The Construction of Complete (kn,n)-Arcs in The Projective Plane PG(2,11) by Geometric Method, with the Related Blocking Sets and Projective Codes Amal SH. Al-Mukhtar Umniyat A. Hassan Dept. of Mathematics/College of Education for Pure Science(Ibn Al-Haitham) /University of Baghdad Received in :17June 2013 , Accepted in: 10October 2013 Abstract In this paper,we construct complete (kn,n)-arcs in the projective plane PG(2,11), n = 2,3,…,10,11 by geometric method, with the related blocking sets and projective codes. Keywords: Complete arcs, blocking set, projective cod. 314 | Mathematics @a@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@ÚÓ‘Ój�n€a@Î@Úœäñ€a@‚Ï‹»‹€@·rÓ:a@Âig@Ú‹©@Ü‹127@@ÖÜ»€a@I1@‚b«@H2014 Ibn Al-Haitham Jour. for Pure & Appl. Sci. Vol. 27 (1) 2014 Introduction A (k,n)-arc is a set of k points of PG(2,q) for some n, but not n + 1 of them, are collinear. A (k,n)-arc is complete if it is not contained in a (k + 1,n)-arc. Let PG(2,q) be the projective plane over Galois field GF(q). The points of PG(2,q) are the non-zero vectors of the vector space V(3,q) with the rule that X(x1,x2,x3) and Y(λx1,λx2,λx3) are the same point, where λ∈ GF(q)\{0}. Similarly, x[x1,x2,x3] and y[(λx1,λx2,λx3] are the same line, where λ∈ GF(q)\{0}. The point X(x1,x2,x3) is on the line Y[y1,y2,y3] if and only if x1y1 + x2y2 + x3y3 = 0. In PG(2,q), there are q2 + q + 1 points and q2 + q + 1 lines, every line contains exactly q + 1 points and every point is on exactly q + 1 lines. Many researcher worked on the construction and classification of the(k,n)-arcs in projective planes PG(2,q),Hirschfeld showed the construction and classification of (k,2)-arcs in PG(2,q),q≤9,Brune showed the relation between the (k,n)-arc and the blocking (b, t) set, 1. Definition: [1] A (kn ,n)-arc K is in PG(2,q) is a set of kn points such that some lines of the plane meet K in n points but no line meets K in more than n points, where n ≥ 2. 2. Definition:[2] A (k,n)-arc is complete if it is not contained in a (k+1,n)-arc. The maximum number of points that (k,2)-arc can have is m(2,q) and this arc is an oval. 3. Theorem:[3] In PG(2,q), q 1 for q odd m(2, q) q 2 for q even + =  + 4. Definition:[1] A line ℓ in PG(2,q) is an i-secant of a (k,n)-arc K if ℓ∩K = i. 5. Definition:[1] 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}. 6. Definition:[1] Let Q(2,q) be the set quadrics in PG(2,q), that is the varieties V(F), where: 2 2 2 11 1 22 2 33 3 12 1 2 13 1 3 23 2 3F a x a x a x a x x a x x a x x= + + + + + …(1) If V(F) is non-singular, then the quadric is a conic, that is, if 1312 11 2312 22 13 23 33 2 2 2 2 2 2 aa a aa a a a a        Α =          Is non singular, then the quadric (1) is a conic. 7. Theorem:[3] In PG(2,q), with q odd, every oval is a conic. 315 | Mathematics @a@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@ÚÓ‘Ój�n€a@Î@Úœäñ€a@‚Ï‹»‹€@·rÓ:a@Âig@Ú‹©@Ü‹127@@ÖÜ»€a@I1@‚b«@H2014 Ibn Al-Haitham Jour. for Pure & Appl. Sci. Vol. 27 (1) 2014 8. Definition:[3] A point N which is not on a (k,n)-arc has index i if there exactly i(n-secants) of the arc through N, the number of the points N of index i is denoted by Ni. 9. Remark:[3] The (k,n)-arc is complete if and only if N0 = 0. Thus the arc is complete if and only if every point of PG(2,q) not on the arc lies on some n-secant of the arc. 10. Definition:[4] An (b,t)-blocking set β in PG(2,q) is a set of b points such that every line of PG(2,q) intersects β in at least t points, and there is a line intersecting β in exactly t points. If β contains a line, it is called trivial, thus β is a subset of PG(2,q) which meets every line but contains no line completely; that is t ≤β ∩ ℓ≤ q for every line ℓ in PG(2,q). So β is a blocking set if and only if PG(2,q)\β is also blocking set. We may note that a blocking set is merely a (k,n)-arc with n ≤ q and no 0-secants. A blocking set β is minimal if β \{p} is not blocking set for every p ∈ β. 11. The Relation Between the Blocking (b,t)-set and the (k,n)-arc: [4] The (k,n)-arcs and the (b,t)-blocking sets 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 find maximal (k,n)-arc in PG(2,q).Blocking sets were studied in details by Di Paola who determined the minimum size of non-trivial blocking set in PG(2,q), q≤ 9 . 12. Definition:[3] In PG(2,q), let β contains a line ℓ minus a point P plus a set of q points one on each of the q lines through P other than ℓ but not all collinear; then β is minimal (2q,1)-blocking set. Blocking sets of this kind are called rédéi-type studied by [Bruen, A.A. and Thas, J.A. (1977)] and in [Blockhuis, A.A. and Brouwer, E. and S.Z. “onyi, T. (1995)]. 13. Definition:[5,6] 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 codewords. 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 distance 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. 