@1a@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@ÚÓ‘Ój�n€a@Î@Úœäñ€a@‚Ï‹»‹€@·rÓ:a@Âig@Ú‹©@Ü‹26@@ÖÜ»€a@I1@‚b«@H2013 Ibn Al-Haitham Jour. for Pure & Appl. Sci. Vol. 26 (1) 2013 Classification and Construction of (k,3)-Arcs on Projective Plane Over Galois Field GF(7) Adil M. Ahmad Aamal SH. Al-Mukhtar Mahmood S. Faiyadh Dept. of Mathematics/College of Education for Pure Science (Ibn AL-Haitham) University of Baghdad Received in : 27 May 2001 , Accepted in : 11 July 2001 Abstract The purpose of this work is to study the classification and construction of (k,3)-arcs in the projective plane PG(2,7). We found that there are two (5,3)-arcs, four (6,3)-arcs, six (7,3)- arcs, six (8,3)-arcs, seven (9,3)-arcs, six (10,3)-arcs and six (11,3)-arcs. All of these arcs are incomplete. The number of distinct (12,3)-arcs are six, two of them are complete. There are four distinct (13,3)-arcs, two of them are complete and one (14,3)-arc which is incomplete. There exists one complete (15,3)-arc. Key words : Arcs,Projective plane, Galois field 259 | Mathematics @1a@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@ÚÓ‘Ój�n€a@Î@Úœäñ€a@‚Ï‹»‹€@·rÓ:a@Âig@Ú‹©@Ü‹26@@ÖÜ»€a@I1@‚b«@H2013 Ibn Al-Haitham Jour. for Pure & Appl. Sci. Vol. 26 (1) 2013 Introduction A projective plane PG(2,P) over GF(P), P is a prime number consists of 1 + P + P2 points and 1 + P + P2 lines, 1 + P points on every line and 1 + P lines through every point. Any point of the plane has the form of a triple (x0,x1,x2), where x0,x1,x2 are elements in GF(P) with the exception of a triple consisting of three zero elements. Two triples (x0,x1,x2), (y0,y1,y2) represent the same point if there exists λ in GF(P) \ {0} such that (y0,y1,y2) = λ (x0,x1,x2). Similarly, any line of the plane has the form of a triple [x0,x1,x2], where x0,x1,x2 are in GF(P) with the exception of a triple consisting three zero elemens. Two triples [x0,x1,x2], [y0,y1,y2] represent the same line if there exists λ in GF(P) \ {0} such that [y0,y1,y2] = λ [x0,x1,x2]. A point P(x0,x1,x2) is incident with the line [y0,y1,y2] iff: x0y0 + x1y1 + x2y2 = 0 Definition 1: [1,2] A (k,3)-arc in PG(2,P) is a set of k points no four of them are collinear. Definition 2: [1,2] A (k,3)-arc is complete if it is not contained in a (k + 1,3)-arc. Definition 3: [3] The i-secant of a (k,3)-arc is a line intersects the arc in exactly i points, for a (k,3)-arc, each line of PG(2,P) is a 3-secant, 2-secant, 1-secant, or 0-secant. A 3-secant is called a trisecant. A (k,3)-arc is complete if every point of PG(2,P) lies on some trisecant of the arc. Let ri be the number of the i-secants of a (k,3)-arc in the plane which are r3, r2, r1, r0. Definition 4: [1,2]: A point N not on a (k,3)-arc has index i, denoted by Ni, if there are exactly i-trisecants of the arc through N. Let Ci = Ni be the number of the points Ni. Thus, a (k,3)-arc is complete iff C0 = 0. Lemma 1: [4] Let ri be the total number of the i-secants of a (k,n)-arc in PG(2,P), then the following equations are hold: n 2 i i 0 n i i 1 n i i 2 r q q 1 ir k(q 1) i(i 1)r k(k 1) = = = = + + = + − = − ∑ ∑ ∑ Definition 5: [4] Let ri be the