Full segments on the triangular grid
Eszter K. Horváth and Zoltán Németh
Dedicated to Gábor Czédli, Ágnes Szendrei and Mária B. Szendrei on their fifty-fifth birthday
Bolyai Institute, University of Szeged, Aradi vértanúk tere 1, 6720 Szeged, Hungary,
horeszt@math.u-szeged.hu, znemeth@math.u-szeged.hu
This research was partially supported by the Hungarian National Foundation for Scientific
Research under Grants # T 049433 and T 046192.
Abstract. Recently the notion of full segments appeared to be useful in information theory
(see [FöSi]); moreover, their maximum number turned out to be an interesting combinatorial question as well. In one dimension, the maximum number of full segments is known (see
[Hä]). Czédli generalized the problem for two dimensional rectangular segments (see[Cz]),
and solved it. Here we investigate the case of the triangular grid. We give upper and lower
estimations, using elementary, lattice theoretical and analitical considerations.
2000 Mathematics Subject Classification. Primary 06D99, Secondary 05A05.
Keywords and phrases: distributive lattice, full segment, weakly independent subset, triangular grid, weak basis, convergent series, digamma function.
1. Introduction
In information theory, in connection with instantaneous (prefix-free) codes, the notion
of the one-dimensional full segments of vectors has appeared, see [FöSi] for further details.
The problem of the maximum number of such one-dimensional full segments is solved first
by Härtel [Hä]. Czédli [Cz] generalized the notion of full segments for two dimensions
(on the square grid), and since in this case the notion might have topographical and
geoinformatical meaning, he called full segments on the square grid as rectangular islands.
In [Cz] the maximum number of the rectangular islands is exactly determined.
After these investigations we raised the naturally appearing case of triangular grid.
Similarly to the rectangular islands [Cz], the notion of full segment on triangular grid has
also some combinatorial interest. However, in the triangular case, the inequality analogous
to [Cz, Eq. (3)], with arbitrary m1 and m2 , is much more complicated, and did not lead us
to the exact formulae of the maximum number of the full segments (“triangular islands”)
on the triangular grid. Nevertheless, we found some interesting inequalities of other kinds,
all of which lead us to lower estimations in elementary ways. We also improved the
lattice-theoretical upper estimation n2 by using lattice-theoretical considerations and by
determining the limits of well known convergent series expressed by the so-called digamma
function.
1
2. Preliminary definitions
On the euclidean plane, given an equilateral triangle of side length n (n ∈ N), we
divide this triangle into small equilateral triangles of side lengths 1 as seen in Figure 1,
and we denote the set of such small triangles by T0 . Those small triangles of side lengths
1 are called cells and denoted by t. (In some of our considerations later, e.g., in Lemmas
5 and 6 the side length corresponding to T0 will be 2n + 2 or 2n + 1 instead of n ; we will
indicate this change explicitly.)
Figure 1.
√
We call two cells neighbouring, if the distance of their center points is at most 2 3/3.
Let T1 and T2 be triangles similar to the big one consisting of small ones. We say that T1
and T2 are far from each other , if they are disjoint and no cell of T1 is neighboring with
any cell of T2 . Obviously, the role of T1 and T2 are symmetric in the definition, i.e., T1
and T2 are interchangeable in it.
Consider the mapping A : T0 → R, t 7→ at . Moreover, consider triangles similar to the
big one consisting of cells. We call such a triangle T a full triangular segment, if for all of
its neighbouring cell tn , atn < min{at : t ∈ T } holds. The set of full triangular segments
of A will be denoted by Str (A).
The set of all equilateral triangles consisting of cells and similar to T0 will be denoted
by T (n). Of course, the empty set does not belong to T (n).
3. Statements in connection with the full segments on the triangular grid
The analogue of [Cz, Lemma 2] is valid for our case as follows in our Lemma 1.
Lemma 1. Let H be a subset of T (n). Then the following two conditions are equivalent:
(i) There exists mapping A : T0 → R, t 7→ at such that H = Str (A).
(ii) T0 ∈ H, and for any T1 , T2 ∈ H either T1 ⊆ T2 , or T2 ⊆ T1 , or T1 and T2 are far from
each other.
Proof. The proof is essentially the same as the proof of [Cz, Lemma 2]. Suppose
H = Str (A) for some mapping A, according to (i). Then T0 ∈ H is evident. If the rest of
2
(ii) fails, then there exists a cell t̂2 ∈ T2 \ T1 which is neighbouring with some cell of T1 .
Since T1 is a full triangular segment, we have
at̂2 < min{at1 : t1 ∈ T1 }.
