396 

© Published by College of Education for Pure Science (Ibn Al-Haitham), University of Baghdad. This 

is an open-access article distributed under the terms of the Creative Commons Attribution 4.0 International License 

 

 

Extension of Size and Degree of (𝒌, 𝒓)-Caps in 𝑷𝑮(𝟑, 𝟏𝟑) 

 

Saja Makki Attook1 and Emad Bakr Al-Zangana2*  

 1,2Department of Mathematics, College of Science, Mustansiriyah University, Baghdad, Iraq. 

*Corresponding Author.  

 

 Received:3 February 2023                     Accepted:30 March 2023                       Published:20 July 2024  

 doi.org/10.30526/37.3.3268 

 
Abstract 

       The aim of this work is an extension of the  (𝑘, 𝑟)-caps (𝑘 is the order, 𝑟 is the degree), 

where 0 < 𝑟 < 14, in the three projective space of dimension over the Galois field of order 

thirteen, 𝑃𝐺(3,13). The extensions have been done on the caps founded by action subgroups of 

projective general linear of order four over the finite field of order thirteen on 𝑃𝐺(3,13).  The 

main condition for the completion of the expansion process on the degree of caps is 14 (number 

of points on the line in 𝑃𝐺(3,13)) and the size  of the cap is points with a zero index zero, as it 

becomes complete when it is equal to zero. In this paper, we present fifteen caps (completes and 

incompletes) in 𝑃𝐺(3,13) of degrees 2,3,4,7 are extended in size until they reach 14, which are 

then complete caps, and then the 𝑐𝑖-distribution are computed for each new cap.  

Keywords: Cap, Complete cap, Finite projective space, Group action. 
 

1. Introduction 

     Let 𝐹𝑞 be the Galois field of 𝑞 elements and 𝑃𝐺(3, 𝑞) be the projective space of dimension 

three. The points [𝑥0, 𝑥1, 𝑥2, 𝑥3] of 𝑃𝐺(3, 𝑞) are the 1-dimensional subspaces of the vector space 

𝐹𝑞
4 over 𝐹𝑞. Subspaces of dimension two are called lines and dimension three planes. The number 

of points and the number of planes in 𝑃𝐺(3, 𝑞) is 𝑞3 + 𝑞2 + 𝑞 + 1. The number of lines is (𝑞𝑛+1 −

1)(𝑞𝑛 − 1)/(𝑞2 − 1)(𝑞 − 1). There are 𝑞 + 1 points on every line, 𝑞 + 1 lines through every 

point, and 𝑞2 + 𝑞 + 1 points on every plane; see [1],[2] and [3]. 

      Many articles have been published about finding (𝑘, 𝑟)-caps in the projective space especially 

caps of degree 2. The classification of caps and determine their completeness are hard tasks 

because of the number of points and lines in the projective space. So, the number of researchers 

focused on the theoretical research method to find a specific case gives complete caps. But no one 

has been able to give a full classification of the caps on specific projective spaces.  In [4], Segre 

constructed complete ((3𝑞 + 2), 2)-caps in 𝑃𝐺(3, 2ℎ). In [5], Alexander gave a generalization of 

Segre’s construction of the complete caps in 𝑃𝐺(3, 2ℎ). In [6], Anbar et. al. used a geometrical 

object arc in the projective plane to produce complete caps of size 𝑘𝑞(𝑁−2)/2 in affine spaces of 

dimension 𝑁 ≡ 0 (𝑚𝑜𝑑 4). A computer search has been used by a number of authors to find new 

caps and complete caps such as Alexander et. al. [7] used a computer search in the finite projective 

spaces 𝑃𝐺(𝑛, 𝑞) for the spectrum of possible sizes 𝑘 of complete 𝑘-caps. Also, in [8-10], a group 

action on the projective space of dimension three have been used to construct special types of caps 

https://creativecommons.org/licenses/by/4.0/
https://orcid.org/0000-0003-0195-2188
mailto:sawsave2233@gmail.com
https://orcid.org/0000-0001-6415-1930
mailto:e.b.abdulkrareem@uomustansiriyah.edu.iq


IHJPAS. 2024, 37( 3 ) 

397 
 

for 𝑞 = 8,11,13,23. As in [9], in this paper, we did an extension of the caps that have been 

constructed in [10] with respect to their size and degree when 𝑞 = 13. All the results obtained 

from this research could be of interest to many researchers because of their relationship to linear 

coding, see[11-18], as well as their relationship to cryptography, see [19-21] and also the network 

system, see [22-24]. Regarding finite projective spaces with a dimension higher than three, some 

studies have appeared recently discussing some projective concepts, as in [25-31]. The algorithms 

in this paper have been implemented by  the GAP program [32]. 

 

2. Preliminary Concepts  

Definition 2.1 [1]: A (𝑘, 𝑟)-cap in 𝑃𝐺(𝑛 ≥ 3, 𝑞) is a set of 𝑘 points such that no 𝑟 + 1   points are 

collinear, but at most 𝑟 points of which lie in any line. Here, 𝑟 is called a degree of the    (𝑘, 𝑟)-

cap. The (𝑘, 𝑟)-cap is called a complete cap if it is not contained in (𝑘 + 1, 𝑟)-cap. 

Definition 2.2 [2]: Let 𝐾 be a cap of degree 𝑟, an 𝑖-secant of a 𝐾 in 𝑃𝐺(𝑛, 𝑞) is a line such that |𝑘 ∩

𝜋| = 𝑖. The number of 𝑖-secants of 𝐾 denoted by 𝜏𝑖. 

Definition 2.3 [1]: Let 𝑄 be a point not on the (𝑘, 𝑟)-cap, 𝐾. The number of 𝑖-secant of 𝐾 passing 

through 𝑄 denoted by 𝜎𝑖(𝑄). The number 𝜎𝑟(𝑄) of 𝑟-secants is called the index of 𝑄 with respect 

to 𝐾.  

Definition 2.4 [2]: The set of all points of index 𝑖 will be denoted by 𝐶𝑖 and the cardinality of 𝐶𝑖 

is denoted by 𝑐𝑖. The sequence (𝑡0, … , 𝑡𝑟) will represent the secant distribution and the sequences 

(𝑐0, … , 𝑐𝑑) refer to the index distribution. 

Definition 2.5 [2,3]: The group of projectivities of  𝑃𝐺(𝑛, 𝑞) is called the projective general linear 

group, and is denoted by 𝑃𝐺𝐿 (𝑛 + 1, 𝑞).  

 

3. (𝒌, 𝒓)-Caps in 𝑷𝑮(𝟑, 𝟏𝟑) 

Let 𝐹13 = 〈𝜏〉 = {0, 1, 𝜏, 𝜏2, 𝜏3, 𝜏4, 𝜏5, 𝜏6, 𝜏7, 𝜏8, 𝜏9, 𝜏10, 𝜏11} be the 13th-order Galois field, where 

𝜏 is the primitive element of 𝐹13. Let 𝑆 be the non-singular 4 × 4 companion matrix over 𝐹13 

𝑆 = [

0  1
0  0

0   0
1   0

0 0
 𝜏5  𝜏3

0  1
0  1

]. 

The cyclic subgroup 〈𝑆〉 of 𝑃𝐺𝐿(4,13) is of order 𝜃3(13) = 2380. The projective space 𝑃𝐺(3,13) 

has 𝜃3(13) = 2380 points and planes, 31110 lines, 14 points on each line and 183 lines passing 

through each point.  

The number 2380 has 22 non-trivial divisors 𝑃 which are: 2,4,5,7,10,14,17,20,28,34,35,68, 

70,85,119,140,170,238,340,476,595,1190. The cyclic subgroups 〈𝑆𝑃〉 of the cyclic group 〈𝑆〉 are 

constructed in [10] and used to formed caps. The following is the summary of these caps results.  

From the action of the subgroups 〈𝑆𝑃〉 on 𝑃𝐺(3,13) deduced 22 equivalence classes (orbits) in 

𝑃𝐺(3,13) formed caps are summarized below: Let 𝜒𝑃 be the first orbit from the action of 〈𝑆𝑃〉 on 

𝑃𝐺(3,13). 

1. 𝜒2 = {1 + 2𝑘|𝑘 = 0, … ,1189} is incomplete (1190,14)-cap. 

2. 𝜒4 = {1 + 4𝑘|𝑘 = 0, … ,594} is complete (595,7)-cap. 

3. 𝜒5 = {1 + 5𝑘|𝑘 = 0, … ,475} is incompletes (476,14)-cap. 

4. 𝜒7 = {1 + 7𝑘|𝑘 = 0, … ,339} is completes (340,4)-cap. 

5. 𝜒10 = {1 + 10𝑘|𝑘 = 0, … ,237} is incompletes (238,14)-cap. 

6. 𝜒14 = {1 + 14𝑘|𝑘 = 0, … ,169} is completes (170,2)-cap. 



IHJPAS. 2024, 37( 3 ) 

398 
 

7. 𝜒17 =  {1 + 17𝑘|𝑘 = 0, … ,139} is incompletes (140,14)-cap. 

8. 𝜒20 = {1 + 20𝑘|𝑘 = 0, … ,118} is incompletes (119,7)-cap. 

9. 𝜒28 = {1 + 28𝑘|𝑘 = 0, … ,84} is incompletes (85,2)-cap. 

10. 𝜒34 = {1 + 34𝑘|𝑘 = 0, … ,69} is incompletes (70,14)-cap. 

11. 𝜒35 = {1 + 35𝑘|𝑘 = 0, … ,67} is incompletes (68,3)-cap. 

12. 𝜒68 = {1 + 68𝑘|𝑘 = 0, … ,34} is incompletes (35,7)-cap. 

13. 𝜒70 =  {1 + 70𝑘|𝑘 = 0, … ,33} is incompletes (34,2)-cap. 

14. 𝜒85 = {1 + 85𝑘|𝑘 = 0, … ,27} is incompletes (28,14)-cap. 

15. 𝜒119 =  {1 + 119𝑘|𝑘 = 0, … ,19} is incompletes (20,2)-cap. 

16. 𝜒140 = {1 + 140𝑘|𝑘 = 0, … ,16} is incompletes (17,2)-cap. 

17. 𝜒170 = {1 + 170𝑘|𝑘 = 0, … ,13} is incompletes (14,14)-cap. 

18. 𝜒238 = {1 + 238𝑘|𝑘 = 0, … ,9} is incompletes (10,2)-cap. 

19. 𝜒340 = {1 + 340𝑘|𝑘 = 0, … ,6} is incompletes (7,7)-cap. 

20. 𝜒476 = {1 + 476𝑘|𝑘 = 0, … ,4} is incompletes (5,2)-cap. 

21. 𝜒595 =  {1 + 595𝑘|𝑘 = 0, … ,3} is incompletes (4,2)-cap. 

22. 𝜒1190 = {1,1191} is incompletes (2,2)-cap. 

 

2. Extension of Caps 

        An exhaustive computer search proves the following theorem: 

Theorem: The caps 𝜒𝑖 , 𝑖 = 4,7,14,20,28,35,68,70,119,140,238,340,476,595,1190 can be 

extended by their sizes and degrees to complete caps. 

Proof: 

Let 𝒟𝑃
𝑝𝑖 = 𝜒𝑃⋃𝑝𝑖, where  𝑝𝑖 is the set of addition points, 1 ≤ 𝑖 ≤ 14. 

1. The line 𝐈(0,0,1,0,0,0) meets the cap 𝜒4 in 7 points, so, seven extension points have been added 

to 𝜒𝑃  in the following orders: 4, 144, 358, 470, 754, 755, 923. 

     Let  𝑝1 = {4}, 𝒟4
𝑝1 = 𝜒4⋃𝑝1. The maximum number of the intersection points between 𝜒4

𝑝1 and 

the lines of 𝑃𝐺(3,13) are eight, so 𝜒4
𝑝1 is (596,8)-cap and 𝑐𝑖 values are  𝑐0 = 1736, 𝑐1 = 48. 

Since 𝑐0 ≠ 0, then 𝜒4
𝑝1 is incomplete. The (596,8)-cap will be complete when you add 196 

points to it.  

    Let 𝑝2 = {4, 144}, 𝒟4
𝑝2 = 𝜒4⋃𝑝2. The maximum number of the intersection points between 𝜒4

𝑝2 

and the lines of 𝑃𝐺(3,13) are nine, so 𝜒4
𝑝2 is (597,9)-cap and 𝑐𝑖 values are  𝑐0 =1778, 𝑐1 = 5. 

Since 𝑐0 ≠ 0, then 𝜒4
𝑝2 is incomplete. The (597,9)-cap will be complete when you add 413 

points to it.   

    Let  𝑝3 = {4, 144,358}, 𝒟4
𝑝3= 𝜒4⋃𝑝3. The maximum number of the intersection points between 

𝜒4
𝑝3 and the lines of 𝑃𝐺(3,13) are ten, so 𝜒4

𝑝3 is (598,10)-cap and 𝑐𝑖 values are  𝑐0 = 1778,

𝑐1 = 4. Since 𝑐0 ≠ 0, then 𝜒4
𝑝3 is incomplete. The (598,10)-cap will be complete when you 

add 600 points to it.   

    Let  𝑝4 = {4, 144,358,470}, 𝒟4
𝑝4 = 𝜒4⋃𝑝4. The maximum number of the intersection points 

between 𝜒4
𝑝4 and lines of 𝑃𝐺(3,13) are eleven, so 𝜒4

𝑝4 is (599,11)-cap and 𝑐𝑖 values are  𝑐0 =

1778, 𝑐1 = 3. Since 𝑐0 ≠ 0, then 𝜒4
𝑝4 is incomplete. The (599,11)-cap will be complete when 

you add 702 points to it.   

