163 | Mathematics 2015) عام 1العدد ( 28مجلة إبن الھيثم للعلوم الصرفة و التطبيقية المجلد Ibn Al-Haitham J. for Pure & Appl. Sci. Vol. 28 (1) 2015 The Construction of Minimal (b,t)-Blocking Sets Containing Conics in PG(2,5) with the Complete Arcs and Projective Codes Related with Them Amal Shihab Al-Mukhtar Hani Sabbar Thumai Dept. of Mathematics, College of Education for pure science University of Baghdad Received in : 28 September 2014 , Accepted in : 21 December 2014 Abstract A (b,t)-blocking set B in PG(2,q) is set of b points such that every line of PG(2,q) intersects B in at least t points and there is a line intersecting B in exactly t points. In this paper we construct a minimal (b,t)-blocking sets, t = 1,2,3,4,5 in PG(2,5) by using conics to obtain complete arcs and projective codes related with them. Keywords: Blocking set, complete arc, projective code. 164 | Mathematics 2015) عام 1العدد ( 28مجلة إبن الھيثم للعلوم الصرفة و التطبيقية المجلد Ibn Al-Haitham J. for Pure & Appl. Sci. Vol. 28 (1) 2015 1- Introduction Let GF(q) denotes the Galois field of q elements and V(3,q) be the vector space of row vectors of length three with entries in GF(q). Let PG(2,q) be the corresponding projective plane. The points of PG(2,q) are the non zero vectors of V(3,q) with the rule that X = (x1,x2,x3) and y = (x1,x2,x3) represent the same point, where   GF(q)\{0}. The number of points of PG(2,q) is q2 +q + 1. If the point P(X) is the equivalence class of the vector X, then we will say that X is a vector representing P(X). A subspace of dimension one is a set of points all of whose representing vectors form a subspace of dimension two of V(3,q), such subspaces are called lines. The number of lines in PG(3,q) is q2 + q + 1. There are q + 1 points on every line and q + 1 lines through every point. The point X(x1,x2,x3) is on the line Y[y1,y2,y3] if and only if x1y1 + x2y2 + x3y3 = 0. Definition (1.1): [1] A (k,n)–arc is a set of k points of a projective plane such that some n but no n + 1 of them are collinear, n  2. Definition (1.2): [2] A (k,n)–arc is complete if it is not contained in a (k + 1,n)-arc. Definition (1.3): [2] A line l in PG(2,q) is an i-secant on a (k,n)-arc K if ℓ  K = i. Definition (1.4): [2] A point N which is not on a (k,n)-arc has index i if there are exactly i (n-secants) of the arc through N, we denote the number of points N of index i by Ni. Remark (1.5): [3] The (k,n)-arc is complete iff N0 = 0. Thus the arc is complete iff every point of PG(2,q) lies on some n-secant of the arc. Definition (1.6): [3] An (b,t)-blocking set B in PG(2,q) is a set of b points such that every line of PG(2,q) intersects B in at least t points, and there is a line intersecting B in exactly t points. If B contains a line, it is called trivial, thus B is a subset of PG(2,q) which meets every line ℓ in PG(2,q), but contains no line completely; that is t  B  ℓ   q for every line ℓ in PG(2,q). So B is a blocking set iff PG(2,q)\B is a blocking set. A blocking set is minimal if B\{P} is not blocking set for every p in B. Lemma (1.7): [4] A (b,1)-blocking set B is minimal in PG(2,q) iff there is a line ℓ in PG(2,q) such that B  ℓ = {Q} for every Q in B. Definition (1.8): [3] A variety V(F) of PG(2,q) is a subset of PG(2,q) such that: V(F) = {P(A)  PG(2,q)  F(A) = 0}. Definition (1.9): [5] Let Q(2,q) be the set of quadrics in PG(2,q); that is the varieties V(F), where: F = a11 2 1x + a22 2 2x + a33 2 3x + a12x1x2 + a13x1x3 + a23x2x3 ...