A note on parabolic and linear one-factorizations of the complete graph +1

Abstract

One-factorizations of the complete graph have wide applications, as an example they are often used for scheduling round-robin tournaments with teams. In this note, we characterize parabolic and linear one-factorizations of complete graphs +1, when is an odd prime. This class of one-factorizations arises from the geometry of conics and lines in the afine plane (2, ). We also include Magma computations for the cases ≤ 19.

Keywords

• eol>One-factorization
• parabolas
• afine lines

Declaration

This work is written in memory of Angelo Sonnino. His

ideas on one-factorizations were very inspiring for this note. We hope that the present work provides something he would find interesting to read. The fourth author is also grateful for his mentorship during the writing of

Master’s thesis, in 2019.

1. Introduction

Some of the most important sport competitions have a fi

nal classification which depends on a round-robin phase.

In such tournaments, each participant plays at least once

against any other team, and the final classification considers all results. Then tournaments as Serie A or NBA need eficient algorithms to compute all the possible match schedules, see [ 1, 2 ]. For example, some tournament uses Berger’s algorithm [ 3 ] (developed by the chess player Johann Berger) which divides the players into two equal sides, from 1 to 2 and from 2 + 1 to ; starts from the first pairing {1, }, {2, − 1}, . . . , {︀ 2 , 2 + 1}︀ ; and ends by giving some combinatorial argument to obtain all the other pairings. In our approach, we use onefactorizations of complete graphs. More precisely, a onefactorization of the complete graph corresponds to a pairing in a round-robin tournament with teams playing. In this note, we characterize parabolic and linear one-factorizations of complete graphs +1, with an odd prime number. In Section 2, we give preliminaries on graph theory and one-factorizations. In Sections 3 and 4, we give characterization results for parabolic and linear one-factorizations, where a one-factor is said to be parabolic or linear if it is represented by a parabola or a line, respectively. This note concludes with an Appendix containing computational results for the parabolic onefactorizations of +1, = 13, 17, 19, and the Magma code used by the authors.

2. Preliminaries

is a partition of its edge set into − 1 disjoint one-factors. sponds to the set of points and are usually denoted by , where is the number of vertices. Let be the complete graph with an even number of vertices. A one-factor is a partition of its vertex set into 2 disjoint edges. A one-factorization of

The vertices of the complete graph +1 correspond to the points of 0 ∪ ∞, while the edges of +1 correspond to the points of type , with = 2 , ∞. Thus the set of edges of +1 corre[ 6 ]. The reader may refer to [14] for notation and termi- line ℓ of 0 is a set consisting of − 2 1 points of ℰ on ℓ, plus = 2 + , represented by a line.

For other notation or definitions on graphs not stated here, we refer the reader to [ 4 ]. Our approach to the problem of constructing one-factorizations of complete graphs is geometric, as in [ 5, 6, 7, 8, 9 ], and is based on techniques that have also been used for multigraphs, see

3. Parabolic one-factorizations

We adopt the same notation and terminology introduced in a paper by Korchmáros, Pace and Sonnino [ 5 ], and then used in subsequent paper by Kiss, Pace and Sonnino nology on finite geometry not explicitly stated here. Let be an odd prime. Fix a projective frame in (2, ) with homogeneous coordinates (0 : 1 : 2), and consider (2, ) as (2, )∪ℓ ∞ where ℓ ∞ has equation 0 = 0. As usual, the points of (2, ) are written as (, ) with = 10 and = 20 . In (2, ), let be the parabola with afine equation where varies in Z, and ∞ = (0 : 0 : 1) the point at infinity of the line

1 = 0. We remark that, in the projective closure of (2, ), any two parabolas and , with ̸= , meet only at the point ∞. Let denote the afine point

( + 2 , 2 + ) and ∞ be the point (0 : 1 : 2) on the line at infinity ℓ . The following ∞ result is easy to check: are on the parabola − 42 .

Lemma 3.1. For a fixed , the points 0, 1, . . . , − 1

Definition 3.2.

