@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@ÚÓ‘Ój�n€a@Î@Úœäñ€a@‚Ï‹»‹€@·rÓ:a@Âig@Ú‹©@Ü‹1a26@@ÖÜ»€a@I2@‚b«@H2013 Ibn Al-Haitham Jour. for Pure & Appl. Sci. Vol. 26 (2) 2013 Complete Arcs in Projective Plane PG (2,11) Over Galois field Mahmood S. Fiadh Department of computer/ College of Education / The Iraqi University Received in: 19 September 2012 , Accepted in:3 February 2013 Abstract In this work, we construct complete (K, n)-arcs in the projective plane over Galois field GF (11), where 122 ≤≤ n ,by using geometrical method (using the union of some maximum(k,2)- Arcs , we found (12,2)-arc, (19,3)-arc , (29,4)-arc, (38,5)-arc , (47,6)-arc, (58,7)-arc, (68,6)-arc, (81,9)-arc, (96,10)-arc, (109,11)-arc, (133,12)-arc, all of them are complete arc in PG(2, 11) over GF(11). Key words : Algebraic geometry, complete arcs, Galois field. 298 | Mathematics @@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@ÚÓ‘Ój�n€a@Î@Úœäñ€a@‚Ï‹»‹€@·rÓ:a@Âig@Ú‹©@Ü‹1a26@@ÖÜ»€a@I2@‚b«@H2013 Ibn Al-Haitham Jour. for Pure & Appl. Sci. Vol. 26 (2) 2013 Introduction Yasin 1986, [8] gave the construction and classification of(k,3)-arcs over the Galois field GF(8), Ahmed (1999) [1] studied the complete arcs in projective plane over Galois field GF (9), also sawan (2001) [2], studied the construction of (k ,n)-arcs form (k , m)-arcs in PG(2,p) The aim of this paper is to find the complete arcs in projective plane PG (2, 11) over Galois field GF (11).This paper is divided into twelve sections, section one consists of the basic theorems and definition of projective plane.From In section two to section twelve, the construction of complete (k, n)-arcs for 122 ≤≤ n in PG (2, 11) is given 1.1Definition [3]: A projective PG(2, p) over Galois field , where p is a prime number , consists of 12 ++ pp points and 12 ++ pp lines, every line contains 1+p points and every point is on 1+p lines. Any point a of the plane has the form of a triple ( ),,( 210 XXX , where 210 ,, XXX are elements in GF(p) with the exception of a triple consisting of three zero elements. Two triples ),,( 210 XXX and ),,( 210 yyy represent the same point if there exists λ in GF (p) \ {0}, s.t. ),,( 210 yyy = λ ),,( 210 XXX . There exists one point of the form (1,0,0) , there exists p points of the form ,)0,1,( X there exists 2p points of the form ,)1,,( YX similarly for the lines . A point P ),,( 210 XXX is incident with the line [ ],, 210 yyy iff : 0221100 =++ yXyXyX The projective plane PG(2, p) satisfying the following axioms : a) Any two distinct lines intersected in a unique point. b) Any two distinct points are contained in a unique line. c) There exists at least four points such that no three of them are collinear. The projective plane PG(2, 11) contains 133 points , 133 lines , every line contains 12 point and every points is on 12 lines. Let ip and iL , i= 1, 2,… ,133 be the points and lines of PG (2, 11) respectively , all the points and lines of PG (2, 11) are given in table (1). 1.2Definition [5]: A (k, n)-arc K in PG (2, P) is a set of k points such that some n, but no n+1 of them are collinear. 1.3Definition [5]: (A k, n)-arc in PG (2, P) k is complete if it is not contained in a (K+1,n)-arc . 