One–factorization of complete graphs induced by certain strong starters

One–factorization of complete graphs induced by certain strong starters

C´ alculo II

´ ´ Vazquez´ Carlos A. Alfaro, Christian Rubio-Montiel and Adrian Avila

Tarea [email protected],[email protected],[email protected] Academia de Ciencias B´asicas

Abstract

In the present poster we give an one–factorization of Kq+1, where q = 2k t + 1 is a odd prime power with k a positive integer and t > 1 an odd integer. Such one–factorization is induced by la figura se Xmuestra gr´ a fica de la funci´ o n f (x). La regi´ o n sombreada A tiene un ∗ a strong starter1.X En satisfying thatdeif abajo {a, b} ∈ then a ∈la QR(q) and b ∈ NQR(q), where QR(q) and NQR(q) are the set of quadratic residues and the set of non-quadratic residues of F R6 q, p r i m i t i v e in GF(2t + 1), x = - 1 and 0 < 2j - 2i < 2t - 2 < p n - 1 f (x) dx = 3.5. Utiliza esta informaci´on para calcular el a´rea de 1.5 unidades se the sabeone que respectively. This strong starter is differentythat given 0 by Mullin and Nemeth [Canad. Math. Bull. 12 (1969), 493–497]. is a r e s i d u e m o d p and h e n c e m u s t be e q u a l to t, a c o n t r a d i c t i o n o c c u r s s i n c e 2j - 2i i s e v e n and t i s odd. The s u m s of c o r r e s p o n d i n g 0 2 2t-2 p a i r s of e l e m e n t s of X a r e x (1 + x ) , x (1 + x), • • • x (1 + x ) . 2i 2j I n d e e d , s u p p o s e t h a t x (1 + x) = x (1 + x ) , t h e n if x f - 1 , (i. e. n -1 2i 2j if p > 3), (1 + x) e x i s t s and we would have x = x for 0 < 2 i , 2j < p - 1 w h i c h c a n only h a p p e n if i = j , and the l e m m a other the following sets f o l l ohand, ws.

valor de las siguientes integrales:

Introduction

Let G be a finite additive abelian group of odd order n, and let G∗ = G \ {0} be the set of Example 3. Z 6 Z A2 starter of G is a set X = {{xi , yZi }6: i = 1, ... , n−1 } such that non–zero elements. 2 1 |f (x)|On dxthe f (x) dx c) (x) dxG∗ and b) y1, ... , yfn−1 }= {x1, ... , x n−1 ,a) 2

2

0

=G . {±(xi − yi ) : i =  n−1 Moreover, if xi + Z yi :2 i = 1, ... , 2 = 2

0

2

∗

1, ... , n−1 2 }

n−1 2 ,

then X isZcalled a strong starter of G. 6




(x){{1, dx 6}, {2, 5}, {3, 4}} e) is a starter 2 + f (x) d) set−2f Example 1. The X = of Zdx 7, whereas that the set 0 X = {{1, 5}, {2, 3}, {4, 6}} is a strong starter of Z7.

0

Strong starters were first introduced by Mullin and Stanton in [2] to construct of Room squares, where a Room square of side r defined on an (r + 1)-set V is an r × r array A in which each cell of the array is either empty or contains a unordered pair of distinct elements from V ; every element of V is contained in precisely one pair of each row and column of A; and each pair determines a unique cell of A. In [3] Kocay and Stinson shows the connection between Room squares and strong starters.

where

are strong

X = {{4, 6}, {5, 2}, {9, 8}, {3, 10}, {1, 7}} 6 T H E O R E M 1. T h e r e e x i s t s a s t r o n g s t a r t e r for G = GF(p n k X7 = {{4, 2}, {5, 8}, {9, 10}, {3, 7}, {1, 6}}

p

= 2 t + 1 for k a p o s i t i v e i n t e g e r and t an odd i n t e g e r > 1. starter for Z11, with α = 4. .k-1 A , and x be a p r i m i t i v e e l e m e n t in Proof. Let 2

t + and 1). T h e n the set On Theorem 3,GF(2 Mullin Nemeth proves in [6] that the following set:

A

/° *

/(x2 A , x 3 A *)

(x(2t.2)Aj

(x , x

, 2A+1 3A+1 (x , x )

, x (2t-2)A+l (2t-l)A+l ( >x );

1

)

A+l. X = J (x , x )

. A-l 2 A - 1 W 3A-1 4A-1 V (x , x )(x , x )

Theorem 1. [3] If a strong starter exists for a finite abelian group G of odd order r , then there exists a Room square.