     Let  𝑝5 = {4, 144,358,470,754}, 𝒟4
𝑝5 = 𝜒4⋃𝑝5. The maximum number of intersection points 

between 𝜒4
𝑝5 and lines of 𝑃𝐺(3,13) are twelve, so 𝜒4

𝑝5 is (600,12)-cap and 𝑐𝑖 values are  𝑐0 =



IHJPAS. 2024, 37( 3 ) 

399 
 

1778, 𝑐1 = 2. Since 𝑐0 ≠ 0, then 𝜒4
𝑝5 is incomplete. The (600,12)-cap will be complete when 

you add 1016 points to it.   

    Let 𝑝6 = {4, 144,358,470,754,755}, 𝒟4
𝑝6 = 𝜒4⋃𝑝6. The maximum number of the intersection 

points between 𝜒4
𝑝6 and lines of 𝑃𝐺(3,13) are thirteen, so 𝜒4

𝑝6 is (601,13)-cap and 𝑐𝑖 values 

are  𝑐0 = 1778, 𝑐1 = 1. Since 𝑐0 ≠ 0, then 𝜒4
𝑝6 is incomplete. The (601,13)-cap will be 

complete when you add 1148 points to it.   

     Let 𝑝7 = {4, 144,358,470,754,755,923}, 𝒟4
𝑝7 = 𝜒4⋃𝑝7. The maximum number of the 

intersection points between 𝜒4
𝑝7 and lines of 𝑃𝐺(3,13) are fourteen, so 𝜒4

𝑝7 is (602,14)-cap and 

𝑐𝑖 values are  𝑐0 = 1778. Since 𝑐0 ≠ 0, then 𝜒4
𝑝7 is incomplete. The (602,14)-cap will be 

complete when you add 1778 points to it.  

  

2. The line 𝐈(0,1,0,0,0,0) meets the cap 𝜒7 in 4 points, so we have ten extension points to be added 

to 𝜒𝑃  in the following orders: 3, 249, 488, 602, 1006, 1075, 1232, 1251, 1433, 2341. 

     Let  𝑝1 = {3}, 𝒟7
𝑝1 = 𝜒7⋃𝑝1. The maximum number of the intersection points between 𝜒7

𝑝1 and 

the lines of 𝑃𝐺(3,13) are five, so 𝜒7
𝑝1 is (341,5)-cap and 𝑐𝑖 values are  𝑐0 = 1751, 𝑐1 = 288. 

Since 𝑐0 ≠ 0, then 𝜒7
𝑝1 is incomplete. The (341,5)-cap will be complete when you add 66 

points to it.   

     Let 𝑝2 = {3, 249},  𝒟7
𝑝2= 𝜒7⋃𝑝2. The maximum number of the intersection points between 

𝜒7
𝑝2 and the lines of 𝑃𝐺(3,13) are six, so 𝜒7

𝑝2 is (342,6)-cap and 𝑐𝑖 values are  𝑐0 =1011, 𝑐1 =

4. Since 𝑐0 ≠ 0, then 𝜒7
𝑝2 is incomplete. The (342,6)-cap will be complete when you add 196 

points to it.   

     Let 𝑝3 = { 3, 249, 488},  𝒟7
𝑝3  =  𝜒7 ⋃ 𝑝3. The maximum number of the intersection points 

between 𝜒7
𝑝3 and the lines of 𝑃𝐺(3,13) are seven, so 𝜒7

𝑝3 is (343,7)-cap and 𝑐𝑖 values are 𝑐0 =

2030, 𝑐1 = 7. Since 𝑐0 ≠ 0, then 𝜒7
𝑝3 is incomplete. The (343,7)-cap will be complete when 

you add 335 points to it.   

    Let 𝑝4 = { 3, 249, 488, 602},  𝒟7
𝑝4= 𝜒7 ⋃ 𝑝4. The maximum number of the intersection points 

between 𝜒7
𝑝4 and the lines of 𝑃𝐺(3,13) are eight, so 𝜒7

𝑝4 is (344,8)-cap and 𝑐𝑖 values are  𝑐0 =

2030, 𝑐1 = 6. Since 𝑐0 ≠ 0, then 𝜒7
𝑝4 is incomplete. The (344,8)-cap will be complete when 

you add 463 points to it.   

    Let 𝑝5 = { 3, 249, 488, 602, 1006},  𝒟7
𝑝5=  𝜒7 ⋃ 𝑝5. The maximum number of the 

intersection points between 𝜒7
𝑝5 and the lines of 𝑃𝐺(3,13) are nine, so 𝜒7

𝑝5 is (345,9)-cap and 

𝑐𝑖 values are  𝑐0 = 2030, 𝑐1 = 5. Since 𝑐0 ≠ 0, then 𝜒7
𝑝5 is incomplete. The (345,9)-cap will 

be complete when you add 685 points to it.   

     Let  𝑝6 = {3, 249, 488, 602, 1006, 1075},  𝒟7
𝑝6 = 𝜒7 ⋃ 𝑝6. The maximum number of the 

intersection points between 𝜒7
𝑝6 and the lines of 𝑃𝐺(3,13) are ten, so 𝜒7

𝑝6 is (346,10)-cap and 

𝑐𝑖 values are  𝑐0 = 2030, 𝑐1 = 4. Since 𝑐0 ≠ 0, then 𝜒7
𝑝6 is incomplete. The (346,10)-cap will 

be complete when you add 816 points to it. 

     Let 𝑝7 = {3, 249, 488, 602, 1006, 1075, 1232},  𝒟7
𝑝7= 𝜒7⋃𝑝7. The maximum number of 

the intersection points between 𝜒7
𝑝7 and the lines of 𝑃𝐺(3,13) are eleven, so 𝜒7

𝑝7 is (347,11)-

cap and 𝑐𝑖 values are  𝑐0 = 2030, 𝑐1 = 3. Since 𝑐0 ≠ 0, then 𝜒7
𝑝7 is incomplete. The (347,11)-

cap will be complete when you add 991 points to it.   



IHJPAS. 2024, 37( 3 ) 

400 
 

     Let 𝑝8 = {3, 249, 488, 602, 1006, 1075, 1232, 1251},  𝒟7
𝑝8 = 𝜒7⋃𝑝8. The maximum number 

of the intersection points between 𝜒7
𝑝8 and the lines of 𝑃𝐺(3,13) are twelve, so 𝜒7

𝑝8 is (348,12)-

cap and 𝑐𝑖 values are  𝑐0 = 2030, 𝑐1 = 2. Since 𝑐0 ≠ 0, then 𝜒7
𝑝8 is incomplete. The (348,12)-

cap will be complete when you add 1180 points to it.   

Let  𝑝9 = {3, 249, 488, 602, 1006, 1075, 1232, 1251, 1433},  𝒟7
𝑝9= 𝜒7⋃𝑝9. The maximum 

number of intersection points between 𝜒7
𝑝9 and the lines of 𝑃𝐺(3,13) are thirteen, so 𝜒7

𝑝9 is 

(349,13)-cap and 𝑐𝑖 values are  𝑐0 = 2030, 𝑐1 = 1. Since 𝑐0 ≠ 0, then 𝜒7
𝑝9 is incomplete. The 

(349,13)-cap will be complete when you add 1395 points to it.   

Let 𝑝10 = {3, 249, 488, 602, 1006, 1075, 1232, 1251, 1433, 2341},  𝒟7
𝑝10 = 𝜒7 ⋃𝑝10. The 

maximum number of the intersection points between 𝜒7
𝑝10  and the lines of 𝑃𝐺(3,13) are 

fourteen, so 𝜒7
𝑝10  is (350,14)-cap and 𝑐𝑖 values are  𝑐0 = 2030. Since 𝑐0 ≠ 0, then 𝜒7

𝑝10  is 

incomplete. The (350,14)-cap will be complete when adding 2030 points to it.   

3. The line 𝐈(0,1,0,0,0,0) meets the cap 𝜒14 in 2 points, so we have twelve extension points that 

can be added to 𝜒𝑃,  as in the following order: 

3,176,249,488,602,638,1006,1075,1232,1251,1433, 2341.  

Let 𝑝1 = {3},  𝒟14
𝑝1= 𝜒14⋃𝑝1. The maximum number of the intersection points between 𝜒14

𝑝1 and 

the lines of 𝑃𝐺(3,13) are three, so 𝜒14
𝑝1 is (171,3)-cap and 𝑐𝑖 values are  𝑐0 = 1351,         𝑐1 =

858. Since 𝑐0 ≠ 0, then 𝜒14
𝑝1 is incomplete. The (171,3)-cap will be complete when you add 20 

points to it.   

Let 𝑝2 = {3, 176},  𝒟14
𝑝2= 𝜒14⋃𝑝2. The maximum number of the intersection points between 𝜒14

𝑝2 

and the lines of 𝑃𝐺(3,13) are four, so 𝜒14
𝑝2 is (172,4)-cap and 𝑐𝑖 values are  𝑐0 =2198, 𝑐1=10. 

Since 𝑐0 ≠ 0, then 𝜒14
𝑝2 is incomplete. The (172,4)-cap will be complete when adding 119 

points to it.   

Let 𝑝3 = {3, 176, 249},  𝒟14
𝑝3= 𝜒14 ⋃ 𝑝3. The maximum number of the intersection points 

between 𝜒14
𝑝3 and the lines of 𝑃𝐺(3,13) are five, so 𝜒14

𝑝3 is (173,5)-cap and 𝑐𝑖 values are  𝑐0 =

2198, 𝑐1 = 9. Since 𝑐0 ≠ 0, then 𝜒14
𝑝3 is incomplete. The (173,5)-cap will be complete when 

you add 254 points to it.   

Let 𝑝4 = {3, 176, 249, 488},  𝒟14
𝑝4  =  𝜒14 ⋃ 𝑝4. The maximum number of the intersection points 

between 𝜒14
𝑝4 and the lines of 𝑃𝐺(3,13) are six, so 𝜒14

𝑝4 is (174,6)-cap and 𝑐𝑖 values are         𝑐0 =

2198, 𝑐1 = 8. Since 𝑐0 ≠ 0, then 𝜒14
𝑝4 is incomplete. The (174,6)-cap will be complete when 

you add 352 points to it.   

Let 𝑝5 = {3, 176, 249, 488, 602},  𝒟14
𝑝5 = 𝜒14⋃𝑝5. The maximum number of the intersection 

points between 𝜒14
𝑝5 and the lines of 𝑃𝐺(3,13) are seven, so 𝜒14

𝑝5 is (175,7)-cap and 𝑐𝑖 values 

are  𝑐0 = 2198, 𝑐1 = 7. Since 𝑐0 ≠ 0, then 𝜒14
𝑝5 is incomplete. The (175,7)-cap will be 

complete when you add 537 points to it. 

Let 𝑝6 = {3, 176, 249, 488, 602, 638 },  𝒟14
𝑝6= 𝜒14⋃𝑝6. The maximum number of the 

intersection points between 𝜒14
𝑝6 and the lines of 𝑃𝐺(3,13) are eight, so 𝜒14

𝑝6 is (176,8)-cap and 

𝑐𝑖 values are  𝑐0 = 2198, 𝑐1 = 6. Since 𝑐0 ≠ 0, then 𝜒14
𝑝6 is incomplete. The (176,8)-cap will 

be complete when you add 629 points to it.   

Let  𝑝7 = { 3, 176, 249, 488, 602, 638, 1006},  𝒟14
𝑝7= 𝜒14⋃𝑝7. The maximum number of 

the intersection points between 𝜒14
𝑝7 and the lines of 𝑃𝐺(3,13) are nine, so 𝜒14

𝑝7  is (177,9)-cap 



IHJPAS. 2024, 37( 3 ) 

401 
 

and 𝑐𝑖 values are  𝑐0 = 2198, 𝑐1 = 5. Since 𝑐0 ≠ 0, then 𝜒14
𝑝7 is incomplete. The (177,9)-cap 

will be complete when you add 818 points to it.   

Let  𝑝8 = {3, 176, 249, 488, 602, 638,1006, 1075},  𝒟14
𝑝8= 𝜒14⋃𝑝8. The maximum number of 

the intersection points between 𝜒14
𝑝8 and the lines of 𝑃𝐺(3,13) are ten, so 𝜒14

𝑝8 is (178,10)-cap 

and 𝑐𝑖 values are  𝑐0 = 2198, 𝑐1 = 4. Since 𝑐0 ≠ 0, then 𝜒14
𝑝8 is incomplete. The (178,10)-cap 

will be complete when you add 970 points to it.   

Let 𝑝9 = { 3, 176, 249, 488, 602, 638, 1006, 1075, 1232},  𝒟14
𝑝9  = 𝜒14 ⋃ 𝑝9. The maximum 

number of the intersection points between 𝜒14
𝑝9 and the lines of 𝑃𝐺(3,13) are eleven, so 𝜒14

𝑝9  is 

(179,11)-cap and 𝑐𝑖 values are  𝑐0 = 2198, 𝑐1 = 3. Since 𝑐0 ≠ 0, then 𝜒14
𝑝9 is incomplete. The 

(179,11)-cap will be complete when you add 1200 points to it.   

Let  𝑝10 = {3, 176, 249, 488, 602, 638, 1006, 1075, 1232, 1251},  𝒟14
𝑝10  =  𝜒14 ⋃ 𝑝10. The 

maximum number of the intersection points between 𝜒14
𝑝10  and the lines of 𝑃𝐺(3,13) are twelve, 

so 𝜒14
𝑝10  is (180,12)-cap and 𝑐𝑖 values are  𝑐0 = 2198, 𝑐1 = 2. Since 𝑐0 ≠ 0, then 𝜒14

𝑝10  is 

incomplete. The (180,12)-cap will be complete when you add 1325 points to it.   

