The Chromatic Index of Projective Triple Systems

The Chromatic Index of Projective TripleSystems

Mariusz MeszkaAGH University of Science and Technology, Krakow, Poland, E-mail:[email protected]

Received July 4, 2012; revised July 4, 2013

Published online 30 July 2013 in Wiley Online Library (wileyonlinelibrary.com).DOI 10.1002/jcd.21368

Abstract: We prove that, for even m, the chromatic index of the projective triple systemSTS(2m+1 − 1) equals 2m + 2. C© 2013 Wiley Periodicals, Inc. J. Combin. Designs 21: 531–540, 2013

Keywords: projective triple system; block coloring; chromatic index

1. INTRODUCTION

A partial triple system of order v, denoted by PTS(v), is a pair (V,B), where V is a set ofv points and B is a collection of 3-element subsets of V called triples (blocks) such thatevery 2-element subset of V is contained in at most one triple. A Steiner triple system,STS(v), is a PTS(v) such that every pair of points is contained in exactly one triple. A setof four triples {a, b, c}, {a, d, e}, {b, d, f }, and {c, e, f } is called a Pasch configuration.

A partial parallel class, PPC, in a PTS(v) is a collection of p pairwise disjoint triples.A PPC is a parallel class if p = v/3 (then v ≡ 3 (mod 6)); similarly a PPC is an almostparallel class if p = (v − 1)/3 (and then v ≡ 1 (mod 6)).

A block coloring of a partial triple system PTS(v) (V,B) is a mapping ψ : B �→C, where C is a set of colors, such that for any two blocks B1, B2 ∈ B, if ψ(B1) =ψ(B2) then B1 ∩ B2 = ∅. If C has cardinality c then ψ is a c-block coloring (or simplyc-coloring). For each c ∈ C, the set of blocks ψ−1(c) makes a partial parallel class that iscalled a color class. The minimum number of colors needed to color blocks in a PTS(v)is called its chromatic index.

Let Wm be an (m + 1)-dimensional vector space over F2. Let ⊕ be the operation ofvector addition in Wm. Any two nonzero (m + 1)-vectors x and y determine uniquely athird vector x ⊕ y in Wm, where addition is performed modulo 2 componentwise. Letevery nonzero vector in Wm+1 be represented by a point in a set V of cardinality 2m+1 − 1.Evidently, every two distinct points, corresponding to x and y, define a unique triple

Journal of Combinatorial DesignsC© 2013 Wiley Periodicals, Inc. 531

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formed by {x, y, x ⊕ y}. The STS(2m+1 − 1) produced in this way is called a projectivetriple system and, by abusing the notation slightly, it is often denoted by PG(m, 2) (justconsider the triples as lines in the projective space over GF(2)). To simplify notation, letevery point in V be labeled by an integer whose binary representation is determined bythe coordinates of its corresponding vector. Thus V (PG(m, 2)) = {1, 2, . . . , 2m+1 − 1}.

In PG(2, 2) every two triples intersect, so its chromatic index is 7. If m is odd, itis known that PG(m, 2) has the chromatic index 2m − 1 [1]. For even m ≥ 4, A. Rosaconjectured [9] that the chromatic index of PG(m, 2) equals 2m + 2. The main result ofthe paper is an affirmative answer to this conjecture.

Theorem 1. For m ≥ 3, the chromatic index of a projective triple system PG(m, 2) is(2m − 1) if m is odd and (2m + 2) otherwise.

This result tends to support another conjecture of A. Rosa [9] that, apart from theFano plane (aka STS(7)), the chromatic index of an STS(v) never exceeds the minimumpossible value (i.e. (v − 1)/2 if v ≡ 3 (mod 6) and (v + 1)/2 otherwise) by more than two.An elementary known upper bound for the chromatic index of an STS(v) is 3(v − 3)/2(for v ≥ 9) [2]. Clearly, the chromatic index attains its minimum in the case of Kirkmantriple systems [8] and Hanani triple systems [10]. Another support for the conjecture canbe found in [4, 6] where the chromatic indices of Steiner triple systems of small ordersv ≤ 19 are determined. Moreover, using probabilistic methods Pippenger and Spencer in[7] proved that the chromatic index of an STS(v) is asymptotic to v/2.