Similarly, there exists a cell t̂1 ∈ T1 \ T2 which is neighbouring with some cell of T2 . Since
T2 is a full triangular segment, we have
at̂1 < min{at2 : t2 ∈ T2 }.
Hence
at̂2 < min{at1 : t1 ∈ T1 } ≤ at̂1 < min{at2 : t2 ∈ T2 }
contradicts t̂2 ∈ T2 . This proves (i) ⇒ (ii).
The converse implication will be proved via induction on n. For n = 1 or |H| = 1
everything is clear. Suppose n > 1, |H| > 1 and (ii) holds for H. Let M1 , . . . , Mk be the
maximal elements of H \ {T0 }. Clearly,
Hi = {T ∈ H : T ⊆ Mi }
satisfies (ii) for the triangle Mi , 1 ≤ i ≤ k. Hence, by the induction hypothesis, there
is a mapping Ai : Mi → R such that Str (Ai ) = Hi for each i ∈ {1, . . . , k}. Now choose
an r ∈ R such that r is strictly less than the minimum of the image values of Ai for all
i ∈ {1, . . . , k}. Then the union of the Ai , 1 ≤ i ≤ k, and the constant mapping
T0 \ (M1 ∪ . . . ∪ Mk ) 7→ {r}
is a T0 → R mapping A. Since the Mi are pairwise far fom each other, obviously H ⊆
Str (A). But those cells that have images r belong only to the full triangular segment T0 .
Subsets H satisfying the equivalent conditions of Lemma 1 will be called systems of
full triangular segments.
With the help of lattice theory — similarly to the case of the two-dimensional rectangular segments — an upper bound can be given for the maximal number of the full
triangular segments. The notion of a distributive lattice can be found e.g. in [Grä]. Let
L = (L; ∨, ∧) be a finite distributive lattice. Following [CzHuSch], a subset H of L is called
weakly independent, if for any k ∈ N and h, h1 , . . . , hk ∈ H which satisfy h ≤ h1 ∨ · · · ∨ hk
there exists an i ∈ {1, . . . , k} such that h ≤ hi . Maximal week independent subsets are
called weak bases of L. We know (see [CzHuSch]) that the set J0 (L) of join-irreducible
elements and all maximal chains are weak bases of L. As for the upper bound, we need
the following result.
Lemma 2 [CzHuSch]. Any two weak bases of a finite distributive lattice have the same
number of elements.
The lattice of all subsets of T0 is a distributive lattice, which will be denoted by
P(T0 ) = (P(T0 ); ∪, ∩).
3
Lemma 3. Let H be a system of triangular segments of T0 . Then H is a weakly independent subset of P(T0 ). Consequently, |H| ≤ n2 .
Proof. The proof is essentially the same as the proof of [Cz, Lemma 3]. Suppose
T ⊆ T1 ∪ · · · ∪ Tk ,
(1)
where T and T1 ∪ · · · ∪ Tk belong to H. We can assume that this inclusion is irredundant,
i.e., there is no i ∈ {1, . . . , k} with
T ⊆ T1 ∪ · · · ∪ Ti−1 ∪ Ti+1 ∪ · · · ∪ Tk .
Then no Ti is disjoint from T and the Ti are pairwise incomparable. If T ⊆ Ti for some
i, then we are done. In the opposite case k ≥ 2 and Ti ⊂ T for all i ∈ {1, . . . , k}. Then
there is a cell t of T neighbouring with Ti . Since the Tj for 1 < j are far from T1 , t does
not belong to T1 ∪ · · · ∪ Tk . This contradicts (1). Consequently, H is a weakly independent
subset of P(T0 ).
Now, we extend H to a weak basis H0 of P(T0 ), and consider a maximal chain C in
P(T0 ). Then C has n2 + 1 elements and it is also a weak basis. Hence we obtain from
Lemma 2 that |H0 | = n2 + 1. On the other hand, the empty set belongs to every weak
basis but not to H, consequently |H| ≤ |H0 | − 1 = n2 . In the sequel we denote by f (n) the maximum number of the full triangular segments
in an equilateral triangle of side length n.
4. Inequalities for the function f (n)
In order to get upper estimates for the function f (n) we need the following inequalities.
Lemma 4. f (n) ≥ f (n − 3) + n + 1.
Proof. Similarly to the proof of formula (3) of [Cz], we draw eqilateral subtriangles of
side lengths n − 3, 2, and 1 into a big equilateral triangle of side length n as seen in Figures
2–5. Clearly, the maximum number of the full triangular segments in the set T0 (of all cells)
is not smaller than the sum of the maximum numbers of the full triangular segments in
the mentioned subtriangles plus one (since T0 itself is obviously a full triangular segment).