1.4Definition [3]: An (n-secant) of a (k,n)-arc is a line intersects K in n points . a o-secant is called an external line of K, 1-secant is called unsecant line, a 2-secant is called a bisecant line and 3-secant is called a trisecant line. 1.5Definition [3]: A point which is not on a (k, n)-arc K has index i denoted by iN , if there are exactly i ( n- secant) of K thought N i . Let iC = iN be the number of the points iN of index i. 1.6Remark [4]: 299 | Mathematics @@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@ÚÓ‘Ój�n€a@Î@Úœäñ€a@‚Ï‹»‹€@·rÓ:a@Âig@Ú‹©@Ü‹1a26@@ÖÜ»€a@I2@‚b«@H2013 Ibn Al-Haitham Jour. for Pure & Appl. Sci. Vol. 26 (2) 2013 The (k, n)-arc K is complete if and only if 00=C , thus K is complete if every point of PG(2, p) lies on some n-secants of K. 1.7Definition [5]: A k, n)-arc K in PG(2. p) is maximal arc if k =(n-1)p+n. 1.8Definition [3]: The maximum number of points that can be a (K, 2)-arc in PG(2, p) is m(2, p)- this arc called an oval. 1.9Definition [5]: A polynomial F in ],...,[ 21 nXXXK is called homogenous or a form of degree d if all its terms have the same degree d. A subset V of PG (n, k) is variety over K if there exists forms RFFF ,...,, 21 in ],...,[ 21 nXXXK such that V = {P(A) in PG(n, k), ).,...,,(}0)(...)()( 2121 RR FFFVAFAFAF ===== 1.10Definition [5]: A variety V(F) in PG(n, k) is called a primal. The order or degree of a primal V(F) is the degree of F. 1.11Definition [5]: A quadric Q in PG(n-1, p) is a primal of order two , so if Q is a quadric then Q=F(V) where F is Quadric form , that is 22112 2 111 1, ... nnnj n ji ji iij XaXXaXaXXa +++=∑ = ≤ 1.12Definition [5]: Let Q(2, p) be the set of quadrics in PG(2, p) that is the varieties V(F), where 322331132112 2 3332 2 22 2 111 XXaXXaXXaXaXaXaF +++++= If                 = 33 3231 23 22 21 1312 11 22 22 22 a aa a a a aa a A is non-singular, then the quadric is a conic. 1.13Theorem [3]: In PG (2, p) , with p odd, every oval is a conic. 1.14Theorem [6]: Let m be a point of a (K, 2)-arc K and let t(m) be the number of unisecants through m in PG(2p) then kpmtt −+== 2)( 1.15Corollary [6]: If k is an oval then t(m) =1 300 | Mathematics @@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@ÚÓ‘Ój�n€a@Î@Úœäñ€a@‚Ï‹»‹€@·rÓ:a@Âig@Ú‹©@Ü‹1a26@@ÖÜ»€a@I2@‚b«@H2013 Ibn Al-Haitham Jour. for Pure & Appl. Sci. Vol. 26 (2) 2013 1.16Theorem [5]: Let k be a (k, 2)-arc in PG (2, p) and let iT be the number of i-secants of k in the plane, that is 2T is the number of bisecants, 1T is the number of unisecant, and 0T is the number of external line kpt −+= 2 , then: a) 2 )1( 2 − = kk T b) ktT =1 c) 2 )1( 2 )1( 0 − + − = ttpp T 1.17Lemma [7]: Let iC be the number of points Q of index i . Then a) kppCi −++=∑ 12 β α b) )1(2 )1( − − =∑ p kk iCi β α where α is smallest I for which 0≠iC , and β is the largest I for which 0≠iC . 1.118Theorem [2]: A (k, n)-arc K is maximal if and only if every line in PG (2, p) is a 0-secant or n-secant, 2. Complete (K, 2)-arc in PG(2, 11) Let A={1,2,13,25} be the set of unit and reference points in PG(2,11) as in the table (1) such that : 1=(1,0,0) , 2=(0,1,0), 13=(0,0,1), 25=(1,1,1), A is (4,2)-arc, since no three points of A are collinear, The general equation of the conic is: 0326315214 2 332 2 2 2 11 =+++++= XXaXXaXXaXaXaXaF … (1) By substituting the points of the arc A in (1), so (1) becomes: 0326315214 =++ XXaXXaXXa … (2) If 04 =a , then the conic is degenerated. Therefore 04 ≠a , similarly 05 ≠a and 06 ≠a Dividing equation (2) by 4a , we get: 0323121 =++ XXXXXX βα …(3) Where 4 5 a a =α , 4 6 a a =β , so that 1+ 0=+ βα (mod.11) β = - (1+ α ), then (3) can be writer as: 0)1( 323121 =+−+ XXXXXX αα Where 0≠α and 10≠α for 0=α or 10=α , we degerenated conics, can be obtaiened thus .9,8,7,6,5,4,3,2,1=α The ovals