),

(x

(2t-l)A-l

x(2t-l)A);

, x

2tA-l

)

i s a s t r o n g s t a r t e r for GF(p ). t X, if c o n s i d e r e d a s an where ∆ = 2k−1 and x is a primitive root ofThe Fq ,s eis a strong starter for Fq , Before showing our a r r a y and r e a d v e r t i c a l l y l i s t s the e l e m e n t s strong starter, we result: 0 need 1 2 the following 2tA-l 2k-l pn-2 n i i

An one–factor in a graph is a set of pairwise edges of the graph such that each vertex is x , x , x , •••, x =x =x in n a t u r a l o r d e r and thus incident with exactly one edge of the set. An one–factorization of a graph is a set of one– k . c[1] o m p Let r i s e s q G=- 2 { 0} Thebe d i fan f e r e nodd c e s bprime e t w e e n power, p a i r s in the s t akr t e> r 1 a positive integer Theorem 4. t + 1 with factors such that each edge occurs in exactly one one–factor. Given a starter X of a finite are i ∆ and t > 1 an odd integer. Then, there exist β ∈ NQR(q) and β ∈ −β α hα i, for some 1 2 1 additive abelian group G of odd order n, it is well known how to construct an one–factorization k−2 A (β1 − 1)(β2 + 1), (β1 +( 21)(β i ∈ {0, ... , 2 − 1}, such that t 2 2 − 1)A ∈ NQR(q). X ( 1 X ) , x°(l x ), , x - >*(l-x ); of the complete graph Kn+1 (see [4]): Let the vertices of Kn+1 be the elements of G with a extra i ∆ 2 ANQR(q) +1 Let α be a generator of QR(q), β ∈ and β ∈ −β α hα some i ∈ {0, ... , ∆1 − 1} fixed point ∞. The set of edges F = X ∪ {∞, 0} is a one–factor of K . Define for each g ∈ G 4 A ( 2 t 2 ) A l _ A )for / „ A 1 + ( 1 xi, 0 j2 X ; X 2. En los siguientes ejercicios, dibuja la regi´onn+1del a´rea signada representada por la integral x (l y - x ), 1 x (1 - x ), the one–factor Fg = F0 +g = {{x +g, y +g} : {x, y} ∈ F0}, where the group operation + of G is with ∆1 = ∆2 , such that calc´ ulala usando geometr´ıa: extended by defining ∞ + g = ∞. Then the set of all such one–factorization F = {Fg : g ∈ G} A - l∈ NQR(q) A. 3A-l (β1 − 1)(β2 + 1) and + 1)(β2 − ( 2 1) t - l )∈ A -NQR(q) ln A. (by Theorem 4). / > ( (β1 A, x (1 - x ), x (1 - x ), , x (1 - x ). is a one–factorization of Kn+1. Z 3 Z 3 Z 1 Then, the following set Example 2. Let consider the following strong starter X = {{1, 5}, {2, 3}, {4, 6}} for Z7, and (1 - x ) is a non z e r o e l e m e n t of G s i n c e the o r d e r of the e l e m e n t x a) 2x dx b) (2x + 4) dx c) 4 dx ) X = (X ∪ · · · ∪ X ∪ (Y ∪ · · · ∪ 2Yi ∆ ' A1+ j') A. β ,β 0 ∆ −1 0 −1 1 1 2 let F0 = X ∪ {∞, 0}. For i = 1, ... , 6, let Fi as it was defined above. Then an one–factorization i s by h y p o t h e s i s 2tA > A- S u p p o s e x (1 - x ) -x (1 - x ) −3 −2 −2 J J of K8 is given by is a strong starter Fqi», where for 0for < i, < t - 1 , and 0 < j , j ! < A - l , t h e n x + x^ 1 = 0. n o Z∞ Z 1 Z 5√ ∞ ∞ ∞ 3 Xj = {α∆i+j , α∆i+j β1} : i = 1. ... , t 6

0

d)

0

5

25 − x26 dx 1

4

6

0

5

2

6 e)

1

4

−2

5

2

4

0 |x| dx

6

1

0

5

2

1

4

3

3

3

F0

F1

F2

F3

∞

∞

∞

5

6 1

4

0

5

2

6 1

4

1 4

1

2

3

3

3

F4

F5

F6

Let q be an odd prime power. An element x ∈ F∗q is called a quadratic residue if there exists an element y ∈ F∗q such that y 2 = x. If there is no such y then x is called a non-quadratic residue. The set of quadratic residues of F∗q is denoted by QR(q) and the set of non-quadratic residues is denoted by NQR(q). It is well known that QR(q) is a cyclic subgroup of F∗q of cardinality q−1 2 (see [5]), that is

Theorem 2. Let q be an odd prime power, then QR(q) is a cyclic subgroup of F∗q . Furthermore, |RQ(q)| = |RNQ(q)| =

(2x − |x|) dx

for j = 0, ... , ∆1 − 1.

Corollary 2. [1] If Xβ1,β2 is a strong starter for Fq given by Theorem 4, then Xβ2,β1, X−β1,−β2 and X−β2,−β1 are strong starters for Fq different from Xβ1,β2 .

q−1 2 .

Example 4. Let F29 = Z29, then k = 2, t = 7, ∆ = 2, ∆1 = 0 and α = 4. We have QR(29) = {1, 4, 5, 6, 7, 9, 13, 16, 20, 22, 23, 24, 25, 28} and NQR(29) = {2, 3, 8, 10, 11, 12, 14, 15, 17, 18, 19, 21, 26, 27}.