2. CONSTRUCTIONS

First we need to describe three recursive constructions.

Construction 1 (i-tuple PG construction). This is a recursive version of the construc-tion of a projective triple system PG(i + 1, 2) (V,B) of order n = 2i+2 − 1 with b =(8 · 4i − 6 · 2i + 1)/3 blocks. Let V = {v1, v2, . . . , vn}. Begin with B0 = {{v1, v2, v3}}.Repeat the following step i times for every j = 1, 2, . . . i:

Let m = 2j+1. For each triple {vx, vy, vz} ∈ Bj−1, construct three new triples{vx, vy+m, vz+m}, {vx+m, vy, vz+m}, {vx+m, vy+m, vz} and put these three new triplestogether with {vx, vy, vz} in Bj . Moreover, add m − 1 triples {vt , vt+m, vm} to Bj ,t = 1, 2, . . . , m − 1. Thus |Bj | = 4|Bj−1| + m − 1.

Finally, let B = Bi . A set U = {v4l : l = 1, 2, . . . , (n − 3)/4} is called an upper set.Notice that none of the triples in B has exactly two points in U . Moreover, every triplewith only one point in U has two remaining points vu and vw such that u − w ≡ 0(mod 4).

Notice that the above recursive method produces the same set of triples as given bydefinition of PG(m, 2) (when vector addition is applied).

Example 1. PG(3, 2).Let V = {1, 2, . . . , 15}. Apply the 2-tuple PG construction; then U = {4, 8, 12}. It iswell known that PG(3, 2) is resolvable and admits 240 distinct resolutions that fall intotwo isomorphism classes [5]. One of those will be used in a general construction:

Journal of Combinatorial Designs DOI 10.1002/jcd

THE CHROMATIC INDEX OF PROJECTIVE TRIPLE SYSTEMS 533

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FIGURE 1. Arrangement of points of PG(3,2).

P1 = {{1, 2, 3}, {4, 8, 12}, {5, 11, 14}, {6, 9, 15}, {7, 10, 13}},P2 = {{1, 4, 5}, {2, 12, 14}, {3, 9, 10}, {6, 11, 13}, {7, 8, 15}},P3 = {{1, 6, 7}, {2, 13, 15}, {3, 8, 11}, {4, 10, 14}, {5, 9, 12}},P4 = {{1, 8, 9}, {2, 4, 6}, {3, 13, 14}, {5, 10, 15}, {7, 11, 12}},P5 = {{1, 10, 11}, {2, 5, 7}, {3, 12, 15}, {4, 9, 13}, {6, 8, 14}},P6 = {{1, 12, 13}, {2, 8, 10}, {3, 5, 6}, {4, 11, 15}, {7, 9, 14}},P7 = {{1, 14, 15}, {2, 9, 11}, {3, 4, 7}, {5, 8, 13}, {6, 10, 12}}.

Figure 1 shows a possible arrangement of points; it has a property that every triplewith exactly one point in U has two remaining points in the same row and vice versa,every pair of points in the same row in V \ U belongs to a triple such that its thirdpoint is in U . So every triple with no point in U has all points in three distinctrows.

Example 2. PG(4, 2).Let V = {1, 2, . . . , 31}. Apply the 3-tuple PG construction; then U = {4, 8, 12, 16,

20, 24, 28}. We need one example of a coloring with 18 colors to be used in a gen-eral construction:

C1 = {{1, 2, 3}, {4, 19, 23}, {5, 9, 12}, {6, 11, 13}, {7, 17, 22}, {8, 20, 28},{10, 18, 24}, {14, 21, 27}, {15, 16, 31}},

C2 = {{1, 4, 5}, {2, 9, 11}, {3, 13, 14}, {6, 17, 23}, {7, 18, 21}, {8, 16, 24},{10, 19, 25}, {12, 22, 26}, {15, 20, 27}},

C3 = {{1, 6, 7}, {2, 8, 10}, {3, 12, 15}, {4, 18, 22}, {5, 17, 20}, {9, 19, 26},{11, 21, 30}, {13, 16, 29}, {14, 23, 25}},

C4 = {{1, 8, 9}, {2, 5, 7}, {3, 17, 18}, {4, 26, 30}, {6, 10, 12}, {11, 20, 31},{13, 22, 27}, {14, 19, 29}, {15, 23, 24}},