in PG(2,11) throught the reference and the unit points has the following points : }116,104,100,87,77,63,53,40,25,13,2,1{1 =C which are the points of )9( 3231211 XXXXXXV ++ }131,110,96,84,78,59,50,42,25,13,2,1{2 =C which are the points of )82( 3231212 XXXXXXV ++ 301 | Mathematics @@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@ÚÓ‘Ój�n€a@Î@Úœäñ€a@‚Ï‹»‹€@·rÓ:a@Âig@Ú‹©@Ü‹1a26@@ÖÜ»€a@I2@‚b«@H2013 Ibn Al-Haitham Jour. for Pure & Appl. Sci. Vol. 26 (2) 2013 }132,115,95,89,76,64,48,41,25,13,2,1{3 =C which are the points of )73( 3231213 XXXXXXV ++ }125,118,108,82,72,65,56,44,25,13,2,1{4 =C which are the points of )64( 3231214 XXXXXXV ++ }127,119,103,99,71,67,51,43,25,13,2,1{5 =C which are the points of )55( 3231215 XXXXXXV ++ }126,114,105,98,88,62,52,45,25,13,2,1{6 =C which are the points of )46( 3231216 XXXXXXV ++ }129,122,106,94,81,75,55,38,25,13,2,1{7 =C which are the points of )37( 3231217 XXXXXXV ++ }128,120,111,92,86,74,60,39,25,13,2,1{8 =C which are the points of )28( 3231218 XXXXXXV ++ }130,117,107,93,83,70,66,54,25,13,2,1{9 =C which are the points of )19( 3231219 XXXXXXV ++ Thus there are nine compelete (12,2)-arcs (conics) in PG(2,11). Hence each arc is a maximum arc, since each line is 0-secant or 2-secant. 3.The construction of compelete (k,3)-arcs in PG(2,11) In this section, we try to get a compelete (k,3)-arc throught following steps. a) We take the union of two maximal (k,2)-arcs, say 1C and 2C denoted by E , }96,78,59{21 −= CCE  , we notice that }131,110,84,50,42,116,104,100,87,77,63,53,40,25,13,2,1{=E is incompelete (k,3)-arc since there exists the points {3,4,5,6,7,8,9,10,11,12,14,15,16,17,18,19,20,21,22,23,24,26,27,28,29,30,31,32,33,34,35,36,3 7,38,39,41,43,44,45,46,47,48,49,51,52,54,55,56,57,58,59,60,61,62,64,65,66, 67,68,69,70,71,72,73,74,75,76,78,79,80,81,82,83,85,86,88,89,90,91,92,93,94,95,96, 97,98,99,101,102,103,105,106,107,108,109,111,112,113,114,115,117,118,119,120, 121,122,123,124,125,126,127,128,129,130,132,133} which are the points of index zero for E . b) We add two points of index zero which are {3,46}. Then }131,116,110,104,100,87,84,77,63,53,50,46,42,40,25,13,3,2,1{=′E is a compelete (19,3)-arc since 00 =C . 4.The construction of compelete (k,4)-arcs in PG(2,11) In this section, we try to get a compelete (k,4)-arc through following steps. a) We take the union of three maximal (k,2)-arcs, say 1C , 2C and 3C denoted by 1E , }132,89,64,48{3211 −= CCCE  , we notice that }115,95,76,41,131,110,96,84,78,59,50,42,116,104,100,87,77,63,53,40,25,13,2,1{1 =E is incompelete (k,4)-arc since there exists the points {3,4,5,6,7,8,9,10,11,12,14,15,16,17,18,19,20,21,22,23,24,26,27,28,29,30,31,32,33,34,35,36,3 7,38,39,43,44,45,46,47,48,49,51,52,54,55,56,57,58,60,61,62,64,65,66,67, 68,69,70,71,72,73,74,75,79,80,81,82,83,85,86,88,89,90,91,92,93,94,97,98,99,101,102, 103,105,106,107,108,109,111,112,113,114,117,118,119,120,121,122,123,124, 125,126,127,128,129,130,132,133} which are the points of index zero for 1E . 302 | Mathematics @@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@ÚÓ‘Ój�n€a@Î@Úœäñ€a@‚Ï‹»‹€@·rÓ:a@Âig@Ú‹©@Ü‹1a26@@ÖÜ»€a@I2@‚b«@H2013 Ibn Al-Haitham Jour. for Pure & Appl. Sci. Vol. 26 (2) 2013 b) We add five points of index zero which are {3,33,34,38,108}. Then }.131,116,115 ,110,108,104,100,96,95,87,84,78,77,76,63,59,53,50,42,41,40,38,34,33,25,13,3,2,1{1 =′E is a compelete (29,4)-arc since 00 =C . 