Hindawi Publishing Corporation
International Journal of Mathematics and Mathematical Sciences
Volume 2012, Article ID 613201, 7 pages
doi:10.1155/2012/613201
Research Article
Translative Packing of Unit Squares into Squares
Janusz Januszewski
Institute of Mathematics and Physics, University of Technology and Life Sciences, Kaliskiego 7,
85-789 Bydgoszcz, Poland
Correspondence should be addressed to Janusz Januszewski, januszew@utp.edu.pl
Received 28 March 2012; Revised 24 May 2012; Accepted 24 May 2012
Academic Editor: Naseer Shahzad
Copyright q 2012 Janusz Januszewski. This is an open access article distributed under the Creative
Commons Attribution License, which permits unrestricted use, distribution, and reproduction in
any medium, provided the original work is properly cited.
Every collection
√ of n arbitrary-oriented unit squares admits a translative packing into any square
of side length 2.5 · n.
1. Introduction
Let i be a positive integer, let 0 ≤ αi < π/2, and let a rectangular coordinate system in the
plane be given. One of the coordinate system’s axes is called x-axis. Denote by Sαi a square
in the plane with sides of unit length and with the angle between the x-axis and a side of Sαi equal to αi . Furthermore, by Is denote a square with side length s and with sides parallel
to the coordinate axes.
We say that a collection of n unit squares Sα1 , . . . , Sαn admits a packing into a set C
if there are rigid motions σ1 , . . . , σn such that the squares σi Sαi are subsets of C and that
they have mutually disjoint interiors. A packing is translative if only translations are allowed
as the rigid motions.
For example, two unit squares can be packed into I2, but they cannot be packed
into I2 − for any > 0. Three and four unit squares can be packed into I2 as well see
Figure 1a. Obviously, two, three, or four squares S0 can be translatively packed into I2.
0 or α2 /
0, then two squares Sα1 and Sα2 cannot be translatively packed into
If either α1 /
I2. The reason is that for every α / 0, the interior of any square Sα translatively packed into
I2 covers the center of I2 see Figure 1b.
The problem of packing of unit squares into squares with possibility of rigid motions
is a well-known problem e.g., see 1–3. The best packings are known for several values of
n. Furthermore, for many values of n, there are good packings that seem to be optimal.
In this paper, we propose the problem of translative packing of squares. Denote by sn
the smallest number s such that any collection of n unit squares Sα1 , . . . , Sαn admits
2
International Journal of Mathematics and Mathematical Sciences
a
b
Figure 1
Q3
λ1
Q2
Q1
√
a
λ2
5
b
Figure 2
a translative packing into Is. The problem is to find sn for n 1, 2, 3, . . .. Obviously, sn >
√
By 4, Theorem 7, we deduce that limn → ∞ sn / n 1. We show that
sn ≤
√
2.5 · n.
√
n.
1.1
2. Packing into Squares
√
√
Example 2.1. We have s1 2. Each unit square can
√ be translatively packed into I 2, but it
is impossible to translatively pack Sπ/4 into I 2 − for any > 0.
√
Example 2.2. We have s2 5 see 5. Here, we only√recall that two squares: Sarctan 1/2
and Sarctan 2 cannot be translatively packed into I 5 − for any > 0 see Figure 2a.
√
√
Example 2.3. We have √
s4 2 2. Four√squares Sπ/4 admit a translative packing into I2 2
2/2 √
≤ λ ≤ 3 2/2. In Figure 3b and Figure 4a, we illustrate the
see Figure 3, where
√
three pictures, we conclude that four
cases when λ 2 and λ 2/2, respectively. By these√
squares√Sπ/4 cannot be translatively packed into I2
√ 2 − , for any > 0. Consequently,
√
s4 ≥ 2 2. On the other hand, four circles of radius 2/2 can be packed into I2 2
√ see
Figure 4b. Since any
√ square Sαi can be translatively packed into a circle of radius 2/2,
it follows that s4 ≤ 2 2.
Lemma 2.4 see 5.
√ Every unit square can be translatively packed into any isosceles right triangle
with legs of length 5.
International Journal of Mathematics and Mathematical Sciences
3
λ
a
b
Figure 3
a
b
Figure 4
√
√
√
√
Theorem 2.5. If n ≥ 3, then sn ≤ 10 5/2 3 · n.
Proof. Let Sα1 , Sα2 , and Sα3 be unit squares and put
λ1 √ 1 √
10 5 .
2
2.1
Three
quadrangles
√ congruent
√
√ Q1 , Q2 , and Q3 , presented in Figure 2b, of side lengths
λ1 , 5, 2.5, and λ2 √ λ1 − 5, are contained in Iλ1 . Since the length of the diagonal of
Sαi is smaller than 2.5, by Lemma 2.4 we deduce that Sαi can be translatively packed
into Qi for i 1, 2, 3. Consequently, the squares Sα1 , Sα2 , and Sα3 can be translatively
packed into Is and
√
s3 ≤ λ1 √
10 5 √
· 3.
√
2 3
2.2
Now assume that 4 ≤ n ≤ 16.
