AbstractThe study of blocking semiovals in finite projective planes was motivated by Batten in connection with cryptography. Dover in studied blocking semiovals in a finite projective plane of order q which meet some line inq− 1 points. In this note, some blocking semiovals in PG(2, q) are considered which admit a homology group, and three new families of blocking semiovals are constructed. Any blocking semioval in the first or the third family meets no line in q− 1 points

Suetake, Chihiro

Elsevier - Publisher Connector 

Some Blocking Semiovals which Admit a Homology Group 

Article No. 10.1006/eujc.2000.0403
Available online at http://www.idealibrary.com on
Europ. J. Combinatorics (2000) 21, 967–972
Some Blocking Semiovals which Admit a Homology Group
CHIHIRO SUETAKE
The study of blocking semiovals in finite projective planes was motivated by Batten [1] in connec-
tion with cryptography. Dover in [4] studied blocking semiovals in a finite projective plane of order q
which meet some line in q − 1 points. In this note, some blocking semiovals in PG(2, q) are consid-
ered which admit a homology group, and three new families of blocking semiovals are constructed.
Any blocking semioval in the first or the third family meets no line in q − 1 points.
c© 2000 Academic Press
1. INTRODUCTION
Let 5 = (P,L) be a projective plane. A blocking set in 5 is a set B of points such that for
every line l ∈ L, l ∩ B 6= φ, but l is not entirely contained in B. A semioval in5 is a set S of
points such that for every point P ∈ S, there exists a unique line l ∈ L such that l ∩ S = {P}.
The idea of a semioval was introduced in [3] and [7]. A blocking semioval in 5 is a set of R
of points which is both a semioval and a blocking set.
At the CMS Special Session in Finite Geometry in 1997, L. M. Batten posed the problem
of classifying all blocking semiovals. The study of the problem was started by Batten and
Dover in [2] and Dover in [4, 5]. Known families of blocking semiovals are few. Dover in [5]
constructed a family of blocking semiovals of size 3q − 4 in a projective plane 5 of order q,
with q ≥ 5, if5 contains a1-configuration. Of course, any desarguesian plane PG(2, q) has
this property. The author in [6] constructed two families of blocking semiovals in PG(2, q).
The first one is a family of blocking semiovals of size 3q − 4, where q ≥ 5. This family
contains the one constructed by Dover as a special case. The second one is a family of blocking
semiovals of size 3q−n−2, where q = ne, n ≥ 3, n is a prime power and e ≥ 2. Any blocking
semioval in these families meets some line in q − 1 points, and admits a nontrivial homology
group if q is odd.
Let S be a blocking semioval in a projective plane 5 of order q . Then let xi denote the
number of lines of 5 which meet S in exactly i points. Clearly x0 = 0 and x1 = |S| by the
definition of S. Dover in [4] proved that xq = 0 for q > 3.
In this note, we consider some blocking semiovals in PG(2, q) which admit a nontrivial
homology group, and construct the following three families of blocking semiovals:
(i) If q = re, r ≥ 3, r is a prime power and e ≥ 2, then there exist blocking semiovals of
size 3q − 4 with xq−1 = 0 (see Theorem 3.4).
(ii) If q = re, r ≥ 3, r is a prime power, e ≥ 2 and 3 ≤ n ≤ r , then there exist blocking
semiovals of size 3q − n − 2 with xq−1 = 1 (see Theorem 4.5).
(iii) If q = re1e2 , r ≥ 3, r is a prime power, e1 6= 1, e2 6= 1 and 3 ≤ n ≤ r , then there exist
blocking semiovals of size 3q − n − 2 with xq−1 = 0 (see Theorem 4.6).
All the other known families of blocking semiovals are unitals and vertexless triangles.