14. Theorem:[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). The Projective Plane PG(2,11) In this paper,we consider the case q = 11 and the elements of GF(11) which are denoted by 0,1,2,3,4,5,6,7,8,9,10 316 | Mathematics @a@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@ÚÓ‘Ój�n€a@Î@Úœäñ€a@‚Ï‹»‹€@·rÓ:a@Âig@Ú‹©@Ü‹127@@ÖÜ»€a@I1@‚b«@H2014 Ibn Al-Haitham Jour. for Pure & Appl. Sci. Vol. 27 (1) 2014 317 | Mathematics @a@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@ÚÓ‘Ój�n€a@Î@Úœäñ€a@‚Ï‹»‹€@·rÓ:a@Âig@Ú‹©@Ü‹127@@ÖÜ»€a@I1@‚b«@H2014 Ibn Al-Haitham Jour. for Pure & Appl. Sci. Vol. 27 (1) 2014 A projective plane π = PG(2,11) over GF(11) consists of 133 points, 133 lines, each line contains 12 points and through each point there are 12 lines. Let Pi and Li be the points and lines of PG(2,11), respectively. i=1,2,……….,133 . Let i stands for the point Pi , and for the line Li whose coordinates are the same coordinates of the point Pi and by using computer program (2-pl- program) [7] the points and the lines of PG(2,11) are given in table (1). 1. The Constructions of (k,n)-Arcs in PG(2,11) Let A={1,2,13,25}be the set of reference and unit points in π=PG(2,11) , where 1≡(1,0,0) , 2≡(0,1,0) , 13≡(0,0,1) , 25≡(1,1,1) A is a(4,2)-arc since no three points of A are collinear , the points of A are the vertices of a quadrangle whose sides are the lines: L1 = [2,13] ={ 2,13,24,35,46,57,68,79,90,101,112,123} L2 = [1,13] ={ 1,13,14,15,16,17,18,19,20,21,22,23} L3=[1,2]= {1,2,3,4,5,6,7,8,9,10,11,12} L4 = [1,25] ={1,24,25,26,27,28,29,30,31,32,33,34} L5 = [2,25] = {2,14,25,36,47,58,69,80,91,102,113,124} L6 = [13,25] = {3,13,25,37,49,61,73,85,97,109,121,133} The diagonal points of A are the points 3,14,24 where. L1∩L4=24 L3∩L6=3 L2∩L5=14 Which are the intersections of the pairs of the opposite sides. Then there are 61 points on the sides of the quadrangle,four of them are the points on the arc A, and three of them are the diagonal points of A. So there are 72 points not on the sides of the quadrangle which are the points of index zero for A, these points are: {38,39,40,41,42,43,44,45,48,50,51,52,53,54,55,56,59,60,62,63,64,65,66,67,70,71,72,74,75,76 ,77, 78,81,82,83,84,86,87,88,89,92,93,94,95,96,98,99,100,103,104,105,106,107,108,110,111,114, 115, 116,117,118,119,120,122,125,126,127,128,129,130,131,132} Hence A is incomplete(4, 2) – arc 2. The Conics in PG (2,11) Through the Reference and Unit Points The general equation of the conic is a11x12 +a22x22 +a33x32+a12x1x2+a13x1x3+a23x2x3 = 0 ...(1) By substituting the points of A in (1), we get: a11=0 , a22=0 , a33=0 a12+a13+a23=0 so the equation (1)becomes: a12x1x2+a13x1x3+a23x2x3 = 0 …(2) If a12=0 then the conic is degenerated . therefore a12≠ 0 similarly a13 ≠ 0 and a23≠ 0 since a12 ≠ 0 we divide equation (2) by a12,we get x1x2 + α x1x3+ βx2x3= 0 …(3) where 13 23 12 12 , a a a a α = β = since 1+α+β = 0 (mod 11) →β = − (1+α) then (3) can be written as x1x2+ αx1x3− (1+α) x2x3= 0…(4) where α ≠ 0 and α ≠ 10 for if α =0 or α=10, we get degenerated conic , that is α = 1,2,3,4,5,6,7,8,9 3. The Equations and the Points of the Conics of PG(2,11 ) Through the Reference and Unite Points. For any value of α , there is a unique conic contains 12 points ,four of them are the reference and unit points 1. If α =1 then the equation of the conic C1 is x1x2+ x1x3 + 9x2x3= 0 The points of C1 are{1,2,13,25,40,53,63,77,87,100,104,116} 2. If α=2 then the equation of the conic C2 is x1x2 + 2x1x3 + 8x2x3= 0 The points of C2 are{1,2,13,25,42,50,59,78,84,96,110,131} 3.If α =3 then the equation of the conic C3 is x1x2+ 3x1x3 + 7x2x3= 0 and the points of C3 are{1,2,13,25,41,48,64,76,89,95,115,132} 4.If α =4 then the equation of the conic C4is x1x2 + 4x1x3 + 6x2x3 = 0 The points of C4 are {1,2,13,25,44,56,65,72,82,108,118,125} 5.If α =5 then the equation of the conic C5 is x1x2+ 5x1x3 + 5x2x3 = 0 The points of C5 are {1,2,13,25,43,51,67,71,99,103,119,127} 6. If α =6 then the equation of the conic C6 is x1x2+ 6 x1x3+ 4x2x3= 0 The points of C6 are {1,2,13,25,45,52,62,88,98,105,114,126} 7.If α =7 then the equation of the conic C7 is x1x2+ 7x1x3+ 3x2x3 = 0 The points of C7 are{1,2,13,25,38,55,75,81,94,106,122,129} 8.If α =8 then the equation of C8 is x1x2+ 8 x1x3+ 2x2x3 =0 The points of C8 are{1,2,13,25,39,60,74,86,92,111,120,128} 9.If α =9 then the equation of C9 is x1x2+ 9x1x3 + x2x3=0 and the points of C9 are{1,2,13,25,54,66,70,83,93,107,117,130} Thus we found eleven conics, two of them are degenerated and the remaining nine conics C1,C2,C3,C4,C5,C6,C7,C8 and C9 are non –degenerated ,which are complete (k,2)-arcs. 318 | Mathematics @a@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@ÚÓ‘Ój�n€a@Î@Úœäñ€a@‚Ï‹»‹€@·rÓ:a@Âig@Ú‹©@Ü‹127@@ÖÜ»€a@I1@‚b«@H2014 Ibn Al-Haitham Jour. for Pure & Appl. Sci. Vol. 27 (1) 2014 4. The Construction of Complete (kn,n)-Arcs in PG(2,11) and the Related Blocking Setsand Projective Codes. The complete (k,n)-arcs in PG(2,11) can be constructed by eliminating the conics given above from PG(2,11) as follows: a. The Construction of Complete (k11,11)-Arc . Let π = PG(2,11) We take a conic say C1 , where C1 ={1,2,13,25,40,53,63,77,87,100,104,116} Let K = π-C1 ={3,4,5,6,7,8,9,10,11,12,14,15,16,17,18,19,20,21,22,23,24,26,27,28,29,30,31,32,33,34,35,36, 37,38,39,41,42,43,44,45,46,47,48,49,50,51,52,54,55,56,57,58,59,60,61,62,64,65,66,67,68,69, 70,71,72,73,74,75,76,78,79,80,81,82,83,84,85,86,88,89,90,91,92,93,94,95,96,97,98,99,101,10 2, 103,105,106,107,108,109,110,111,112,113,114,115,117,118,119,120,121,122,123,124,125,12 6, 127,128,129,130,131,132,133} This points in table(3.1) represent the elements of the projective plane PG(2,11)=π ,some points of π are deleted to satisfy the definition of the arc which is mentioned in definition 1.1 ,and some of these deleted points are of index zero therefore we must add them to the remained points in π to make the arc K11 complete as mentioned in remark 1.9 . The construction of complete (k11,11) –arc must satisfy the following : 1.Any line of π must intersect the arc in at most 11 points . 2. Every point not in the arc is on at least one (11-secant) of the arc. We eliminate twenty points from K which are : 7,8,10,12,17,24,26,27,28,31,35,46,47,57,68,79,90,101,112,124 to satisfy (1) since the points {1,116,104,100,87,77,63,53,40,25} are of index zero therefore we add them to K to satisfy (2) Then K11={1,3,4,5,6,9,11,14,15,16,18,19,20,21,22,23,25,29,30,32,33,34,36,37,38,39,40,41,42,43,4 4,45,48,49,50,51,52,53,54,55,56,58,59,60,61,62,63,64,65,66,67,69,70,71,72,73,74,75,76,77,7 8,80,81,82,83,84,85,86,87,88,89,91,92,93,94,95,96,97,98,99,100,102,103,104,105,106,107,10 8,109,110,111,113,114,115,116,117,118,119,120,121,122,123,125,126,127,128,129,130,131, 132,133}. K11 is a complete (111,11)-arc . By theorem 1.14, there exists a projective [111,3,100] code which is equivalent to the complete (111,11)-arc K11 Let β1= π – K11 β1 = {2,7,8,10,12,13,17,24,26,27,28,31,35,46,47,57,68,79,90,101,112,124} β1 is (22,1)–blocking set of size (2q)which is of Redei-type ,(figure 1) contains the line L1,where L1\{123} = {2,13,24,35,46,57,68,79,90,101,112} and one point on each Line through the point 123 other than L1 which are non-collinear points: 7,8,10,12,17,26,27,28,31,47,124. Note that each line in π intersect β1 in at least one point . b. The Construction of Complete (k10,10)-Arc Let π =PG(2,11) In this section we construct (k10,10)-arc by eliminating the union of two conics, say C1 and C2where C1= {1,2,13,25,40,53,63,77,87,100,104,116} 319 | Mathematics @a@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@ÚÓ‘Ój�n€a@Î@Úœäñ€a@‚Ï‹»‹€@·rÓ:a@Âig@Ú‹©@Ü‹127@@ÖÜ»€a@I1@‚b«@H2014 Ibn Al-Haitham Jour. for Pure & Appl. Sci. Vol. 27 (1) 2014 C2= {1,2,13,25,42,50,59,78,84,96,110,131} Let K = π –C1∪C2 ={3,4,5,6,7,8,9,10,11,12,14,15,16,17,18,19,20,21,22,23,24,26,27,28,29,30,31,32,33,34,35,36, 37,38, 39,41,43,44,45,46,47,48,49,51,52,54,55,56,57,58,60,61,62,64,65,66,67,68,69,70,71,72,73,74, 75,76, 79,80,81,82,83,85,86,88,89,90,91,92,93,94,95,97,98,99,101,102,103,105,106,107,108,109,11 1,112,113,114,115,117,118,119,120,121,122,123,124,125,126,127,128,129,130,132,133} The construction of a complete (k10,10)-arc must satisfy the following : 1.Any line of π intersects K in at most 10 points . 2. Every point not in K is on at least one 10 -secant . we have to eliminate twenty points of K which are 5,7,21,26,35,46,47,54,57,68,79,83,90,99,101,105,112,115,123,126 to satisfy(1). there are six points of index zero which are 1,2,50,77,87,116 therefore we add them to K to satisfy (2) then K10={1,2,3,4,6,8,9,10,11,12,14,15,16,17,18,19,20,22,23,24,27,28,29,30,31,32,33,34,36,37,38, 39,41,43,44,45,48,49,50,51,52,55,56,58,60,61,62,64,65,66,67,69,70,71,72,73,74,75,76,77,80, 81,82,85,8687,88,89,91,92,93,94,95,97,98,102,103,106,107,108,109,111,113,114,116,117,11 8,119,120,121,122, 124,125, 127,128,129,130,132,133} then K10 is a complete (99,10) –arc By theorem 1.14, there exists a projective [99,3,89] code which is equivalent to the complete (99,10) –arc K10 Let β2 = π – K10 ={5,7,13,21,25,26,35,40,42,46,47,53,54,57,59,63,68,78,79,83,84,90,96,99,100,101,104,105,1 10, 112,115,123,126,131} β2 is (34,2)-blocking set. Note that each line in π intersects β2 in at least two points . c. The Construction of Complete (k9,9)-Arc In this section, we eliminate the union of three conics, say C1,C2, and C3 where C1= {1,2,13,25,40,53,63,77,87,100,104,116} C2= {1,2,13,25,42,50,59,78,84,96,110,131} C3= {1,2,13,25,41,48,64,76,89,95,115,132} LetK= π –C1∪C2∪C3 ={3,4,5,6,7,8,9,10,11,12,14,15,16,17,18,19,20,21,22,23,24,26,27,28,29,30,31,32,33,34,35,36, 37,38,39,43,44,45,46,47,49,51,52,54,55,56,57,58,60,61,62,65,66,67,68,69,70,71,72,73,74,75, 79,80,81,82,83,85,86,88,90,91,92,93,94,97,98,99,101,102,103,105,106,107,108,109,111,112, ,113,114,117,118,119,120,121,122,123,124,125,126,127,128,129,130,133}. The construction of a complete (k9,9)-arc must satisfy the following : 1. Any line in π intersects K in at most 9 point. 