C5 = {{1, 10, 11}, {2, 4, 6}, {3, 16, 19}, {5, 8, 13}, {7, 27, 28}, {9, 23, 30},{12, 21, 25}, {14, 22, 24}, {15, 18, 29}},

C6 = {{1, 12, 13}, {2, 17, 19}, {3, 5, 6}, {4, 25, 29}, {7, 9, 14}, {8, 23, 31},{10, 22, 28}, {11, 16, 27}, {15, 21, 26}},


534 MESZKA

C7 = {{1, 14, 15}, {2, 16, 18}, {3, 4, 7}, {5, 19, 22}, {6, 25, 31}, {10, 20, 30},{11, 23, 28}, {12, 17, 29}, {13, 21, 24}},

C8 = {{1, 16, 17}, {2, 12, 14}, {3, 8, 11}, {4, 9, 13}, {5, 26, 31}, {6, 18, 20},{7, 25, 30}, {10, 23, 29}, {15, 19, 28}},

C9 = {{1, 18, 19}, {2, 13, 15}, {3, 9, 10}, {4, 8, 12}, {5, 25, 28}, {6, 27, 29},{7, 24, 31}, {11, 17, 26}, {14, 16, 30}},

C10 = {{1, 20, 21}, {2, 24, 26}, {3, 28, 31}, {4, 10, 14}, {7, 16, 23}, {8, 17, 25},{9, 18, 27}, {11, 22, 29}, {13, 19, 30}},

C11 = {{1, 22, 23}, {2, 25, 27}, {3, 29, 30}, {4, 11, 15}, {6, 19, 21}, {9, 17, 24},{12, 16, 28}, {13, 18, 31}, {14, 20, 26}},

C12 = {{1, 24, 25}, {2, 20, 22}, {4, 17, 21}, {5, 27, 30}, {6, 9, 15}, {7, 26, 29},{12, 19, 31}, {14, 18, 28}},

C13 = {{1, 26, 27}, {2, 21, 23}, {4, 16, 20}, {5, 10, 15}, {6, 8, 14}, {9, 22, 31},{11, 19, 24}, {12, 18, 30}, {13, 17, 28}},

C14 = {{1, 28, 29}, {3, 20, 23}, {4, 27, 31}, {5, 16, 21}, {6, 24, 30}, {7, 11, 12},{8, 18, 26}, {15, 22, 25}},

C15 = {{1, 30, 31}, {3, 21, 22}, {4, 24, 28}, {7, 8, 15}, {9, 20, 29}, {10, 17, 27},{11, 18, 25}, {13, 23, 26}},

C16 = {{2, 28, 30}, {3, 24, 27}, {5, 18, 23}, {8, 21, 29}, {10, 16, 26}, {13, 20, 25},{14, 17, 31}},

C17 = {{2, 29, 31}, {3, 25, 26}, {5, 11, 14}, {6, 16, 22}, {7, 10, 13}, {8, 19, 27},{9, 21, 28}, {12, 20, 24}, {15, 17, 30}},

C18 = {{5, 24, 29}, {6, 26, 28}, {7, 19, 20}, {8, 22, 30}, {9, 16, 25}, {10, 21, 31},{12, 23, 27}}.

Construction 2 (i-tuple PG construction with a hole). A projective triple system(V,B′) of order n = 2i+2 − 1 with a hole U of size 2i − 1 and b = 10 · 4i−1 − 3 · 2i−1

blocks is constructed. Let V = {v1, v2, . . . , vn} and U = {v4l : l = 1, 2, . . . , (n − 3)/4}.Apply the i-tuple PG construction and remove triples with all points in U , i.e every triple{vx, vy, vz} such that x ≡ y ≡ z ≡ 0 (mod 4). Thus every triple in B′ has either onepoint in U or none.