In the case of n − 3 = 4k we obtain
f (n) ≥ f (n − 3) + f (2) + 2f (2)k + 1 + 1
= f (n − 3) + 4k + 3 + 1 = f (n − 3) + n + 1,
see also Figure 2.
4
Figure 2.
In the case of n − 3 = 4k + 1 we obtain
f (n) ≥ f (n − 3) + f (2) + 2f (2)k + 2 + 1
= f (n − 3) + 4k + 4 + 1 = f (n − 3) + n + 1,
see also Figure 3.
Figure 3.
In the case of n − 3 = 4k + 2 we obtain
f (n) ≥ f (n − 3) + f (2) + 2f (2)k + f (2) + 1 + 1
= f (n − 3) + 4k + 5 + 1 = f (n − 3) + n + 1,
see also Figure 4.
Figure 4.
5
Finally, in the case of n − 3 = 4k + 3 we obtain
f (n) ≥ f (n − 3) + f (2) + 2f (2)k + 4 + 1
= f (n − 3) + 4k + 6 + 1 = f (n − 3) + n + 1,
see also Figure 5.
Figure 5.
For all cases the promised inequality holds.
Lemma 5. f (2n + 2) ≥ 3f (n) + f (n + 1) + 1.
Proof. Now we start with a big equilateral triangle of side length 2n + 2. In this
proof we use the notation T0 for the set of all cells in this equilateral triangle of side length
2n+2. Into the big triangle we draw three equilateral subtriangles of side lengths n and one
equilateral subtriangle of side length n + 1 into the center as shown in Figure 6. Similarly
to the former cases, if we add 1 to the maximal number of segments corresponding to the
mentioned four triangles (since T0 itself is also a full triangular segment), then this sum is
clearly not greater than the maximum number of full triangular segments in T0 . Figure 6.
6
Lemma 6. f (2n + 1) ≥ 3f (n) + f (n − 1) + 1.
Proof. Now we start with a big equilateral triangle of side length 2n + 1. In this
proof we use the notation T0 for the set of all cells in this eqilateral triangle of side length
2n + 1. Into the big triangle, we draw three subtriangles of side lengths n and one triangle
of side length n − 1 into the center, as shown in Figure 7. Similarly to the former two
cases, if we add 1 to the maximal number of segments corresponding to the mentioned
four triangles (since T0 itself is also a full triangular segment), then this sum is clearly not
greater than the maximum number of full triangular segments in T0 . Figure 7.
5. Lower estimate
As we mentioned, we do not have inequality similar to [Cz, Eq. (3)], with arbitrarily
chosen parameters m1 and m2 . Here we present two approaches (based on the above
mentioned three Lemmas), both of which yielding lower estimates.
First, if we draw subtriangles the sizes of which are “as different as possible” into
the big triangle (see Figures 2–5), then we may use the recursive relation of Lemma 4 as
follows
f (n) ≥ f (n − 3) + n + 1.
It is clear that the functions
gc (n) :=
1 2 5
n + n+c
6
6
satisfy the recursive relation
gc (n) = gc (n − 3) + n + 1.
Moreover, if we choose c := − 13 , then the relations for the initial values
gc (1) ≤ 1,
gc (2) ≤ 2,
7
gc (3) ≤ 4
are fulfilled. So, it is clear that
f (n) ≥
1 2 5
1
n + n− ,
6
6
3
which yields
1
f (n)
≥ .
2
n
6
Remark. Finding inequalities between f (n) and f (n − k) where k > 3 instead of the
one in Lemma 4 would probably result in a better estimation, but the obtaining inequality
analogous to the one in Lemma 4 seems very complicated if possible at all.
In the second case we draw subtriangles the sizes of which are “as same as possible”
into the big triangle (see Figures 6–7). This way we obtained the relations of Lemmas 5
and 6, which are our starting points now:
f (2n + 2) ≥ 3f (n) + f (n + 1) + 1,
f (2n + 1) ≥ 3f (n) + f (n − 1) + 1.
Now we define the quadratic functions
gc (n) :=
3c + 1 2 3c + 1
n +
n + c.
6
2
Easy calculation gives that the equalities
gc (2n + 2) = 3gc (n) + gc (n + 1) + 1,
and
gc (2n + 1) = 3gc (n) + gc (n − 1) + 1
hold; moreover, if we choose c :=
1
18 ,
then the initial conditions
g(1) ≤ 1,
g(2) ≤ 2
are fulfilled. Finally, it is clear, that
f (n) ≥
7
1
7 2
n + n+ .