2. Every point not on K is on at least one (9-secant). we eliminate twenty five points of K which are 4,23,24,27,28,33,35,46,52,58,60,61,68,69,74,79,88,90,93,101,106,112,123,126,129, to satisfy (1) we add the points50,58 which are of index zero to K to satisfy (2), then K9={3,5,6,7,8,9,10,11,12,14,15,16,17,18,19,20,21,22,26,29,30,31,32,34,36,37,38,39,43,44,45 ,47,49,50,51,54,55,56,57,58,62,65,66,67,70,71,72,73,75,80,81,82,83,85,86,91,92,94,97,98,99, 102,103, 105,107,108,109,111,113,114,117,118,119,120,121,122,124,125,127,128,130,133} Then K9 is a complete (82,9)-arc . By theorem 1.14, there exists a projective [82,3,73] code which is equivalent to the complete (82,9)-arc K9. Let β3=π−K9 320 | Mathematics @a@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@ÚÓ‘Ój�n€a@Î@Úœäñ€a@‚Ï‹»‹€@·rÓ:a@Âig@Ú‹©@Ü‹127@@ÖÜ»€a@I1@‚b«@H2014 Ibn Al-Haitham Jour. for Pure & Appl. Sci. Vol. 27 (1) 2014 ={1,2,4,13,23,24,25,27,28,33,35,40,41,42,46,48,52,53,59,60,61,63,64,68,69,74,76,77,78,79,8 4,87, 88,89,90,93,95,96,100,101,104,106,110,112,115,116,123,126,129,131,132} Then β3 is a(51,3)-blocking set ,note that each line intersects β3 in at least three points . d. The Construction of Complete (k8,8)-Arc In this section ,we take the union of four conics, say C1, C2, C3 ,and C4 where C1={1,2,13,25,40,53,63,77,87,100,104,116} C2={1,2,13,25,42,50,59,78,84,96,110,131} C3={1,2,13,25, 41,48,64,76,89,95,115,132} C4={1,2,13,25,44,56,65,72,82,108,118,125} Let K=π−C1∪C2∪C3∪C4 ={3,4,5,6,7,8,9,10,11,12,14,15,16,17,18,19,20,21,22,23,24,26,27,28,29,30,31,32,33,34,35,36, 37,38,39,43,45,46,47,49,51,52,54,55,57,58,60,61,62,66,67.68,69,70,71,73,74,75,79,80,81,83, ,85,86,88,90,91,92,93,94,97,98,99,101,102,103,105,106,107,109,111,112,113,114,117,119,12 0,121,122,123,124,126,127,128,129,130,133}. The construction of complete(k8,8)-arc must satisfy the following: 1.Any line in π intersects K in at most 8 points. 2. Every point not in K is on at least one (8-secant). we eliminate thirty points from K which are 3,5,6,7,17,23,24,27,28,29,35,36,45,46,57,61,68,70,86,88,90,97,101,105,112,113,114,123,126, 129 to satisfy (1). K is incomplete arc since there are points of index zero which are 7,44 we add them to K to satisfy(2),then K8={4,7,8,9,10,11,12,14,15,16,18,19,20,21,22,26,30,31,32,33,34,37,38,39,43,44,47,49,51,52, 54,55,58,60,62,66,67,69,71,73,74,75,79,80,81,83,85,91,92,93,94,98,99,102,103,106,107,109, 111,117,119120,121,122,124,127,128,130,133} K8 is a complete(69,8)-arc . By theorem 1.14 there exists a projective [69,3,61]code which is equivalent to the complete(69,8)-arc K8 Let β4=π−K8 ={1,2,3,5,6,13,17,23,24,25,27,28,29,35,36,40,41,42,45,46,48,50,53,56,57,59,61,63,64,65,68, 70,72, 76,77,78,82,84,86,87,88,89,90,95,96,97,100,101,104,105,108,110,112,113,114,115,116,118,1 23, 125,126,129,131,132} Then β4 is a(64,4)-blocking set .Note that each line in π intersects β4 in at least four points . e. The Construction of Complete (k7,7)-Arc In this section ,we take the union of five conics, say C1, C2, C3, C4, and C5where C1={1,2,13,25,40,53,63,77,87,100,104,116} C2={1,2,13,25,42,50,59,78,84,96,110,131} C3={1,2,13,25,41,48,64,76,89,95,115,132} C4={1,2,13,25,44,56,65,72,82,108,118,125} C5={1,2,13,25,43,51,67,71,99,103,119,127} Let K=π−C1∪C2∪C3∪C4∪C5 ={3,4,5,6,7,8,9,10,11,12,14,15,16,17,18,19,20,21,22,23,24,26,27,28,29,30,31,32,33,34,35,36, 37,38 ,39,45,46,47,49,52,54,55,57,58,60,61,62,66,68,69,70,73,74,75,79,80,81,83,85,86,88,90,91,92, 93,94 321 | Mathematics @a@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@ÚÓ‘Ój�n€a@Î@Úœäñ€a@‚Ï‹»‹€@·rÓ:a@Âig@Ú‹©@Ü‹127@@ÖÜ»€a@I1@‚b«@H2014 Ibn Al-Haitham Jour. for Pure & Appl. Sci. Vol. 27 (1) 2014 ,97,98,101,102,105,106,107,109,111,112,113,114,117,120,121,122,123,124,126,128,129,130, 133} The construction of complete(k7,7)-arc must satisfy the following: 1. Any line in π intersects K in at most 7 points. 2. Every point not in K is on at least one(7-secant). We eliminate thirty-seven points of K which are 3,8,12,14,19,20,21,27,31,34,35,38,39,46,49,55,57,58,60,68,69,74,79,86,90,92,93,98,101,109, 111,112 ,117,120,123,126,128, to satisfy(1) K is incomplete arc since there are the points 53,56,63,82,87,95 of index zero therefore we add them to K to satisfy (2), then K7={4,5,6,7,9,10,11,15,16,17,18,22,23,24,26,28,29,30,32,33,36,37,45,47,52,53,54,56,61,62,6 3,66,70,73,75,80,81,82,83,85,87,88,91,94,95,97,102,105,106,107,113,114,121,122,124,129,1 30,133}. Then K7 is a complete(58,7)-arc . By theorem 1.14, there exists a projective [58,3,51]code which is equivalent to the complete (58,7)-arc K7 Let β5=π−K7 ={1,2,3,8,12,13,14,19,20,21,25,27,31,34,35,38,39,40,41,42,43,44,46,48,49,50,51,55,57,58,59 , 60,64,65,67,68,69,71,72,74,76,77,78,79,84,86,89,90,92,93,96,98,99,100,101,103,104,108,109 ,110, 111,112,115,116,117,118,119,120,123,125,126,127,128,131,132}. β5 is a (75,5)-blocking set. Note that each line in π intersects β5 in at least five points . f. The Construction of Complete (k6,6)-Arc In this section, we take the union of six conics say C1,C2,C3,C4,C5, and C6where C1={1,2,13,25,40,53,63,77,87,100,104,116} C2={1,2,13,25,42,50,59,78,84,96,110,131} C3={1,2,13,25, 41,48,64,76,89,95,115,132} C4={1,2,13,25,44,56,65,72,82,108,118,125} C5={1,2,13,25, 43,51,67,71,99,103,119,127} C6={1,2,13,25,45,52,62,88,98,105,114,126} Let K=π−C1∪C2∪C3∪C4∪C5∪C6 ={3,4,5,6,7,8,9,10,11,12,14,15,16,17,18,19,20,21,22,23,24,26,27,28,29,30,31,32,33,34,35,36, 37 38,39,46,47,49,54,55,57,58,60,61,66,68,69,70,73,74,75,79,80,81,83,85,86,90,91,92,93,94,97, 101,102,106,107,109,111,112,113,117,120,121,122,123,124,128,129,130,133}. The construction of complete (k6,6)-arc must satisfy the following: 1. Any line in π intersects K in at most 6 points. 