Let 𝑝11 = {3, 176, 249, 488, 602, 638, 1006, 1075, 1232, 1251, 1433},  𝒟14
𝑝11 = 𝜒14⋃𝑝11. The 

maximum number of the intersection points between 𝜒14
𝑝11  and the lines of 𝑃𝐺(3,13) are 

thirteen, so 𝜒14
𝑝11  is (181,13)-cap and 𝑐𝑖 values are  𝑐0 = 2198, 𝑐1 = 1. Since 𝑐0 ≠ 0, then 

𝜒14
𝑝11  is incomplete. The (181,13)-cap will be complete when you add 1516 points to it.   

Let  𝑝12 = {3, 176, 249, 488, 602, 638, 1006, 1075, 1232, 1251, 1433, 2341} .                         

Put 𝒟14
𝑝12= 𝜒14⋃𝑝12. The maximum number of the intersection points between 𝜒14

𝑝12  and the lines 

of 𝑃𝐺(3,13) are fourteen, so 𝜒14
𝑝12 is (182,14)-cap and 𝑐𝑖 values are  𝑐0 = 2198. Since 𝑐0 ≠0, 

then 𝜒14
𝑝12  is incomplete. The (182,14)-cap will be complete when you add 2198 points to it. 

 4. The line 𝐈(𝜏11, 1,0,0,0,0) meets the cap 𝜒20 in 7 points, so we have seven extension points that 

can be added to 𝜒𝑃, as in the following orders: 171, 511, 851, 1191, 1531, 1871, 2211. 

Let 𝑝1 = {171},  𝒟20
𝑝1= 𝜒20⋃𝑝1. The maximum number of the intersection points between 𝜒20

𝑝1 

and the lines of 𝑃𝐺(3 13) are eight, so 𝜒20
𝑝1 is (120,8)-cap and 𝑐𝑖 values are  𝑐0 = 2254, 𝑐1 =

6. Since 𝑐0 ≠ 0, then 𝜒20
𝑝1 is incomplete. The (120,8)-cap will be complete when you add 658 

points to it. 

Let 𝑝2 = {171, 511},  𝒟20
𝑝2= 𝜒20⋃𝑝2. The maximum number of the intersection points between 

𝜒20
𝑝2 and the lines of 𝑃𝐺(3,13) are nine, so 𝜒20

𝑝2 is (121,9)-cap and 𝑐𝑖 values are  𝑐0 = 2254,

𝑐1 = 5. Since 𝑐0 ≠ 0, then 𝜒20
𝑝2 is incomplete. The (121,9)-cap will be complete when you add 

827 points to it.   

Let 𝑝3 = { 171, 511, 851},  𝒟20
𝑝3 =  𝜒20  ⋃ 𝑝3. The maximum number of the intersection 

points between 𝜒20
𝑝3 and the lines of 𝑃𝐺(3,13) are ten, so 𝜒20

𝑝3 is (122,10)-cap and 𝑐𝑖 values 

are  𝑐0 = 2254, 𝑐1 = 4. Since 𝑐0 ≠ 0, then 𝜒20
𝑝3 is incomplete. The (122,10)-cap will be 

complete when you add 1033 points to it.  

Let 𝑝4 = {171, 511, 851, 1191},  𝒟20
𝑝4 = 𝜒20⋃𝑝4. The maximum number of intersection points 

between 𝜒20
𝑝4  and the lines of 𝑃𝐺(3,13) are eleven, so 𝜒20

𝑝4 is (123,11)-cap and 𝑐𝑖 values are 

 𝑐0 = 2254, 𝑐1 = 3. Since 𝑐0 ≠ 0, then 𝜒20
𝑝4 is incomplete. The (123,11)-cap will be 

complete when you add 1250 points to it.  

Let 𝑝5 = {171, 511, 851, 1191, 1531 },  𝒟20
𝑝5= 𝜒20⋃𝑝5. The maximum number of the 

intersection points between 𝜒20
𝑝5 and the lines of 𝑃𝐺(3,13) are twelve, so 𝜒20

𝑝5 is (124,12)-cap 



IHJPAS. 2024, 37( 3 ) 

402 
 

and 𝑐𝑖 values are  𝑐0 = 2254, 𝑐1 = 2. Since 𝑐0 ≠ 0, then 𝜒20
𝑝5 is incomplete. The (124,12)-

cap will be complete when you add 1383 points to it.  

Let  𝑝6 = {171, 511, 851, 1191, 1531, 1871},  𝒟20
𝑝6= 𝜒20⋃𝑝6. The maximum number of 

the intersection points between 𝜒20
𝑝6 and the lines of 𝑃𝐺(3,13) are thirteen, so 𝜒20

𝑝6 is (125,13)-

cap and 𝑐𝑖 values are  𝑐0 = 2254, 𝑐1 = 1. Since 𝑐0 ≠ 0, then 𝜒20
𝑝6 is incomplete. The 

(125,13)-cap will be complete when you add 1555 points to it.  

Let  𝑝7 = {171, 511, 851, 1191,1531, 1871, 2211},  𝒟20
𝑝7=  𝜒20⋃ 𝑝7. The maximum number 

of the intersection points between 𝜒20
𝑝7 and the lines of 𝑃𝐺(3,13) are fourteen, so 𝜒20

𝑝7 is        

(126,14)-cap and 𝑐𝑖 values are  𝑐0 = 2254. Since 𝑐0 ≠ 0, then 𝜒20
𝑝7 is incomplete. The 

(126,14)-cap will be complete when you add 2254 points to it. 

5. The line 𝐈(0,0,1,0,0,0) meets the cap 𝜒28 in 2 points, so we have twelve extension points that 

can be added to 𝜒𝑃, as in the following order: 

 4,144,358,470,593,754,755,923,1213,2105,2165, 2369.  

Let 𝑝1 = {4},  𝒟28
𝑝1= 𝜒28⋃𝑝1. The maximum number of the intersection points between 𝜒28

𝑝1 and 

the lines of 𝑃𝐺(3,13) are three so 𝜒28
𝑝1 is (86,3)-cap and 𝑐𝑖 values are  𝑐0 = 2096, 𝑐1 = 198. 

Since 𝑐0 ≠ 0, then 𝜒28
𝑝1 is incomplete. The (86,3)-cap will be complete when you add 74 points 

to it.   

Let 𝑝2 = {4, 144},  𝒟28
𝑝2 = 𝜒28⋃𝑝2. The maximum number of the intersection points between 

𝜒28
𝑝2 and the lines of 𝑃𝐺(3,13) are four so 𝜒28

𝑝2 is (87,4)-cap and 𝑐𝑖 values are  𝑐0 =2283, 𝑐1 =

10. Since 𝑐0 ≠ 0, then 𝜒28
𝑝2 is incomplete. The (87,4)-cap will be complete when you add 176 

points to it.  

Let 𝑝3 = {4, 144, 358},  𝒟28
𝑝3= 𝜒28⋃𝑝3. The maximum number of the intersection points 

between 𝜒28
𝑝3 and the lines of 𝑃𝐺(3,13) are five, so 𝜒28

𝑝3 is (88,5)-cap and 𝑐𝑖 values are  𝑐0 =

2283, 𝑐1 = 9. Since 𝑐0 ≠ 0, then 𝜒28
𝑝3 is incomplete. The (88,5)-cap will be complete when 

you add 319 points to it.  

Let 𝑝4 = {4, 144, 358, 470}. Put  𝒟28
𝑝4= 𝜒28⋃𝑝4. The maximum number of the intersection 

points between 𝜒28
𝑝4 and the lines of 𝑃𝐺(3,13) are six, so 𝜒28

𝑝4 is (89,6)-cap and 𝑐𝑖 values are       

𝑐0 = 2283, 𝑐1 = 8. Since 𝑐0 ≠ 0, then 𝜒28
𝑝4 is incomplete. The (89,6)-cap will be complete 

when adding 433 points to it. Let 𝑝5 = {4, 144, 358, 470, 593},  𝒟28
𝑝5= 𝜒28⋃𝑝5. The maximum 

number of the intersection points between 𝜒28
𝑝5 and the lines of 𝑃𝐺(3,13) are seven, so 𝜒28

𝑝5  is 

(90,7)-cap and 𝑐𝑖 values are  𝑐0 = 2283, 𝑐1 = 7. Since 𝑐0 ≠ 0, then 𝜒28
𝑝5 is incomplete. The 

(90,7)-cap will be complete when you add 623 points to it.  

Let 𝑝6 = { 4, 144, 358, 470, 593, 754},  𝒟28
𝑝6  =  𝜒28 ⋃ 𝑝6. The maximum number of the 

intersection points between 𝜒28
𝑝6 and the lines of 𝑃𝐺(3,13) are eight, so 𝜒28

𝑝6 is (91,8)-cap and 

𝑐𝑖 values are  𝑐0 = 2283, 𝑐1 = 6. Since 𝑐0 ≠ 0, then 𝜒28
𝑝6 is incomplete. The (91,8)-cap will 

be complete when you add 671 points to it.  

Let 𝑝7 = {4, 144, 358, 470, 593, 754, 755},  𝒟28
𝑝7 = 𝜒28⋃𝑝7. The maximum number of the 

intersection points between 𝜒28
𝑝7 and the lines of 𝑃𝐺(3,13) are nine, so 𝜒28

𝑝7 is (92,9)-cap and 

𝑐𝑖 values are  𝑐0 = 2283, 𝑐1 = 5. Since 𝑐0 ≠ 0, then 𝜒28
𝑝7 is incomplete. The (92,9)-cap will 

be complete when you add 892 points to it.  

Let 𝑝8 = {4, 144, 358, 470, 593, 754, 755, 923},  𝒟28
𝑝8 = 𝜒28⋃𝑝8. The maximum number of the 

intersection points between 𝜒28
𝑝8 and the lines of 𝑃𝐺(3,13) are ten, so 𝜒28

𝑝8 is (93,10)-cap and 



IHJPAS. 2024, 37( 3 ) 

403 
 

𝑐𝑖 values are  𝑐0 = 2283, 𝑐1 = 4. Since 𝑐0 ≠ 0, then 𝜒28
𝑝8 is incomplete. The (93,10)-cap will 

be complete when you add 1047 points to it.  

Let 𝑝9 = {4, 144, 358, 470, 593, 754, 755, 923, 1213, }. Put  𝒟28
𝑝9= 𝜒28 ⋃ 𝑝9. The maximum 

number of the intersection points between 𝜒28
𝑝9 and the lines of 𝑃𝐺(3,13) are eleven, so 𝜒28

𝑝9  is 

(94,11)-cap and 𝑐𝑖 values are  𝑐0 = 2283, 𝑐1 = 3. Since 𝑐0 ≠ 0, then 𝜒28
𝑝9 is incomplete. The 

(94,11)-cap will be complete when you add 1299 points to it.  

Let 𝑝10 = { 4, 144, 358, 470, 593, 754, 755, 923, 1213, 2105}. Put  𝒟28
𝑝10  =  𝜒28⋃ 𝑝10. The 

maximum number of the intersection points between 𝜒28
𝑝10  and the lines of 𝑃𝐺(3,13) are 

twelve, so 𝜒28
𝑝10  is (95,12)-cap and 𝑐𝑖 values are  𝑐0 = 2283, 𝑐1 = 2. Since 𝑐0 ≠ 0, then 𝜒28

𝑝10  

is incomplete. The (95,12)-cap will be complete when you add 1367 points to it.  

Let 𝑝11 = {4, 144, 358, 470, 593, 754, 755, 923, 1213, 2105, 2165},  𝒟28
𝑝11  =  𝜒28⋃ 𝑝11. The 

maximum number of the intersection points between 𝜒28
𝑝11  and the lines of 𝑃𝐺(3,13) are 

thirteen, so 𝜒28
𝑝11  is (96,13)-cap and 𝑐𝑖 values are  𝑐0 = 2283, 𝑐1 = 1. Since 𝑐0 ≠ 0, then 𝜒28

𝑝11  

is incomplete. The (96,13)-cap will be complete when you add 1577 points to it.  

Let 𝑝12 = {4, 144, 358, 470, 593, 754, 755, 923, 1213, 2105, 2165, 2369},  𝒟28
𝑝12 = 𝜒28⋃𝑝12. 

The maximum number of the intersection points between 𝜒28
𝑝12  and the lines of 𝑃𝐺(3,13) are 

fourteen,  so 𝜒28
𝑝12  is (97,14)-cap and 𝑐𝑖 values are  𝑐0 = 2283. Since 𝑐0 ≠ 0, then 𝜒28

𝑝12  is 

incomplete. The (97,14)-cap will be complete when you add 2283 points to it.  

6. The line 𝐈(𝜏9, 𝜏10, 1,0,0,0) meets the cap 𝜒35 in 3 points, so we have eleven extension points 

that can be added to 𝜒𝑃 , as in the following order: 

356,565,718,1049,1054,1314,1377,1756, 1818,2253,2331.  

Let 𝑝1 = {356},  𝒟35
𝑝1= 𝜒35⋃𝑝1. The maximum number of the intersection points between 𝜒35

𝑝1 

and the lines of 𝑃𝐺(3,13) are four, so 𝜒35
𝑝1 is (69,4)-cap and 𝑐𝑖 values are  𝑐0 = 2291, 𝑐1 =

20. Since 𝑐0 ≠ 0, then 𝜒35
𝑝1 is incomplete. The (69,4)-cap will be complete when you add 189 

points to it.  

Let 𝑝2 = {356, 565},  𝒟35
𝑝2= 𝜒35⋃𝑝2. The maximum number of the intersection points between 

𝜒35
𝑝2 and the lines of 𝑃𝐺(3,13) are five, so 𝜒35

𝑝2 is (70,5)-cap and 𝑐𝑖 values are  𝑐0 = 2301,

𝑐1 = 9. Since 𝑐0 ≠ 0, then 𝜒35
𝑝2 is incomplete. The (70,5)-cap will be complete when you add 

334 points to it.  