(1) If V(F) is non-singular, then the quadric is a conic. 165 | Mathematics 2015) عام 1العدد ( 28مجلة إبن الھيثم للعلوم الصرفة و التطبيقية المجلد Ibn Al-Haitham J. for Pure & Appl. Sci. Vol. 28 (1) 2015 That is, if 1312 11 2312 22 13 23 33 aa a 2 2 aa a 2 2 a a a 2 2                   is nonsingular, then the quadric (1) is a conic. 1.10 The Relation Between The Blocking (b,t)-Set and The (k,n)-arc [5] The (k,n)-arc and the (b,t)-blocking set are each complement to the other in the projective plane PG(2,q), that is, n + t = q + 1 and k + b = q2 + q + 1. Thus the complement of the (b,t)- blocking set is the set of points that intersects every line in at most n points which represents the (k,n)-arc. Also finding minimal (b,t)-blocking set is equivalent to finding maximal (k,n)- arc in PG(2,q). Lemma (1.11): [4] Let  = C  ℓ  {P} \ {P1,P2}, where C is a conic, ℓ is a (2-secant) of C such that C  ℓ = {P1,P2}, P is the point of intersection of the two tangents to C at P1 and P2, then  is a minimal (2p – 1,1)-blocking set. Definition (1.12): [5] Let V(n,q) denote the vector space of all ordered n-tuples over GF(q). A linear code C over GF(q) of length n and dimension k is a k-dimensional subspace of V(n,q). The vectors of C are called code words. The Hamming distance between two codewords is defined to be the number of coordinate places in which they differ. The minimum distance of a code is the smallest distances between distinct codewords. Such a code is called an [n,k,d]q code if its minimum hamming distance is d. There exists a relationship between complete (n,r)-arcs in PG(2,q) and [n,3,d]q codes given by the next theorem. Theorem (1.13): [5] There exists a projective [n,3,d]q code if and only if there exists an (n,n – d)-arc in PG(2,q). Theorem (1.14): [6] Let 2 be a double blocking set in PG(2,q): (1) If q < 9, then 2 has at least 3q points. (2) If q = 11, 13, 17 or 19, then 2  (5q + 7)/2. Theorem (1.15): [6] Let 3 be a trible blocking set in PG(2,q): (1) If q = 5, 7,9, then 3 has at least 4q points and if q = 8, then 3 has at least 31 points. (2) If q = 11, 13 or 17, then 3  (7q + 9)/2. Now, we prove the following theorem: Theorem (1.16): A (b,t)-blocking set B is minimal in PG(2,q) then every point P in B there is a t-secant of B containing P. Proof: Suppose B is minimal blocking set, let P be any point in B. Let K be the complement of B, then K is complete (k,n)-arc in PG(2,q) and P is not K., then P is an (n-secant) of K, but q + 1 = t + n and so t = q + 1 – n. Thus P is on an (t-secant) of B. 166 | Mathematics 2015) عام 1العدد ( 28مجلة إبن الھيثم للعلوم الصرفة و التطبيقية المجلد Ibn Al-Haitham J. for Pure & Appl. Sci. Vol. 28 (1) 2015 2- The Projective Plane PG(2,5) In this paper we consider the case q = 5 and the elements of GF(5) are denoted by 0,1,2,3,4. A projective plane  = PG(2,5) over GF(5) consists of 31 points, 31 lines each line contains 6 points and through every point there is 6 lines. Let Pi and ℓi be the points and lines of PG(2,5) respectively. Let i stands for the point Pi, i = 1,2,…,31. The points and lines of PG(2,5) are given in the table (1). 