2. BLOCKING SEMIOVALS OF HOMOLOGY TYPE
Let P and L be the set of points and the set of lines in a desarguesian plane PG(2, q) with
q ≥ 7. PG(2, q) is defined on the three-dimensional vector space V = {(a, b, c)|a, b, c ∈
0195–6698/00/070967 + 06 $35.00/0 c© 2000 Academic Press
968 C. Suetake
G F(q)} over G F(q). For (a, b, c) ∈ V , set [a, b, c] = {(ax, bx, cx)|x ∈ G F(q)}. Then
P = {[a, b, c]|a, b, c ∈ G F(q) and (a, b, c) 6= (0, 0, 0)}. The lines of PG(2, q) are the two-
dimensional subspaces of V . Choose a line l0 from L. Let P0, P1, . . . , Pn be n + 1 distinct
points on l0, where 2 ≤ n ≤ q − 2. Choose a point Q0 off l0. Let, for i ∈ {0, 1, . . . , n}, i be
a subset of Pi Q0 − {Q0, Pi } with |i | ≥ 2. Set
S = (l0) ∪0 ∪1 ∪ · · · ∪n − {P0, P1, . . . , Pn}.
We will derive a necessary and sufficient condition for S to be a blocking semioval. We may
assume that Q0 = [1, 0, 0], P0 = [0, 0, 1], Pi = [0, 1, ai ] (i = 1, 2, . . . , n). We define
subsets 1i of G F(q)∗ = G F(q)− {0} (i = 0, 1, . . . , n) as follows:
10 = {x ∈ G F(q)|[1, 0, x] ∈ 0},
1i = {x ∈ G F(q)|[1, x, ai x] ∈ i } (i = 1, 2, . . . , n).
Clearly, |1i | ≥ 2 (i = 0, 1, . . . , n). For i, j, k ∈ {1, 2, . . . , n} with i 6= j , set 8 jk =
(ak − a j )1k and 8i jk = ak−a jai−a j 1k .
In the rest of this section we assume the following.
ASSUMPTION 2.1. If q is odd, −1i = 1i for all i ∈ {0, 1, . . . , n}.
LEMMA 2.2. S is a blocking set if and only if the following hold.
(i) G F(q)∗ = 11 ∪12 ∪ · · · ∪1n .
(ii) For any i ∈ {1, 2, . . . , n}, G F(q)∗ = 10 ∪⋃1≤ j (6=i)≤n 8i j .
PROOF. S is a blocking set if and only if for any i ∈ {0, 1, . . . , n}, any line through the
point Pi except Q0 Pi and l0 intersects 0 ∪1 ∪ · · · ∪ ̂i ∪ · · · ∪n .
Any line {[1, b, x]|x ∈ G F(q)} ∪ {P0}(b ∈ G F(q)∗) through the point P0 except Q0 P0
and l0 intersects 1 ∪2 ∪ · · · ∪n .
⇔ If b ∈ G F(q)∗, then there exists j ∈ {1, 2, . . . , n} such that b ∈ 1 j .
⇔ G F(q)∗ = 11 ∪12 ∪ · · · ∪1n .
Next let i ∈ {1, 2, . . . , n}.
Any line {[1, x, ai x + b]|x ∈ G F(q)} ∪ {Pi }(b ∈ G F(q)∗) through the point Pi except
Q0 Pi and l0 intersects 0 ∪1 ∪ · · · ∪ ̂i ∪ · · · ∪n .
⇔ If b ∈ G F(q)∗−10, then there exists j ∈ {1, 2, . . . iˆ, . . . , n} such that ai x + b = a j x .
⇔ If b ∈ G F(q)∗ − 10, then there exists j ∈ {1, 2, . . . , iˆ, . . . , n} such that b ∈ (a j −
ai )1 j = 8i j .
⇔ G F(q)∗ = 10 ∪⋃1≤ j (6=i)≤n 8i j . 2
LEMMA 2.3. S is a semioval if and only if the following hold.
(i) If a ∈ G F(q)− {a1, a2, . . . , an}, then G F(q)∗ = 10 ∪⋃1≤i≤n(ai − a)1i .
(ii) If a ∈ 10, then there exists a unique element j ∈ {1, 2, . . . , n} such that
a /∈⋃1≤k(6= j)≤n 8 jk , but a ∈⋃1≤k(6=i)≤n 8ik for all i ∈ {1, 2, . . . , jˆ, . . . , n}.