Let 𝑝3 = {356, 565, 718},  𝒟35
𝑝3  =  𝜒35 ⋃ 𝑝3. The maximum number of the intersection points 

between 𝜒35
𝑝3 and the lines of 𝑃𝐺(3,13) are six, so 𝜒35

𝑝3 is (71,6)-cap and 𝑐𝑖 values are         𝑐0 =

2301, 𝑐1 = 8. Since 𝑐0 ≠ 0, then 𝜒35
𝑝3 is incomplete. The (71,6)-cap will be complete when 

you add 486 points to it.  

Let 𝑝4 = {356, 565, 718, 1049},  𝒟35
𝑝4 = 𝜒35⋃𝑝4. The maximum number of the intersection 

points between 𝜒35
𝑝4 and the lines of 𝑃𝐺(3,13) are seven, so 𝜒35

𝑝4 is (72,7)-cap and 𝑐𝑖 values are 

 𝑐0 = 2301, 𝑐1 = 7. Since 𝑐0 ≠ 0, then 𝜒35
𝑝4 is incomplete. The (72,7)-cap will be complete 

when you add 627 points to it.  

Let 𝑝5 = {356, 565, 718, 1049, 1054},  𝒟35
𝑝5= 𝜒35⋃𝑝5. The maximum number of the 

intersection points between 𝜒35
𝑝5 and the lines of 𝑃𝐺(3,13) are eight, so 𝜒35

𝑝5 is (73,8)-cap and 

𝑐𝑖 values are  𝑐0 = 2301, 𝑐1 = 6. Since 𝑐0 ≠ 0, then 𝜒35
𝑝5 is incomplete. The (73,8)-cap will 

be complete when you add 692 points to it.  



IHJPAS. 2024, 37( 3 ) 

404 
 

Let 𝑝6 = {356, 565, 718, 1049, 1054, 1314},  𝒟35
𝑝6 = 𝜒35⋃𝑝6. The maximum number of the 

intersection points between 𝜒35
𝑝6 and the lines of 𝑃𝐺(3,13) are nine, so 𝜒35

𝑝6 is (74,9)-cap and 

𝑐𝑖 values are  𝑐0 = 2301, 𝑐1 = 5. Since 𝑐0 ≠ 0, then 𝜒35
𝑝6 is incomplete. The (74,9)-cap will 

be complete when you add 876 points to it.  

Let 𝑝7 = {356, 565, 718, 1049, 1054, 1314, 1377},  𝒟35
𝑝7= 𝜒35⋃𝑝7. The maximum number of 

the intersection points between 𝜒35
𝑝7 and the lines of 𝑃𝐺(3,13) are ten, so 𝜒35

𝑝7 is (75,10)-cap 

and 𝑐𝑖 values are  𝑐0 = 2301, 𝑐1 = 4. Since 𝑐0 ≠ 0, then 𝜒35
𝑝7 is incomplete. The (75,10)-cap 

will be complete when you add 1037 points to it.  

Let  𝑝8 = { 356, 565, 718, 1049, 1054, 1314, 1377, 1756},  𝒟35
𝑝8  = 𝜒35⋃𝑝8. The 

maximum number of the intersection points between 𝜒35
𝑝8 and the lines of 𝑃𝐺(3,13) are eleven, 

so 𝜒35
𝑝8 is (76,11)-cap and 𝑐𝑖 values are  𝑐0 = 2301, 𝑐1 = 3. Since 𝑐0 ≠ 0, then 𝜒35

𝑝8 is 

incomplete. The (76,11)-cap will be complete when you add 1312 points to it.  

Let 𝑝9 = {356, 565, 718, 1049, 1054, 1314, 1377, 1756, 1818}. Put  𝒟35
𝑝9= 𝜒35⋃𝑝9. The 

maximum number of intersection points between 𝜒35
𝑝9 and the lines of 𝑃𝐺(3,13) are twelve, so 

𝜒35
𝑝9 is (77,12)-cap and 𝑐𝑖 values are  𝑐0 = 2301, 𝑐1 = 2. Since 𝑐0 ≠ 0, then 𝜒35

𝑝9 is 

incomplete. The (77,12)-cap will be complete when you add 1401 points to it.  

Let 𝑝10 = {356, 565, 718, 1049, 1054, 1314, 1377, 1756, 1818, 2253},  𝒟35
𝑝10 = 𝜒35⋃𝑝10. 

The maximum number of the intersection points between 𝜒35
𝑝10  and the lines of 𝑃𝐺(3,13) are 

thirteen, so 𝜒35
𝑝10  is (78,13)-cap and 𝑐𝑖 values are  𝑐0 = 2301, 𝑐1 = 1. Since 𝑐0 ≠ 0, then 𝜒35

𝑝10  

is incomplete. The (78,13)-cap will be complete when you add 1591 points to it.  

Let  𝑝11 = {356, 565, 718, 1049, 1054, 1314, 1377, 1756, 1818, 2253, 2331}.                       

Put  𝒟35
𝑝11  =  𝜒35 ⋃ 𝑝11. The maximum number of the intersection points between 𝜒35

𝑝11  and the 

lines of 𝑃𝐺(3,13) are fourteen, so 𝜒35
𝑝11  is (79,14)-cap and 𝑐𝑖 values are  𝑐0 = 2301. Since 

𝑐0 ≠ 0, then 𝜒35
𝑝11  is incomplete. The (79,14)-cap will be complete when you add 2301 points 

to it. 

7. The line 𝐈(𝜏11, 1,0,0,0,0) meets the cap 𝜒68 in 7 points, so we have seven extension points that 

can be added to 𝜒𝑃, as in the following order: 171, 511, 851, 1191, 1531, 1871, 2211. 

Let 𝑝1 = {171},  𝒟68
𝑝1= 𝜒68⋃𝑝1. The maximum number of the intersection points between 𝜒68

𝑝1 

and the lines of 𝑃𝐺(3,13) are eight, so 𝜒68
𝑝1 is (36,8)-cap and 𝑐𝑖 values are  𝑐0 = 2338, 𝑐1 =

6. Since 𝑐0 ≠ 0, then 𝜒68
𝑝1  is incomplete. The (36,8)-cap will be complete when you add 709 

points to it.  

Let 𝑝2 = {171, 511},  𝒟68
𝑝2= 𝜒68⋃𝑝2. The maximum number of the intersection points between 

𝜒68
𝑝2 and the lines of 𝑃𝐺(3,13) are nine, so 𝜒68

𝑝2 is (37,9)-cap and 𝑐𝑖 values are  𝑐0 = 2338,

𝑐1 = 5. Since 𝑐0 ≠ 0, then 𝜒68
𝑝2 is incomplete. The (37,9)-cap will be complete when you add 

901 points to it. 

Let 𝑝3 = {171, 511, 851},  𝒟68
𝑝3 = 𝜒68⋃𝑝3. The maximum number of the intersection points 

between 𝜒68
𝑝3 and the lines of 𝑃𝐺(3,13) are ten, so 𝜒68

𝑝3 is (38,10)-cap and 𝑐𝑖 values are      𝑐0 =

2338, 𝑐1 = 4. Since 𝑐0 ≠ 0, then 𝜒68
𝑝3 is incomplete. The (38,10)-cap will be complete when 

you add 1065 points to it.  

Let 𝑝4 = {171, 511, 851, 1191},  𝒟68
𝑝4 = 𝜒68⋃𝑝4. The maximum number of the intersection 

points between 𝜒68
𝑝4 and the lines of 𝑃𝐺(3,13) are eleven, so 𝜒68

𝑝4 is (39,11)-cap and 𝑐𝑖 values 



IHJPAS. 2024, 37( 3 ) 

405 
 

are  𝑐0 = 2338, 𝑐1 = 3. Since 𝑐0 ≠ 0, then 𝜒68
𝑝4 is incomplete. The (39,11)-cap will be 

complete when you add 1395 points to it.  

Let 𝑝5 = {171, 511, 851, 1191, 1531},  𝒟68
𝑝5= 𝜒68⋃𝑝5. The maximum number of the 

intersection points between 𝜒68
𝑝5 and the lines of 𝑃𝐺(3,13) are twelve, so 𝜒68

𝑝5 is (40,12)-cap 

and 𝑐𝑖 values are  𝑐0 = 2338, 𝑐1 = 2. Since 𝑐0 ≠ 0, then 𝜒68
𝑝5 is incomplete. The (40,12)-cap 

will be complete when you add 1427 points to it.  

Let 𝑝6 = {171, 511, 851, 1191, 1531, 1871},  𝒟68
𝑝6 = 𝜒68⋃𝑝6. The maximum number of the 

intersection points between 𝜒68
𝑝6 and the lines of 𝑃𝐺(3,13) are thirteen, so 𝜒68

𝑝6  is (41,13)-cap 

and 𝑐𝑖 values are  𝑐0 = 2338, 𝑐1 = 1. Since 𝑐0 ≠ 0, then 𝜒68
𝑝6 is incomplete. The (41,13)-cap 

will be complete when you add 1615 points to it.  

Let 𝑝7 = {171, 511, 851, 1191, 1531, 1871, 2211},  𝒟68
𝑝7= 𝜒68⋃𝑝7. The maximum number of 

the intersection points between 𝜒68
𝑝7 and the lines of 𝑃𝐺(3,13) are fourteen, so 𝜒68

𝑝7 is (42,14)-

cap and 𝑐𝑖 values are  𝑐0 = 2338. Since 𝑐0 ≠ 0, then 𝜒68
𝑝7 is incomplete. The (42,14)-cap will 

be complete when you add 2338 points to it.  

8. The line 𝐈(𝜏5, 𝜏5, 1,0,0,0) meets the cap 𝜒70 in 2 points, so we have twelve extension points that 

can be added to 𝜒𝑃, as in the following order: 20,202,772,1110,1150,1152,1325,1398, 

1637,1787,2155,2224.  

Let 𝑝1 = {20},  𝒟70
𝑝1 = 𝜒70⋃𝑝1. The maximum number of the intersection points between 𝜒70

𝑝1 

and the lines of 𝑃𝐺(3,13) are three, so 𝜒70
𝑝1 is (35,3)-cap and 𝑐𝑖 values are  𝑐0 = 2323, 𝑐1 =

22. Since 𝑐0 ≠ 0, then 𝜒70
𝑝1 is incomplete. The (35,3)-cap will be complete when you add 109 

points to it.  

Let 𝑝2 = {20, 202, },  𝒟70
𝑝2= 𝜒70⋃𝑝2. The maximum number of the intersection points between 

𝜒70
𝑝2 and the lines of 𝑃𝐺(3,13) are four, so 𝜒70

𝑝2 is (36,4)-cap and 𝑐𝑖 values are  𝑐0 = 2334,

𝑐1 = 10. Since 𝑐0 ≠ 0, then 𝜒70
𝑝2 is incomplete. The (36,4)-cap will be complete when you add 

226 points to it.  

Let 𝑝3 = {20, 202, 772},  𝒟70
𝑝3= 𝜒70⋃𝑝3. The maximum number of the intersection points 

between 𝜒70
𝑝3 and the lines of 𝑃𝐺(3,13) are five, so 𝜒70

𝑝3 is (37,5)-cap and 𝑐𝑖 values are  𝑐0 =

2334, 𝑐1 = 9. Since 𝑐0 ≠ 0, then 𝜒70
𝑝3 is incomplete. The (37,5)-cap will be complete when 

you add 368 points to it. 

Let 𝑝4 = {20, 202, 772, 1110},  𝒟70
𝑝4= 𝜒70⋃𝑝4. The maximum number of the intersection 

points between 𝜒70
𝑝4 and the lines of 𝑃𝐺(3,13) are six, so 𝜒70

𝑝4 is (38,6)-cap and 𝑐𝑖 values are        

𝑐0 = 2334, 𝑐1 = 8. Since 𝑐0 ≠ 0, then 𝜒70
𝑝4 is incomplete. The (38,6)-cap will be complete 

when you add 468 points to it.  

Let 𝑝5 = {20, 202, 772, 1110, 1150},  𝒟70
𝑝5 = 𝜒70⋃𝑝5. The maximum number of the 

intersection points between 𝜒70
𝑝5 and the lines of 𝑃𝐺(3,13) are seven, so 𝜒70

𝑝5 is (39,7)-cap and 

𝑐𝑖 values are  𝑐0 = 2334, 𝑐1 = 7. Since 𝑐0 ≠ 0, then 𝜒70
𝑝5 is incomplete. The (39,7)-cap will 

be complete when you add 639 points to it.  

Let 𝑝6 = { 20, 202, 772, 1110, 1150, 1152},  𝒟70
𝑝6  = 𝜒70⋃ 𝑝6. The maximum number of the 

intersection points between 𝜒70
𝑝6 and the lines of 𝑃𝐺(3,13) are eight, so 𝜒70

𝑝6 is (40,8)-cap and 

𝑐𝑖 values are  𝑐0 = 2334, 𝑐1 = 6. Since 𝑐0 ≠ 0, then 𝜒70
𝑝6 is incomplete. The (40,8)-cap will 

be complete when you add 697 points to it.  



IHJPAS. 2024, 37( 3 ) 

406 
 

Let 𝑝7 = {20, 202, 772, 1110, 1150, 1152, 1325},  𝒟70
𝑝7= 𝜒70⋃𝑝7. The maximum number of 

the intersection points between 𝜒70
𝑝7 and the lines of 𝑃𝐺(3,13) are nine, so 𝜒70

𝑝7 is (41,9)-cap 

and 𝑐𝑖 values are  𝑐0 = 2334, 𝑐1 = 5. Since 𝑐0 ≠ 0, then 𝜒70
𝑝7 is incomplete. The (41,9)-cap 

will be complete when you add 989 points to it.  

Let 𝑝8 = { 20, 202, 772, 1110, 1150, 1152, 1325, 1398},  𝒟70
𝑝8= 𝜒70⋃ 𝑝8. The maximum 

number of the intersection points between 𝜒70
𝑝8 and the lines of 𝑃𝐺(3,13) are ten, so 𝜒70

𝑝8 is 

(42,10)-cap and 𝑐𝑖 values are  𝑐0 = 2334, 𝑐1 = 4. Since 𝑐0 ≠ 0, then 𝜒70
𝑝8 is incomplete. The 

(42,10)-cap will be complete when you add 1056 points to it.  