Hence hα2i = {1, 7, 16, 20, 23, 24, 25} and −hα2i = {4, 5, 6, 9, 13, 22, 28}. If β1 = 2 and β2 = 26 ∈ −β1α0hα2i = {2, 10, 11, 15, 17, 21, 26}, then (β1 − 1)(β2 + 1) ∈ NQR(29) and (β1 + 1)(β2 − 1) ∈ NQR(29). Therefore X2,26 = X0 ∪ Y0, where X0 = {{16, 3}, {24, 19}, {7, 14}, {25, 21}, {23, 17}, {20, 11}, {1, 2}}, Y0 = {{13, 10}, {5, 15}, {22, 8}, {4, 12}, {6, 18}, {9, 27}, {28, 26}}.

is a strong starter for F29. Moreover, by Corollary 2 the following X26,2, X27,3 and X3,27 are strong starters for F29 differents from X2,26: X26,2 = X0 ∪ Y0, where X0 = {{16, 10}, {24, 15}, {7, 8}, {25, 12}, {23, 18}, {20, 27}, {1, 26}}, Y0 = {{13, 3}, {5, 19}, {22, 14}, {4, 21}, {6, 17}, {9, 11}, {28, 2}}.

Corollary 1. Let q be an odd prime power, then 1 2

if either x, y ∈ QR(q) or x, y ∈ NQR(q), then xy ∈ QR(q), if x ∈ QR(q) and y ∈ NQR(q), then xy ∈ NQR(q).

Results

n o 495 Yj = {−α∆i+j , α∆i+j β2} : i = 1 ... , t ,

To end, the strong starter given by Mullin and Nemeth is different to the strong starter Xβ1,β2 . Since if x is a primitive root then x ∈ NQR(q). Hence {x 2i∆+j , x (2i+1)∆+j } ⊆ QR(q), for i = 1, ... , t, if j is even, and {x 2i∆+j , x (2i+1)∆+j } ⊆ NQR(q), for i = 1, ... , t, if j is odd.

0

5

2

−1

2

3

0

f)

X27,3 = X0 ∪ Y0, where

X0 = {{16, 26}, {24, 10}, {7, 15}, {25, 8}, {23, 12}, {20, 18}, {1, 27}}, Y0 = {{13, 19}, {5, 14}, {22, 21}, {4, 17}, {6, 11}, {9, 2}, {28, 3}}.

1

Let k be a positive integer and t be an odd integer greater than 1. In this work we prove that if q = 2k t + 1 is an odd prime power, then there exists a strong starter X for Fq such that {a, b} ∈ X satisfy a ∈ QR(q) and b ∈ NQR(q). A distinct strong starter was proposed by Mullin and Nemeth in [6] to demonstrate the following:

X3,27 = X0 ∪ Y0, where

X0 = {{16, 19}, {24, 14}, {7, 21}, {25, 17}, {23, 11}, {20, 2}, {1, 3}}, Y0 = {{13, 26}, {5, 10}, {22, 15}, {4, 8}, {6, 12}, {9, 18}, {28, 27}}.

Lemma 1. If q ≡ 3 mod 4 is an odd prime power with q 6= 3, then there exists a strong starter X for Fq .

It can be verify that the followings are strong starters for F29:

Theorem 3. If q = 2k t + 1 is an odd prime power with k a positive integer and t > 1 an odd integer, then there exists a strong starter X for Fq .

Xβ1,β2 X2,26 X2,10 X3,15 X3,12 X8,11 X10,14 X10,17

On Lemma 1, Mullin and Nemeth proves that the following set XM = {{r 0, r 1}, {r 2, r 3}, ... , {r q−2, r q−1}}

is a strong starter for Fq , where r is a primitive root of Fq . This strong starter is equivalent to the following set: n o p−1 p−1 Xβ = {α, αβ}, {α2, α2β}, ... , {α 2 , α 2 β} , where α is a generator of QR(q) and β is any element of NQR(q) such that 1 + β 6= 0.

Xβ2,β1 X26,2 X10,2 X15,3 X12,3 X11,8 X14,10 X17,10

X−β1,−β2 X27,3 X27,19 X26,14 X26,17 X21,18 X19,15 X19,12

X−β2,−β1 X3,27 X19,27 X14,26 X17,26 X18,21 X15,19 X12,19

Bibliography ´ ´ [1] C. A. Alfaro, C. Rubio-Montiel and A. VazquezAvila. On strong starters for Fq . submitted

[3] W. L. Kocay, D. R. Stinson and S. A. Vanstone On strong starters in cyclic groups, Discrete Math., 56(1):45–60,1985.

[2] R.G. Stanton and R.C. Mullin. Construction of Room squares. Ann. Math. Statist., 39:1540–1548,1968

[4] Horton, J. D. Orthogonal starters in finite abelian groups, Combinatorica, 8(3):293–295, 1988.

Carlos A. Alfaro, Christian Rubio-Montiel and Adri´ One--factorization an V´ azquez-´ Avila

[5] G. H. Hardy and E.W. Wright. An introduction to the theory of numbers, Oxford University Press, Oxford, 2008. [6] R.C. Mullin and E. Nemeth An existence theorem for room squares, Canad. Math. Bull., 12:493–497, 1969.

of complete graphs induced by certain April 8, 2017 strong 1 / sta 1