Let = (, ) be a point. The symptome of is defined as = 2 − .

The symptome of ℓ is defined as ℓ = 2 + 4.

It is straightforward to check the following lemma:

We recall some basic properties of the parabolas . Lemma 3.3. Let = (, ) be a point and ℓ : = + be a non-vertical line. Then ( ,

2 − ). only if ℓ − element in GF()). • is an external (internal) point of 0 if and only if is a square (non-square) element in GF(). • ℓ is a tangent (secant or external line) to if and 4 is 0 (a square or a non-square • The pole of ℓ with respect to 0 is the point = These points are called external w.r.t. 0.

Definition 3.4.

A one-factor represented by a parabola is a set of − 2 1 points of type on , together with a suitable point at infinity. A one-factor so defined is referred to as a parabolic one-factor.

Definition 3.5.

A one-factor represented by a secant

the pole of ℓ with respect to 0. a set consisting of +1 points of ℰ on ℓ.

2

A one-factor represented by an external line ℓ of 0 is

Definition 3.6.

A one-factorization of +1 is called a parabolic one-factorization if − are represented by parabolas and one of its one-factors is 1 of its one-factors

In [6] the authors proved the existence of an infinite family of parabolic one-factorization.

Theorem 3.7. [6, Theorem 3.4] Let be an odd prime. Then the complete graph +1 has a parabolic onefactorization.

Proof. The proof is constructive. The set 0 = ︂{ − 2 : = 1, 2, . . . , − 1 }︂ 2 ∪ {0∞} is a one-factor represented by the line secant line of 0 the sets = = ︂{ ︂{ 2 +2 : = 0, 1, . . . , − 3 }︂

2 2 +(2+1) : = 0, 1, . . . , − 3 }︂ 2 ∪ {︁− ∞2 }︁ , ∪ {︁ ∞}︁ 2 .

By Lemma 3.1 ∖ {︁− ∞2 }︁ and ∖{︁ ∞}︁ are disjoint 2 one-factors represented by the parabola − 42 . subsets of the parabola − 2 , and both and are 4

Definition 3.8.

A one-factorization of +1 is called an almost parabolic one-factorization if at least one of its onefactors is represented by − 42 for all ∈ {1, 2, . . . − 2 1 }, and all other of its one-factors are represented by lines.

Let ℓ be a non-vertical line with equation = + . of equation = 0, and 0 is the pole. Now we define For < 11 exhaustive computer search shows is +21 . If 412 +41 = 422 +42, then (1 − 2)(1 + 2 + 11. because the sum in the second factor is not 0.

1) = 0. So for 1 ̸= 2 we have 412 + 41 = 422 + 42, again.

The set obviously contains 0. Moreover, we claim

that contains both square and non-square elements. Suppose to the contrary, that all of the non-zero products ( +1) are either squares or non-squares. If 2 is a square, then 1 · 2 is a square, hence 2 · 3 is also a square which implies that 3 is a square. In the same way, step by step, we get that all of the elements 4, . . . , − 1 2 , +1 are squares.

2 Hence there are at least +1 square elements, a contradiction. If 2 is a non-square, then 1 hence 2

· 3 is also a non-square, so 3 is a square. But 4 is also a square, hence 3 · 4 is a square, a contradiction · 2 is a non-square, 2 that each almost parabolic one-factorization of +1 is parabolic. In the rest of the paper we will assume ≥ Lemma 3.9. The number of one-factors represented by lines in an almost parabolic one-factorization is either one, or at least ⌈ +41 ⌉.

Proof. If more than one one-factors are represented by

lines, then there exist parabolas − 2 which represent only one one-factor. Hence − at least ⌈ +41 ⌉ lines to cover these points. are covered by the lines represented the other one-factors. Any line meets − 2 in at most two points, so we need 4

2 2 − 1 =4 +1 of its points of [6, Theorem 3.5].

Lemma 3.10. Let ℒ denote the set of points of type belonging to the one-factors represented by lines in an almost parabolic one-factorization. Suppose that a onefactor is represented by the line = 0. Then for each ∈ {1, 2, . . . − 1

2 } either ℒ ⊂ or ℒ ⊂ .