Let 𝑝9 = {20, 202, 772, 1110, 1150, 1152, 1325, 1398, 1637},  𝒟70
𝑝9= 𝜒70⋃𝑝9. The 

maximum number of the intersection points between 𝜒70
𝑝9 and the lines of 𝑃𝐺(3,13) are eleven, 

so 𝜒70
𝑝9 is (43,11)-cap and 𝑐𝑖 values are  𝑐0 = 2334, 𝑐1 = 3. Since 𝑐0 ≠ 0, then 𝜒70

𝑝9 is 

incomplete. The (43,11)-cap will be complete when you add 1330 points to it.  

Let 𝑝10 = {20, 202, 772, 1110, 1150, 1152, 1325, 1398, 1637, 1787},  𝒟70
𝑝10 = 𝜒70⋃𝑝10. 

The maximum number of the intersection points between 𝜒70
𝑝10  and the lines of 𝑃𝐺(3,13) are 

twelve, so 𝜒70
𝑝10  is (44,12)-cap and 𝑐𝑖 values are  𝑐0 = 2334, 𝑐1 = 2. Since 𝑐0 ≠ 0, then 𝜒70

𝑝10  

is incomplete. The (44,12)-cap will be complete when you add 1421 points to it.  

Let  𝑝11 = {20, 202, 772, 1110, 1150, 1152, 1325, 1398, 1637, 1787,2155}.              

Put  𝒟70
𝑝11 = 𝜒70⋃𝑝11. The maximum number of the intersection points between 𝜒70

𝑝11  and the 

lines of 𝑃𝐺(3,13) are thirteen, so 𝜒70
𝑝11  is (45,13)-cap and 𝑐𝑖 values are  𝑐0 = 2334, 𝑐1 = 1. 

Since 𝑐0 ≠ 0, then 𝜒70
𝑝11  is incomplete. The (45,13)-cap will be complete when you add 1612 

points to it.  

Let  𝑝12 = {20, 202, 772, 1110, 1150, 1152, 1325, 1398, 1637, 1787, 2155, 2224}.             

Put  𝒟70
𝑝12= 𝜒70⋃𝑝12. The maximum number of the intersection points between 𝜒70

𝑝12  and the 

lines of 𝑃𝐺(3,13) are fourteen, so 𝜒70
𝑝12  is (46,14)-cap and 𝑐𝑖 values are  𝑐0 = 2334. Since 

𝑐0 ≠ 0, then 𝜒70
𝑝12  is incomplete. The (46,14)-cap will be complete when you add 2334 points 

to it.  

9. The line 𝐈(0,0,1,0,0,0) meets the cap 𝜒119  in 2 points, so we have twelve extension points that 

can be added to 𝜒𝑃, as in the following order: 4, 144, 470, 593, 754, 755,923,1213,2017, 

2105, 2165, 2369.  

Let 𝑝1 = {4},  𝒟119
𝑝1 = 𝜒119⋃𝑝1. The maximum number of the intersection points between 𝜒119

𝑝1  

and the lines of 𝑃𝐺(3,13) are three, so 𝜒119
𝑝1  is (21,3)-cap and 𝑐𝑖 values are  𝑐0 = 2337, 𝑐1 =

22. Since 𝑐0 ≠ 0, then 𝜒119
𝑝1  is incomplete. The (21,3)-cap will be complete when you add 117 

points to it. 

Let 𝑝2 = {4, 144},  𝒟119
𝑝2 = 𝜒119⋃𝑝2. The maximum number of the intersection points between 

𝜒119
𝑝2  and the lines of 𝑃𝐺(3,13) four, so 𝜒119

𝑝2  is (22,4)-cap and 𝑐𝑖 values are  𝑐0 = 2348, 𝑐1 =

10. Since 𝑐0 ≠ 0, then 𝜒119
𝑝2  is incomplete. The (22,4)-cap will be complete when you add 241 

points to it.  

Let 𝑝3 = {4, 144, 470},  𝒟119
𝑝3 = 𝜒119⋃𝑝3. The maximum number of the intersection points 

between 𝜒119
𝑝3  and the lines of 𝑃𝐺(3,13) are five, so 𝜒119

𝑝3  is (23,5)-cap and 𝑐𝑖 values are   𝑐0 =

2348, 𝑐1 = 9. Since 𝑐0 ≠ 0, then 𝜒119
𝑝3  is incomplete. The (23,5)-cap will be complete when 

you add 392 points to it.  



IHJPAS. 2024, 37( 3 ) 

407 
 

Let 𝑝4 = {4, 144, 470, 593},  𝒟119
𝑝4 = 𝜒119⋃𝑝4. The maximum number of the intersection points 

between 𝜒119
𝑝4  and the lines of 𝑃𝐺(3,13) are six, so 𝜒119

𝑝4  is (24,6)-cap and 𝑐𝑖 values are     𝑐0 =

2348, 𝑐1 = 8. Since 𝑐0 ≠ 0, then 𝜒119
𝑝4  is incomplete. The (24,6)-cap will be complete when 

you add 507 points to it. 

Let 𝑝5 = {4, 144, 470, 593, 754},  𝒟119
𝑝5 = 𝜒119⋃𝑝5. The maximum number of the intersection 

points between 𝜒119
𝑝5  and the lines of 𝑃𝐺(3,13) are seven, so 𝜒119

𝑝5  is (25,7)-cap and 𝑐𝑖 values 

are  𝑐0 = 2348, 𝑐1 = 7. Since 𝑐0 ≠ 0, then 𝜒119
𝑝5  is incomplete. The (25,7)-cap will be 

complete when you add 639 points to it.  

Let 𝑝6 = {4, 144, 470, 593, 754, 755},  𝒟119
𝑝6 = 𝜒119⋃𝑝6. The maximum number of the 

intersection points between 𝜒119
𝑝6  and the lines of 𝑃𝐺(3,13) are eight, so 𝜒119

𝑝6  is (26,8)-cap and 

𝑐𝑖 values are  𝑐0 = 2348, 𝑐1 = 6. Since 𝑐0 ≠ 0, then 𝜒119
𝑝6  is incomplete. The (26,8)-cap will 

be complete when you add 734 points to it.  

Let 𝑝7 = {4, 144, 470, 593, 754, 755, 923},  𝒟119
𝑝7 = 𝜒119⋃𝑝7. The maximum number of the 

intersection points between 𝜒119
4,144,470,593,754,755,923

 and the lines of 𝑃𝐺(3,13) are nine, so 𝜒119
𝑝7  

is (27,9)-cap and 𝑐𝑖 values are  𝑐0 = 2348, 𝑐1 = 5. Since 𝑐0 ≠ 0, then 𝜒119
𝑝7  is incomplete. 

The (27,9)-cap will be complete when you add 891 points to it.  

Let 𝑝8 = {4, 144, 470, 593, 754, 755, 923, 1213},  𝒟119
𝑝8 = 𝜒119⋃𝑝8. The maximum number of 

the intersection points between 𝜒119
𝑝8  and the lines of 𝑃𝐺(3,13) are ten, so 𝜒119

𝑝8  is (28,10)-cap 

and 𝑐𝑖 values are  𝑐0 = 2348, 𝑐1 = 4. Since 𝑐0 ≠ 0, then 𝜒119
𝑝8  is incomplete. The (28,10)-

cap will be complete when you add 1066 points to it.  

Let 𝑝9 = {4, 144, 470, 593, 754, 755, 923, 1213, 2017},  𝒟119
𝑝9  =  𝜒119  ⋃ 𝑝9. The maximum 

number of the intersection points between 𝜒119
𝑝9  and the lines of 𝑃𝐺(3,13) are eleven, so 𝜒119

𝑝9  

is (29,11)-cap and 𝑐𝑖 values are  𝑐0 = 2348, 𝑐1 = 3. Since 𝑐0 ≠ 0, then 𝜒119
𝑝9  is incomplete. 

The (29,11)-cap will be complete when you add 1335 points to it.  

Let  𝑝10 = {4, 144, 470, 593, 754, 755, 923, 1213, 2017, 2105},  𝒟119
𝑝10  = 𝜒119⋃ 𝑝10. 

The maximum number of the intersection points between 𝜒119
𝑝10  and the lines of 𝑃𝐺(3,13) are 

twelve, so 𝜒119
𝑝10  is (30,12)-cap and 𝑐𝑖 values are  𝑐0 = 2348, 𝑐1 = 2. Since 𝑐0 ≠ 0, then 𝜒119

𝑝10  

is incomplete. The (30,12)-cap will be complete when you add 1444 points to it.  

Let 𝑝11 = {4, 144, 470, 593, 754, 755, 923, 1213, 2017, 2105, 2165},  𝒟119
𝑝11 = 𝜒119⋃𝑝11. The 

maximum number of the intersection points between 𝜒119
𝑝11  and the lines of 𝑃𝐺(3,13) are 

thirteen, so 𝜒119
𝑝11  is (31,13)-cap and 𝑐𝑖 values are  𝑐0 = 2348, 𝑐1 = 1. Since 𝑐0 ≠ 0, then 𝜒119

𝑝11  

is incomplete. The (31,13)-cap will be complete when you add 1623 points to it.  

Let  𝑝12 = {4, 144, 470, 593, 754, 755, 923, 1213, 2017, 2105, 2165, 2369}.           

Put  𝒟119
𝑝12 = 𝜒119⋃𝑝12. The maximum number of the intersection points between 𝜒119

𝑝12  and the 

lines of 𝑃𝐺(3,13) are fourteen, so 𝜒119
𝑝12  is (32,14)-cap and 𝑐𝑖 values are  𝑐0 = 2348. Since 

𝑐0 ≠ 0, then 𝜒119
𝑝12  is incomplete. The (32,14)-cap will be complete when you add 2348 points 

to it.  

10. The line 𝐈(𝜏9, 𝜏10, 1,0,0,0) meets the cap 𝜒140 in 2 points, so we have twelve extension points 

that can be added to 𝜒𝑃, as in the following order: 

 36,356,565,718,1049,1054,1314,1377, 1756,1818,2253,2331. 

Let 𝑝1 = {36},  𝒟140
𝑝1 = 𝜒140⋃𝑝1. The maximum number of the intersection points between 𝜒140

𝑝1  

and the lines of 𝑃𝐺(3,13) are three, so 𝜒140
𝑝1  is (18,3)-cap and 𝑐𝑖 values are  𝑐0 = 2340,    𝑐1 =



IHJPAS. 2024, 37( 3 ) 

408 
 

22. Since 𝑐0 ≠ 0, then 𝜒140
𝑝1  is incomplete. The (18,3)-cap will be complete when you add 128 

points to it.  

Let 𝑝2 = {36, 356},  𝒟140
𝑝2  =  𝜒140  ⋃ 𝑝2. The maximum number of the intersection points 

between 𝜒140
𝑝2  and the lines of 𝑃𝐺(3,13) are four, so 𝜒140

𝑝2  is (19,4)-cap and 𝑐𝑖 values are                

𝑐0 = 2351, 𝑐1 = 10. Since 𝑐0 ≠ 0, then 𝜒140
𝑝2  is incomplete. The (19,4)-cap will be complete 

when you add 248 points to it.  

Let 𝑝3 = {36, 356, 565},  𝒟140
𝑝3 = 𝜒140⋃𝑝3. The maximum number of the intersection points 

between 𝜒140
𝑝3  and the lines of 𝑃𝐺(3,13) are five, so 𝜒140

𝑝3  is (20,5)-cap and 𝑐𝑖 values are     𝑐0 =

2351, 𝑐1 = 9. Since 𝑐0 ≠ 0, then 𝜒140
𝑝3  is incomplete. The (20,5)-cap will be complete when 

you add 388 points to it. 

Let 𝑝4 = {36, 356, 565, 718},  𝒟140
𝑝4 = 𝜒140⋃𝑝4. The maximum number of the intersection 

points between 𝜒140
𝑝4  and the lines of 𝑃𝐺(3,13) are six, so 𝜒140

𝑝4  is (21,6)-cap and 𝑐𝑖 values are     

𝑐0 = 2351, 𝑐1 = 8. Since 𝑐0 ≠ 0, then 𝜒140
𝑝4  is incomplete. The (21,6)-cap will be complete 

when you add 489 points to it. 

Let 𝑝5 = {36, 356, 565, 718, 1049},  𝒟140
𝑝5 = 𝜒140⋃𝑝5. The maximum number of the 

intersection points between 𝜒140
𝑝5  and the lines of 𝑃𝐺(3,13) are seven, so 𝜒140

𝑝5  is (22,7)-cap 

and 𝑐𝑖 values are  𝑐0 = 2351, 𝑐1 = 7. Since 𝑐0 ≠ 0, then 𝜒140
𝑝5  is incomplete. The (22,7)-cap 

will be complete when you add 647 points to it. 

Let 𝑝6 = { 36, 356, 565, 718, 1049, 1054},  𝒟140
𝑝6 =  𝜒140 ⋃ 𝑝6. The maximum number of the 

intersection points between 𝜒140
𝑝6  and the lines of 𝑃𝐺(3,13) are eight, so 𝜒140

𝑝6  is (23,8)-cap and 

𝑐𝑖 values are  𝑐0 = 2351, 𝑐1 = 6. Since 𝑐0 ≠ 0, then 𝜒140
𝑝6  is incomplete. The (23,8)-cap will 

be complete when you add 707 points to it. 

Let 𝑝7 = {36, 356, 565, 718, 1049, 1054, 1314},  𝒟140
𝑝7 = 𝜒140⋃𝑝7. The maximum number of 

the intersection points between 𝜒140
𝑝7  and the lines of 𝑃𝐺(3,13) are nine, so 𝜒140

𝑝7  is (24,9)-cap 

and 𝑐𝑖 values are  𝑐0 = 2351, 𝑐1 = 5. Since 𝑐0 ≠ 0, then 𝜒140
𝑝7  is incomplete. The (24,9)-cap 

will be complete when you add 904 points to it.  