(iii) Let i ∈ {1, 2, . . . , n} and a ∈ 1i . Then one of the following occurs.
(a) If a /∈ ⋃1≤k(6=i)≤n 1k , then a ∈ 1ai−a j 10 ∪ ⋃1≤k(6=i, j)≤n 8i jk for all j ∈
{1, 2, . . . , iˆ, . . . , n}.
Some blocking semiovals 969
(b) If a ∈⋃1≤k(6=i)≤n 1k , then there exists a unique element j ∈ {1, 2, . . . , iˆ, . . . , n}
such that a /∈ 1
ai−a j 10∪
⋃
1≤k(6=i, j)≤n 8i jk , but a ∈ 1ai−as10∪
⋃
1≤k(6=i,s)≤n 8isk
for all s ∈ {1, 2, . . . , iˆ, . . . , jˆ, . . . , n}.
PROOF. S is a semioval if and only if the following hold.
(α) Let P ∈ (l0) − {P0, P1, . . . , Pn}. If l ∈ L, P ∈ l, Q0 /∈ l and l 6= l0, then (l) ∩ (0 ∪
1 ∪ · · · ∪n) 6= φ.
(β) For any i ∈ {0, 1, . . . , n} and for any P ∈ i , there exists a unique element j ∈
{0, 1, . . . , iˆ, . . . , n} such that the line P Pj does not intersect0 ∪1 ∪ · · · ∪ ̂i ∪ · · · ∪ ̂ j ∪
· · · ∪n .
Let P ∈ (l0)−{P0, P1, . . . , Pn}. Then P = [0, 1, a] for some a ∈ G F(q)−{a1, a2, . . . , an}.
Let l be a line such that P ∈ l, Q0 /∈ l and l 6= l0. Then l = {[1, x, ax+b]|x ∈ G F(q)}∪{P}
for some b ∈ G F(q)∗. (l) ∩0 6= φ if and only if b ∈ 10.
Let 1 ≤ i ≤ n.
(l)∩i 6= φ.⇔ There exists x ∈ 1i such that ax+b = ai x .⇔ b ∈ (ai−a)1i . Therefore,
(α) is equivalent to (i).
Let i = 0 and P ∈ 0. Then P = [1, 0, a] for some a ∈ 10. Let 1 ≤ j ≤ n. P Pj =
{[1, x, a j x + a]|x ∈ G F(q)} ∪ {Pj }. Let k ∈ {0ˆ, 1, . . . , jˆ, . . . , n}.
The line P Pj does not intersect k .
⇔ For any x ∈ 1k a j x + a 6= ak x .
⇔ a
ak−a j /∈ 1k .⇔ a /∈ (ak − a j )1k = 8 jk .
Therefore, when i = 0, (β) is equivalent to (ii).
Let 1 ≤ i ≤ n and P ∈ i . Then P = [1, a, ai a] for some a ∈ 1i . P P0 = {[1, a, x]|x ∈
G F(q)} ∪ {P0}. Let k ∈ {0ˆ, 1, . . . , iˆ, . . . , n}.
The line PP0 does not intersect k .
⇔ For any x ∈ 1k x 6= a.
⇔ a /∈ 1k .
Let j ∈ {1, 2, . . . , iˆ, . . . , n}. Then P Pj = {[1, x + a, a j x + ai a]|x ∈ G F(q)} ∪ {Pj } =
{[1, x, a j (x − a)+ ai a]|x ∈ G F(q)} ∪ {Pj }. Let k′ ∈ {1, . . . , iˆ, . . . , jˆ, . . . , n}.
The line P Pj intersects 0.
⇔ ai a − a j a ∈ 10.
⇔ a ∈ 1
ai−a j 10.
The line P Pj intersects k′ .
⇔ a j (x − a)+ ai a = ak′x for some x ∈ 1k′ .