5.The construction of compelete (k,5)-arcs in PG(2,11) In this section, we try to get a compelete (k,5)-arc throught following steps. a) We take the union of four maximal (k,2)-arcs, say 1C , 2C , 3C and 4C denoted by 2E , }118,108,82,56,44{43212 −= CCCCE  , }125,72,65,132,115 ,95,89,76,64,48,41,131,110,96,84,78,59,50,42,116,104,100,87,77,63,53,40,25,13,2,1{2 =E 2E is incompelete (k,5)-arc since there exists the points {3,4,5,6,7,8,9,10,11,12,14,15,16,17,18,19,20,21,22,23,24,26,27,28,29,30,31,32,33,34,35,36,3 7,38,39,43,44,45,46,47,49,51,52,54,55,56,57,58,60,61,62,66,67,68,69,70 ,71,73,74,75,79,80,81,82,83,85,86,88,89,90,91,92,93,94,97,98,99,101,102,103,105 ,106,107,108,109,111,112,113,114,117,118,119,120,121,122,123,124,126,127,128 ,129,130,133} which are the points of index zero for 2E . b) We add seven points of index zero which are {9,10,12,19,31,39,68}. Then }.132,131,125,116,115,110,104,100,96,95 ,89,87,84,78,77,76,72,68,65,64,63,59,53,50,48,42,41,40,39,31,25,19,13,12,10,9,2,1{2 =′E 2E′ is a compelete (38,5)-arc since 00 =C . 6.The construction of compelete (k,6)-arcs in PG(2,11) In this section, we try to get a compelete (k,6)-arc through following steps. a) We take the union of five maximal (k,2)-arcs, say 1C , 2C , 3C , 4C and 5C denoted by 3E , , }118,82{543213 −= CCCCCE  , }.127,119,103,99,71,67,51,43,125,108,72,65,56,44,132,115,95 ,89,76,64,48,41,131,110,96,84,78,59,50,42,116,104,100,87,77,63,53,40,25,13,2,1{3 =E 3E is incompelete (k,6)-arc since there exists the points {3,4,5,6,7,8,9,10,11,12,14,15,16,17,18,19,20,21,22,23,24,26,27,28,29,30,31,32,33,34,35,36,3 7,38,39,45,46,47,49,52,54,55,57,58,60,61,62,66,68,69,70,73,74,75,79,80,81, 82,83,85,86,88,90,91,92,93,94,97,98,101,102,105,106,107,109,111,112,113,114,117,118,120, 121,122,123,124,126,128,129,130,133} which are the points of index zero for 3E . b) We add six points of index zero which are {9,10,12,15,16,85}. Then }.132,131,127,125,119,116,115,110,108,104,103,100,99,96,95,89,87,85,84,78 ,77,76,72,71,67,65,64,63,59,56,53,51,50,48,44,43,42,41,40,25,16,15,13,12,10,9,2,1{3 =′E 3E′ is a compelete (48,6)-arc since 00 =C . 7.The construction of compelete (k,7)-arcs in PG(2,11) In this section, we try to get a compelete (k,6)-arc through following steps. a)We take the union of six maximal (k,2)-arcs, say 1C , 2C , 3C , 4C , 5C and 6C denoted by 4E , 6543214 CCCCCCE = , 303 | Mathematics @@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@ÚÓ‘Ój�n€a@Î@Úœäñ€a@‚Ï‹»‹€@·rÓ:a@Âig@Ú‹©@Ü‹1a26@@ÖÜ»€a@I2@‚b«@H2013 Ibn Al-Haitham Jour. for Pure & Appl. Sci. Vol. 26 (2) 2013 }.126,114 ,105,98,88,62,52,45,127,119,103,99,71,67,51,43,125,118,108,82,72,65,56,44,132,115,95 ,89,76,64,48,41,131,110,96,84,78,59,50,42,116,104,100,87,77,63,53,40,25,13,2,1{4 =E 4E is incompelete (k,6)-arc since there exists the points {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} which are the points of index zero for 4E . b) We add six points of index zero which are {4,19,27,31,34,81}. Then }.132,131,127,126,125 ,119,118,116,115,114,110,108,105,104,103,100,99,98,96,95,89,88,87,84,82,81,78,77,76,72 ,71,67,65,64,63,62,59,56,53,52,51,50,48,45,44,43,42,41,40,34,31,27,25,19,13,4,2,1{4 =′E 4E′ is a compelete (58,7)-arc since 00 =C 8.The construction of compelete (k,8)-arcs in PG(2,11) In this section, we try to get a compelete (k,8)-arc through following steps. a) We take the union of seven maximal (k,2)-arcs, say 1C , 2C , 3C , 4C , 5C 6C ,and 7C denoted by 5E , 76543215 CCCCCCCE = , }129,122,106,94,81,75,55,38,126,114 ,105,98,88,62,52,45,127,119,103,99,71,67,51,43,125,118,108,82,72,65,56,44,132,115,95 ,89,76,64,48,41,131,110,96,84,78,59,50,42,116,104,100,87,77,63,53,40,25,13,2,1{5 =E 5E is incompelete (k,8)-arc since there exists the points {3,4,5,6,7,8,9,10,11,12,14,15,16,17,18,19,20,21,22,23,24,26,27,28,29,30,31,32,33,34,35,36,3 7,39,46,47,49,54,57,58,60,61,66,68,69,70,73,74,79,80,83,85,86,90,91,92,93,97,101,102,107,1 09,111,112,113,117,120,121,123,124,,128,130,133} which are the points of index zero for 5E . b) We add eight points of index zero which are {3,4,5,6,7,11,19,31}. Then }132,131,129,127,126,125,122,119,118,116,115,114,110 ,108,106,105,104,103,100,99,98,96,95,94,89,88,87,84,82,81,78,77,76,75,72,71,67,65,64 ,63,62,59,56,55,53,52,51,50,48,45,44,43,42,41,40,38,31,25,19,13,11,7,6,5,4,3,2,1{5 =′E ′ 5E is a compelete (68,8)-arc since 00 =C . 9.The construction of compelete (k,9)-arcs in PG(2,11) In this section, we try to get a compelete (k,9)-arc through following steps. a) We take the union of eight maximal (k,2)-arcs, say 1C , 2C , 3C , 4C , 5C 6C , 7C and 8C denoted by 6E , 876543216 CCCCCCCCE = , 6E ={1,2,13,25,40,53,63,77,87,100,,104,116,42,50,59,78,84,96,110,131,41,48,64,76, 89,95,115,132,44,56,65,72,82,108,118,125,43,51,67,71,99,103,119,127,45,52,62,88, 98,105,114,126,38,55,75,81,94,106,122,129,39,60,74,86,92,111,120,128} 6E is incompelete (k,9)-arc since there exist the points {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, 61,66,68,69,70,73,79,80,83,85,90,91,93,97,101,102,107,109,112,113,117,121,123, 124,130,133} which are the points of index zero for 6E . 304 | Mathematics @@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@ÚÓ‘Ój�n€a@Î@Úœäñ€a@‚Ï‹»‹€@·rÓ:a@Âig@Ú‹©@Ü‹1a26@@ÖÜ»€a@I2@‚b«@H2013 Ibn Al-Haitham Jour. for Pure & Appl. Sci. Vol. 26 (2) 2013 b) We add thirteen points of index zero whichare{4,5,6,7,10,11,12,18,22,37,49,61, 109}Then 6E′ ={1,2,4,5,6,7,10,11,12,13,18,22,25,37,38,39,40,41,42,43,44,45,48,49,50,51,52,53,55,56,59,6 0,61,62,63,64,65,67,71,72,74,75,76,77,78,81,82,84,86,87,88,89, 92,94,95,96,98,99,100,103,104,105,106,108,109,110,111,114,115,116,118,119,120, 122,125,126,127,128,129,131,132}is complete (81,9) –arc,since 00 =C . 10.The construction of compelete (k,10)-arcs in PG(2,11) In this section, we try to get a compelete (k,10)-arc through following steps. a)We take the union of nine maximal (k,2)-arcs, say 1C , 2C , 3C , 4C , 5C , 7C 8C and 9C denotedby 7E , 9876543217 CCCCCCCCCE = , 7E ={1,2,13,25,40,53,63,77,87,100,,104,116,42,50,59,78,84,96,110,131, 41,48,64,76,89,95,115,132,44,56,65,72,82,108,118,125,43,51,67,71,99,103,119,127, 45,52,62,88,98,105,114,126,38,55,75,81,94,106,122,129,39,60,74,86,92,111,120,128,54,66,7 0,83,93,107,117,130} 7E is incompelete (k,10)-arc since there exists the points {3,4,5,6,7,8,9,10,11,12,14,15,16,17,18,19,20,21,22,23,24,26,27,28,29,30,31,32,33,34,35,36,3 7,46,47,49,57,58,61,68,69,73,79,80,85,90,91,97,101,102,109,112,113,121,123,124,133} which are the points of index zero for 7E . b) We add twenty points of index zero which are{3,4,5,6,7,8,9,10,14,15,16,22,23,61 69,73,85,97,109,124}Then 7E′ ={1,2,3,4,5,6,7,8,9,10,13,14,15,16,22,23,25,38,39,40, 41,42,43,44,45,48,50,51,52,53,54,55,56,59,60,61,62,63,64,65,66,67,69,70,71,72,73, 74,75,76,77,78,81,82,83,84,85,86,87,88,89,92,93,94,95,96,97,98,99,100,103,104,105, 106,107,108,109,110,111,114,115,116,117,118,119,120,122,124,125,126,127,128, 129,130,131,132}is complete (96,10) –arc,since 00 =C . 