Let  𝑝8 = {36, 356, 565, 718, 1049, 1054, 1314, 1377},  𝒟140
𝑝8  =  𝜒140  ⋃ 𝑝8. The maximum 

number of the intersection points between 𝜒140
𝑝8  and the lines of 𝑃𝐺(3,13) are ten, so 𝜒140

𝑝8  is 

(25,10)-cap and 𝑐𝑖 values are  𝑐0 = 2351, 𝑐1 = 4. Since 𝑐0 ≠ 0, then 𝜒140
𝑝8  is incomplete. The 

(25,10)-cap will be complete when you add 1062 points to it.  

Let  𝑝9 = {36, 356, 565, 718, 1049, 1054, 1314, 1377, 1756}. Put  𝒟140
𝑝9  =  𝜒140 ⋃ 𝑝9. The 

maximum number of the intersection points between 𝜒140
𝑝9  and the lines of 𝑃𝐺(3,13) are eleven, 

so 𝜒140
𝑝9  is (26,11)-cap and 𝑐𝑖 values are  𝑐0 = 2351, 𝑐1 = 3. Since 𝑐0 ≠ 0, then 𝜒140

𝑝9  is 

incomplete. The (26,11)-cap will be complete when you add 1340 points to it.  

Let 𝑝10 = {36, 356, 565, 718, 1049, 1054, 1314, 1377, 1756, 1818},  𝒟140
𝑝10  =  𝜒140  ⋃ 𝑝10. 

The maximum number of the intersection points between 𝜒140
𝑝10  and the lines of 𝑃𝐺(3,13) are 

twelve, so 𝜒140
𝑝10  is (27,12)-cap and 𝑐𝑖 values are  𝑐0 = 2351, 𝑐1 = 2. Since 𝑐0 ≠ 0, then 𝜒140

𝑝10  

is incomplete. The (27,12)-cap will be complete when you add 1383 points to it.  

Let  𝑝11 = {36, 356, 565, 718, 1049, 1054, 1314, 1377, 1756, 1818, 2253}.           

Put 𝒟140
𝑝11 = 𝜒140⋃𝑝11. The maximum number of the intersection points between 𝜒140

𝑝11  and the 

lines of 𝑃𝐺(3,13) are thirteen, so 𝜒140
𝑝11  is (28,13)-cap and 𝑐𝑖 values are  𝑐0 = 2351, 𝑐1 = 1. 



IHJPAS. 2024, 37( 3 ) 

409 
 

Since 𝑐0 ≠ 0, then 𝜒140
𝑝11  is incomplete. The (28,13)-cap will be complete when you add 1625 

points to it.  

Let  𝑝12 = {36, 356, 565, 718, 1049, 1054, 1314, 1377, 1756, 1818, 2253, 2331}.  

Put  𝒟140
𝑝12=  𝜒140⋃𝑝12. The maximum number of the intersection points between 𝜒140

𝑝12  and the 

lines of 𝑃𝐺(3,13) are fourteen, so 𝜒140
𝑝12  is (29,14)-cap and 𝑐𝑖 values are  𝑐0 = 2351. Since 

𝑐0 ≠ 0, then 𝜒140
𝑝12  is incomplete. The (29,14)-cap will be complete when you add 2351 points 

to it.  

11. The line 𝐈(𝜏7, 𝜏9, 1,0,0,0) meets the cap 𝜒238 in 2 points, so we have twelve extension points 

that can be added to 𝜒𝑃, as in the following order: 24, 315, 453, 794, 860, 870, 968,    

1324, 1534, 1658, 1895, 2373. 

Let 𝑝1 = {24},  𝒟238
𝑝1 = 𝜒238⋃𝑝1. The maximum number of the intersection points between 𝜒238

𝑝1  

and the lines of 𝑃𝐺(3,13) are three, so 𝜒238
𝑝1  is (11,3)-cap and 𝑐𝑖 values are  𝑐0 = 2358,     𝑐1 =

11. Since 𝑐0 ≠ 0, then 𝜒238
𝑝1  is incomplete. The (11,3)-cap will be complete when you add 119 

points to it.  

Let 𝑝2 = {24, 315},  𝒟238
𝑝2 = 𝜒238⋃𝑝2. The maximum number of the intersection points 

between 𝜒238
𝑝2  and the lines of 𝑃𝐺(3,13) are four, so 𝜒238

𝑝2  is (12,4)-cap and 𝑐𝑖 values are  𝑐0 =

2358, 𝑐1 = 10. Since 𝑐0 ≠ 0, then 𝜒238
𝑝2  is incomplete. The (12,4)-cap will be complete when 

you add 243 points to it.  

Let 𝑝3 = {24, 315, 453},  𝒟238
𝑝3 = 𝜒238⋃𝑝3. The maximum number of the intersection points 

between 𝜒238
𝑝3  and the lines of 𝑃𝐺(3,13) are five, so 𝜒238

𝑝3  is (13,5)-cap and 𝑐𝑖 values are  𝑐0 =

2358, 𝑐1 = 9. Since 𝑐0 ≠ 0, then 𝜒238
𝑝3  is incomplete. The (13,5)-cap will be complete when 

you add 393 points to it. 

Let 𝑝4 = {24, 315, 453, 794},  𝒟238
𝑝4 = 𝜒238⋃𝑝4. The maximum number of the intersection 

points between 𝜒238
𝑝4  and the lines of 𝑃𝐺(3,13) are six, so 𝜒238

𝑝4  is (14,6)-cap and 𝑐𝑖 values are     

𝑐0 = 2358, 𝑐1 = 8. Since 𝑐0 ≠ 0, then 𝜒238
𝑝4  is incomplete. The (14,6)-cap will be complete 

when you add 509 points to it.  

Let 𝑝5 = {24, 315, 453, 794, 860},  𝒟238
𝑝5 = 𝜒238⋃𝑝5. The maximum number of the intersection 

points between 𝜒238
𝑝5  and the lines of 𝑃𝐺(3,13) are seven, so 𝜒238

𝑝5  is (15,7)-cap and 𝑐𝑖 values 

are  𝑐0 = 2358, 𝑐1 = 7. Since 𝑐0 ≠ 0, then 𝜒238
𝑝5  is incomplete. The (15,7)-cap will be 

complete when you add 643 points to it.  

Let 𝑝6 = {24, 315, 453, 794, 860, 870},  𝒟238
𝑝6  = 𝜒238 ⋃ 𝑝6. The maximum number of the 

intersection points between 𝜒238
𝑝6  and the lines of 𝑃𝐺(3,13) are eight, so 𝜒238

𝑝6  is (16,8)-cap 

and 𝑐𝑖 values are  𝑐0 = 2358, 𝑐1 = 6. Since 𝑐0 ≠ 0, then 𝜒238
𝑝6  is incomplete. The (16,8)-cap 

will be complete when you add 707 points to it.  

Let 𝑝7 = {24, 315, 453, 794, 860, 870, 968},  𝒟238
𝑝7 = 𝜒238⋃𝑝7. The maximum number of the 

intersection points between 𝜒238
𝑝7  and the lines of 𝑃𝐺(3,13) are nine, so 𝜒238

𝑝7  is (17,9)-cap and 

𝑐𝑖 values are  𝑐0 = 2358, 𝑐1 = 5. Since 𝑐0 ≠ 0, then 𝜒238
𝑝7  is incomplete. The (17,9)-cap will 

be complete when you add 896 points to it.  

Let 𝑝8 = {24, 315, 453, 794, 860, 870, 968, 1324},  𝒟238
𝑝8 = 𝜒238⋃𝑝8. The maximum number 

of the intersection points between 𝜒238
𝑝8  and the lines of 𝑃𝐺(3,13) are ten, so 𝜒238

𝑝8  is (18,10)-

cap and 𝑐𝑖 values are  𝑐0 = 2358, 𝑐1 = 4. Since 𝑐0 ≠ 0, then 𝜒238
𝑝8  is incomplete. The        

(18,10)-cap will be complete when you add 1065 points to it. 



IHJPAS. 2024, 37( 3 ) 

410 
 

Let 𝑝9 = {24, 315, 453, 794, 860, 870, 968, 1324, 1534},  𝒟238
𝑝9 = 𝜒238⋃𝑝9. The maximum 

number of the intersection points between 𝜒238
𝑝9  and the lines of 𝑃𝐺(3,13) are eleven, so 𝜒238

𝑝9  

is (19,11)-cap and 𝑐𝑖 values are  𝑐0 = 2358, 𝑐1 = 3. Since 𝑐0 ≠ 0, then 𝜒238
𝑝9  is incomplete. 

The (19,11)-cap will be complete when you add 1406 points to it.  

Let  𝑝10 = {24, 315, 453, 794, 860, 870, 968, 1324, 1534, 1658},  𝒟238
𝑝10  =  𝜒238 ⋃  𝑝10. The 

maximum number of the intersection points between 𝜒238
𝑝10  and the lines of 𝑃𝐺(3,13) are 

twelve, so 𝜒238
𝑝10  is (20,12)-cap and 𝑐𝑖 values are  𝑐0 = 2358, 𝑐1 = 2. Since 𝑐0 ≠ 0, then 𝜒238

𝑝10  

is incomplete. The (20,12)-cap will be complete when you add 1450 points to it.  

Let 𝑝11 = {24, 315, 453, 794, 860, 870, 968, 1324, 1534, 1658, 1895},  𝒟238
𝑝11  = 𝜒238  ⋃ 𝑝11. 

The maximum number of the intersection points between 𝜒238
𝑝11  and the lines of 𝑃𝐺(3,13) are 

thirteen, so 𝜒238
𝑝11  is (21,13)-cap and 𝑐𝑖 values are  𝑐0 = 2358, 𝑐1 = 1. Since 𝑐0 ≠ 0, then 𝜒238

𝑝11  

is incomplete. The (21,13)-cap will be complete when you add 1629 points to it.  

Let  𝑝12 = {24, 315, 453, 794, 860, 870, 968, 1324, 1534, 1658, 1895, 2373}.  

Put  𝒟238
𝑝12 = 𝜒238⋃𝑝12. The maximum number of the intersection points between 𝜒238

𝑝12  and the 

lines of 𝑃𝐺(3,13) are fourteen, so 𝜒238
𝑝12  is (22,14)-cap and 𝑐𝑖 values are  𝑐0 = 2358. Since 

𝑐0 ≠ 0, then 𝜒238
𝑝12  is incomplete. The (22,14)-cap will be complete when you add 2358 points 

to it.  

12. The line 𝐈(𝜏11, 1,0,0,0,0) meets the cap 𝜒340 in 7 points, so we have seven extension points 

that can be added to 𝜒𝑃, as in the following order: 

 171, 511, 851, 1191, 1531, 1871, 2211. 

Let 𝑝1 = {171},  𝒟340
𝑝1 = 𝜒340⋃𝑝1. The maximum number of intersection points between 𝜒340

𝑝1  

and the lines of 𝑃𝐺(3,13) are eight, so 𝜒340
𝑝1  is (8,8)-cap and 𝑐𝑖 values are  𝑐0 = 2366,     𝑐1 =

6. Since 𝑐0 ≠ 0, then 𝜒340
𝑝1  is incomplete. The (8,8)-cap will be complete when you add 702 

points to it.  

Let 𝑝2 = {171, 511},  𝒟340
𝑝2 = 𝜒340⋃𝑝2. The maximum number of the intersection points 

between 𝜒340
𝑝2  and the lines of 𝑃𝐺(3,13) are nine, so 𝜒340

𝑝2  is (9,9)-cap and 𝑐𝑖 values are  𝑐0 =

2366, 𝑐1 = 5. Since 𝑐0 ≠ 0, then 𝜒340
𝑝2  is incomplete. The (9,9)-cap will be complete when 

you add 912 points to it. 

Let 𝑝3 = {171, 511, 851},  𝒟340
𝑝3  =  𝜒340  ⋃ 𝑝3. The maximum number of the intersection points 

between 𝜒340
𝑝3  and the lines of 𝑃𝐺(3,13) are ten, so 𝜒340

𝑝3  is (10,10)-cap and 𝑐𝑖 values are  𝑐0 =

2366, 𝑐1 = 4. Since 𝑐0 ≠ 0, then 𝜒340
𝑝3  is incomplete. The (10,10)-cap will be complete when 

you add 1077 points to it.  

Let 𝑝4 = {171, 511, 851, 1191},  𝒟340
𝑝4 = 𝜒340⋃𝑝4. The maximum number of the intersection 

points between 𝜒340
𝑝4  and the lines of 𝑃𝐺(3,13) are eleven, so 𝜒340

𝑝4  is (11,11)-cap and 𝑐𝑖 values 

are  𝑐0 = 2366, 𝑐1 = 3. Since 𝑐0 ≠ 0, then 𝜒340
𝑝4  is incomplete. The (11,11)-cap will be 

complete when you add 1408 points to it. 

Let 𝑝5 = {171, 511, 851, 1191, 1531},  𝒟340 
𝑝5 =  𝜒340 ⋃ 𝑝5. The maximum number of the 

intersection points between 𝜒340
𝑝5  and the lines of 𝑃𝐺(3,13) are twelve, so 𝜒340

𝑝5  is (12,12)-cap 

and 𝑐𝑖 values are  𝑐0 = 2366, 𝑐1 = 2. Since 𝑐0 ≠ 0, then 𝜒340
𝑝5  is incomplete. The       (12,12)-

cap will be complete when you add 1451 points to it.  

