The Lower Bounds of Eight and Fourth Blocking Sets and Existence of Minimal Blocking Sets

This paper contains two main results relating to the size of eight and fourth blocking set in PG(2,16). First gives new example for (129,9)complete arc. The second result we prove that there exists (k,13)complete arc in PG(2,16), k≤197. We classify the minimal blocking sets of size eight in PG(2,4).We show that Rédei –type minimal blocking sets of size eight exist in PG(2, 4). Also we classify the minimal blocking sets of size ten in PG(2, 5), We obtain an example of a minimal blocking set of size ten with at most 4-secants.We show that Rédei – type minimal blocking sets of size ten exists in PG(2, 5). ةظحلام : ةحورطلأا نم لتسم ثحبلا The Lower Bounds of Eight and Fourth ... 100 1.


Eight blocking sets in PG
Let B be an eight blocking set in PG(2, q) , q is square such that through each of its points there areq+1 lines, each lines contains at least q+8 points of B and forming a dual Baer subline .Then ( 1 ) For q>64 , B has at least 8q+2q+8 points.

Proof.
( 1 )Call the lines meeting B in q+8 or more points long lines . If two long lines meet out side of B , then B has at least 2(q+8 )+8(q-1)= 8q+2q+8 points and the desired bound is obtained . Hence |B|8q+2q+8 . So to assume that two long lines meet in B .Take l , a long line ,and p ,a point of B not on l. Then the long lines through p contain a dual Baer subline and meet l in a Baer subline. Let Q be a point on this Baer subline. Consider long lines through a point on an 8-secant to Q. These meet l in another Baer subline not containing Q. Two Baer sublines meet in at most two points and so l has at least 2q points . Since l was arbitrary every long line has at least 2q points and it follows that B has at least (q+1)( 2q-1)+1+7(q-q)=9q-6q points. Since 9q-6q≥ 8q+2q+8 so that |B|≥ 8q+2q+8 points.
Let pB, through B , since there are q+1 long lines through p. B has at least (q+1) (q+7)+1+7(q+1-(q+1) )=8q+q+8 points. Now |B|≥140 . If this bound is a chafed then (k ,9)-arc has k=133 and that impossible .See Table(3)from [2]. If |B|=8q+q+9 then k=132, that impossible. Since k≤131, hence |B|≥ 8q+q+10.    (2, 16) The object of this section is to obtain good lower bounds for the size of a fourth blocking sets in PG(2, q) , q is square. Theorem ( 2.2.1) (q>9, q is a square) Let B be a fourth blocking set in PG(2, q) , q is square, such that through each of its points there are q+1 lines, each containing at least q+4 points of B and forming a dual Baer subline . Then B has at least 4q+2q+4 points. Proof. Call the lines meeting B in q+4 or more points long lines .If two long lines meet out side of B ,then B has at least 2(q+4)+4(q-1)= 4q+2q+4 points and the desired bound is obtained .So assume that two long lines meet in B . let l be a long line and p a point of B not on l.Then the long lines through p contain a dual Baer subline and meet l in a Baer subline. Let Q be a point on this Baer subline. Consider long lines through a point on a 4-secant to Q. These meet l in another Baer subline not containing Q. Two Baer subline meets in at most two points and so l has at least 2q points . Since l was arbitrary every long line has at least 2q points and it follows that B has at least (q+1)( 2q-1)+1+3(q-q)=5q-2q points.

On Blocking sets:
In this section we have given the following information on the structure of such blocking sets. Definition (3.1.1) ( unital): [3] Points, that every line 1 qq  A unital in PG(2,q) is a set U of 1 q  joining two points of U intersects U in precisely points.
again straight forward counting gives that all other lines of the plane intersect U in precisely one point, and in fact at each point of U there is a unique tangent. So a unital is a minimal blocking set. In fact it turns out to be the largest one. Theorem(3.1.2) : [3] let B be a minimal blocking set in PG(2, q). Then |B|≤ q +1 with equality if and only if B is a unital in PG(2, q) ,q is square . Theorem (3.1.3) : [7] In PG(2,q),q square , q≥25 or q=9, there is no minimal blocking set of size q . qq Yahya Theorem(3.1.4) : [7] For q square , q≥16, there is no minimal blocking k-set B k  1 qq  in PG(2,q) with Theorem(3.1.5) : [7] In a Desargusian plane of order at least 4 there exists a blocking set of order k if 2q-1≤ k ≤3q-3. From these equations, we get d=3 . The first solution (8,10,0,3) and fourth solution (11,1,9, (11,1,9,0).Since c>0,let be a 3-secant. Now any two 3-secant must be intersect in a point of B.

Minimal
Since if two3-secant intersect in a point p  B, then |B|≥2*3+3*1=9 which is impossible. On every p  B there are at most three 3-secants passing through p. Now since T 3 =9 then the remaining eight 3-secants pass through the three points of  B , So we have a point of  B with at least four 3-secants, and that is impossible. Hence (11,1,9,0) does not exist.
Minimal Blocking sets in PG( 2 , 5 ): 3.3 The following lemmas give the properties of minimal blocking sets of size ten. Lemma( 3.3.1): Every blocking set of size ten in PG (2,5)    3.3. 9 Minimal blocking sets of size ten with at most 4secants: We find an example of minimal blocking sets of size ten with ten points. 3.3.11 Minimal blocking sets of size ten with 5-secants: The following theorems prove that the existence of minimal blocking sets of size ten, T 5 >0, T 4 ≠0. Theorem(3.3.12): Let B have at most 5 points on a line. Let the numbers of 1-, 2-, 3-,4-and 5-secants be denoted by a, b, c, d ,e resp . Then these numbers satisfy one of the following possibilities: a b c d e ( 2 ) 2b+ 6c+ 12d+20e =90 ...