2.1 The Conic in PG(2,5) Through The Reference and Unit Points The general equation of the conic is: 2 2 2 11 1 22 2 33 3 12 1 2 13 1 3 23 2 3a x + a x + a x + a x x + a x x + a x x = 0 …(1) By substituting the reference points: 1(1,0,0), 2(0,1,0), 7(0,0,1) and the unit point 13 (1,1,1), which are four points no three of them are collinear, in (1), we get: a12 + a13 + a23 = 0 and a11 = a22 = a33 = 0, so (1) becomes: 12 1 2 13 1 3 23 2 3a x x + a x x + a x x = 0 …(2) If a12 = 0, then the conic is degenerated, therefore a12  0, similarly, a13  0 and a23  0. Dividing equation (2) by a12, we get: 1 2 1 3 2 3x x + α x x + β x x = 0 ,where 13 23 12 12 a a α = , β = a a , then  = – (1 + ) since 1 +  +  = 0 (mod 5). Then x1 x2+α x1x3−(1+α ) x2x3 =0, where   0 and   4, for if  = 0 or  = 4 we get a degenerated conic, that is,  = 1,2,3. 2.2 The Equations and the Points of the Conics in PG(2,5) Through the Reference and Unit Points For any value of , there is a unique conic contains 6 points, 4 of them are the reference and unit points 1. If  = 1, then the equation of the conic C1 is 1 2 1 3 2 3x x + x x + 3 x x = 0 The points of C1 are : 1,2,7,13,20,26. 2. If  = 2, then the equation of the conic C2 is 1 2 1 3 2 3x x + 2 x x + 2 x x = 0 The points of C2 are : 1,2,7,13,21,29. 3. If  = 3, then the equation of the conic C3 is 1 2 1 3 2 3x x + 3x x + x x = 0 The points of C3 are : 1,2,7,13,24,30. Thus we found five conics two of them are degenerated and the remaining three conics C1, C3, C3 are non-degenerated. Table (1) i Pi Li 1 1 0 0 2 7 12 17 22 27 2 0 1 0 1 7 8 9 10 11 3 1 1 0 6 7 16 20 24 28 4 2 1 0 4 7 14 21 23 30 5 3 1 0 5 7 15 18 26 29 6 4 1 0 3 7 13 19 25 31 7 0 0 1 1 2 3 4 5 6 8 1 0 1 2 11 16 21 26 31 167 | Mathematics 2015) عام 1العدد ( 28مجلة إبن الھيثم للعلوم الصرفة و التطبيقية المجلد Ibn Al-Haitham J. for Pure & Appl. Sci. Vol. 28 (1) 2015 9 2 0 1 2 9 14 19 24 29 10 3 0 1 2 10 15 20 25 30 11 4 0 1 2 8 13 18 23 28 12 0 1 1 1 27 28 29 30 31 13 1 1 1 6 11 15 19 23 27 14 2 1 1 4 9 16 18 25 27 15 3 1 1 5 10 13 21 24 27 16 4 1 1 3 8 14 20 26 27 17 0 2 1 1 17 18 19 20 21 18 1 2 1 5 11 14 17 25 28 19 2 2 1 6 9 13 17 26 30 20 3 2 1 3 10 16 17 23 29 21 4 2 1 4 8 15 17 24 31 22 0 3 1 1 22 23 24 25 26 23 1 3 1 4 11 13 20 22 29 24 2 3 1 3 9 15 21 22 28 25 3 3 1 6 10 14 18 22 31 26 4 3 1 5 8 16 19 22 30 27 0 4 1 1 12 13 14 15 16 28 1 4 1 3 11 12 18 24 30 29 2 4 1 5 9 12 20 23 31 30 3 4 1 4 10 12 19 26 28 31 4 4 1 6 8 12 21 25 29 2.3 The Construction of Minimal (b,t)-Blocking Sets By Using Conic-Type Blocking Sets We construct minimal (b,t)-blocking set in PG(2,5) from the minimal blocking (9,1)-sets of lemma (1.15) by using conic. 2.3.1 The Construction of Minimal (9,1)-Blocking Set by Lemma (1.11) We take the conic C1 in section 2. Let 1 = C1  L1 \ {P1,P2}  {P}, C1 ={1,2,7,13,20,26}, L1 = {2,7,12,17,22,27}, C1  L1 = {2,7}, L4 and L9 are the two tangents to C1 at the points 7 and 2 respectively. L4  L9 = {14}, then 1 = {1,12,13,14,17,20,22,26,27}, 1 is a (9,1)-blocking set in PG(2,5). Since each point of 1 is on line ℓ in PG(2,9) such that 1  ℓ = {P} (lemma 1.7), 1 satisfies the following conditions: (a) 1 intersects every line in PG(2,5) in at least one point. (b) Every point in 1, there is a line ℓ in PG(2,5) such that 1  ℓ = {P}. The complement of 1 is the complete (22,5)-arc K5, by theorem (1.13) there exists a projective [22,3,17] code. 2.3.2 The Construction of Minimal (b,2)-Blocking Set In PG(2,5) We construct two (9,1)-blocking sets. Let 1 = {1,12,13,14,17,20,22,26,27} be the minimal (9,1)-blocking set of section (2.3.1). We construct another (9,1)-blocking set 1 = C2  L8 \ {C2  L8}  {15}, where C2 = {1,2,7,13,21,29}, L8 = {2,11,16,21,26,31}, C2  L8 = {2,21}, L10  L24 = {15} and L10 and L24 are tangents to C2 at the points 2 and 21 respectively. 