Construction 3 (i-tuple Pasch construction). This construction produces apartial triple system (V ′′,B′′) of order n = 3 · 2i with b = 4i blocks. Let V ′′ ={v1, v2, . . . , v4·2i−1} \ {v4, v8, v12 . . . , v4·2i−4}. Repeat the following recursive step i timesfor every j = 1, 2, . . . i, starting with B0 = {{v1, v2, v3}}:

Let m = 2j+1. For each triple {vx, vy, vz} ∈ Bj−1, construct three new triples{vx, vy+m, vz+m}, {vx+m, vy, vz+m}, {vx+m, vy+m, vz} and put these three new triples to-gether with {vx, vy, vz} in Bj . Finally, let B′′ = Bi . Thus every step of this construc-tion consists of consecutive replacements of triples by Pasch configurations. Hence,



for every triple {vx, vy, vz} ∈ B′′ the following holds: x − y ≡ 0 (mod 4), x − z ≡ 0(mod 4), and y − z ≡ 0 (mod 4). Notice that the same partial triple system (V ′′,B′′)may be obtained by applying the i-tuple PG construction and then by removal of the setU = {v4l : l = 1, 2, . . . , 2i − 1} together with all triples with at least one point in U .

3. PROOFS

PG(3, 2) has the following useful property.

Lemma 2. The projective triple system PG(3, 2) contains a parallel class P suchthat, for every triple T ∈ P , there exist three other parallel classes PT

2 , PT3 and PT

4which, together with P , pairwise intersect in T , i.e. PT

i ∩ PTj = T , where PT

1 = P and1 ≤ i < j ≤ 4. Moreover, every triple T ′ ∈ P belongs to exactly two of the parallelclasses in the set {PS

i : S ∈ P, i = 2, 3, 4}.

Proof. Take the PG(3, 2) constructed in Example 1. Let P consist of triples TA ={1, 2, 3}, TB = {4, 8, 12}, TC = {5, 11, 14}, TD = {6, 9, 15}, and TE = {7, 10, 13}. Therequired parallel classes are:

PTA

2 = {{1, 2, 3}, {4, 9, 13}, {5, 10, 15}, {6, 8, 14}, {7, 11, 12}},PTA

3 = {{1, 2, 3}, {4, 10, 14}, {5, 9, 12}, {6, 11, 13}, {7, 8, 15}},PTA

4 = {{1, 2, 3}, {4, 11, 15}, {5, 8, 13}, {6, 10, 12}, {7, 9, 14}},PTB

2 = {{1, 6, 7}, {2, 9, 11}, {3, 13, 14}, {4, 8, 12}, {5, 10, 15}},PTB

3 = {{1, 10, 11}, {2, 13, 15}, {3, 5, 6}, {4, 8, 12}, {7, 9, 14}},PTB

4 = {{1, 14, 15}, {2, 5, 7}, {3, 9, 10}, {4, 8, 12}, {6, 11, 13}},PTC

2 = {{1, 6, 7}, {2, 8, 10}, {3, 12, 15}, {4, 9, 13}, {5, 11, 14}},PTC

3 = {{1, 8, 9}, {2, 13, 15}, {3, 4, 7}, {5, 11, 14}, {6, 10, 12}},PTC

4 = {{1, 12, 13}, {2, 4, 6}, {3, 9, 10}, {5, 11, 14}, {7, 8, 15}},PTD

2 = {{1, 4, 5}, {2, 8, 10}, {3, 13, 14}, {6, 9, 15}, {7, 11, 12}},PTD

3 = {{1, 10, 11}, {2, 12, 14}, {3, 4, 7}, {5, 8, 13}, {6, 9, 15}},PTD

4 = {{1, 12, 13}, {2, 5, 7}, {3, 8, 11}, {4, 10, 14}, {6, 9, 15}},PTE

2 = {{1, 4, 5}, {2, 9, 11}, {3, 12, 15}, {6, 8, 14}, {7, 10, 13}},PTE

3 = {{1, 8, 9}, {2, 12, 14}, {3, 5, 6}, {4, 11, 15}, {7, 10, 13}},PTE

4 = {{1, 14, 15}, {2, 4, 6}, {3, 8, 11}, {5, 9, 12}, {7, 10, 13}}.

It is straightforward to check that the number of occurrences of every triple T ′ ∈ B \ Pin the above 15 sets is exactly two. �

Lemma 3. For every i ≥ 2, a partial triple system of order 3 · 2i produced by thei-tuple Pasch construction is 2i-colorable.