total number of i-secants of a (k,n)-arc in PG(2,P), then the type of a (k,n)- arc w.r.t. its lines is denoted by (rn,rn – 1,…,r0). Lemma 2: [4] Let (k1,n)-arc be of type (rn,rn – 1,…,r0) and a (k2,n)-arc be of type (tn,tn – 1,…,t0), then (k1,n) and (k2,n) have the same type iff ri = ti for all i. Definition 6: [2] Two arcs are projectively equivalent under type of lines iff they have the same type. Definition 7: [5] 260 | Mathematics @1a@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@ÚÓ‘Ój�n€a@Î@Úœäñ€a@‚Ï‹»‹€@·rÓ:a@Âig@Ú‹©@Ü‹26@@ÖÜ»€a@I1@‚b«@H2013 Ibn Al-Haitham Jour. for Pure & Appl. Sci. Vol. 26 (1) 2013 Let Q1, Q2 be two points in PG(2,P) which are not on a (k,n)-arc and let K1 = K∪{Q1}, K2 = K∪{Q2}, then Q1, Q2 are in the same set iff (k1,n) and (k2,n) are projectively equivalent under type of lines. Lemma 3: [5] Let Q1, Q2 be two points in PG(2,P) which are not on a (k,n)-arc, then (i) Q1, Q2 are in the same set if they have the same type. (ii) Q1, Q2 are in the different sets if they have the different types.’ Let Pi and ℓi, i = 1, 2, …, 57 be the points and lines of PG(2,7), respectively. Let i stands for the points Pi, all the points and the lines of PG(2,7) are given in the table. Definition 8: [2] A projectively of (k,3)-arc is a non-singular (3×3) matrix which keeps the k-points of the arc point wise or globally invariant. The Construction of the Projectively Distinct (5,3)-arcs in PG(2,7): Let A = {1,2,9,17} be a set of reference points in PG(2,7) no three of them are collinear. The distinct (5,3)-arcs can be constructed by adding to A in each time a point from the remaining 53 points of the plane. By definition 6, there are only two projectively distinct (5,3)-arcs, which are: B1 = {1,2,9,17,3} and B2 = {1,2,9,17,4}. The set of projectivities fixing a (k,3)-arc B forms the group G(B). The group G(B1) has eight projectivities: 1 2 3 4 5 1 0 0 6 0 0 0 1 0 0 6 0 0 0 6 T 0 1 0 , T 0 6 0 , T 1 0 0 , T 6 0 0 , T 1 1 1 , 0 0 1 1 1 1 0 0 1 1 1 1 6 0 0                    = = = = =                             6 7 8 0 0 6 1 1 1 1 1 1 T 1 1 1 , T 0 0 6 , T 0 0 6 0 6 0 6 0 0 0 6 0            = = =                 So G(B1) ≅ D4, while G(B2) has four projectivities: 1 2 3 4 6 0 0 6 0 0 1 0 0 T , T 0 6 0 , T 2 1 0 , T 5 6 0 1 1 1 0 0 1 1 1 1            = Ι = = =                 G(B2) ≅ Z2×Z2. The Classification and Construction of (6,3)-Arcs: There are 42 points of index zero for B1. Then G(B1) partitions these points into 9 orbits. So, we have nine (6,3)-arc to be constructed by adding one point from each of these nine orbits to B1. We have 47 points of index zero for B2. G(B2) partitions these points into 15 orbits. So we have 15 (6,3)-arcs to be constructed by adding one point from each of these orbits to B2. By definition 6, we have four (6,3)-arcs which are projectivity distinct : C1 = {1,2,9,17,3,10}, C2 = {1,2,9,17,3,11}, C3 = {1,2,9,17,3,26}, C4 = {1,2,9,17,4,28}. G(C1) ≅ S4, G(C2) ≅ Z3, G(C3) ≅ Z2, G(C4) ≅ I. 261 | Mathematics @1a@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@ÚÓ‘Ój�n€a@Î@Úœäñ€a@‚Ï‹»‹€@·rÓ:a@Âig@Ú‹©@Ü‹26@@ÖÜ»€a@I1@‚b«@H2013 Ibn Al-Haitham Jour. for