11.The construction of compelete (k,11)-arcs in PG(2,11) In this section, we take (96,10)-arc 7E′ which is incomplete (k,11)-arc, since there exist points of index zero for 7E′ which are{11,12,17,18,19,20,21,24,26,27,28 ,29,30,31,32,33,34,35,36,37,46,47,49,57,58,68,79,80,90,91,101,102,112,113,121,123, 133} ,i.e. 00 ≠C . Now ,we add to 7E′ thirteen points of index zero which are {11,17,18,19,20,26,27,35, 36,47,102,121,133}. Then 8E′ ={1,2,3,4,5,6,7,8,9,10,11,13,14,15,16,17,18,19,20,22,23,25,26,27,35,36,38,39,40,41,42,43,44 ,45,47,48,50,51,52,53,54,55,56,59,60,61,62,63,64,65,66,67,69,70,71, 72,73,74,75,76,77,78,81,82,83,84,85,86,87,88,89,92,93,94,95,96,97,98,99,100,102, 103,104,105,106,107,108,109,110,111,114,115,116,117,118,119,120,121,122,124, 125,126,127,128,129,130,131,132,133}is a complete (109,11) –arc,since 00 =C . 12.The construction of compelete (k,12)-arcs in PG(2,11) In this section, we take from 8E′ the complete (109,11)-arc which is incomplete (k,12)-arc 8E′ since there exist points of index zero for 8E′ which are{12, ,21,24,28,29, 30,31,32,33,34,37,46,49,57,58,68,79,80,90,91,101,112,113, 123,} ,i.e. 00 ≠C . 305 | Mathematics @@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@ÚÓ‘Ój�n€a@Î@Úœäñ€a@‚Ï‹»‹€@·rÓ:a@Âig@Ú‹©@Ü‹1a26@@ÖÜ»€a@I2@‚b«@H2013 Ibn Al-Haitham Jour. for Pure & Appl. Sci. Vol. 26 (2) 2013 We add the points of index zero to 8E′ denoted by 9E′ then 9E′ contains all the points of the plane i.e., 9E′ ={1,2,3,…,131,132,133} is a complete (133,12)-arc since 00 =C . This arc is the whole plane PG(2,11), since each line in it contains 12 points . Hence this arc is maximal. References 1. Ahmed, A.m (1999). "Complete Arcs in Projective Plane Over GF(9)", Accepted to puplish , Ibn Al-Hitham, Iraq. 2. Swsan,J.K (2001). M.SC.Thesis,University of Baghdad, Ibn-Al-Haitham College, Iraq. 3. Hirschfeld, J.W.P (1979). "Projective Geometries Over Finite Fields", Oxford Press. 4. Saleh,R.A (1999). "Complete Arcs in Projective Plane Over Galois filed",M.S.C.Thesis,College of Education Ibn-Al-Haitham, University of Baghdad,Iraq. 5. Hughes,D.R. and Piper, F.C (1973)."Projective Plane", Springer Village,New York, Inc. 6. Al-Mukhtar, A.S (2001). Mathematics and Physics.J.Vol.17. رسالة ماجستیر ، جامعة الموصل. ،arc-(k,n)) القید األعلى لالقواس 2001. بان عبد الكریم ، (7 8. Yasin, A. L (1986). "Cubic Arcs in the Projective Plane of Order Eight", Ph.D. Thesis, Uinversity of Sussex, England. Table (1):Points and lines of PG(2,11) i Pi Li 1 (1,0,0) 2 13 24 35 46 57 68 79 90 101 112 123 2 (0,1,0) 1 13 14 15 16 17 18 19 20 21 22 23 3 (1,1,0) 12 13 34 44 54 64 74 84 94 104 114 124 4 (2,1,0) 7 13 29 45 50 66 71 87 92 108 113 129 5 (3,1,0) 9 13 31 38 56 63 70 88 95 102 120 127 6 (4,1,0) 10 13 32 40 48 67 75 83 91 110 118 126 7 (5,1,0) 4 13 26 39 52 65 78 80 93 106 119 132 8 (6,1,0) 11 13 33 42 51 60 69 89 98 107 116 125 9 (7,1,0) 5 13 27 41 55 58 72 86 100 103 117 131 10 (8,1,0) 6 13 28 43 47 62 77 81 96 111 115 130 11 (9,1,0) 8 13 30 36 53 59 76 82 99 105 122 128 12 (10,1,0) 3 13 25 37 49 61 73 85 97 109 121 133 13 (0,0,1) 1 2 3 4 5 6 7 8 9 10 11 12 14 (1,0,1) 2 23 34 45 56 67 78 89 100 111 122 133 15 (2,0,1) 2 18 29 40 51 62 73 84 95 106 117 128 16 (3,0,1) 2 20 31 42 53 64 75 86 97 108 119 130 17 (4,0,1) 2 21 32 43 54 65 76 87 98 109 120 131 18 (5,0,1) 2 15 26 37 48 59 70 81 92 103 114 125 19 (6,0,1) 2 22 33 44 55 66 77 88 99 110 121 132 20 (7,0,1) 2 16 27 38 49 60 71 82 93 104 115 126 21 (8,0,1) 2 17 28 39 50 61 72 83 94 105 116 127 22 (9,0,1) 2 19 30 41 52 63 74 85 96 107 118 129 23 (10,0,1) 2 14 25 36 47 58 69 80 91 102 113 124 24 (0,1,1) 1 123 124 125 126 127 128 129 130 131 132 133 25 (1,1,1) 12 23 33 43 53 63 73 83 93 103 113 123 26 (2,1,1) 7 18 34 39 55 60 76 81 97 102 118 123 27 (3,1,1) 9 20 27 45 52 59 77 84 91 109 116 123 306 | Mathematics @@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@ÚÓ‘Ój�n€a@Î@Úœäñ€a@‚Ï‹»‹€@·rÓ:a@Âig@Ú‹©@Ü‹1a26@@ÖÜ»€a@I2@‚b«@H2013 Ibn Al-Haitham Jour. for Pure & Appl. 