536 MESZKA

Proof. Let n′ = 2i+2 − 1 and V ′′ = {1, 2, . . . , n′} \ {4, 8, 12, . . . , n′ − 3}. Apply thei-tuple Pasch construction to get a partial triple system (V ′′,B′′).For i = 2, note that blocks in B′′ can be partitioned into four parallel classes:

P21 = {{1, 2, 3}, {5, 10, 15}, {6, 11, 13}, {7, 9, 14}},

P22 = {{1, 6, 7}, {2, 13, 15}, {3, 9, 10}, {5, 11, 14}},

P23 = {{1, 10, 11}, {2, 5, 7}, {3, 13, 14}, {6, 9, 15}},

P24 = {{1, 14, 15}, {2, 9, 11}, {3, 5, 6}, {7, 10, 13}}.

For i = 3, blocks in B′′ can be partitioned into eight parallel classes:

P31 = {{1, 2, 3}, {5, 10, 15}, {6, 17, 23}, {7, 25, 30}, {9, 19, 26}, {11, 22, 29},

{13, 18, 31}, {14, 21, 27}},P3

2 = {{1, 6, 7}, {2, 9, 11}, {3, 25, 26}, {5, 18, 23}, {10, 21, 31}, {13, 22, 27},{14, 19, 29}, {15, 17, 30}},

P33 = {{1, 10, 11}, {2, 5, 7}, {3, 29, 30}, {6, 19, 21}, {9, 18, 27}, {13, 23, 26},

{14, 17, 31}, {15, 22, 25}},P3

4 = {{1, 14, 15}, {2, 17, 19}, {3, 21, 22}, {5, 26, 31}, {6, 27, 29}, {7, 10, 13},{9, 23, 30}, {11, 18, 25}},

P35 = {{1, 18, 19}, {2, 13, 15}, {3, 5, 6}, {7, 26, 29}, {9, 22, 31}, {10, 17, 27},

{11, 21, 30}, {14, 23, 25}},P3

6 = {{1, 22, 23}, {2, 29, 31}, {3, 17, 18}, {5, 27, 30}, {6, 11, 13}, {7, 9, 14},{10, 19, 25}, {15, 21, 26}},

P37 = {{1, 26, 27}, {2, 21, 23}, {3, 9, 10}, {5, 11, 14}, {6, 25, 31}, {7, 17, 22},

{13, 19, 30}, {15, 18, 29}},P3

8 = {{1, 30, 31}, {2, 25, 27}, {3, 13, 14}, {5, 19, 22}, {6, 9, 15}, {7, 18, 21},{10, 23, 29}, {11, 17, 26}}.

For i ≥ 4 we proceed by induction. Suppose that the partial triple system (V ′′j ,B′′

j )of order 3 · 2j is 2j -colorable with colors c1, c2, . . . , c2j , where j = i − 2. To colorthe blocks in (V ′′,B′′) we split the set of 2i colors into 2j pairwise disjoint sets Ck ={ck, c2j +k, c2·2j +k, c3·2j +k}, each of cardinality four, k = 1, 2, . . . , 2j . In the last twoiterations of the i-tuple Pasch construction, each triple T = {x, y, z} in B′′

j is replacedby a set DT of 16 triples on the set {x + t · 2i , y + t · 2i , z + t · 2i : t = 0, 1, 2, 3} of 12points in V ′′. If ck is the color of a triple T then the triples in DT are colored using fourcolors from Ck , a coloring is given in the first step of the proof (for i = 2). �

Lemma 4.

(a) PG(3, 2) with a hole U of size 3 is 7-colorable in such way that 6 colors are used tocolor all triples incident with points in U .



(b) PG(4, 2) with a hole U of size 7 is 18-colorable in such way that 12 colors are usedto color all triples incident with points in U .