⇔ a(ai−a j )
ak′−a j ∈ 1k′ .
⇔ a ∈ ak′−a j
ai−a j 1k′ = 8i jk′ .
Therefore, when 1 ≤ i ≤ n, (β) is equivalent to (iii). Thus the lemma is proved. 2
3. n = 2
In this section, under Assumption 2.1 we consider the point set S of PG(2, q) defined in
Section 2, when n = 2. We remark that from the definition of 1’s, |1i | ≥ 2(i = 0, 1, 2).
Then we may assume that a1 = 1, a2 = 0.
Lemmas 2.2 and 2.3 yield the following theorem.
970 C. Suetake
THEOREM 3.1. Let n = 2 and a1 = 1, a2 = 0. Then, S is a blocking semioval if and only
if the following hold.
(i) If i 6= j ∈ {0, 1, 2}, then G F(q)∗ = 1i ∪1 j .
(ii) If a ∈ G F(q)− {0, 1}, then G F(q)∗ = 10 ∪ (1− a)11 ∪ a12.
(iii) If {i, j, k} = {0, 1, 2}, then 1i = (1 j −1k) ∪ (1k −1 j ).
From Theorem 3.1 the next result follows.
COROLLARY 3.2. If S is a blocking semioval, then |S| = 3q − 4 and, furthermore, when
q is odd, an involutory homology
(
1 0 0
0 −1 0
0 0 −1
)
acts on S.
PROOF. Let S be a blocking semiovals. Since Theorem 3.1(iii) implies that the sets10,11
and12 exactly double cover G F(q)∗, |S| = |10|+|11|+|12|+(q−2) = 2(q−1)+(q−2) =
3q − 4. The existence of the involutory homology follows from Assumption 2.1. 2
EXAMPLE 3.3. Set 10 = G F(q)∗. If q is odd, let 11 6= φ be a subset of G F(q)∗ such
that −11 = 11 and 11 6= G F(q)∗. If q is even, let 11 be a subset of G F(q)∗ such that
2 ≤ |11| ≤ q − 3. Set 12 = G F(q)∗ −11. Then, 10, 11 and 12 satisfy (i), (ii) and (iii) of
Theorem 3.1. Therefore, the point set S corresponding to10,11 and12 is a blocking semio-
val of size 3q − 4 such that xq−1 = 1. This is a blocking semioval as defined in Theorem 4.2
of [5].
THEOREM 3.4. Let q = re, r ≥ 3, r be a prime power and e ≥ 2. Set 12 = G F(r)∗. Let
8 be a nonempty subset of G F(r)∗ such that −8 = 8 and 8 6= G F(r)∗. Here, if r is even,
let 2 ≤ |8| ≤ r − 3. Set 11 = (G F(q)∗ − G F(r)∗ ∪8 and 10 = G F(q)∗ −8. Then 10,
11 and 12 satisfy (i), (ii) and (iii) of Theorem 3.1. Therefore, the point set S corresponding
to 10, 11 and 12 is a blocking semioval of size 3q − 4 such that xq−1 = 0 and xq−2 6= 0.
PROOF. It is clear that 10, 11 and 12 satisfy (i) and (iii) of Theorem 3.1. We will show
that 10, 11, and 12 satisfy (ii) of Theorem 3.1. Let a ∈ G F(q) − {0, 1}. We want to show
that G F(q)∗ = 10 ∪ (1 − a)11 ∪ a12. Let x ∈ G F(q)∗ − 10. Then x ∈ 8 ⊆ G F(r)∗.
If a /∈ G F(r)∗, then x1−a /∈ G F(r)∗, and hence x ∈ (1 − a)11. If a ∈ G F(r)∗, then
x
a
∈ G F(r)∗ = 12 and therefore x ∈ a12. Thus G F(q)∗ = 10 ∪ (1− a)11 ∪ a12. 2
The author proved that the following conjecture is true for q ∈ {7, 11, 13, 17} using a
computer.
CONJECTURE 3.5. If q is a prime with q ≥ 7, then any blocking semioval S defined in
Theorem 3.1 is one of the blocking semiovals constructed in Example 3.3.