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Vol. 26 (2) 2013 28 (4,1,1) 10 21 29 37 56 64 72 80 99 107 115 123 29 (5,1,1) 4 15 28 41 54 67 69 82 95 108 121 123 30 (6,1,1) 11 22 31 40 49 58 78 87 96 105 114 123 31 (7,1,1) 5 16 30 44 47 61 75 89 92 106 120 123 32 (8,1,1) 6 17 32 36 51 66 70 85 100 104 119 123 33 (9,1,1) 8 19 25 42 48 65 71 88 94 111 117 123 34 (10,1,1) 3 14 26 38 50 62 74 86 98 110 122 123 35 (0,2,1) 1 68 69 70 71 72 73 74 75 76 77 78 36 (1,2,1) 11 23 32 41 50 59 68 88 97 106 115 124 37 (2,2,1) 12 18 28 38 48 58 68 89 99 109 119 129 38 (3,2,1) 5 20 34 37 51 65 68 82 96 110 113 127 39 (4,2,1) 7 21 26 42 47 63 68 84 100 105 121 126 40 (5,2,1) 6 15 30 45 49 64 68 83 98 102 117 132 41 (6,2,1) 9 22 29 36 54 61 68 86 93 111 118 125 42 (7,2,1) 8 16 33 39 56 62 68 85 91 108 114 131 43 (8,2,1) 10 17 25 44 52 60 68 87 95 103 122 103 44 (9,2,1) 3 19 31 43 55 67 68 80 92 104 116 128 45 (10,2,1) 4 14 27 40 53 66 68 81 94 107 120 133 46 (0,3,1) 1 90 91 92 93 94 95 96 97 98 99 100 47 (1,3,1) 10 23 31 39 47 66 74 82 90 109 117 125 48 (2,3,1) 6 18 33 37 52 67 71 86 90 105 120 124 49 (3,3,1) 12 20 30 40 50 60 70 80 90 111 121 131 50 (4,3,1) 4 21 34 36 49 62 75 88 90 103 116 129 51 (5,3,1) 8 15 32 38 55 61 78 84 90 107 113 130 52 (6,3,1) 7 22 27 43 48 64 69 85 90 106 122 127 53 (7,3,1) 11 16 25 45 54 63 72 81 90 110 119 128 54 (8,3,1) 3 17 29 41 53 65 77 89 90 102 114 126 55 (9,3,1) 9 19 26 44 51 58 76 83 90 108 115 133 56 (10,3,1) 5 14 28 42 56 59 73 87 90 104 114 132 57 (0,4,1) 1 101 102 103 104 105 106 107 108 109 110 111 58 (1,4,1) 9 23 0 37 55 62 69 87 94 101 119 126 59 (2,4,1) 11 18 27 36 56 65 74 83 92 101 121 130 60 (3,4,1) 8 20 26 43 49 66 72 89 95 101 118 124 61 (4,4,1) 12 21 31 41 51 61 71 81 91 101 121 132 62 (5,4,1) 10 15 34 42 50 58 77 85 93 101 120 128 63 (6,4,1) 5 22 25 39 53 67 70 84 98 101 115 129 64 (7,4,1) 3 16 28 40 52 64 76 88 100 101 113 125 65 (8,4,1) 7 17 33 38 54 59 75 80 96 101 117 133 66 (9,4,1) 4 19 32 45 47 60 73 86 99 101 114 127 67 (10,4,1) 6 14 29 44 48 63 78 82 97 101 116 131 68 (0,5,1) 1 35 36 37 38 39 40 41 42 43 44 45 69 (1,5,1) 8 23 29 35 52 58 75 81 98 104 121 127 70 (2,5,1) 5 18 32 35 49 63 77 80 94 108 122 125 71 (3,5,1) 4 20 33 35 48 61 74 87 100 102 115 128 72 (4,5,1) 9 21 28 35 53 60 78 85 92 110 117 124 73 (5,5,1) 12 15 25 35 56 66 76 86 96 106 116 126 74 (6,5,1) 3 22 34 35 47 59 71 83 95 107 119 131 75 (7,5,1) 6 16 31 35 50 65 69 84 99 103 118 133 76 (8,5,1) 11 17 26 35 55 64 73 82 91 111 120 129 307 | Mathematics @@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@ÚÓ‘Ój�n€a@Î@Úœäñ€a@‚Ï‹»‹€@·rÓ:a@Âig@Ú‹©@Ü‹1a26@@ÖÜ»€a@I2@‚b«@H2013 Ibn Al-Haitham Jour. for Pure & Appl. 