2. Every point not in K is on at least one (6-secant). We eliminate thirty seven points from K which are 3,8,10,11,14,17,18,20,24,26,28,29,37,39,46,47,57,60,61,66,68,70,73,79,83,90,91,101,111,112 ,113,117,120,122,123,129,133 to satisfy(1). since the points 73,132 of index zero therefore we add them to K to satisfy(2) ,then K6={4,5,6,7,9,12,15,16,19,21,22,23,27,30,31,32,33,34,35,36,38,49,54,55,58,69,73,74,75,80,8 1,85,86,92,93,94,97,102,106,107,109,121,124,128,130,132}. then K6 is a complete(46,6)-arc . 322 | Mathematics @a@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@ÚÓ‘Ój�n€a@Î@Úœäñ€a@‚Ï‹»‹€@·rÓ:a@Âig@Ú‹©@Ü‹127@@ÖÜ»€a@I1@‚b«@H2014 Ibn Al-Haitham Jour. for Pure & Appl. Sci. Vol. 27 (1) 2014 By theorem 1.14, there exists a projective[46,3,40]code which is equivalent to the complete(46,6)-arc K6 Let β6=π−K6 ={1,2,3,8,10,11,13,14,17,18,20,24,25,26,28,29,37,39,40,41,42,43,44,45,46,47,48,50,51,52,53 ,56 57,59,60,61,62,63,64,65,66,67,68,70,71,72,76,77,78,79,82,83,84,87,88,89,90,91,95,96,98,99, 100, 101,103,104,105,108,110,111,112,113,114,115,116,117,118,119,120,122,123,125,126,127,12 9,131, 133} then β6 is a(87,6)-blocking set. Note that each line in π intersects β6 in at least six points. g. The Construction of Complete(k5,5)-Arc In this section, we take the union of seven conics, say C1,C2,C3,C4,C5,C6 and C7 C1={1,2,13,25,40,53,63,77,87,100,104,116} C2={1,2,13,25,42,50,59,78,84,96,110,131} C3={1,2,13,25,41,48,64,76,89,95,115,132} C4={1,2,13,25,44,56,65,72,82,108,118,125} C5={1,2,13,25,43,51,67,71,99,103,119,127} C6={1,2,13,25,45,52,62,88,98,105,114,126} C7={1,2,13,25,38,55,75,81,94,106,122,129} Let K=π−C1∪C2∪C3∪C4∪C5∪C6∪C7 ={3,4,5,6,7,8,9,10,11,12,14,15,16,17,18,19,20,21,22,23,24,26,27,28,29,30,31,32,33,34,35,36, 37, 39,46,47,49,54,57,58,60,61,66,68,69,70,73,74,79,80,83,85,86,90,91,92,93,97,101,102,107,10 9,111,112,113,117,120,121,123,124,128,130,133} The construction of complete (k5,5)-must satisfy the following: 1. Any line in π intersects K in at most 5 points. 2. Every point not in K is on at least one (5-secant). We eliminate forty points from K which are 3,5,10,11,12,14,15,20,22,23,28,29,31,32,33,36,39,46,49,58,60,66,68,70,73,74,86,90,92,111,1 12, 113,117,120,121,123,124,128,130,133 to satisfy(1) K is incomplete arc since the points43,45,65are of index zero, therefore we add themt to K to satisfy(2), then K5={4,6,7,8,9,16,17,18,19,21,24,26,27,30,34,35,37,43,45,47,54,57,61,65,69,79,80,83,85,91,9 3,97, 101,102,107,109} Then K5 is a complete(36,5)-arc . By theorem 1.14, there exists a projective[36,3,31]code which is equivalent to the complete(36,5)-arc K5 Let β7=π−K5 ={1,2,3,5,10,11,12,13,14,15,20,22,23,25,28,29,31,32,33,36,38,39,40,41,42,44,46,48,49,50,51 , 52,53,55,56,58,59,60,62,63,64,66,67,68,70,71,72,73,74,75,76,77,78,81,82,84,86,87,88,89,90, 92,94,95,96,98,99,100,103,104,105,106,108,110,111,112,113,114,115,116,117,118,119,120,1 21,122,123, 124,125,126,127,128,129,130,131,132,133} Then β7 is a (97,7)-blocking set , which is trivial since β7 contains some lines completely. Note that each line in π intersects β7 in at least seven points . h. The Construction of Complete (k4,4)-Arc 323 | Mathematics @a@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@ÚÓ‘Ój�n€a@Î@Úœäñ€a@‚Ï‹»‹€@·rÓ:a@Âig@Ú‹©@Ü‹127@@ÖÜ»€a@I1@‚b«@H2014 Ibn Al-Haitham Jour. for Pure & Appl. Sci. Vol. 27 (1) 2014 In this section, we take the union of eight conics say C1, C2, C3, C4, C5, C6, C7, and C8 where C1={1,2,13,25,40,53,63,77,87,100,104,116} C2={1,2,13,25,42,50,59,78,84,96,110,131} C3={1,2,13,25,41,48,64,76,89,95,115,132} C4={1,2,13,25,44,56,65,72,82,108,118,125} C5={1,2,13,25,43,51,67,71,99,103,119,127} C6={1,2,13,25,45,52,62,88,98,105,114,126} C7={1,2,13,25,38,55,75,81,94,106,122,129} C8={1,2,13,25,39,60,74,86,92,111,120,128} Let K=π−C1∪C2∪C3∪C4∪C5∪C6∪C7∪C8 ={3,4,5,6,7,8,9,10,11,12,14,15,16,17,18,19,20,21,22,23,24,26,27,28,29,30,31,32,33,34,35,36, 37,46,47,49,54,57,58,61,66,68,69,70,73,79,80,83,85,90,91,93,97,101,102,107,109,112,113,11 7,121,123,124,130,133}. The construction of complete(k4,4)-arc must satisfy the following: 1. Any line in π intersects K in at least 4 points. 