4. n ≥ 3
In this section we consider the set S of points of PG(2, q) defined in Section 2 for n ≥ 3.
ASSUMPTION 4.1. (i) q = re, r ≥ 3, r is a prime power and e ≥ 2.
(ii) 3 ≤ n ≤ r and G F(r) ⊇ {a1, a2, . . . , an}.
(iii) For any i ∈ {0, 1, . . . , n} and for any x ∈ G F(r)∗, x1i = 1i .
Lemmas 2.2 and 2.3 yield the following theorem.
Some blocking semiovals 971
THEOREM 4.2. Under Assumption 4.1, S is a blocking semioval if and only if the following
hold.
(i) For any i ∈ {0, 1, . . . , n}, G F(q)∗ = ∪0≤ j (6=i)≤n1 j .
(ii) For any a ∈ G F(q)− {a1, a2, . . . , an}, G F(q)∗ = 10 ∪⋃1≤i≤n(ai − a)1i .
(iii) For any i ∈ {0, 1, . . . , n} and for any a ∈ 1i , there exists a unique element j ∈
{0, 1, . . . , iˆ, . . . , n} such that a ∈ 1 j .
PROOF. Assume that S is a blocking semioval. It is clear that (i) and (ii) hold. Let i ∈
{0, 1, . . . , n} and a ∈ 1i . Then, by (ii) and (iii) of Lemma 2.3, there exists a unique ele-
ment j ∈ {0, 1, . . . , iˆ, . . . , n} such that a /∈ ⋃0≤k(6=i, j)≤n 1k . However, since G F(q)∗ =⋃
0≤k(6=i)≤n 1n , a ∈ 1 j . Thus we obtain (iii).
The converse is clear. 2
THEOREM 4.3. Under Assumption 4.1, S is a blocking semioval if and only if for all dis-
tinct i , j ∈ {0, 1, . . . , n} there exists a subset 1i j of G F(q)∗ which satisfies the following.
(i) Each 1i j is closed under multiplication by G F(r)∗.
(ii) For all distinct i, j ∈ {0, 1, . . . , n}, 1i j = 1 j i .
(iii) For any i ∈ {0, 1, . . . , n}, 1i =⋃0≤ j (6=i)≤n 1i j 6= φ.
(iv) G F(q)∗ =⋃0≤i< j≤n 1i j is a disjoint union.
(v) For any a ∈ G F(q)− {a1, . . . , an}, G F(q)∗ = 10 ∪⋃1≤i≤n(ai − a)1i .
PROOF. Assume that S is a blocking semioval. For all distinct i , j ∈ {0, 1, . . . , n}, set
1i j = 1i ∩1 j . Then clearly (i), (ii) and (v) hold.
Let 0 ≤ i ≤ n. From Theorem 4.2(i) it follows that 1i = 1i ∩ G F(q)∗ = 1i ∩(⋃
0≤ j (6=i)≤n 1 j
) =⋃0≤ j (6=i)≤n(1i ∩1i ) =⋃0≤ j (6=i)≤n 1i j . Therefore we obtain (iii).
Let 0 ≤ i 6= j ≤ n, 0 ≤ i 6= j ′ ≤ n, and j 6= j ′. If b ∈ 1i j ∩1i j ′ , then b ∈ 1i , b ∈ 1 j
and b ∈ 1 j ′ which are contrary to (i) and (iii) of Theorem 4.2. Therefore 1i j ∩ 1i j ′ = φ.
Furthermore, let 0 ≤ i ′ 6= i ≤ n and i ′ ≤ j ′. Since 1i j and 1i j ′ are mutually disjoint, and
1i j ′ = 1i ∩ 1 j ′ = 1 j ′i and 1i ′ j ′ are mutually disjoint, 1i j ∩ 1i ′ j ′ = φ. Thus, by (i) of
Theorem 4.2, we have (iv).