36
12
18
Summarizing our estimates gives the following result.
Theorem 1. For the function f (n) the inequalities
f (n) ≥
and
f (n) ≥
1 2 5
1
n + n−
6
6
3
7
1
7 2
n + n+
36
12
18
hold.
We observe that the second estimate is slightly closer than the first one.
8
6. Upper estimate
We say that C is a companion set of H if C ∩ H = ∅ and H ∪ C ∪ ∅ is a weak basis.
A cell t ∈ T0 is called an outer cell of H, if t is not in any maximal element of H \ {T0 }.
The set of the maximal elements of H \ {T0 } will be denoted by max(H). The set of outer
cells of H will be denoted by out(H). For a cell t let t+ = {t}. For a set S of cells, let
S + = {t+ : t ∈ S}. For T ∈ H, let HT denote {T ∈ H : S ⊆ T }.
Lemma 7. Let H be a system of triangular segments of T0 . Then H has a companion
set C. Moreover, for each companion set C of H there is an unique outer cell t of H such
that t+ does not belong to C.
Proof. The proof is essentially the same as the proof of [Cz, Lemma 4]. We prove the
lemma by induction on |H|. For |H| = 1 we can choose any cell t and let C = ((T0 ) \ {t})+.
Then Lemma 2 and |H ∪ C ∪ ∅| = n2 + 1 easily give that H ∪ C ∪ ∅ is a weak basis, whence
C is a companion set of H.
Suppose now that |H| > 1 and max(H) = {T1 , . . . Tk }. We denote the system of
triangular segments of Ti by HTi . Suppose that for 1 ≤ i ≤ k, Ci is a companion set of
HTi in the triangle Ti . Fix a cell t ∈ out(H) arbitrarily. We show that
C = C1 ∪ · · · ∪ Ck ∪ (out(H)+ \ {t+ })
(2)
is a companion set of H. It is straightforward to see that H ∪ C ∪ ∅ is weakly independent.
Since
.
[
.
.
|H ∪ C ∪ {∅}| = (HTi ∪ Ci ) ∪ out(H)+ \ {t+ } ∪ {T0 , ∅}
1≤i≤k
X
= 1 + | out(H)+ | +
1≤i≤k
= 1 + | out(H)| +
X
1≤i≤k
|(HTi ∪ Ci )|
|Ti |
.
. [
= 1 + out(H) ∪
Ti = 1 + |T0 | = 1 + n2 ,
1≤i≤k
H ∪ C ∪ {∅} is a weak basis by Lemma 1. This proves (2).
Now we show the remaining part of Lemma 7. Case n = 1 is trivial. If out(H)+ ⊆ C,
then
[
T0 ⊆ out(H)+ ∪
Ti
1≤i≤k
would contradict the weak independence of H ∪ C. If
c, d ∈ out(H)+ \ C
with
c 6= d,
then H ∪ C ∪ {∅, d} would be still independent, contradicting the maximality of H ∪ C ∪ {∅}.
9
Lemma 8. |H| + |C| = n2 .
Proof. Easily follows from |H ∪ C ∪ {∅}| = 1 + n2 , which we obtained and used in the
proof of Lemma 7. Now, for a system H of triangular segments of T0 , we define the deficiency of H as
d(H) = n2 − |H|. Now we give a lower estimate for d(H) in order to obtain upper estimate
for f (n). We define d(n) = inf d(H).
9
Now we define the series en in the following way: let e1 := 11
, and let
n
X
9
1
−
.
en =
11
k 2 + 3k + 23
k=2
Theorem 2. en n2 + (en − 1)(3n + 23 ) ≤ d(n).
Proof. We prove the statement by induction on n. In case of n = 1, |H| = 1, |C| = 0,
and d(1) = 0,
3
e1 12 + (e1 − 1) 3 +
≤ 0.
2
In the case of n > 1, there exists a system of triangular segments H such that d(n) =
d(H).
First, we magnify T0 as well as the elements of max(H) as seen in Figure 8.
Figure 8.
We denote the magnified segments (which are now hexagons) by T0∗ and Ti∗ . We define
G := T0∗ \
G,
[
Ti .
i
Let us denote the area of the cell t by µ(t). Since µ(t) = 1 for all t ∈ T0 , the area of
µ(G) = | out(H)| + 3n +
10
3
.
2
We denote the side length of Ti by ai .