2. Every point not in K is on at least one 4-secant. We eliminate thirty nine points from K which are 3,5,9,10,11,12,14,15,17,20,22,23,24,27,28,29,32,33,35,36,37,46,47,49,57,66,83,85,91,102,10 7,112,113,117,121,123,124,130,133 to satisfy(1). K is incomplete arc since the point 82 is of index zero therefore we add it to K to satisfy(2), then K4={4,6,7,8,16,18,19,21,26,30,31,34,54,58,61,68,69,70,73,79,80,82,90,93,97,101,109}. Then K4 is a complete (27,4)-arc . By theorem 1.14, there exists a projective [27,3,23]code which is equivalent to the complete(27,4)-arc K4 Let β8=π−K4 ={1,2,3,5,9,10,11,12,13,14,15,17,20,22,23,24,25,27,28,29,32,33,35,36,37,38,39,40,41,42,43, 44,45,46,47,48,49,50,51,52,53,55,56,57,59,60,62,63,64,65,66,67,71,72,74,75,76,77,78,81,83, 84,85,86,87,88,89,91,92,94,95,96,98,99,100,102,103,104,105,106,107,108,110,111,112,113,1 14,115,116,117,118,119,120,121,122,123,124,125,126,127,128,129,130,131,132,133}. Then β8 is a (106,8)-blocking set which is trivial since β8 contains some lines completely. Note that each line in π intersects β8 in at least eight points . i. The Construction of a Complete (k3,3)-Arc In this section, we take the union of nine conics C1,C2,C3,C4,C5,C6,C7,C8 and C9,where C1={1,2,13,25,40,53,63,77,87,100,104,116} C2={1,2,13,25,42,50,59,78,84,96 ,110,131 } C3={1,2,13,25,41,48,64,76,89,95,115,132 } C4={1,2,13,25,44,56,65,72,82,108,118,125} C5={1,2,13,25,43,51,67,71,99,103,119,127} C6={1,2,13,25,45,52,62,88,98,105,114,126 } C7={1,2,13,25,38,55,75,81,94,106,122,129 } C8={1,2,13,25,39,60,74,86,92,111,120,128 } C9={1,2,13,25,54,66,70,83,93,107,117,130 } Let K=π −C1∪C2∪C3∪C4∪C5∪C6∪C7∪C8∪C9 K={3,4,5,6,7,8,9,10,11,12,14,15,16,17,18,19,20,21,22,23,24,26,27,28,29,30,31,32,33,34,35,3 6,37,46,47,49,57,58,61,68,69,73,79,80,85, 90,91,97,101,102,109,112,113,121,123,124,133} The construction of a complete (k3,3)-arc must satisfy the following: 324 | Mathematics @a@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@ÚÓ‘Ój�n€a@Î@Úœäñ€a@‚Ï‹»‹€@·rÓ:a@Âig@Ú‹©@Ü‹127@@ÖÜ»€a@I1@‚b«@H2014 Ibn Al-Haitham Jour. for Pure & Appl. Sci. Vol. 27 (1) 2014 1. Any line in π intersects the arc in at most three points. 2. Every point not in the arc is on at least one (3-secant). We eliminate forty two points from K which are 3,5,8,9,10,11,12,14,15,16,17,18,20,22,23,24,27,28,29,30,32,33,35,36,46,47,49,57,58,61,68,73 ,79,85,90,91,102,112,113, 121,123,133 to satisfy(1). K is incomplete arc since there are three points of index zero which are 15, 74, 130 therefore we add them to K to satisfy (2). Then K3={4,6,7,15,19,21,26,31,34,37,69,74,80,97,101,109,124,130} K3 is a complete (18,3)-arc . By theorem 1.14, there exists a projective [18,3,15]code which is equivalent to the complete (18,3)-arc K3 Let β9=π-K3 ={1,2,3,5,8,9,10,11,12,13,14,16,17,18,20,22,23,24,25,27,28,29,30,32,33,35,36,38,39,40,41,4 2,43,44,45,46,47,48,49,50,51,52,53,54,55,56,57,58,59,60,61,62,63,64,65,66,67,68,70,71,72,7 3,75,76,77,78,79,81,82,83,84,85,86,87,88,89,90,91,92,93,94,95,96,98,99,100,102,103,104,10 5,106,107,108,110,111,112,113,114,115,116,117,118,119,120,121,122,123,125,126,127,128, 129,131,132,133}. β9 is (115,9)-blocking set which is trivial since β9 contains some line completely, each line in π intersects β9 in at least nine points . j. The Construction of Complete (k2,2)-Arc In this section, we construct a complete (k2,2)-arc by using the arc K3 which is constructed above . The construction of complete (k2,2)-arc must satisfies the following: 1.Any line in π intersects the arc in at most 2 points. 2.Every point not in the arc is on at least one (2-secant). We eliminate seventeen points from K3 which are 4,6,7,15,19,21,26,31,34,37,69,74,80,97,101,124,130 to satisfy (1) Let K2=K3\{4,6,7,15,19,21,26,31,34,37,69,74,80,97,101,124,130}. K2={109} K is incomplete arc since there are eleven points of index zero which are 133,132,122,120,110,95,90,75,35,20,15 , therefore we add them to K to satisfy (2),then K2={15,20,35,75,90,95,109,110,120,122,132,133} K2 is a complete (12,2)-arc . By theorem 1.14 there exists a projective [12,3,10]code which is equivalent to the complete (12,2)-arc ,K2. Let β10=π─K2 ={1,2,3,4,5,6,7,8,9,10,11,12,13,14,16,17,18,19,21,22,23,24,25,26,27,28,29,30,31,32,33,34, 36,37,38,39,40,41,42,43,44,45,46,47,48,49,50,51,52,53,54,55,56,57,58,59,60,61,62,63,64,65, 66,67,68,69,70,71,72,73,74,76,77,78,79,80,81,82,83,84,85,86,87,88,89,91,92,93,94,96,97,98, 99,100,101,102,103,104,105,106,107,108,111,112,113,114,115,116,117,118,119,121,123,124 ,125,126,127,128,129,130,131},thenβ10 is (121,10)-blocking set which is trivial since β10 contains some lines completely, we notice that each line in π intersects β10 in at least ten points. Conclusion 1.We can construct complete(kn,n)-arcs by eliminating some conics, wheren=11,10,…,3. 