Proof. A coloring in the case (a) follows immediately from Example 1; in fact, letU = {4, 8, 12}. The remaining triples in P1 are not incident with U .In the case (b), let V = {1, 2, . . . , 31} and U = {4, 8, 12, . . . , 28}. The following 12color classes are used to color all triples incident with U :

C1 = {{1, 2, 3}, {4, 17, 21}, {5, 9, 12}, {6, 8, 14}, {7, 19, 20}, {10, 18, 24},{11, 23, 28}, {13, 22, 27}, {15, 16, 31}},

C2 = {{1, 4, 5}, {2, 9, 11}, {3, 12, 15}, {6, 17, 23}, {7, 24, 31}, {8, 21, 29},{10, 22, 28}, {13, 20, 25}, {14, 16, 30}},

C3 = {{1, 6, 7}, {2, 8, 10}, {3, 16, 19}, {4, 26, 30}, {5, 17, 20}, {9, 21, 28},{12, 23, 27}, {13, 18, 31}, {14, 22, 24}},

C4 = {{1, 8, 9}, {2, 4, 6}, {3, 20, 23}, {5, 10, 15}, {7, 27, 28}, {11, 19, 24},{12, 21, 25}, {13, 16, 29}},

C5 = {{1, 12, 13}, {2, 16, 18}, {3, 4, 7}, {5, 25, 28}, {6, 24, 30}, {8, 23, 31},{9, 19, 26}, {11, 22, 29}, {15, 20, 27}},

C6 = {{1, 16, 17}, {2, 12, 14}, {3, 8, 11}, {4, 27, 31}, {5, 18, 23}, {6, 26, 28},{7, 25, 30}, {9, 20, 29}, {13, 21, 24}},

C7 = {{1, 20, 21}, {2, 24, 26}, {3, 28, 31}, {4, 18, 22}, {5, 8, 13}, {6, 10, 12},{9, 16, 25}, {14, 19, 29}, {15, 17, 30}},

C8 = {{1, 24, 25}, {2, 20, 22}, {4, 10, 14}, {5, 16, 21}, {7, 11, 12}, {8, 18, 26},{15, 19, 28}},

C9 = {{1, 28, 29}, {3, 25, 26}, {4, 11, 15}, {6, 18, 20}, {7, 16, 23}, {8, 22, 30},{9, 17, 24}, {12, 19, 31}, {14, 21, 27}},

C10 = {{2, 28, 30}, {3, 24, 27}, {4, 9, 13}, {6, 16, 22}, {7, 8, 15}, {10, 21, 31},{12, 17, 29}, {14, 20, 26}},

C11 = {{4, 19, 23}, {5, 24, 29}, {8, 17, 25}, {10, 20, 30}, {11, 16, 27}, {12, 22, 26},{14, 18, 28}},

C12 = {{4, 25, 29}, {8, 19, 27}, {10, 16, 26}, {11, 20, 31}, {12, 18, 30},{13, 17, 28}, {15, 23, 24}},

The remaining six color classes are the same as P33 ,P3

4 , . . . ,P38 in the proof of

Lemma 3. �

If m is odd, it is known that PG(m, 2) has a coloring with 2m − 1 colors [1]. For thecompletion of a construction, we present a uniform method to get optimum colorings forall m’s.

Lemma 5. For m ≥ 3, a projective triple system PG(m, 2) is (2m − 1)-colorable if m

is odd and (2m + 2)-colorable otherwise.


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FIGURE 2. Arrangement of points of PG(m,2).

Proof. For m ≤ 4 the assertion follows from Examples 1 and 2. Let m ≥ 5. LetV = {1, 2, . . . , 2m+1 − 1} and U = {4, 8, 12, . . . , 2m+1 − 4}. In fact, we are going toconstruct a coloring of PG(m, 2) that has an additional property (E) that exactly 3 · 2m−2

colors are used to color all triples with just one point in the upper set U . Notice thatpoints of V may be arranged as in Figure 2; each unlabeled point in a row denoted byp and column +q gets a label p + q. On the set {1, 2, . . . , 15} (i.e. on points in the firstcolumn) we put the PG(3, 2) given in Example 1. Then P1 = {TA, TB, TC, TD, TE} isa parallel class (cf. Example 1 and Lemma 2). For each Tα = {x, y, z} ∈ P1, let Vα ={x + 16k, y + 16k, z + 16k : k = 0, 1, . . . , 2m−3 − 1}, where α = A,B,C, D, E. LetU ′ = {16k : k = 1, 2, . . . , 2m−3 − 1}.