We denote by E those parts of T0∗ that are not covered by the hexagons Ti∗ where Ti
are the elements of max(H). Now we may estimate as follows:
d(n) = d(H) = |C|
=
k
X
|(Ci )| + | out(H)| − 1
k
X
|(Ci )| + µ(G) − 3n −
i=1
=
i=1
5
2
k
k
X
5 X
|(Ci )| +
µ(Ti∗ \ Ti ) + µ(E)
= −3n − +
2 i=1
i=1
k
X
5
∗
|(Ci )| + µ(Ti \ Ti )
= −3n − + µ(E) +
2
i=1
k
X
5
3
≥ −3n − + µ(E) +
d(ai ) + 3ai +
2
2
i=1
k h
X
5
3i
3
2
≥ −3n − + µ(E) +
+ 3ai + .
eai ai + (eai − 1) 3ai +
2
2
2
i=1
Keeping in mind the monotonicity of en and observing the fact that the formula in the
square brackets is equal to eai (a2i + 3ai + 23 ), we have
k
X
5
d(n) ≥ −3n − + µ(E) + en−1
µ(Ti∗ )
2
i=1
k
≥ −3n −
X
5
µ(Ti∗ ) =: (∗).
+ (1 − en−1 )µ(E) + en−1 µ(E) +
2
i=1
Observing the facts that 1 − en−1 > 0, µ(E) ≥ 0 and
µ(E) +
k
X
3
µ(Ti∗ ) = µ(T0∗ ) = n2 + 3n + ,
2
i=1
we have
5
3
+ en−1 n2 + 3n +
2
2
3
2
= en−1 n + (en−1 − 1) 3n +
− 1.
2
Now, using the recursive relation
(∗) ≥ −3n −
en = en−1 −
n2
11
1
,
+ 3n + 23
we arrive at
dn ≥ en n2 + (en − 1) 3n +
3
.
2
The series
∞
X
1
k 2 + 3k +
k=2
3
2
converges, moreover, it is well known (see, e.g., [PBM]) that
∞
∞
X
X
2
1
2
1
3 = − 3 − 11 +
2
2
k + 3k + 2
k + 3k + 23
k=2
2
2
=− −
+
3 11
k=0
∞
X
1
√ √ 3
3
3
3
k=0 k + ( 2 − 2 ) k + ( 2 + 2 )
√
√ i
3
3
2
1 h 3
3
2
+ √ Ψ +
−Ψ −
,
=− −
3 11 − 3
2
2
2
2
here Ψ(x) is the so-called digamma function, see, for example, [PBM]. Calculating the
value lim en numerically gives
lim en = 0.47 . . . ,
which leads us to the upper estimate
f (n)
≤ 0, 52 . . .
n2
for all n.
Now we define the series ên with the following recursion:
9
(2n2 + 6n)ên−1 + 1
ê1 :=
,
ên :=
.
11
2n2 + 6n + 3
Theorem 3. ên n2 + (ên − 1)(3n + 23 ) ≤ d(n).
Proof. We use µ(E) ≥ 32 instead of µ(E) ≥ 0 in the proof of Theorem 1; it is possible
if we consider the small grey triangles in Figure 9.
Figure 9.
12
This is a better estimation than the previous one, so this must be also convergent.
Unfortunately, we cannot exactly determine the limit of ên , but we can calculate e.g. ê5000
by Maple, and if we put
e∗n
= ê5000 −
n
X
k=5001
k2
1
,
+ 3k + 23
then with the help of the digamma function, we can determine
lim e∗n = 0.62 . . . ,
which leads us to the improved upper estimate
f (n)
≤ 0.38 . . .
n2
for all n.
We can collect our estimates and state our final numerical result in the following form.
Theorem 4. For the function f (n) the asymptotical inequalities
0.19 . . . ≈
7
f (n)
≤ 2 ≤ 0.38 . . .
37
n
hold.
References
[Cz] G. Czédli, The number of rectangular islands by means of distributive lattices, submitted.
[CzHuSch] G. Czédli, A. P. Huhn and E. T. Schmidt: Weakly independent subsets in
lattices, Algebra Universalis, 20 (1985), 194–196.
[FöSi] S. Földes and N. M. Singhi, On instantaneous codes, J. Combin. Inform. System
Sci., 31 (2006), pp. 307–316.
[Grä] G. Grätzer, General Lattice Theory, Birkhauser Verlag, Basel–Stuttgart, 1978.
[Hä] Gernot Härtel, Unpublished result on one-dimensional full segments, mentioned by
Stephan Földes, in the University of Szeged Algebra Seminar , Szeged, Hungary, 2007.
[PBM] A. P. Prudnikov, Yu. A. Brychkov and O. I. Marichev, Integrals and series, Nauka,
Moscow, 1981 (in Russian).
13