Proposition 3.11. An almost parabolic one-factorization contains at most two one-factors which are represented by vertical lines.

The following Lemma is a straightforward corollary

Now we show that the set Hence

= +′ = 3.10.

′ 2 +(2+1)′ where = 0, 1, . . . , − 3 2 So the line = intersects ′ contradicting Lemma Lemma 3.12. In GF() let denote the set = ︂{ 2 : = 1, 2, . . . , − 1 }︂ 2 equals to the set of square elements of GF(). The cardinality of is − 1 2 , because 12 = 22 implies (1 − 2)(1 + 2) = 0, and the second factor is never 0, because 1 + 2 ≤ − 1.

Choose elements 1, 2 ∈ such that 1 is a square and 2 is a non-square. Then 1 ∩ 2 = ∅, hence |{0} ∪ 1 ∪ 2 | = 1 + − 1 2 + − 1 2 = ,

Proposition

3.13. If an almost parabolic onefactorization contains a one-factors which is represented by a vertical line, then it cannot contain one-factors represented by non-vertical lines.

Proof. We may assume that the equation of the corresponding vertical line is = 0. Suppose to the contrary . that it contains the line ℓ : = + . We claim that such that ℓ = (42 + 4)2. Then there exists at least one ∈ {1, 2, . . . , − 1 2 } such that ℓ ∩ ̸= ∅ ̸= ℓ ∩ . By Lemma 3.12, there exist and ℓ + 2 = (42 + 4)2 + 2 = ((2 + 1))2, so, by Lemma 3.3, ℓ intersects − 2 and the diference of 4 the first coordinates of the two intersections is (2 + 1).

Hence one of the two points belongs to and the other

one belongs to . The statement follows from Lemma 3.10.

The main result of this section is the following theorem, which derives from Propositions 3.9, 3.11 and 3.13.

Theorem 3.14. Let be an odd prime. If an almost parabolic one-factorization ℱ of +1 contains a onefactor which is represented by a vertical line, then the onefactorization is parabolic.

Proof. Suppose to the contrary that it contains at least

three one-factors which are represented by vertical lines. the statement is proved. {1, 2, . . . − 2 1 }.

We may assume that the equations of the corresponding

lines are = 0, = , = and − = ′ ∈

We also may assume that the line = intersects ′ . Then = 2′ + 2′ where = 0, 1, . . . , − 2 3 . 2 2 = {︁ (42 + 4)2 : = 0, 1, . . . , − 1 , = 1, 2, . . . , − 1 }︁

Proof. For < 11 the statement follows from an exhaus

tive computer search, see [ 6 ]. conIftai n≥s e1i1th,ethreonne⌈,o+4r1a⌉t l≥ea3st. tHhereneceo,nbey-fLacetmorms ath3a.t9,aℱre TbyhethreesHeaurncghaorifatnheNfirasttioanuatlhRoresweaarscpha, rDtieavlleylospumppeonrtteadnd represented by lines. In the former case, we are done. Innovation Ofice OTKA grant no. SNN 132625. The The latter case leads to a contradiction, since, by Lemma research of the fourth author was partially supported by 3.11, the number of vertical lines is at most two, and by the Italian National Group for Algebraic and Geometric Lemma 3.13, there is no non-vertical line among the lines Structures and their Applications (GNSAGA - INdAM) representing the one-factors. and by the INdAM - GNSAGA Project Tensors over finite ifelds and their applications , number E53C23001670001.

Acknowledgments

4. Linear one-factorizations

In this section, we consider one-factorizations whose all one-factors are represented by lines.

Let Ω be the set of points of an irreducible conic in

(2, ) with ≥ 5 odd.

Theorem 4.1. Let +1 be a one-factorization on Ω whose one-factors are represented by lines. Then some of those lines are a chord of Ω .

Proof. Let +1 be represented by the lines ℓ1, . . . , ℓ.

