = 1 , c51c15 = 1 , c60c06 = 1 .

The equations for I 115 , I 117 and I 124 are very complicated. For simplication, let ussuppose, that the values of the invariants are such, that

c42c212 = 1 , c60c324 = 1 .

Then they become64I 115 = 1 + (c51c224)16I 117 = 3 3 (c51c224)16I 124 = (c51c224)

From it, the corresponding values of the invariants are

I 117 = 12I 115I 124 = 12 I 115 (1 + 32I 115 )

and we can choose I 115 freely in the interval < 132 , 0 > , the set is independent. We willuse the values I 115 = 0, I 117 = 0, I 124 = 0 and c51c224 = 1 for future computations of theother set of invariants c33, c42c24, c51c15 , c60c06, (c42c212), (c60c324) and (c51c224).

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In the 7th order, we choose I 205 , I 206 , I 210 , I 211 , I 212 , I 213 , I 214 , I 215 . We obtain equations

(2i)8I 205 = c11c70c07 + 5 c11c61c16 9c11c52c25 + 5 c11c43c34(2i)10I 206 = 4c311c61c16 + 12 c311c52c25 8c311c43c34(2i)7I 210 = 4c211c30c25 + 12 c211c21c34 12c211c12c43 + 4 c211c03c52(2i)8I 211 = 4c311c30c25 + 4 c311c21c34 + 4 c311c12c43 4c311c03c52(2i)

9

I 212 = 2 c11c230c21c07 2c11c

230c12c16 12c11 c30c

221c16 + 24 c11c30c21c12c25

4c11c30c21c03c34 12c11c30c212c34 + 4 c11c30c12c03c43 + 18 c11c321c2554c11c221c12c34 + 12 c11c221c03c43 + 54 c11 c21c212c43 24c11c21c12c03c52+2 c11c21c203c61 18c11c312c52 + 12 c11c212c03c61 2c11c12c203c70(2i)9I 213 = 2 c11c230c12c16 2c11c230c03c25 2c11c30c221c16 4c11c30c21c12c25+8 c11c30c21c03c34 + 4 c11c30c212c34 8c11c30c12c03c43 + 2 c11c30c203c52+6 c11c321c25 12c11c221c12c34 4c11 c221c03c43 + 12 c11c21c212c43+4 c11c21c12c03c52 2c11c21c203c61 6c11c312c52 + 2 c11c212c03c61(2i)9I 214 = 16c411c21c34 + 16 c411c12c43(2i)10I 215 = 2 c211c230c21c07 2c211c230c12c16 8c211c30c221c16 + 12 c211c30c21c12c25

4c211c30c212c34 + 6 c211c321c25

6c211c221c12c34

4c211c221c03c43

6c211c21c212c43 + 12 c211c21c12c03c52 2c211c21c203c61 + 6 c211c312c528c211c212c03c61 + 2 c211c12c203c70 (127)From I 214 = 0 , I 210 = 0 , I 211 = 0 , I 206 = 0 and I 205 = 0 we have

(c43c12) = 0 , c52c03 = c43c12, i.e. c52c25 = c43c34 , c61c16 = c43c34, c70c07 = c43c34 .

If we substitute it into the equation for I 213 , we obtain

(c61c212c03) = 64I 213 ,i.e. (c61c212c03) = 0 for I 213 = 0 and we obtain from the equation for I 212

(c70c12c203) = 128I 212 ,

i.e. (c70c12c203) = 1 for I 212 = 1128 . The last equation becomes

256I 215 = (c70c12c203) 3 (c61c212c03) + 3 (c52c03) (c43c12)or

256I 215 =

(c43c34)2

1

c43c12

andc43c12 =

65536I 2215 + 1512I 215

.

We can choose I 215 freely, the set is independent. We will use the values I 215 = 1256 ,c43c12 = 1, c52c03 = 1, c61c212c03 = 1 and c70c12c203 = i in the future of the other set of invariants c43c12, c52c03, c61c212c03 and c70c12c203 (It is 8 real values).

The 8th order is an example, where the rst choice is dependent, in spite of it satisesthe rule of thumb. If we choose the set I 280 , I 281 , I 282 , I 283 , I 284 , I 288 , I 289 , I 290 , I 291 , then

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the equation for I 284 does not bring any new information, the value of the I 284 cannot bechosen, while the equation for relation between c62c26 and c53c35 vanish. If we choose theset I 280 , I 281 , I 282 , I 283 , I 288 , I 289 , I 290 , I 291 , I 294 instead, we obtain equations

(2i)8I 280 = c80c08 8c71c17 + 28 c62c26 56c53c35 + 35 c244(

2i)8I 281 = 16c411c44(2i)10I 282 = 4 c211c71c17 24c211 c62c26 + 60c211c53c35 40c211c244(2i)10I 283 = c330c12c08 c330c03c17 c230c221c08 5c230c21c12c17+7 c230c21c03c26 + 5 c230c212c26 8c230c12c03c35 + 2 c230c203c44+6 c30c321c17 3c30c221c12c26 13c30c221c03c35 6c30c21c212c35+28 c30c21c12c03c44 8c30c21c203c53 + 3 c30c312c44 13c30c212c03c53+7 c30c12c203c62 c30c303c71 9c421c26 + 27 c321c12c35 + 3 c321c03c44