The converse is clear. 2
COROLLARY 4.4. If S is a blocking semioval, then |S| = 3q − n − 2 and the homology(
1 0 0
0 s 0
0 0 s
)
(s ∈ G F(r)∗) acts on S.
PROOF. |S| = ∑0≤i≤n |1i | + (q − n) = ∑0≤i≤n ∑0≤ j (6=i)≤n |1i j | + (q − n) =
2
∑
0≤i< j≤n |1i j | + (q − n) = 2(q − 1) + (q − n) = 3q − n − 2. Since each 1i is closed
under multiplication by G F(r)∗, the homologies do exist. 2
THEOREM 4.5. Assume Assumption 4.1. Set 10 = G F(q)∗ and let G F(q)∗ = 11 ∪ · · · ∪
1n be a mutually disjoint union. Then10,11, . . . ,1n satisfy (i), (ii) and (iii) of Theorem 4.2,
and the size of the blocking semioval corresponding to 10,11, . . . ,1n is 3q − n − 2 and
xq−1 = 1. In particular, when n = r , the blocking semioval is one of the blocking semiovals
defined in Theorem 5.2 of [6].
972 C. Suetake
THEOREM 4.6. Let q = re = re1e2 , r be a prime power, e1 6= 1, e2 6= 1 and 3 ≤ n ≤ r .
Set 101 = G F(q) − G F(re1). Let G F(re1)∗ = ⋃0≤i< j≤n,(i, j)6=(0,1)1i j be a partition such
that each 1i j is closed under multiplication by G F(r)∗. For distinct i, j ∈ {0, 1, . . . , n} with
i < j , set1 j i = 1i j . Then,1i j (0 ≤ i 6= j ≤ n) satisfy (i)–(v) of Theorem 4.3. Therefore, the
set S of points corresponding to1i j (0 ≤ i 6= j ≤ n) is a blocking semioval of size 3q−n−2
with xq−1 = 0 and xq−n 6= 0.
PROOF. We must show that 1i j (0 ≤ i 6= j ≤ n) satisfy (i)–(v) of Theorem 4.3. It is clear
that (i), (ii), (iii) and (iv) hold. We will show that (v) holds. Let a ∈ G F(q)−{a1, a2, . . . , an}.
Let x ∈ G F(q)−10. Then x ∈ G F(re1)∗.
Assume that a /∈ G F(re1)∗. Then for any i ∈ {1, 2, . . . , n}, x
ai−a /∈ G F(re1)∗ and hence
x
ai−a ∈ 101 ⊆ 11. In particular, let i = 1. Then x ∈ (a1 − a)11.
Assume that a ∈ G F(re1)∗. Then, for any i ∈ {1, . . . , n}, x
ai−a ∈ G F(re1)∗ =⋃
0≤i< j≤n,(i,k)6=(0,1)1i j . Therefore there exists j ∈ {1, 2, . . . , n} such that xai−a ∈ 1 j . In
particular, let i = j . Then x ∈ (a j − a)1 j . 2
REFERENCES
1. L. M. Batten, Determining sets, submitted.
2. L. M. Batten and J. M. Dover, Blocking semiovals with three intersection sizes, preprint.
3. F. Buekenhout, Characterizations of semiquadrics. A survey, Teorie Combinatorie (Rome, 1973),
vol. I, Atti Acc. Naz. Lincei, 17 (1976), 393–421.
4. J. M. Dover, Semiovals with large collinear subsets, J. Geom., to appear.
5. J. M. Dover, A lower bound on blocking semiovals, Europ. J. Combinatorics, to appear.
6. C. Suetake, Two families of blocking semiovals, Europ. J. Combinatorics, to appear.
7. J. A. Thas, On semiovals and semiovoids, Geom. Ded., 3 (1974), 229–231.
Received 6 September 1999 and accepted 15 February 2000
CHIHIRO SUETAKE
Himeji-kita High School,
Himeji,
Hyogo 670-0012,
Japan
E-mail: c-suetake@mvh.biglobe.ne.jp


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Some Blocking Semiovals which Admit a Homology Group

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