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Vol. 26 (2) 2013 77 (9,5,1) 10 19 27 35 54 62 70 89 97 105 113 132 78 (10,5,1) 7 14 30 35 51 67 72 88 93 109 114 130 79 (0,6,1) 1 112 113 114 115 116 117 118 119 120 121 122 80 (1,6,1) 7 23 28 44 49 65 70 86 91 107 112 128 81 (2,6,1) 10 18 26 45 53 61 69 88 96 104 112 131 82 (3,6,1) 11 20 29 38 47 67 76 85 94 103 112 132 83 (4,6,1) 6 21 25 40 55 59 74 89 93 108 112 127 84 (5,6,1) 3 15 27 39 51 63 75 87 99 111 112 124 85 (6,6,1) 12 22 32 42 52 62 72 82 92 102 112 133 86 (7,6,1) 9 16 34 41 48 66 73 80 98 109 112 130 87 (8,6,1) 4 17 30 43 56 58 71 84 97 110 112 125 88 (9,6,1) 5 19 33 36 50 64 78 81 95 109 112 126 89 (10,6,1) 8 14 31 37 54 60 77 83 100 106 112 129 90 (0,7,1) 1 46 47 48 49 50 51 52 53 54 55 56 91 (1,7,1) 6 23 27 42 46 61 76 80 95 110 114 129 92 (2,7,1) 4 18 31 44 46 59 72 85 98 111 113 126 93 (3,7,1) 7 20 25 41 46 62 78 83 99 104 120 125 94 (4,7,1) 3 21 33 45 46 58 70 82 94 106 118 130 95 (5,7,1) 5 15 29 43 46 60 74 88 91 105 119 133 96 (6,7,1) 10 22 30 38 46 65 73 81 100 108 116 124 97 (7,7,1) 12 16 26 36 46 67 77 87 97 107 117 127 98 (8,7,1) 8 17 34 40 46 63 69 86 92 109 115 132 99 (9,7,1) 11 19 28 37 46 66 75 84 93 102 122 131 100 (10,7,1) 9 14 32 39 46 64 71 89 96 103 121 128 101 (0,8,1) 1 57 58 59 60 61 62 63 64 65 66 67 102 (1,8,1) 5 23 26 40 54 57 71 85 99 102 116 130 103 (2,8,1) 9 18 25 43 50 57 75 82 100 107 114 132 104 (3,8,1) 3 20 32 44 56 57 69 81 93 105 117 129 105 (4,8,1) 11 21 30 39 48 57 77 86 95 104 113 133 106 (5,8,1) 7 15 31 36 52 57 73 89 94 110 115 131 107 (6,8,1) 8 22 28 45 51 57 74 80 97 103 120 126 108 (7,8,1) 4 16 29 42 55 57 70 83 96 109 122 124 109 (8,8,1) 12 17 27 37 47 57 78 88 98 108 118 128 110 (9,8,1) 6 19 34 38 53 57 72 87 91 106 121 125 111 (10,8,1) 10 14 33 41 49 57 76 84 92 111 119 127 112 (0,9,1) 1 79 80 81 82 83 84 85 86 87 88 89 113 (1,9,1) 4 23 25 38 51 64 77 79 92 105 118 131 114 (2,9,1) 3 18 30 42 54 66 78 79 91 103 115 127 115 (3,9,1) 10 20 28 36 55 63 71 79 98 106 114 133 116 (4,9,1) 8 21 27 44 50 67 73 79 96 102 119 125 117 (5,9,1) 9 115 33 40 47 65 72 79 97 105 122 129 118 (6,9,1) 66 22 26 41 56 60 75 79 94 109 113 128 119 (7,9,1) 7 16 32 37 53 58 74 79 95 111 116 132 120 (8,9,1) 5 17 31 45 48 62 76 79 93 107 121 124 121 (9,9,1) 12 19 29 39 49 59 69 79 100 110 120 130 122 (10,9,1) 11 14 34 43 52 61 70 79 99 108 117 126 123 (0,10,1) 1 24 25 26 27 28 29 30 31 32 33 34 124 (1,10,1) 3 23 24 36 48 60 72 84 96 108 120 132 125 (2,10,1) 8 18 24 41 47 64 70 87 93 110 116 133 308 | Mathematics @@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@ÚÓ‘Ój�n€a@Î@Úœäñ€a@‚Ï‹»‹€@·rÓ:a@Âig@Ú‹©@Ü‹1a26@@ÖÜ»€a@I2@‚b«@H2013 Ibn Al-Haitham Jour. for Pure & Appl. Sci. Vol. 26 (2) 2013 126 (3,10,1) 6 20 24 39 54 58 73 88 92 102 122 126 127 (4,10,1) 5 21 24 38 52 66 69 83 97 111 114 128 128 (5,10,1) 11 15 24 44 53 62 71 80 100 109 118 127 129 (6,10,1) 4 22 24 37 50 63 76 89 91 104 117 130 130 (7,10,1) 10 16 24 43 51 59 78 86 94 102 121 129 131 (8,10,1) 9 17 24 42 49 67 74 81 99 106 113 131 132 (9,10,1) 7 19 24 40 56 61 77 82 89 103 119 124 133 (10,10,1) 12 14 24 45 55 65 75 85 95 105 115 125 309 | Mathematics @@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@ÚÓ‘Ój�n€a@Î@Úœäñ€a@‚Ï‹»‹€@·rÓ:a@Âig@Ú‹©@Ü‹1a26@@ÖÜ»€a@I2@‚b«@H2013 Ibn Al-Haitham Jour. for Pure & Appl. Sci. Vol. 26 (2) 2013 حول حقل كالو PG(2,11) االسقاطيى األقواس الكاملة في المستو محمود سالم فیاض الجامعة العراقیة / كلیة التربیة / قسم الحاسوب 2013شباط 3قبل البحث في: ، 2012ایلول 19استلم البحث في: الخالصة ≥n≥2 12عندما GF(11)األسقاطي على حقل كالوا ىفي المستو (k,n)في ھذا البحث قمنا بإنشاء األقواس الكاملة ) و (19,3و (12,2) . وجدنا األقواس(12,2)باستخدام الطریقة الھندسیة عن طریق اتحاد بعض األقواس العظمى ) أقواس (133,12) و (109,11) و(96,10) و (81,9) و(68,8) و(58,7) و (47,6) و(38,5) و (29,4 PG(2,11)االسقاطي ىكاملة في المستو الھندسة الجبریة ، االقواس الكاملة ، حقل كالوا .: الكلمات المفتاحیة 310 | Mathematics