1 = {1,7,11,13,15,16,26,29,31} is (9,1)-blocking set. Now, we construct (b,2)-blocking set as follows: 168 | Mathematics 2015) عام 1العدد ( 28مجلة إبن الھيثم للعلوم الصرفة و التطبيقية المجلد Ibn Al-Haitham J. for Pure & Appl. Sci. Vol. 28 (1) 2015 Let A = 1  1 = {1,7,11,12,13,14,15,16,17,20,22,26,27,29,31}. A must satisfies the following conditions: (a) A intersects every line of PG(2,5) in at least two points. (b) Every point in A is on at least one 2-secant of A. We add three points 3,10 and 18 to A and eliminate the points 15 and 26 from A to satisfy these conditions, then: 2 = A{3,10,18}\ {15,26} = {1,3,7,10,11,12,13,14,16,17,18,20,22,27,29,31} is a minimal (16,2)-blocking set. The complement of 2 is the complete (15,4)-arc K4. By theorem (1.13) there exists a projective [15,3,11] code. 2.3.3 The Construction of Minimal (b,3)-Blocking Set In PG(2,5) We take the (9,1)-blocking sets in section (2.3.2) 1 = {1,7,11,13,15,16,26,29,31}, 1 = {1,12,13,14,17,20,22,26,27}, Let 1 = C3  L28  {8} \ {C3  L28}, C3 ={1,2,7,13,24,30}, L28 = {3,11,12,18,24,30}, C3  L28 = {24,30} and L21  L26 = {8}, where L21 and L26 are tangents to C3 at the points 24 and 30 respectively. 1 = {1,2,3,7,8,11,12,13,18} is a minimal (9,1)-blocking set. We must construct a minimal (b,3)-blocking set from 1, 1 and 1 as follows:. Let B= 1  1  1 ={1,2,3,7,8,11,12,13,14,15,16,17,18,20,22,26,27,29,31}. B must satisfy the following conditions: (a) B intersects every line in PG(2,5) in at least three points. (b) Every point in B is on at least one 3-secant of B. We add two points 4 and 5 to B and eliminate the point 31 from B to satisfy these conditions, then: 3 = B{4,5}\ {31} = {1,2,3,4,5,7,8,11,12,13,14,15,16,17,18,20,22,26,27,29} is a minimal (20,3)-blocking set which is trivial since 3 contains some lines completely. The complement of 3 is the complete (11,3)-arc K3. By theorem (1.13) there exists a projective [11,3,8] code in PG(2,5). 2.3.4 The Construction of Minimal (b,4)-Blocking Set In PG(2,5) We take three minimal (9,1)-blocking sets in section (2.3.3) which are: 1 = {1,7,11,13,15,16,26,29,31}, 1 = {1,12,13,14,17,20,22,26,27}, 1 = {1,2,3,7,8,11,12,13,18}. Let 1 = C1  L2{30}\ {C1  L2}, where C1 is the conic C1 = {1,2,7,13,20,26}, L2 = {1,7,8,9,10,11}, C1  L2 = {1,7}, L4  L12 = {30}, L4 and L12 are tangents to C1 at the points 7 and 1 respectively, then. 1 = {2,8,9,10,11,13,20,26,30} is a minimal (9,1)-blocking set. We construct a minimal (b,4)-blocking set from 1, 1,  1 and 1 as follows:. Let C = 1  1  1  1 = {1,2,3,7,…,14,15,16,17,18,20,22, 26,27,29,30,31}. C must satisfy the following conditions: (a) C intersects every line in at least four points. (b) Every point in C is on at least one 4-secant of C. We add the points 6,45,21,24,28 to C, and eliminate one point 29 from C to satisfy these conditions, then: 4=C{6,21,24,28}\{29}={1,2,3,6,7,…,18,20,21,22,24,26,27,28,30,31} is a minimal (25,4)- blocking set which is trivial since 4 contains some lines completely. The complement of 4 is the complete (6,2)-arc K2. By theorem (1.13) there exists a projective [6,3,4] code. 2.3.5 The Construction of Minimal (b,5)-Blocking Set In PG(2,5) We take four minimal (9,1)-blocking sets of section (2.3.4) which are 1 = {1,7,11,13,15,16,26,29,31}, 1 = {1,12,13,14,17,20,22,26,27}, 1 = {1,2,3,7,8,11,12,13,18}, 1 = {2,8,9,10,11,13,20,26,30}. We construct another minimal (9,1)-blocking set. 169 | Mathematics 2015) عام 1العدد ( 28مجلة إبن الھيثم للعلوم الصرفة و التطبيقية المجلد Ibn Al-Haitham J. for Pure & Appl. Sci. Vol. 28 (1) 2015 