Assume on the contrary that each those lines is an external line to Ω . Let 1, . . . , be the poles of ℓ1, . . . , ℓ with respect to the orthogonal polarity associated with Ω . For = 1, . . . , , let be the involutory perspectivity with center and axis ℓ which preserves Ω . Let ∼= (2, ) be the orthogonal group of Ω , i.e. the subgroup of (3, ) which commutes with . Then preserves Ω and acts on its points as (2, ) on the projective line over F. Furthermore, ∈ . Let ℱ be the set consisting of , = 1, . . . , together with the identity of . Since +1 is a one-factorization, for any two points , ∈ Ω there exists a unique ∈ ℱ such that ( ) = . Therefore, ℱ is a sharply transitive permutation set on Ω containing the identity. From the classification of sharply transitive subsets of (2, ) [15], it turns out ℱ is a subgroup of (2, ) of order + 1. On the other hand, from Dickson’s classification of subgroups of (2, ), the subgroups of (2, ) entirely consisting of involutions together with the identity, have order either 2 or 4. But then ≤ 3, a contradiction.

We conclude by reporting a conjecture that is sup

ported by computer-aided searches. With the aid of

Magma [16] we veriefid that the conjecture holds for

≤ 23.

Conjecture 4.2. [6, Conjecture 3.6] Let > 7 be an odd prime, ℱ be a one-factorization of the complete graph +1 such that each one-factor of ℱ is either represented by a line or a parabola. Then ℱ is either a parabolic onefactorization or each one-factor of ℱ is represented by a line. (2, 2ℎ), Discrete Math. 231 (2001) 447–451. 4,9 : (1 : 11 : 0), 4,4 : (1 : 6 : 6), 4,12 : (1 : 1 : 10), doi:10.1016/S0012-365X(00)00337-X. 4,7 : (1 : 9 : 12)} [14] Gy. Kiss, T. Szőnyi, Finite Geometries, CRC Press, 5 : {4,∞ : (0 : 1 : 8), 5,9 : (1 : 5 : 9), 5,6 : (1 : 2 : 1), [15] TS.ayElboerr&haFrrda,ncSishGarrpoluyp,trBaoncsaitRivaetosne,tsFLin,20192.(), 5,3 : (1 : 12 : 11), 5,0 : (1 : 9 : 0), 5,10 : (1 : 6 : 7), 5,7 : (1 : 3 : 6)} Adv. Geom. 21 (2021) 611–612. doi:10.1515/ 5 : {9,∞ : (0 : 1 : 5), 5,1 : (1 : 10 : 6), 5,11 : (1 : 7 : 7), advgeom-2021-0029. [16] W. Bosma, J. Cannon, C. Playoust, The magma alge- 5,8 : (1 : 4 : 0), 5,5 : (1 : 1 : 11), 5,2 : (1 : 11 : 1), bra system. i. the user language., J. Symbolic Com- 5,12 : (1 : 8 : 9)} put. 24 (1997) 235–265. doi:10.1006/jsco.1996. 6 : {10,∞ : (0 : 1 : 7), 6,3 : (1 : 6 : 1), 6,2 : (1 : 5 : 3), 0125. 6,1 : (1 : 4 : 7), 6,0 : (1 : 3 : 0), 6,12 : (1 : 2 : 8), 6,11 : (1 : 1 : 5)} 6 : {3,∞ : (0 : 1 : 6), 6,9 : (1 : 12 : 5), 6,8 : (1 : 11 : 8), A. Tables 6,7 : (1 : 10 : 0), 6,6 : (1 : 9 : 7), 6,5 : (1 : 8 : 3), 6,4 : (1 : 7 : 1)}

Corollary 3.14 states that almost parabolic one

factorization of +1, odd prime, containing a vertical line are parabolic. In this Appendix, we report exam- A.2. = 17 ples of such parabolic one-factorizations of the complete graphs 14, 18 and 20, found by computations on 0 : {0,∞ : (0 : 1 : 0), 1,8 : (1 : 0 : 4), 2,16 : (1 : 0 : 16),

Magma, [16]. 4 admits only 1 one-factorization, and

are well-known the 6 examples of 1-factorizations of 6.