An equivalent way to construct blocks in B, instead of applying the (m − 1)-tuple PGconstruction directly, is to apply the following 3 steps:



1. On the set VB ∪ U ′ apply the (m − 3)-tuple PG construction, where U ′ in an upperset.

2. Separately for each Tα ∈ P1 \ TB , on the set Tα ∪ U ′ apply the (m − 3)-tuple PGconstruction with a hole U ′.

3. For each triple {x, y, z} ∈ P1, on the set {x + 16k, y + 16k, z + 16k : k =0, 1, . . . , 2m−3 − 1} apply the (m − 3)-tuple Pasch construction.

Properties of those three constructions guarantee that every pair of points u, w belongsto exactly one triple in B.

To color the triples of PG(m, 2) we proceed by induction. Suppose that, for l = m − 2,PG(l, 2) is c-colorable where c = 2l − 1 if l is odd and c = 2l + 2 otherwise. Moreover,if l ≥ 5, a given coloring satisfies the property E. We color separately triples constructedin steps 1–3. A set of triples obtained in step 1 is a set of triples of a PG(l, 2) and, byan inductive step, may be colored using a set C of c colors. We split the set C into twodisjoint subsets CB and C2 of cardinalities c1 = 3 · 2l−2 and c2 = c − c1, respectively.Moreover, each of the four sets of triples built in step 2 has a coloring with c colors, whichis given by Lemma 4 for l = 3, 4 and by an inductive step for l ≥ 5. Among these colors,c1 are used to color triples incident with points in U ′ (U ′ is a hole for this subsystem). Foreach of those four sets, we take a set Cα , where α = A,C, D, E, respectively, of c1 colorsthat were not previously used, and we color triples incident with U ′; to color remainingtriples we use additionally c2 colors from the set C2. Thus the total number of colorsis 5 · c1 + c2 = 2l+2 − 1 if l is odd and 5 · c1 + c2 = 2l+2 + 2 otherwise. It remains tocolor triples constructed in step 3 using colors from the set CA ∪ CB ∪ CC ∪ CD ∪ CE

(notice that colors from Cα were used so far to color triples on the set Vα ∪ U ′ only,where α = A, B,C, D, E). First, we are going to distribute colors into parallel classesgiven in the proof of Lemma 2. In fact, we split every set Cα into 3 disjoint subsets C2

α ,C3

α, and C4α , each of cardinality 2l−2, and assign each subset Ck

α to a parallel class PTα

k ,k = 2, 3, 4. By Lemma 2, each triple T ∈ P1 of the PG(3, 2) gets two such subsets. Sothe number of colors assigned to T is 2l−1. By Lemma 3, we may color every set oftriples constructed in step 3 from T using just that set of 2l−1 colors.

Notice that the above coloring satisfies the property E. Namely, each triple with exactlyone point in the set U = VB ∪ U ′ has a color from the set CA ∪ CC ∪ CD ∪ CE of size4c1 = 3 · 2m−2. �

Proof of Theorem 1. For odd m, a set of triples of PG(m, 2) can be partitioned into atleast 2m − 1 partial parallel classes (which actually need to be parallel classes to satisfythis bound). For even m, R. Wilson proved (cf. [3, Th. 19.6]) that PG(m, 2) does notcontain an almost parallel class. Thus the maximum size of a partial parallel class is(2m+1 − 5)/3 and then the number of partial parallel classes is at least 2m + 2. On theother hand, a coloring with 2m + 2 colors is given in Lemma 5. �

Notice that the proof of Lemma 5 provides a polynomial time algorithm for coloringtriples of a projective triple system. In fact, a PG(m, 2) of order n = 2m+1 − 1 can becolored in time �(n2 log n),

ACKNOWLEDGMENTS

The author would like to thank Alexander Rosa for useful discussions, many helpful commentsand suggestions. The research was supported by the NCN Grant No. 2011/01/B/ST1/04056.


540 MESZKA

REFERENCES

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[6] R. A. Mathon, K. T. Phelps, and A. Rosa, Small Steiner triple systems and their properties, ArsCombin 15 (1983), 3–110. Errata 16 (1983), 286.

[7] N. Pippenger and J. Spencer, Asymptotic behavior of the chromatic index for hypergraphs, JCombin Theory A 51 (1989), 24–42.

[8] D. K. Ray-Chaudhuri and R. M. Wilson, Solution of Kirkman’s schoolgirl problem, ProcSympos Pure Math 19 (1971), 187–204.

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The Chromatic Index of Projective Triple Systems