Pure & Appl. Sci. Vol. 26 (1) 2013 The Classification and Construction of (7,3)-Arcs: The groups G(C1), G(C2), G(C3) and G(C4) partition the points of index zero for C1, C2, C3 and C4 into 4, 12, 24, 46 orbits, respectively. By definition 6, we have only six (7,3)-arcs which are projectively distinct. These are: D1 = {1,2,9,17,3,10,16}, D2 = {1,2,9,17,3,11,16}, D3 = {1,2,9,17,3,11,20}, D4 = {1,2,9,17,3,11,26}, D5 = {1,2,9,17,3,26,32}, D6 = {1,2,9,17,4,28,50}. The groups G(D2), G(D3), G(D4) and G(D6) are isomorphic to the identity group, G(D1) ≅ S4 and G(D5) ≅ Z3. We have six distinct (7,3)-arcs, each one of them is incomplete. The Classification and Construction of (8,3)-Arcs: The groups G(D1), G(D2), …, G(D6) partition the points of index zero for D1, D2, …, D6 into 2, 25, 30, 33, 20, 45 orbits, respectively. By definition 6, we have only six (8,3)-arcs which are projectively distinct: E1 = {1,2,9,17,3,10,16,26}, E2 = {1,2,9,17,3,10,16,27}, E3 = {1,2,9,17,3,11,16,26}, E4 = {1,2,9,17,3,11,26,20}, E5 = {1,2,9,17,3,26,32,22}, E6 = {1,2,9,17,4,28,50,31}. The groups G(E3), G(E4), G(E5) and G(E6) are isomorphic to the identity group, G(E1) ≅ Z2 and G(E2) ≅ Z3, we have six distinct (8,3)-arcs, each one of them is incomplete. The Classification and Construction of (9,3)-Arcs: The groups G(E1), G(E2), …, G(E6) partition the points of index zero for E1, E2, …, E6 into 8, 7, 24, 29, 34, 39 orbits, respectively. By definition 6, we have only seven (9,3)-arcs which are projectively distinct, these are: F1 = {1,2,9,17,3,10,16,26,29}, F2 = {1,2,9,17,3,10,16,26,27}, F3 = {1,2,9,17,3,10,16,27,28}, F4 = {1,2,9,17,3,10,16,27,39}, F5 = {1,2,9,17,3,11,26,20,47}, F6 = {1,2,9,17,4,28,50,31,12}, F7 = {1,2,9,17,4,28,50,31,42} The groups G(F3), G(F4), G(F5), G(F6) and G(E7) are isomorphic to the identity group, G(F1) ≅ Z3 and G(F2) ≅ Z2. The Classification and Construction of (10,3)-Arcs: The groups G(F1), G(F2), …, G(F7) partition the points of index zero for F1, F2, …, F7 into 5, 13, 8, 18, 26, 30, 33 orbits, respectively. By definition 6, we have only six (10,3)-arcs which are projectively distinct, these arcs are: H1 = {1,2,9,17,3,10,16,26,29,32}, H2 = {1,2,9,17,3,10,16,26,27,32}, H3 = {1,2,9,17,3,10,16,26,29,39}, H4 = {1,2,9,17,3,10,16,27,28,39}, H5 = {1,2,9,17,4,28,31,50,12,42}, H6 = {1,2,9,17,4,28,31,50,12,42}. The groups G(H1), G(H2), G(H3), G(H4), G(H5) and G(E6) are isomorphic to the identity group. We have only six distinct (10,3)-arcs, each one of them is in complete. The Classification and Construction of (11,3)-Arcs: The groups G(H1), …, G(H6) partition the points of index zero for H1, …, H6 into 5, 8, 11, 14, 20, 22 orbits. By definition 6, we have only six (11,3)-arcs which are projectively distinct, these arcs are: I1 = {1,2,9,17,3,10,16,26,29,32,36}, I2 = {1,2,9,17,3,10,16,26,27,32,36}, I3 = {1,2,9,17,3,10,16,26,27,32,39}, I4 = {1,2,9,17,3,10,27,28,39,36}, I5 = {1,2,9,17,3,10,16,27,28,39,54}, I6 = {1,2,9,17,4,28,31,50,12,42,25}. The groups G(Ii), i = 1,2,…,6 are isomorphic to the identity group. The Classification and Construction of (12,3)-Arcs: 262 | Mathematics @1a@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@ÚÓ‘Ój�n€a@Î@Úœäñ€a@‚Ï‹»‹€@·rÓ:a@Âig@Ú‹©@Ü‹26@@ÖÜ»€a@I1@‚b«@H2013 Ibn Al-Haitham Jour. for Pure & Appl. Sci. Vol. 26 (1) 2013 The groups G(I1), …, G(I6) partition the points of index zero for I1, …, I6 into 2, 4, 4, 10, 10, 16 orbits, respectrively. By definition 6, we have only six (12,3)-arcs which are projectively distinct. These arcs are: L1 = {1,2,9,17,3,10,16,26,29,32,36,54}, L2 = {1,2,9,17,3,10,16,26,29,32,36,46}, L3 = {1,2,9,17,3,10,16,26,27,32,36,46}, L4 = {1,2,9,17,3,10,16,27,28,39,36,43}, L5 = {1,2,9,17,3,10,16,27,28,39,36,46}, L6 = {1,2,9,17,4,28,31,50,12,42,25,55}. The groups G(Li), i = 1,2,…,6 are isomorphic to the identity group. We have six distinct arcs, two of them are complete and the others are incomplete. The Classification and Construction of (13,3)-Arcs: The groups G(L1), …, G(L6) partition the points of index zero for L1, …, L6 into 1, 4, 5, 4, 5, 15 orbits, respectrively. By definition 6, we have only four (13,3)-arcs which are projectively distinct: M1 = {1,2,9,17,3,10,16,26,27,32,36,46,43}, M2 = {1,2,9,17,3,10,16,27,28,39,36,43,34}, M3 = {1,2,9,17,3,10,16,27,28,39,36,46,43}, M4 = {1,2,9,17,4,28,31,50,12,42,25,55,18}. The groups G(M1), G(M2) and G(M4) are isomorphic to the identity group, G(M3) ≅ Z2. M1 and M2 are complete arcs, while M3 and M4 are incomplete arcs. The Classification and Construction of (14,3)-Arcs: The groups G(M3) and G(M4) partition the points of index zero for M3 and M4 into 1 and 2 orbits, respectively. So there exist three (14,3)-arcs to be constructed. By definition 6, we have only one (14,3)-arcs which is incomplete: N = {1,2,9,17,3,10,16,27,28,39,36,46,43,54}. G(N) is isomprphic to the identity group. The Classification and Construction of (15,3)-Arcs: The point 55 is the only point of index zero for N. We construct (15,3)-arc by adding the point 55 to N, then: Q = {1,2,9,17,3,10,16,27,28,39,36,46,43,54,55} is a complete (15,3)-arc. G(Q) is isomorphic to the identity group. References 1. Faiyad, M.S. (2000) Classification and construction of (k,3)Arcs on Projective Plane Over Galois Field GF(7), M.Sc. Thesis, University of Baghdad, Iraq. 2. Kareem, F.F. (2000) Classification and Construction of (k,3)-arcs in PG(2,9), M.Sc. Thesis, University of Baghdad, Iraq. 3. Kwaam, A.A. (1999) Classification and Construction of (k,3)-arcs in PG(2,11), M.Sc. Thesis, University of Baghdad, Iraq. 4. Hirschfeld, J. W. P. (1979) Projective Geometries Over Finite Fields, Second Edition, Oxford University Press. 5. Abood, H.M. (1997) Classification of (k,4)-Arc in PG(2,3), for k ≥ 4, J.Basrah Research, Vol.13, Part 1. 263 | Mathematics @1a@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@ÚÓ‘Ój�n€a@Î@Úœäñ€a@‚Ï‹»‹€@·rÓ:a@Âig@Ú‹©@Ü‹26@@ÖÜ»€a@I1@‚b«@H2013 Ibn Al-Haitham Jour. for Pure & Appl. Sci. Vol. 26 (1) 2013 Table 1 : Points and Lines of PG(2,7) i Pi ℓi 1 1 0 0 2 9 16 23 30 37 44 51 2 0 1 0 1 9 10 11 12 13 14 15 3 1 1 0 8 9 22 28 34 40 46 52 4 2 1 0 5 9 19 29 32 42 45 55 5 3 1 0 4 9 18 27 36 38 47 56 6 4 1 0 7 9 21 26 31 43 48 53 7 5 1 0 6 9 20 24 35 39 50 54 8 6 1 0 3 9 17 25 33 41 49 57 9 0 0 1 1 2 3 4 5 6 7 8 10 