Let 1 = C2  L6 \ {7,13}  {24}, where C2 is a conic, C2 = {1,2,7,13,21,29}, L6 = {3,7,13,19,25,31}, C2  L6 = {7,13}, L3  L22 = {24}, where L3 and L22 are tangents to C2 at the points 7 and 13 respectively, then. 1 = {1,2,3,19,21,24,25,29,31} is a minimal (9,1)-blocking set. Now, we must construct a minimal (b,5)-blocking set from 1, 1,  1, 1 and 1 as follows:. Let D=111 1 1={1,2,3,7,…,22, 24,…,27,29,30,31}. D must satisfy the following conditions: (a) D intersects every line in at least five points. (b) Every point of D is on at least one 5-secant of D. We add four points 5,6,23,28 to D to satisfy these conditions, then: 5 = D  {5,6,23,28} = {1,2,3,5,…,31} is a minimal (30,5)-blocking set which is trivial since 5 contains some lines completely. The complement of 5 is not arc since every (k,n) cannot exist when n < 2. Conclusion 1. We construct a minimal (9,1)-blocking set, which is containing a conic as in lemma (1.12). Also we construct minimal (16,2)-blocking by taking the union of two blocking (9,1)-sets of type in lemma (1.12). We construct minimal (20,3)-blocking set, by taking the union of three (9,1)- blocking sets of type in lemma (1.12). We construct minimal (25,4)-blocking set by taking the union of four (9,1)-blocking sets of type in lemma (1.12) and finally we construct minimal (30,5)-blocking set B5 by taking the union five (9,1)-blocking sets of type in lemma (1.12). 2. The minimal (9,1)-blocking set B1 and the minimal (16,2)-blocking set B2 are non-trivial, but the minimal (20,3)-blocking set B3, the minimal (25,4)-blocking set B4 and the minimal (30,5)-blocking set B5 are trivial References 1. Al-Mukhtar, A.S.,Ahmed,A.M. and Faiyadh, M.S., (2013), The Construction of (k,3)-arcs on Projective Plane Over Galois Field GF(7),Ibn-Al-Haitham Journal For Pure and Applied Science,(26),(2),259-265. 2. Al-Mukhtar, A.S., Ahmed, and Kareem, F.F., (2013), The Construction of (k,3)-arcs in PG(2,9) by using Geometric Method,Ibn-Al-Haitham Journal For Pure and Applied Science,(26),(2),239-248. 3. Hassan, U.A., (2013), The Reverse Construction for the Complete Arcs in the Projective Plane PG(2,p) Over Galois Field GF(p) by Using Geometric Methods, M.Sc. Thesis, University of Baghdad, Iraq. 4. Hirschfeld, J. W. P., (1998), Projective Geometries Over Finite Fields, Second Edition, Oxford University Press. 5. Rumen Daskalov, (2008), A Geometric Construction of (38,2)-Bloking Set in PG(2,13) and the Related [145,3,133]13 Cod, Discrete Mathematics Technical University of Gabrovo, Bulgaria, 308 (1341-1345). 6. Ball, S., (1995), Multiple Blocking Sets and arcs in Finite Plane, School of Mathematical and Physical Sciences, University of Sussex, Brighton BN, QH,U,K,1-16. 170 | Mathematics 2015) عام 1العدد ( 28مجلة إبن الھيثم للعلوم الصرفة و التطبيقية المجلد Ibn Al-Haitham J. for Pure & Appl. Sci. Vol. 28 (1) 2015 PG(2,5)صغرى تحتوي على مخروطيات في (b,t)–بناء مجموعات قالبية واالقواس الكاملة والشفرات االسقاطية المرتبطة بھا آمال شھاب المختار ثميل ھاني صبار ، جامعة بغداد كلية التربية للعلوم الصرفة، قسم الرياضيات 2014الول اكانون 21 , قبل البحث في :2014ايلول 28أستلم البحث في : الخالصه يقطع PG(2,q)من النقاط بحيث ان كل مستقيم في bھي مجموعة من PG(2,q)في B(b,t) - المجموعة القالبية B فيt من النقاط في االقل ويوجد مستقيم يقطعB فيt .من النقاط فقط ، باعتماد مخروطيات وحصلنا PG(2,5) ،t = 1,2,3,4,5صغرى في (b,t) –بية في ھذا البحث قمنا ببناء مجموعات قال على أقواس كاملة وشفرات إسقاطية مرتبطة بھا. مجموعة قالبية ، قوس كامل ، شفرة إسقاطية. الكلمات المفتاحية :