We refer to [6] for the cases 8 and 12.

3,7 : (1 : 0 : 2), 4,15 : (1 : 0 : 13), 5,6 : (1 : 0 : 15), 6,14 : (1 : 0 : 8), 7,5 : (1 : 0 : 9), 8,13 : (1 : 0 : 1)} 1 : {8,∞ : (0 : 1 : 16), 1,9 : (1 : 1 : 5), 1,11 : (1 : 3 : 13), 1,13 : (1 : 5 : 12), 1,15 : (1 : 7 : 2), 1,0 : (1 : 9 : 0), 1,2 : (1 : 11 : 6), 1,4 : (1 : 13 : 3), 1,6 : (1 : 15 : 8)} A.1. = 13 0 : {0,∞ : (0 : 1 : 0), 1,6 : (1 : 0 : 3), 2,12 : (1 : 0 : 12), 1 : {9,∞ : (0 : 1 : 1), 1,10 : (1 : 2 : 8), 1,12 : (1 : 4 : 3), 3,5 : (1 : 0 : 1), 4,11 : (1 : 0 : 9), 5,4 : (1 : 0 : 10), 1,14 : (1 : 6 : 6), 1,16 : (1 : 8 : 0), 1,1 : (1 : 10 : 2), 6,10 : (1 : 0 : 4)} 1,3 : (1 : 12 : 12), 1,5 : (1 : 14 : 13), 1,7 : (1 : 16 : 5)} 1 : {6,∞ : (0 : 1 : 12), 1,7 : (1 : 1 : 4), 1,9 : (1 : 3 : 12), 2 : {16,∞ : (0 : 1 : 15), 2,1 : (1 : 2 : 3), 2,5 : (1 : 6 : 1), 1,11 : (1 : 5 : 2), 1,0 : (1 : 7 : 0), 1,2 : (1 : 9 : 6), 2,9 : (1 : 10 : 14), 2,13 : (1 : 14 : 8), 2,0 : (1 : 1 : 0), 1,4 : (1 : 11 : 7)} 2,4 : (1 : 5 : 7), 2,8 : (1 : 9 : 12), 2,12 : (1 : 13 : 15)} 1 : {7,∞ : (0 : 1 : 1), 1,8 : (1 : 2 : 7), 1,10 : (1 : 4 : 6), 2 : {1,∞ : (0 : 1 : 2), 2,3 : (1 : 4 : 15), 2,7 : (1 : 8 : 12), 1,12 : (1 : 6 : 0), 1,1 : (1 : 8 : 2), 1,3 : (1 : 10 : 12), 2,11 : (1 : 12 : 7), 2,15 : (1 : 16 : 0), 2,2 : (1 : 3 : 8), 1,5 : (1 : 12 : 4)} 2,6 : (1 : 7 : 14), 2,10 : (1 : 11 : 1), 2,14 : (1 : 15 : 3)} 2 : {12,∞ : (0 : 1 : 11), 2,1 : (1 : 2 : 3), 2,5 : (1 : 6 : 9), 3 : {7,∞ : (0 : 1 : 14), 3,10 : (1 : 3 : 11), 3,16 : (1 : 9 : 15), 2,9 : (1 : 10 : 8), 2,0 : (1 : 1 : 0), 2,4 : (1 : 5 : 11), 3,5 : (1 : 15 : 6), 3,11 : (1 : 4 : 1), 3,0 : (1 : 10 : 0), 2,8 : (1 : 9 : 2)} 3,6 : (1 : 16 : 3), 3,12 : (1 : 5 : 10), 3,1 : (1 : 11 : 4)} 2 : {1,∞ : (0 : 1 : 2), 2,3 : (1 : 4 : 2), 2,7 : (1 : 8 : 11), 3 : {10,∞ : (0 : 1 : 3), 3,13 : (1 : 6 : 4), 3,2 : (1 : 12 : 10), 2,11 : (1 : 12 : 0), 2,2 : (1 : 3 : 8), 2,6 : (1 : 7 : 9), 3,8 : (1 : 1 : 3), 3,14 : (1 : 7 : 0), 3,3 : (1 : 13 : 1), 2,10 : (1 : 11 : 3)} 3,9 : (1 : 2 : 6), 3,15 : (1 : 8 : 15), 3,4 : (1 : 14 : 11)} 3 : {5,∞ : (0 : 1 : 10), 3,8 : (1 : 3 : 10), 3,1 : (1 : 9 : 4), 4 : {15,∞ : (0 : 1 : 13), 4,2 : (1 : 4 : 12), 4,10 : (1 : 12 : 4), 3,7 : (1 : 2 : 5), 3,0 : (1 : 8 : 0), 3,6 : (1 : 1 : 2), 4,1 : (1 : 3 : 5), 4,9 : (1 : 11 : 15), 4,0 : (1 : 2 : 0), 3,12 : (1 : 7 : 11)} 4,8 : (1 : 10 : 11), 4,16 : (1 : 1 : 14), 4,7 : (1 : 9 : 9)} 3 : {8,∞ : (0 : 1 : 3), 3,11 : (1 : 6 : 11), 3,4 : (1 : 12 : 2),4 : {2,∞ : (0 : 1 : 4), 4,6 : (1 : 8 : 9), 4,14 : (1 : 16 : 14), 3,10 : (1 : 5 : 0), 3,3 : (1 : 11 : 5), 3,9 : (1 : 4 : 4), 4,5 : (1 : 7 : 11), 4,13 : (1 : 15 : 0), 4,4 : (1 : 6 : 15), 3,2 : (1 : 10 : 10)} 4,12 : (1 : 14 : 5), 4,3 : (1 : 5 : 4), 4,11 : (1 : 13 : 12)} 4 : {11,∞ : (0 : 1 : 9), 4,2 : (1 : 4 : 12), 4,10 : (1 : 12 : 10),5 : {6,∞ : (0 : 1 : 12), 5,11 : (1 : 5 : 6), 5,4 : (1 : 15 : 2), 4,5 : (1 : 7 : 6), 4,0 : (1 : 2 : 0), 4,8 : (1 : 10 : 5), 5,14 : (1 : 8 : 11), 5,7 : (1 : 1 : 16), 5,0 : (1 : 11 : 0), 4,3 : (1 : 5 : 8)} 5,10 : (1 : 4 : 14), 5,3 : (1 : 14 : 7), 5,13 : (1 : 7 : 13)} 4 : {2,∞ : (0 : 1 : 4), 4,6 : (1 : 8 : 8), 4,1 : (1 : 3 : 5), 5 : {11,∞ : (0 : 1 : 5), 5,16 : (1 : 10 : 13), 5,9 : (1 : 3 : 7), 5,2 : (1 : 13 : 14), 5,12 : (1 : 6 : 0), 5,5 : (1 : 16 : 16),

Below, we report the code used to find the parabolic one-factorizations.

p:=19; F:=GF(p); AG:=AffineSpace(F,3); r:=(p-1)/2; rr:=(p-3)/2; F0:={AG![ 0,1,0 ]}; for k in [1..r] do kk:=F!k; l:=AG![1,0,(-kk/2)^2+(-kk/2)*kk]; F0:=F0 join {l}; end for; printf "F 0 : %o\n" ,F0; for k in [1..r] do kk:=F!k; G:={AG![0,1,-kk]}; H:={AG![0,1,kk]}; for j in [0..rr] do jj:=F!j; i:=F!(kk/2+2*jj*kk); ii:=F!(kk/2+(2*jj+1)*kk); m:=AG![1,i+kk/2,i^2+i*kk]; n:=AG![1,ii+kk/2,ii^2+ii*kk]; G:=G join {m}; H:=H join {n}; end for; printf "G %o : %o\n" ,kk,G; printf "H %o : %o\n" ,kk,H; end for;

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