Let 𝑝6 = {171, 511, 851, 1191, 1531, 1871},  𝒟340
𝑝6 = 𝜒340⋃𝑝6. The maximum number of the 

intersection points between 𝜒340
𝑝6  and the lines of 𝑃𝐺(3,13) are thirteen, so 𝜒340

𝑝6  is       (13,13)-



IHJPAS. 2024, 37( 3 ) 

411 
 

cap and 𝑐𝑖 values are  𝑐0 = 2366, 𝑐1 = 1. Since 𝑐0 ≠ 0, then 𝜒340
𝑝6  is incomplete. The 

(13,13)-cap will be complete when you add 1635 points to it.  

Let 𝑝7 = {171, 511, 851, 1191, 1531, 1871, 2211},  𝒟340
𝑝7 = 𝜒340  ⋃ 𝑝7. The maximum number 

of the intersection points between 𝜒340
𝑝7  and the lines of 𝑃𝐺(3,13) are fourteen, so 𝜒340

𝑝7  is 

(14,14)-cap and 𝑐𝑖 values are  𝑐0 = 2366. Since 𝑐0 ≠ 0, then 𝜒340
𝑝7  is incomplete. The 

(14,14)-cap will be complete when you add 2366 points to it. 

13. The line 𝐈(𝜏, 𝜏6, 1,0,0,0) meets the cap 𝜒476 in 2 points, so we have twelve extension points 

that can be added to 𝜒𝑃 , as in the following order: 205, 217, 220, 360, 574, 686,

809, 970, 971, 1139, 2233, 2321. 

Let 𝑝1 = {205},  𝒟476
𝑝1 = 𝜒476⋃𝑝1. The maximum number of the intersection points between 

𝜒476
𝑝1  and the lines of 𝑃𝐺(3,13) are three, so 𝜒476

𝑝1  is (6,3)-cap and 𝑐𝑖 values are  𝑐0 = 2363,      

𝑐1 = 11. Since 𝑐0 ≠ 0, then 𝜒476
𝑝1  is incomplete. The (6,3)-cap will be complete when you add 

137 points to it.  

Let 𝑝2 = {205, 217},  𝒟476
𝑝2 = 𝜒476⋃𝑝2. The maximum number of the intersection points 

between 𝜒476
𝑝2  and the lines of 𝑃𝐺(3,13) are four, so 𝜒476

𝑝2  is (7,4)-cap and 𝑐𝑖 values are  𝑐0 =

2363, 𝑐1 = 10. Since 𝑐0 ≠ 0, then 𝜒476
𝑝2  is incomplete. The (7,4)-cap will be complete when 

you add 243 points to it.  

Let 𝑝3 = {205, 217, 220},  𝒟476
𝑝3 = 𝜒476⋃𝑝3. The maximum number of the intersection points 

between 𝜒476
𝑝3  and the lines of 𝑃𝐺(3,13) are five, so 𝜒476

𝑝3  is (8,5)-cap and 𝑐𝑖 values are     𝑐0 =

2363, 𝑐1 = 9. Since 𝑐0 ≠ 0, then 𝜒476
𝑝3  is incomplete. The (8,5)-cap will be complete when 

you add 399 points to it.  

Let 𝑝4 = {205, 217, 220, 360},  𝒟476
𝑝4 = 𝜒476⋃𝑝4. The maximum number of the intersection 

points between 𝜒476
𝑝4  and the lines of 𝑃𝐺(3,13) are six, so 𝜒476

𝑝4  is (9,6)-cap and 𝑐𝑖 values are 

 𝑐0 = 2363, 𝑐1 = 8. Since 𝑐0 ≠ 0, then 𝜒476
𝑝4  is incomplete. The (9,6)-cap will be complete 

when you add 509 points to it.  

Let 𝑝5 = {205, 217, 220, 360, 574},  𝒟476
𝑝5 = 𝜒476⋃𝑝5. The maximum number of the 

intersection points between 𝜒476
𝑝5  and the lines of 𝑃𝐺(3,13) are seven, so 𝜒476

𝑝5  is (10,7)-cap 

and 𝑐𝑖 values are  𝑐0 = 2363, 𝑐1 = 7. Since 𝑐0 ≠ 0, then 𝜒476
𝑝5  is incomplete. The (10,7)-cap 

will be complete when you add 632 points to it.  

Let 𝑝6 = {205, 217, 220, 360, 574, 686},  𝒟476
𝑝6 = 𝜒476  ⋃ 𝑝6. The maximum number of the 

intersection points between 𝜒476
𝑝6  and the lines of 𝑃𝐺(3,13) are eight, so 𝜒476

𝑝6  is (11,8)-cap 

and 𝑐𝑖 values are  𝑐0 = 2363, 𝑐1 = 6. Since 𝑐0 ≠ 0, then 𝜒476
𝑝6  is incomplete. The (11,8)-cap 

will be complete when you add 677 points to it.  

Let 𝑝7 = {205, 217, 220, 360, 574, 686, 809},  𝒟476
𝑝7 = 𝜒476⋃𝑝7. The maximum number of 

the intersection points between 𝜒476
𝑝7  and the lines of 𝑃𝐺(3,13) are nine, so 𝜒476

𝑝7  is (12,9)-cap 

and 𝑐𝑖 values are  𝑐0 = 2363, 𝑐1 = 5. Since 𝑐0 ≠ 0, then 𝜒476
𝑝7  is incomplete. The (12,9)-cap 

will be complete when you add 896 points to it.  

Let 𝑝8 = {205, 217, 220, 360, 574, 686, 809, 970},  𝒟476
𝑝8 = 𝜒476⋃𝑝8. The maximum 

number of the intersection points between 𝜒476
𝑝8  and the lines of 𝑃𝐺(3,13) ten, so 𝜒476

𝑝8  is 

(13,10)-cap and 𝑐𝑖 values are  𝑐0 = 2363, 𝑐1 = 4. Since 𝑐0 ≠ 0, then 𝜒476
𝑝8  is incomplete. The 

(13,10)-cap will be complete when you add 1069 points to it.  



IHJPAS. 2024, 37( 3 ) 

412 
 

Let 𝑝9 = {205, 217, 220, 360, 574, 686, 809, 970, 971},  𝒟476
𝑝9 = 𝜒476⋃𝑝9. The 

maximum number of the intersection points between 𝜒476
𝑝9  and the lines of 𝑃𝐺(3,13) are eleven, 

so 𝜒476
𝑝9  is (14,11)-cap and 𝑐𝑖 values are  𝑐0 = 2363, 𝑐1 = 3. Since 𝑐0 ≠ 0, then 𝜒476

𝑝9  is 

incomplete. The (14,11)-cap will be complete when you add 1407 points to it.  

Let  𝑝10 = {205, 217, 220, 360, 574, 686, 809, 970, 971, 1139},  𝒟476
𝑝10  

 =  𝜒476  ⋃ 𝑝10. The 

maximum number of the intersection points between 𝜒476
𝑝10  and the lines of 𝑃𝐺(3,13) are 

twelve, so 𝜒476
𝑝10  is (15,12)-cap and 𝑐𝑖 values are  𝑐0 = 2363, 𝑐1 = 2. Since 𝑐0 ≠ 0, then 𝜒476

𝑝10  

is incomplete. The (15,12)-cap will be complete when you add 1389 points to it.  

Let 𝑝11 = {205, 217, 220, 360, 574, 686, 809, 970, 971, 1139, 2233},  𝒟476
𝑝11 = 𝜒476⋃ 𝑝11. 

The maximum number of the intersection points between 𝜒476
𝑝11  and the lines of 𝑃𝐺(3,13) are 

thirteen, so 𝜒476
𝑝11  is (16,13)-cap and 𝑐𝑖 values are  𝑐0 = 2363, 𝑐1 = 1. Since 𝑐0 ≠ 0, then 𝜒476

𝑝11  

is incomplete. The (16,13)-cap will be complete when you add 1630 points to it.  

Let 𝑝12 = {205, 217, 220, 360, 574, 686, 809, 970, 971, 1139, 2233, 2321}. 

Put 𝒟476
𝑝12 = 𝜒476⋃𝑝12. The maximum number of the intersection points between 𝜒476

𝑝12  and the 

lines of 𝑃𝐺(3,13) are fourteen, so 𝜒476
𝑝12  is (17,14)-cap and 𝑐𝑖 values are  𝑐0 = 2363. Since 

𝑐0 ≠ 0, then 𝜒476
𝑝12  is incomplete. The (17,14)-cap will be complete when you add 2363 points 

to it. 

14. The line 𝐈(𝜏8, 𝜏2, 1,0,0,0) meets the cap 𝜒595 in 2 points, so we have twelve extension points 

that can be added to 𝜒𝑃, as in the following order: 108, 483, 501, 539, 1201, 1321,

1392, 1424, 1507, 1741, 1840, 2235. 

Let 𝑝1 = {108},  𝒟595
𝑝1 = 𝜒595  ⋃ 𝑝1. The maximum number of the intersection points between 

𝜒595
𝑝1  and the lines of 𝑃𝐺(3,13) are three, so 𝜒595

𝑝1  is (5,3)-cap and 𝑐𝑖 values are  𝑐0 = 2364,     

𝑐1 = 11. Since 𝑐0 ≠ 0, then 𝜒595
𝑝1  is incomplete. The (5,3)-cap will be complete when you add 

118 points to it.  

Let 𝑝2 = {108, 483},  𝒟595
𝑝2 = 𝜒595⋃𝑝2. The maximum number of the intersection points 

between 𝜒595
𝑝2  and the lines of 𝑃𝐺(3,13) are four, so 𝜒595

𝑝2  is (6,4)-cap and 𝑐𝑖 values are  𝑐0 =

2364, 𝑐1 = 10. Since 𝑐0 ≠ 0, then 𝜒595
𝑝2  is incomplete. The (6,4)-cap will be complete when 

you add 221 points to it. 

Let 𝑝3 = {108, 483, 501},  𝒟595
𝑝3 = 𝜒595⋃𝑝3. The maximum number of the intersection points 

between 𝜒595
𝑝3  and the lines of 𝑃𝐺(3,13) are five, so 𝜒595

𝑝3  is (7,5)-cap and 𝑐𝑖 values are     𝑐0 =

2364, 𝑐1 = 9. Since 𝑐0 ≠ 0, then 𝜒595
𝑝3  is incomplete. The (7,5)-cap will be complete when 

you add 404 points to it.  

Let 𝑝4 = {108, 483, 501, 539},  𝒟595
𝑝4 = 𝜒595⋃𝑝4. The maximum number of the intersection 

points between 𝜒595
𝑝4  and the lines of 𝑃𝐺(3,13) are six, so 𝜒595

𝑝4  is (8,6)-cap and 𝑐𝑖 values are 

 𝑐0 = 2364, 𝑐1 = 8. Since 𝑐0 ≠ 0, then 𝜒595
𝑝4  is incomplete. The (8,6)-cap will be complete 

when you add 491 points to it.  

Let 𝑝5 = {108, 483, 501, 539, 1201},  𝒟595
𝑝5 = 𝜒595⋃𝑝5. The maximum number of the 

intersection points between 𝜒595
𝑝5  and the lines of 𝑃𝐺(3,13) are seven, so 𝜒595

𝑝5  is (9,7)-cap and 

𝑐𝑖 values are  𝑐0 = 2364, 𝑐1 = 7. Since 𝑐0 ≠ 0, then 𝜒595
108,483,501,539,1201

 is incomplete. The 

(9,7)-cap will be complete when you add 633 points to it.  

Let 𝑝6 = {108, 483, 501, 539, 1201, 1321},  𝒟595
𝑝6 = 𝜒595⋃𝑝6. The maximum number of the 

intersection points between 𝜒595
𝑝6  and the lines of 𝑃𝐺(3,13) are eight, so 𝜒595

𝑝6  is (10,8)-cap and 



IHJPAS. 2024, 37( 3 ) 

413 
 

𝑐𝑖 values are  𝑐0 = 2364, 𝑐1 = 6. Since 𝑐0 ≠ 0, then 𝜒595
𝑝6  is incomplete. The (10,8)-cap will 

be complete when you add 695 points to it.  

Let 𝑝7 = {108, 483, 501, 539, 1201, 1321, 1392},  𝒟595
𝑝7 = 𝜒595⋃𝑝7. The maximum number 

of the intersection points between 𝜒595
𝑝7  and the lines of 𝑃𝐺(3,13) are nine, so 𝜒595

𝑝7  is (11,9)-

cap and 𝑐𝑖 values are  𝑐0 = 2364, 𝑐1 = 5. Since 𝑐0 ≠ 0, then 𝜒595
𝑝7  is incomplete. The (11,9)-

cap will be complete when you add 902 points to it.  

Let 𝑝8 = {108, 483, 501, 539, 1201, 1321, 1392, 1424},  𝒟595
𝑝8 = 𝜒595⋃𝑝8. The maximum 

number of the intersection points between 𝜒595
𝑝8  and the lines of 𝑃𝐺(3,13) are ten, so 𝜒595

𝑝8  is 

(12,10)-cap and 𝑐𝑖 values are  𝑐0 = 2364, 𝑐1 = 4. Since 𝑐0 ≠ 0, then 𝜒595
𝑝8  is incomplete. The 

(12,10)-cap will be complete when you add 1073 points to it.  

Let  𝑝9 = {108, 483, 501, 539, 1201, 1321, 1392, 1424, 1507},  𝒟595
𝑝9  =  𝜒595 ⋃ 𝑝9. The 

maximum number of the intersection points between 𝜒595
𝑝9  and the lines of 𝑃𝐺(3,13) are eleven, 

so 𝜒595
𝑝9  is (13,11)-cap and 𝑐𝑖 values are  𝑐0 = 2364, 𝑐1 = 3. Since 𝑐0 ≠ 0, then 𝜒595

𝑝9  is 

incomplete. The (13,11)-cap will be complete when you add 1346 points to it.  

Let  𝑝10 = { 108, 483, 501, 539, 1201, 1321, 1392, 1424, 1507, 1741}.       