1 0 1 2 15 22 29 36 43 50 57 11 2 0 1 2 12 19 26 33 40 47 54 12 3 0 1 2 11 18 25 32 39 46 53 13 4 0 1 2 14 21 28 35 42 49 56 14 5 0 1 2 13 20 27 34 41 48 55 15 6 0 1 2 10 17 24 31 38 45 52 16 0 1 1 1 51 52 53 54 55 56 57 17 1 1 1 8 15 21 27 33 39 45 51 18 2 1 1 5 12 22 25 35 38 48 51 19 3 1 1 4 11 20 29 31 40 49 51 20 4 1 1 7 14 19 24 36 41 46 51 21 5 1 1 6 13 17 28 32 43 47 51 22 6 1 1 3 10 18 26 34 42 50 51 23 0 2 1 1 30 31 32 33 34 35 36 24 1 2 1 7 15 20 25 30 42 47 52 25 2 2 1 8 12 18 24 30 43 49 55 26 3 2 1 6 11 22 26 30 41 45 56 27 4 2 1 5 14 17 27 30 40 50 53 28 5 2 1 3 13 21 29 30 38 46 54 29 6 2 1 4 10 19 28 30 39 48 57 30 0 3 1 1 23 24 25 26 27 28 29 31 1 3 1 6 15 19 23 34 38 49 53 32 2 3 1 4 12 21 23 32 41 50 52 33 3 3 1 8 11 17 23 36 42 48 54 34 4 3 1 3 14 22 23 31 39 47 55 35 5 3 1 7 13 18 23 35 40 45 57 36 6 3 1 5 10 20 23 33 43 46 56 37 0 4 1 1 44 45 46 47 48 49 50 38 1 4 1 5 15 18 28 31 41 44 54 39 2 4 1 7 12 17 29 34 39 44 56 40 3 4 1 3 11 19 27 35 43 44 52 41 4 4 1 8 14 20 26 32 38 44 57 42 5 4 1 4 13 22 24 33 42 44 52 43 6 4 1 6 10 21 25 36 40 44 55 44 0 5 1 1 37 38 39 40 41 42 43 45 1 5 1 4 15 17 26 35 37 46 55 46 2 5 1 3 12 20 28 36 37 45 53 47 3 5 1 5 11 21 24 34 37 47 57 48 4 5 1 6 14 18 29 33 37 48 52 49 5 5 1 8 13 19 25 31 37 50 56 50 6 5 1 7 10 22 27 32 37 49 54 51 0 6 1 1 16 17 18 19 20 21 22 52 1 6 1 3 15 16 24 32 40 48 56 53 2 6 1 6 12 16 27 31 42 46 57 54 3 6 1 7 11 16 28 33 38 50 55 55 4 6 1 4 14 16 25 34 43 45 54 56 5 6 1 5 13 16 26 36 39 49 52 57 6 6 1 8 10 16 29 35 41 47 53 264 | Mathematics @1a@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@ÚÓ‘Ój�n€a@Î@Úœäñ€a@‚Ï‹»‹€@·rÓ:a@Âig@Ú‹©@Ü‹26@@ÖÜ»€a@I1@‚b«@H2013 Ibn Al-Haitham Jour. for Pure & Appl. Sci. Vol. 26 (1) 2013 GF(7)في مستوي إسقاطي حول حقل كالوا (k,3)تصنیف وبناء االقواس عادل محمود أحمد آمال شھاب المختار محمود سالم فیاض قسم علوم الریاضیات / كلیة التربیة للعلوم الصرفة (إبن الھیثم) / جامعة بغداد 2001تموز 11قبل البحث في: ، 2001 یارأ 27استلم البحث في : الخالصة . لقد وجدنا قوسین G(2,7)في المستوي االسقاطي (k,3)الغرض من ھذا البحث ھو لدراسة تصنیف وبناء االقواس –ستة أقواس ،(9,3) –سبعة أقواس (8,3) –، ستة أقواس (7,3) –، ستة أقواس (6,3) –، اربعة اقواس (5,3) – ، كل ھذه االقواس تكون غیر كاملة.(11,3) –، وستة أقواس (10,3) اثنان منھا كاملة وقوس (13,3) –تكون ستة، اثنان منھا تكون كاملة. توجد أربعة أقواس (12,3) –عدد االقواس المختلفة غیر كامل. (14,3) –واحد یكون كامال ً. (15,3) –یوجد قوس واحد ستوي اسقاطي ، حقل كالوا: أقواس ، م الكلمات المفتاحیة 265 | Mathematics Definition 1: [1,2] Definition 2: [1,2] Definition 4: [1,2]: Lemma 1: [4] Definition 5: [4] Lemma 2: [4] Definition 6: [2] Definition 7: [5] Lemma 3: [5] Definition 8: [2] قسم علوم الرياضيات / كلية التربية للعلوم الصرفة (إبن الهيثم) / جامعة بغداد الغرض من هذا البحث هو لدراسة تصنيف وبناء الاقواس (k,3) في المستوي الاسقاطي G(2,7) . لقد وجدنا قوسين – (5,3)، اربعة اقواس – (6,3)، ستة أقواس – (7,3)، ستة أقواس – (8,3) سبعة أقواس – (9,3)، ستة أقواس –(10,3)، وستة أقواس – (11,3)، كل هذه الاقواس ت... عدد الاقواس المختلفة – (12,3) تكون ستة، اثنان منها تكون كاملة. توجد أربعة أقواس – (13,3) اثنان منها كاملة وقوس واحد – (14,3) غير كامل. يوجد قوس واحد – (15,3) يكون كاملا ً.