36c221c212c44 6c221c12c03c53 + 5 c221c203c62 + 27 c21c312c533c21c212c03c62 5c21c12c203c71 + c21c303c80 9c412c62 + 6 c312c03c71c212c203c80(2i)8I 288 = 2c11 c230c17 + 12c11c30c21c2612c11c30c12c35 + 4 c11c30c03c44 18c11c

221c35 + 36c11c21c12c4412c11c21c03c53 18c11c212c53 + 12 c11 c12c03c62 2c11c203c71(2i)9I 289 = 2c211 c230c17 + 8 c211 c30c21c26 4c211c30c12c35

6c211 c221c35 + 4 c211 c21c03c53 + 6 c211c212c53 8c211c12c03c62+2 c211c203c71(2i)10I 290 = 8c311 c30c21c26 + 8 c311c30c12c35 + 24 c311c221c35 48c311 c21c12c44+8 c311c21c03c53 + 24 c311c212c53 8c311c12c03c62(2i)10I 291 = 8c311 c30c12c35 + 8 c311c30c03c44 + 8 c311c221c35

8c311 c21c12c44 8c311c21c03c53 + 8 c311c212c53(2i)9I 294 = 4 c311c40c26 8c311c31c35 + 8 c311c13c53 4c311c04c62 .

(128)

From I 281 =1

16 we have c44 = 1, from I 282 = 0 we have c71c17 = 6 c62c26 15c53c35 + 10,and from I 280 = 0 we have c80c08 = 20c62c26 64c53c35 + 45. From I 291 = 132 we obtain(c53c212) = 1, from I 290 = 0 we obtain (c62c12c03) = 1, from I 288 = 0 we obtain(c71c203) = 1 and from I 283 = 1512 we obtain (c80c212c203) = 1. The last two equations

128I 289 = (c71c203) 4 (c62c12c03)) + (c53c212)64I 294 = 2 (c53c212) + (c62c12c03) .

If we substitute the expression for c71c17 from I 282 = 0 and (c62c12c03) = 2 (c53c212)from I 294 = 0 into the equation for I 289 , we obtain

128I 289 = (9 3) (c53c212) ,I 289 can be chosen arbitrarily and this set is independent. We choose I 289 = 0, c44 = 1,c53c212 = 1, c62c12c03 = 1, c71c203 = 1 and c80c212c203 = 1 as the values of the other set of invariants c44, c53c212, c62c12c03, c71c203 and c80c212c203 (It is 9 real values) for the future.

In the 9th order, we choose I 328 , I 329 , I 330 , I 331 , I 332 , I 334 , I 335 , I 340 , I 341 , I 344 . Weobtain equations

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(2i)10I 328 = c11c90c09 + 7 c11c81c18 20c11c72c27 + 28 c11c63c36 14c11c54c45(2i)9I 329 = c330c09 + 9 c230c21c18 9c230c12c27 + 3 c230c03c36 27c30c221c27+54 c30c21c12c36 18c30c21c03c45 27c30c212c45 + 18 c30c12c03c543c30c203c63 + 27c321c36 81c221c12c45 + 27 c221c03c54 + 81 c21c212c54

54c21c12c03c63 + 9 c21c2

03c72

27c3

12c63 + 27 c2

12c03c72

9c12c2

03c81

+ c303c90(2i)9I 330 = 8 c311c30c36 24c311 c21c45 + 24c311c12c54 8c311c03c63(2i)10I 331 = c11c330c09 + 7 c11c230c21c18 5c11c230c12c27 + c11c230c03c36

15c11c30c221c27 + 18 c11c30c21c12c36 2c11c30c21c03c45 3c11c30c212c452c11 c30c12c03c54 + c11c30c203c63 + 9 c11c321c36 9c11c221c12c453c11 c221c03c54 9c11c21c212c54 + 18 c11c21c12c03c63 5c11c21c203c72+9 c11c312c63 15c11c212c03c72 + 7 c11c12c203c81 c11c303c90(2i)10I 332 = 8 c411c30c36 8c411c21c45 8c411c12c54 + 8 c411c03c63(2i)9I 334 = 4c211 c50c27 + 20c211c41c36 40c211c32c45 + 40 c211c23c5420c211c14c63 + 4 c211c05c72

(2i)10

I 335 = 4c311 c50c27 + 12c

311c41c36 8c

311c32c45 8c

311c23c54+12 c311c14c63 4c311c05c72(2i)9I 340 = 2 c11c70c18 14c11 c61c27 + 42c11c52c36 70c11c43c45 + 70 c11c34c54

42c11c25c63 + 14 c11c16c72 2c11c07c81(2i)10I 341 = 2 c211c70c18 10c211 c61c27 + 18c211c52c36 10c211c43c45 10c211c34c54+18 c211c25c63 10c211c16c72 + 2 c211c07c81(2i)9I 344 = 2 c11c30c40c18 8c11c30c31c27 + 12 c11c30c22c36 8c11c30c13c45+2 c11c30c04c54 6c11c21c40c27 + 24 c11c21c31c36 36c11c21c22c45+24 c11c21c13c54 6c11c21c04c63 + 6 c11c12c40c36 24c11c12c31c45+36 c11c12c22c54 24c11c12c13c63 + 6 c11c12c04c72 2c11c03c40c45+8 c11c03c31c54

12c11c03c22c63 + 8 c11c03c13c72

2c11c03c04c81

(129)From I 330 = 0 we have (c63c03) = 3 (c54c12), from I 332 = 0 we have