Put  𝒟595
𝑝10 = 𝜒595⋃𝑝10. The maximum number of the intersection points between 𝜒595

𝑝10  and the 

lines of 𝑃𝐺(3,13) are twelve, so 𝜒595
𝑝10  is (14,12)-cap and 𝑐𝑖 values are  𝑐0 = 2364, 𝑐1 = 2. 

Since 𝑐0 ≠ 0, then 𝜒595
𝑝10  is incomplete. The (14,12)-cap will be complete when you add 1411 

points to it.  

Let  𝑝11 = { 108, 483, 501, 539, 1201, 1321, 1392, 1424, 1507, 1741, 1840}.  

Put  𝒟595
𝑝11= 𝜒595⋃𝑝11. The maximum number of the intersection points between 𝜒595

𝑝11  and the 

lines of 𝑃𝐺(3,13) are thirteen, so 𝜒595
𝑝11  is (15,13)-cap and 𝑐𝑖 values are  𝑐0 = 2364, 𝑐1 = 1. 

Since 𝑐0 ≠ 0, then 𝜒595
𝑝11  is incomplete. The (15,13)-cap will be complete when you add 

 1632 points to it.  

Let  𝑝12 = {108, 483, 501, 539, 1201, 1321, 1392, 1424, 1507, 1741, 1840, 2235}.  

Put 𝒟595
𝑝12= 𝜒595⋃𝑝12. The maximum number of the intersection points between 𝜒595

𝑝12  and the 

lines of 𝑃𝐺(3,13) are fourteen, so 𝜒595
𝑝12  is (16,14)-cap and 𝑐𝑖 values are  𝑐0 = 2334. Since 

𝑐0 ≠ 0, then 𝜒595
𝑝12  is incomplete. The (16,14)-cap will be complete when you add 2364 points 

to it.  

15. The line 𝐈(𝜏11, 1,0,0,0,0) meets the cap 𝜒1190 in 2 points, so we have twelve extension points 

that can be added to 𝜒𝑃, as in the following order: 171, 341, 511, 681, 851, 1021,

1361, 1531, 1701, 1871, 2041, 2211. 

Let 𝑝1 = {171},  𝒟1190
𝑝1 = 𝜒1190⋃𝑝1. The maximum number of the intersection points 

between 𝜒1190
𝑝1  and the lines of 𝑃𝐺(3,13) are three, so 𝜒1190

𝑝1  is (3,3)-cap and 𝑐𝑖 values are 

 𝑐0 = 2366, 𝑐1 = 11. Since 𝑐0 ≠ 0, then 𝜒1190
𝑝1  is incomplete. The (3,3)-cap will be complete 

when you add 119 points to it.  

Let 𝑝2 = {171, 341},  𝒟1190
𝑝2 = 𝜒1190⋃𝑝2. The maximum number of the intersection points 

between 𝜒1190
𝑝2  and the lines of 𝑃𝐺(3,13) are four, so 𝜒1190

𝑝2  is (4,4)-cap and 𝑐𝑖 values are  𝑐0 =

2366, 𝑐1 = 10. Since 𝑐0 ≠ 0, then 𝜒1190
𝑝2  is incomplete. The (4,4)-cap will be complete when 

you add 222 points to it.  

Let 𝑝3 = {171, 341, 511},  𝒟1190
𝑝3 = 𝜒1190⋃𝑝3. The maximum number of the intersection points 

between 𝜒1190
𝑝3  and the lines of 𝑃𝐺(3,13) are five, so 𝜒1190

𝑝3  is (5,5)-cap and 𝑐𝑖 values are  𝑐0 =



IHJPAS. 2024, 37( 3 ) 

414 
 

2366, 𝑐1 = 9. Since 𝑐0 ≠ 0, then 𝜒1190
𝑝3  is incomplete. The (5,5)-cap will be complete when 

you add 401 points to it.  

Let 𝑝4 = {171, 341, 511, 681},  𝒟1190
𝑝4 = 𝜒1190⋃𝑝4. The maximum number of the intersection 

points between 𝜒1190
𝑝4  and the lines of 𝑃𝐺(3,13) are six, so 𝜒1190

𝑝4  is (6,6)-cap and 𝑐𝑖 values are 

 𝑐0 = 2366, 𝑐1 = 8. Since 𝑐0 ≠ 0, then 𝜒1190
𝑝4  is incomplete. The (6,6)-cap will be complete 

when you add 491 points to it.  

Let 𝑝5 = {171, 341, 511, 681, 851},  𝒟1190
𝑝5  = 𝜒1190 ⋃ 𝑝5. The maximum number of the 

intersection points between 𝜒1190
𝑝5  and the lines of 𝑃𝐺(3,13) are seven, so 𝜒1190

𝑝5  is (7,7)-cap 

and 𝑐𝑖 values are  𝑐0 = 2366, 𝑐1 = 7. Since 𝑐0 ≠ 0, then 𝜒1190
𝑝5  is incomplete. The (7,7)-cap 

will be complete when you add 603 points to it.  

Let 𝑝6 = {171, 341, 511, 681, 851, 1021},  𝒟1190
𝑝6 = 𝜒1190⋃𝑝6. The maximum number of the 

intersection points between 𝜒1190
𝑝6  and the lines of 𝑃𝐺(3,13) are eight, so 𝜒1190

𝑝6  is (8,8)-cap 

and 𝑐𝑖 values are  𝑐0 = 2366, 𝑐1 = 6. Since 𝑐0 ≠ 0, then 𝜒1190
𝑝6  is incomplete. The (8,8)-cap 

will be complete when you add 681 points to it.  

Let 𝑝7 = {171, 341, 511, 681, 851, 1021, 1361},  𝒟1190
𝑝7 = 𝜒1190⋃𝑝7. The maximum number 

of the intersection points between 𝜒1190
𝑝7  and the lines of 𝑃𝐺(3,13) are nine, so 𝜒1190

𝑝7  is (9,9)-

cap and 𝑐𝑖 values are  𝑐0 = 2366, 𝑐1 = 5. Since 𝑐0 ≠ 0, then 𝜒1190
𝑝7  is incomplete. The (9,9)-

cap will be complete when you add 899 points to it.  

Let 𝑝8 = {171, 341, 511, 681, 851, 1021, 1361, 1531},  𝒟1190
𝑝8 = 𝜒1190⋃𝑝8. The maximum 

number of the intersection points between 𝜒1190
𝑝8  and the lines of 𝑃𝐺(3,13) are ten, so 𝜒1190

𝑝8  is 

(10,10)-cap and 𝑐𝑖 values are  𝑐0 = 2366, 𝑐1 = 4. Since 𝑐0 ≠ 0, then 𝜒1190
𝑝8  is incomplete. 

The (10,10)-cap will be complete when you add 1069 points to it.  

Let 𝑝9 = { 171, 341, 511, 681, 851, 1021, 1361, 1531, 1701},  𝒟1190
𝑝9  =  𝜒1190  ⋃ 𝑝9. 

The maximum number of the intersection points between 𝜒1190
𝑝9  and the lines of 𝑃𝐺(3,13) are 

eleven, so 𝜒1190
𝑝9  is (11,11)-cap and 𝑐𝑖 values are  𝑐0 = 2366, 𝑐1 = 3. Since 𝑐0 ≠ 0, then 

𝜒1190
𝑝9  is incomplete. The (11,11)-cap will be complete when you add 1347 points to it. 

Let 𝑝10 = {171, 341, 511, 681, 851, 1021, 1361, 1531, 1701, 1871},  𝒟1190
𝑝10 = 𝜒1190⋃𝑝10. 

The maximum number of the intersection points between 𝜒1190
𝑝10  and the lines of 𝑃𝐺(3,13) are 

twelve, so 𝜒1190
𝑝10  is (12,12)-cap and 𝑐𝑖 values are  𝑐0 = 2366, 𝑐1 = 2. Since 𝑐0 ≠ 0, then 

𝜒1190
𝑝10  is incomplete. The (12,12)-cap will be complete when you add 1434 points to it. 

Let 𝑝11 = {171, 341, 511, 681, 851, 1021, 1361, 1531, 1701, 1871,

2041},  𝒟1190
𝑝11 = 𝜒1190⋃𝑝11. The maximum number of the intersection points between 𝜒1190

𝑝11  

and the lines of 𝑃𝐺(3,13) are thirteen, so 𝜒1190
𝑝11  is (13,13)-cap and 𝑐𝑖 values are  𝑐0 = 2366,

𝑐1 = 1. Since 𝑐0 ≠ 0, then 𝜒1190
𝑝11  is incomplete. The (13,13)-cap will be complete when you 

add 1635 points to it. 

Let  𝑝12 = {171, 341, 511, 681, 851, 1021, 1361, 1531, 1701, 1871, 2041, 2211}.  

Put  𝒟1190
𝑝12 = 𝜒1190⋃𝑝12. The maximum number of the intersection points between 𝜒1190

𝑝12  and 

the lines of 𝑃𝐺(3,13) are fourteen, so 𝜒1190
𝑝12  is (14,14)-cap and 𝑐𝑖 values are  𝑐0 = 2366. 

Since 𝑐0 ≠ 0, then 𝜒1190
𝑝12  is incomplete. The (14,14)-cap will be complete when you add 2366 

points to it.  

 

 



IHJPAS. 2024, 37( 3 ) 

415 
 

4. Results and discussions 

The order pairs (𝑠 + 𝑖, 𝑑 + 𝑖) refer to the size and degree of the new extension cap from (𝑠, 𝑑)-

cap, 𝒟𝑃
𝑝𝑖 . Let # denote the number of added points to the orbit to be complete. Table 1 contains a 

summary of the getting results. 

 
Table 1. Details about the extension complete caps. 

 𝒟𝑃
𝑝𝑖   (size, degree) of the cap # 

1 𝒟4
𝑝𝑖  (595,7) → (595 + 𝑖, 7 + 𝑖), 𝑖 = 1, … ,7 196, 413, 600, 702, 1016, 1148, 1778 

2 𝒟7
𝑝𝑖   (340,4) → (340 + 𝑖, 4 + 𝑖), 𝑖 = 1, … ,10 

66, 196, 335, 463, 685, 816, 991, 

1180, 1395, 2030 

3 𝒟14
𝑝𝑖  (170,2) → (170 + 𝑖, 2 + 𝑖), 𝑖 = 1, … ,12 

20, 119, 254, 352, 537, 629, 818, 970, 

1200, 1325, 1516, 2198 

4 𝒟20
𝑝𝑖   (119,7) → (119 + 𝑖, 7 + 𝑖), 𝑖 = 1, … ,7 

658, 827, 1033, 1250, 1383, 1555, 

2254 

5 𝒟28
𝑝𝑖  (85,2) → (85 + 𝑖, 2 + 𝑖), 𝑖 = 1, … ,12 

74, 176, 319, 433, 623, 671, 892, 

1047, 1299, 1367, 1577, 2283 

6 𝒟35
𝑝𝑖  (68,3) → (68 + 𝑖, 3 + 𝑖), 𝑖 = 1, … ,11 

189, 334, 486, 627, 692, 876, 1037, 

1312, 1401, 1591, 2301 

7 𝒟68
𝑝𝑖  (35,7) → (35 + 𝑖, 7 + 𝑖), 𝑖 = 1, … ,7 

709, 901, 1065, 1395, 1427, 1615, 

2338 

8 𝒟70
𝑝𝑖  (34,2) → (34 + 𝑖, 2 + 𝑖), 𝑖 = 1, … ,12 

109, 226, 368, 468, 639, 697, 989, 

1056, 1330, 1421, 1612, 2334 

9 𝒟119
𝑝𝑖  (20,2) → (20 + 𝑖, 2 + 𝑖), 𝑖 = 1, … ,12 

117, 241, 392, 507, 639, 734, 891, 

1066, 1335, 1444, 1623, 2348 

10 𝒟140
𝑝𝑖  (19,2) → (19 + 𝑖, 2 + 𝑖), 𝑖 = 1, … ,12 

128, 248, 388, 489, 647, 707, 904, 

1062, 1340, 1383, 1625, 2351 

11 𝒟238
𝑝𝑖  (10,2) → (10 + 𝑖, 2 + 𝑖), 𝑖 = 1, … ,12 

119, 243, 393, 509, 643, 707, 896, 

1065, 1406, 1450, 1629, 2358 

12 𝒟340
𝑝𝑖  (7,7) → (7 + 𝑖, 7 + 𝑖), 𝑖 = 1, … ,7 

702, 912, 1077, 1408, 1451, 1635, 

2366 

13 𝒟476
𝑝𝑖  (5,2) → (5 + 𝑖, 2 + 𝑖), 𝑖 = 1, … ,12 

137, 243, 399, 509, 632, 677, 896, 

1069, 1407, 1389, 1630, 2363 

14 𝒟595
𝑝𝑖  (4,2) → (4 + 𝑖, 2 + 𝑖), 𝑖 = 1, … ,12 

118, 221, 404, 491, 633, 695, 902, 

1073, 1346, 1411, 1632, 2364 

15 𝒟1190
𝑝𝑖  (2,2) → (2 + 𝑖, 2 + 𝑖), 𝑖 = 1, … ,12 

119, 222, 401, 491, 603, 681, 899, 

1069, 1347, 1434, 1635, 2366 

 

5. Conclusion 

Theory of Group action is very helpful to introduce new caps in the finite projective space of 

dimension higher than two. All caps that are founded in this paper can be used to construct linear 

codes. Also, all these caps can be used to construct more caps which will deal with it in the next 

paper.     

 

Acknowledgment  

         The authors would like to thank the University of Mustansiriyah 

(https://uomustansiriyah.edu.iq/), Department of Mathematics in the College of Sciences for their 

motivation and support. 



IHJPAS. 2024, 37( 3 ) 

416 
 

 

Conflicts of Interest   

The authors declare no conflict of interest. 

 

Funding 

Non 

 

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