2.We can construct a complete(k2,2)-arc by eliminating some points from a complete(k3,3)- arc 325 | Mathematics @a@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@ÚÓ‘Ój�n€a@Î@Úœäñ€a@‚Ï‹»‹€@·rÓ:a@Âig@Ú‹©@Ü‹127@@ÖÜ»€a@I1@‚b«@H2014 Ibn Al-Haitham Jour. for Pure & Appl. Sci. Vol. 27 (1) 2014 Table (1) : Points and Lines of PG(2,11) Li Pi i 123 112 101 90 79 68 57 46 35 24 13 2 (1,0,0) 1 23 22 21 20 19 18 17 16 15 14 13 1 (0,1,0) 2 124 114 104 94 84 74 64 54 44 34 13 12 (1,1,0) 3 129 113 108 92 87 71 66 50 45 29 13 7 (2,1,0) 4 127 120 102 95 88 70 63 56 38 31 13 9 (3,1,0) 5 126 118 110 91 83 75 67 48 40 32 13 10 (4,1,0) 6 132 119 106 93 80 78 65 52 39 26 13 4 (5,1,0) 7 125 116 107 98 89 69 60 51 42 33 13 11 (6,1,0) 8 131 117 103 100 86 72 58 55 41 27 13 5 (7,1,0) 9 130 115 111 96 81 77 62 47 43 28 13 6 (8,1,0) 10 128 122 105 99 82 76 59 53 36 30 13 8 (9,1,0) 11 133 121 109 97 85 73 61 49 37 25 13 3 (10,1,0) 12 12 11 10 9 8 7 6 5 4 3 2 1 (0,0,1) 13 133 122 111 100 89 78 67 56 45 34 23 2 (1,0,1) 14 128 117 106 95 84 73 62 51 40 29 18 2 (2,0,1) 15 130 119 108 97 86 75 64 53 42 31 20 2 (3,0,1) 16 131 120 109 98 87 76 65 54 43 32 21 2 (4,0,1) 17 125 114 103 92 81 70 59 48 37 26 15 2 (5,0,1) 18 132 121 110 99 88 77 66 55 44 33 22 2 (6,0,1) 19 126 115 104 93 82 71 60 49 38 27 16 2 (7,0,1) 20 127 116 105 94 83 72 61 50 39 28 17 2 (8,0,1) 21 129 118 107 96 85 74 63 52 41 30 19 2 (9,0,1) 22 124 113 102 91 80 69 58 47 36 25 14 2 (10,0,1) 23 133 132 131 130 129 128 127 126 125 124 123 1 (0,1,1) 24 123 113 103 93 83 73 63 53 43 33 23 12 (1,1,1) 25 123 118 102 97 81 76 60 55 39 34 18 7 (2,1,1) 26 123 116 109 91 84 77 59 52 45 27 20 9 (3,1,1) 27 123 115 107 99 80 72 64 56 37 29 21 10 (4,1,1) 28 123 121 108 95 82 69 67 54 41 28 15 4 5,1,1) 29 123 114 105 96 87 78 58 49 40 31 22 11 (6,1,1) 30 123 120 106 92 89 75 61 47 44 30 16 5 (7,1,1) 31 123 119 104 100 85 70 66 51 36 32 17 6 (8,1,1) 32 123 117 111 94 88 71 65 48 42 25 19 8 (9,1,1) 33 123 122 110 98 86 74 62 50 38 26 14 3 (10,1,1) 34 78 77 76 75 74 73 72 71 70 69 68 1 (0,2,1) 35 124 115 106 97 88 68 59 50 41 32 23 11 (1,2,1) 36 129 119 109 99 89 68 58 48 38 28 18 12 (2,2,1) 37 127 113 110 96 82 68 65 51 37 34 20 5 (3,2,1) 38 126 121 105 100 84 68 63 47 42 26 21 7 (4,2,1) 39 132 117 102 98 83 68 64 49 45 30 15 6 (5,2,1) 40 125 118 111 93 86 68 61 54 36 29 22 9 (6,2,1) 41 131 114 108 91 85 68 62 56 39 33 16 8 (7,2,1) 42 130 122 103 95 87 68 60 52 44 25 17 10 (8,2,1) 43 128 116 104 92 80 68 67 55 43 31 19 3 (9,2,1) 44 133 120 107 94 81 68 66 53 40 27 14 4 (10,2,1) 45 100 99 98 97 96 95 94 93 92 91 90 1 (0,3,1) 46 125 117 109 90 82 74 66 47 39 31 23 10 (1,3,1) 47 326 | Mathematics @a@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@ÚÓ‘Ój�n€a@Î@Úœäñ€a@‚Ï‹»‹€@·rÓ:a@Âig@Ú‹©@Ü‹127@@ÖÜ»€a@I1@‚b«@H2014 Ibn Al-Haitham Jour. for Pure & Appl. Sci. Vol. 27 (1) 2014 Li Pi i 124 120 105 90 86 71 67 52 37 33 18 6 (2,3,1) 48 131 121 111 90 80 70 60 50 40 30 20 12 (3,3,1) 49 129 116 103 90 88 75 62 49 36 34 21 4 (4,3,1) 50 130 113 107 90 84 78 61 55 38 32 15 8 (5,3,1) 51 127 122 106 90 85 69 64 48 43 27 22 7 (6,3,1) 52 128 119 110 90 81 72 63 54 45 25 16 11 (7,3,1) 53 126 114 102 90 89 77 65 53 41 29 17 3 (8,3,1) 54 133 115 108 90 83 76 58 51 44 26 19 9 (9,3,1) 55 132 118 104 90 87 73 59 56 42 28 14 5 (10,3,1) 56 111 110 109 108 107 106 105 104 103 102 101 1 (0,4,1) 57 126 119 101 94 87 69 62 55 37 30 23 9 (1,4,1) 58 130 121 101 92 83 74 65 56 36 27 18 11 (2,4,1) 59 124 118 101 95 89 72 66 49 43 26 20 8 (3,4,1) 60 132 122 101 91 81 71 61 51 41 31 21 12 (4,4,1) 61 128 120 101 93 85 77 58 50 42 34 15 10 (5,4,1) 62 129 115 101 98 84 70 67 53 39 25 22 5 (6,4,1) 63 125 113 101 100 88 76 64 52 40 28 16 3 (7,4,1) 64 133 117 101 96 80 75 59 54 38 33 17 7 (8,4,1) 65 127 114 101 99 86 73 60 47 45 32 19 4 (9,4,1) 66 131 116 101 97 82 78 63 48 44 29 14 6 (10,4,1) 67 45 44 43 42 41 40 39 38 37 36 35 1 (0,5,1) 68 127 121 104 98 81 75 58 52 35 29 23 8 (1,5,1) 69 125 122 108 94 80 77 63 49 35 32 18 5 (2,5,1) 70 128 115 102 100 87 74 61 48 35 33 20 4 (3,5,1) 71 124 117 110 92 85 78 60 53 35 28 21 9 (4,5,1) 72 126 116 106 96 86 76 66 56 35 25 15 12 (5,5,1) 73 131 119 107 95 83 71 59 47 35 34 22 3 (6,5,1) 74 133 118 103 99 84 69 65 50 35 31 16 6 (7,5,1) 75 129 120 111 91 82 73 64 55 35 26 17 11 (8,5,1) 76 132 113 105 97 89 70 62 54 35 27 19 10 (9,5,1) 77 130 114 109 93 88 72 67 51 35 30 14 7 (10,5,1) 78 122 121 120 119 118 117 116 115 114 113 112 1 (0,6,1) 79 128 112 107 91 86 70 65 49 44 28 23 7 (1,6,1) 80 131 112 104 96 88 69 61 53 45 26 18 10 (2,6,1) 81 132 112 103 94 85 76 67 47 38 29 20 11 (3,6,1) 82 127 112 108 93 89 74 59 55 40 25 21 6 (4,6,1) 83 124 112 111 99 87 75 63 51 39 27 15 3 (5,6,1) 84 133 112 102 92 82 72 62 52 42 32 22 12 (6,6,1) 85 130 112 105 98 80 73 66 48 41 34 16 9 (7,6,1) 86 125 112 110 97 84 71 58 56 43 30 17 4 (8,6,1) 87 126 112 109 95 81 78 64 50 36 33 19 5 (9,6,1) 88 129 112 106 100 83 77 60 54 37 31 14 8 (10,6,1) 89 56 55 54 53 52 51 50 49 48 47 46 1 (0,7,1) 90 129 114 110 95 80 76 61 46 42 27 23 6 (1,7,1) 91 126 113 111 98 85 72 59 46 44 31 18 4 (2,7,1) 92 125 120 104 99 83 78 62 46 41 25 20 7 (3,7,1) 93 130 118 106 94 82 70 58 46 45 33 21 3 (4,7,1) 94 133 119 105 91 88 74 60 46 43 29 15 5 (5,7,1) 95 327 | Mathematics @a@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@ÚÓ‘Ój�n€a@Î@Úœäñ€a@‚Ï‹»‹€@·rÓ:a@Âig@Ú‹©@Ü‹127@@ÖÜ»€a@I1@‚b«@H2014 Ibn Al-Haitham Jour. for Pure & Appl. Sci. Vol. 27 (1) 2014 Li Pi i 124 116 108 100 81 73 65 46 38 30 22 10 (6,7,1) 96 127 117 107 97 87 77 67 46 36 26 16 12 (7,7,1) 97 132 115 109 92 86 69 63 46 40 34 17 8 (8,7,1) 98 131 122 102 93 84 75 66 46 37 28 19 11 (9,7,1) 99 128 121 103 96 89 71 64 46 39 32 14 9 (10,7,1) 100 67 66 65 64 63 62 61 60 59 58 57 1 (0,8,1) 101 130 116 102 99 85 71 57 54 40 26 23 5 (1,8,1) 102 132 114 107 100 82 75 57 50 43 25 18 9 (2,8,1) 103 129 117 105 93 81 69 57 56 44 32 20 3 (3,8,1) 104 133 113 104 95 86 77 57 48 39 30 21 11 (4,8,1) 105 131 115 110 94 89 73 57 52 36 31 15 7 (5,8,1) 106 126 120 103 97 80 74 57 51 45 28 22 8 (6,8,1) 107 124 122 109 96 83 70 57 55 42 29 16 4 (7,8,1) 108 128 118 108 98 88 78 57 47 37 27 17 12 (8,8,1) 109 125 121 106 91 87 72 57 53 38 34 19 6 (9,8,1) 110 127 119 111 92 84 76 57 49 41 33 14 10 (10,8,1) 111 89 88 87 86 85 84 83 82 81 80 79 1 (0,9,1) 112 131 118 105 92 79 77 64 51 38 25 23 4 (1,9,1) 113 127 115 103 91 79 78 66 54 42 30 18 3 (2,9,1) 114 133 114 106 98 79 71 63 55 36 28 20 10 (3,9,1) 115 125 119 102 96 79 73 67 50 44 27 21 8 (4,9,1) 116 129 122 104 97 79 72 65 47 40 33 15 9 (5,9,1) 117 128 113 109 94 79 75 60 56 41 26 22 6 (6,9,1) 118 132 116 111 95 79 74 58 53 37 32 16 7 (7,9,1) 119 124 121 107 93 79 76 62 48 45 31 17 5 (8,9,1) 120 130 120 110 100 79 69 59 49 39 29 19 12 (9,9,1) 121 126 117 108 99 79 70 61 52 43 34 14 11 (10,9,1) 122 34 33 32 31 30 29 28 27 26 25 24 1 (0,10,1) 123 132 120 108 96 84 72 60 48 36 24 23 3 (1,10,1) 124 133 116 110 93 87 70 64 47 41 24 18 8 (2,10,1) 125 126 122 107 92 88 73 58 54 39 24 20 6 (3,10,1) 126 128 114 111 97 83 69 66 52 38 24 21 5 (4,10,1) 127 127 118 109 100 80 71 62 53 44 24 15 11 (5,10,1) 128 130 117 104 91 89 76 63 50 37 24 22 4 (6,10,1) 129 129 121 102 94 86 78 59 51 43 24 16 10 (7,10,1) 130 131 113 106 99 81 74 67 49 42 24 17 9 (8,10,1) 131 124 119 103 98 82 77 61 56 40 24 19 7 (9,10,1) 132 125 115 105 95 85 75 65 55 45 24 14 12 (10,10,1) 133 328 | Mathematics @a@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@ÚÓ‘Ój�n€a@Î@Úœäñ€a@‚Ï‹»‹€@·rÓ:a@Âig@Ú‹©@Ü‹127@@ÖÜ»€a@I1@‚b«@H2014 Ibn Al-Haitham Jour. for Pure & Appl. Sci. Vol. 27 (1) 2014 329 | Mathematics @a@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@ÚÓ‘Ój�n€a@Î@Úœäñ€a@‚Ï‹»‹€@·rÓ:a@Âig@Ú‹©@Ü‹127@@ÖÜ»€a@I1@‚b«@H2014 Ibn Al-Haitham Jour. for Pure & Appl. Sci. Vol. 27 (1) 2014 Figure (1) References 1. Kadhum, S.J., (2001), Construction of (k,n)-arcs from (k,m)-arcs in PG(2,p) for 2 ≤ m < n, M.Sc. Thesis, University of Baghdad, Iraq. 2. Hirschfeld, J. W. P., (2001), Complete arcs Discrete Math., North Holland Mathematics Studies 123, North-Holland, Amesterdam, 243-250. 3. Hirschfeld, J. W. P., (1979), Projective Geometries Over Finite Fields, Second Edition, Oxford University Press, Oxford. 4. Hirschfeld, J. W. P. and Storme, L.,(1998), The Packing Problem in Statistics, Coding Theory and Finite Projective Spaces, J.Statistical Planning and Inference, 72, pp.355-380. 5. R.Hill, (1992), Optimal Linear Codes in: C.Mitichell (Ed.) Crytography and Coding, Oxford University ,Discrete Mathematics, Press: Oxford, pp.75-104. 6. Rumen Daskalov, (2008), A GeometricConstruction of (38,2)-Blocking Set in PG(2,13) and the Related [14,5,3,133]13 Code, Discrete Mathematics,Technical University of Gabrovo, Bulgaria, 308 (1341-1345). 7. Hassan, H. K., (2002), "Reverse Construction of (k,n)-arcs in PG(2,q) over some GF(q)" , M.Sc.Thesis, College of Education Ibn-Al Haitham, University of Baghdad . بطریقة ھندسیة مع PG(2,11)في المستوي إالسقاطي (kn,n)بناء االقواس المجموعات القالبیة والشفرات االسقاطیة المرتبطة بھا آمال شھاب المختار امنیات عدنان حسن جامعة بغداد / )ابن الھیثم(كلیة التربیة للعلوم الصرفة /قسم الریاضیات 2013تشرین االول 10، قبل للنشرفي : 2013حزیران 17أستلم البحث في : الخالصة منھا على استقامة واحدة، n + 1وال توجد nأذ توجد PG(2,q),من النقاط في kھو مجموعة من (k,n)القوس .(k + 1,n)-كامال ً اذا لم یكن محتوى في قوس (k,n) –فیكون القوس PG(2,11)قاطي في المستوي االس n = 2,3,…,10,11كاملة ،إذ (kn,n) –في ھذا البحث نقوم ببناء اقواس بطریقة ھندسیة، مع المجموعات القالبیة والشفرات االسقاطیة المرتبطة بھا. أقواس كاملة ، مجموعة قالبیة، شفرة أسقاطیة كلمات مفتاحیة : 330 | Mathematics @a@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@ÚÓ‘Ój�n€a@Î@Úœäñ€a@‚Ï‹»‹€@·rÓ:a@Âig@Ú‹©@Ü‹127@@ÖÜ»€a@I1@‚b«@H2014 Ibn Al-Haitham Jour. for Pure & Appl. Sci. Vol. 27 (1) 2014 قسم الرياضيات / كلية التربية للعلوم الصرفة (ابن الهيثم) / جامعة بغداد أستلم البحث في : 17 حزيران 2013 , قبل للنشرفي : 10 تشرين الاول 2013 القوس (k,n)هو مجموعة من k من النقاط في ,PG(2,q)أذ توجد n ولا توجد n + 1 منها على استقامة واحدة، فيكون القوس – (k,n) كاملا ً اذا لم يكن محتوى في قوس-(k + 1,n). في هذا البحث نقوم ببناء اقواس – (kn,n) كاملة ,إذ n = 2,3,…,10,11 في المستوي الاسقاطي PG(2,11) بطريقة هندسية، مع المجموعات القالبية والشفرات الاسقاطية المرتبطة بها.