Denote by mn the smallest number s such that n circles of unit radius can be packed
into Is. The problem of minimizing the side of a square into which n congruent circles can
be packed is a well-known question. The values of mn are known, among others, for n ≤ 16
see Table 2.2.1 in 6 or 7. We know that
m4 4,
m10 < 6.75,
m5 < 4.83,
m11 < 7.03,
m6 < 5.33,
m7 < 5.74,
m12 < m13 < 7.47,
m8 < m9 6,
m14 < m15 < m16 8.
2.3
4
International Journal of Mathematics and Mathematical Sciences
a
b
Figure 5
√
Since each unit square is contained√in a circle of radius 2/2, it follows that n unit
squares can be translatively packed into I 2mn /2. It is easy to verify that
√
sn ≤
2
mn <
2
√
√
10 5 √
· n,
√
2 3
2.4
for n 4, 5, . . . , 16.
Finally, assume that n > 16. We use two lattice arrangements of circles.
There exists an integer m ≥ 4 such that either
m2 < n ≤ m 2 m
or m2 m < n ≤ m 12 .
2.5
√
√
2/2 can be packed into
I √2m see Figure 5a.
Obviously, m2 circles of radius
√
√
Moreover, m2 m circles of radius 2/2 can be packed into I 2m 2/2 see Figure 5b;
it is easy to check that
√
√2 √
√
2
< 2m ,
3m 2 ·
2
2
2.6
provided m ≥ 4.
If m2 1 ≤ n ≤ m2 m, then
sn ≤
√
√
√
√
√
2 √ √
2
10 5 √
≤ 2· n−1
<
· n.
2m √
2
2
2 3
If m√2 m 1 ≤ n ≤ m 12 , then sn ≤
m ≤ 1/2 4n − 3 − 1/2. Thus,
√
2.7
2m 1. Since m2 m 1 ≤ n, it follows that
√
√
√ 1√
10 5 √
1
4n − 3 <
· n.
sn ≤ 2
√
2
2
2 3
By Theorem 2.5, Examples 2.1 and 2.2, we conclude the following result.
2.8
International Journal of Mathematics and Mathematical Sciences
5
Corollary 2.6. Every
√ collection of n unit squares
√ admits a translative packing into any square of area
2.5 n, that is, sn ≤ 2.5 · n. Furthermore, s2 5.
For n ≥ 1980, the following upper bound is better than the bound presented in
Theorem 2.5.
Lemma 2.7. Let n be a positive integer, and let k be the greatest integer not over
sn ≤
π √
1
21 k n 2k2 − 4k 2 .
2k
√
4
n. Then
2.9
Proof. Assume that Sα1 , . . . , Sαn is a collection of n unit squares. Let k be the greatest
√
integer not over 4 n, let η π/2k, and put
√
ζ 1η
21 k n 2k2 − 4k 2 .
2.10
For each i ∈ {1, . . . , n}, there exists j ∈ {1, 2, . . . , k} such that
j − 1 η ≤ αi < jη.
2.11
Put ϕ αi − j − 1η. We have 0 ≤ ϕ < η. Moreover, let λ3 and λ4 denote the lengths of
the segments presented in Figure 6a. Since
λ3 λ4 cos ϕ sin ϕ ≤ 1 ϕ < 1 η,
2.12
it follows that Sαi is contained in a square Pi with side length 1 η and with the angle
between the x-axis and a side of Pi equal to j − 1η. We say that Pi is a j-square.
To prove Lemma 2.7, it suffices to show that P1 , . . . , Pn can be translatively packed into
Iζ. Denote by Aj the total area of the j-squares, for j 1, . . . , k. Obviously,
k
2
Aj n 1 η .
j1
2.13
Put
hj √ Aj
√ 2 2 1η ,
ζ−2 2 1η
2.14
for j 1, . . . , k. Observe that
k
j1
hj √ A1 · · · Ak
√ 2k 2 1 η .
ζ−2 2 1η
2.15
6
International Journal of Mathematics and Mathematical Sciences
1
λ3
λ4
ϕ
a
b
Figure 6
Thus,
2
√ n 1η
hj √ 2k 2 1 η .
ζ−2 2 1η
j1
2.16
2
√ n 1η
√ 2k 2 1 η x
x−2 2 1η
2.17
√ 2
x2 − x · 2 2 1 η k 1 8k − n 1 η 0.
2.18
k
The equation
is equivalent to
It is easy to verify that x ζ is a solution of this equation. Consequently,
2
k
√ n 1η
hj √ 2k 2 1 η ζ.
ζ−2 2 1η
j1
2.19
We divide Iζ into k rectangles R1 , . . . , Rk , where Rj is a rectangle of side lengths ζ
√
and hj . Since the diagonal of each Pi equals 21 η and
√ √ Aj ζ − 2 2 1 η hj − 2 2 1 η ,
2.20
it follows that all j-squares admit a translative packing into Rj for j 1, . . . , k see
Figure 6b. Hence, P1 , . . . , Pn , and consequently Sα1 , . . . , Sαn can be translatively packed
into Iζ. This implies that sn ≤ ζ.
Theorem 2.8. Let n be a positive integer. Then
√
√
π √
4
2
n O1,
sn ≤ n 2
2.21
as n → ∞.
√
4
n. Since
√
√
n 2k2 − 4k 2 < n 2k2 ≤ n 2 n < n 1,
Proof. Let k be the greatest integer not over
2.22
International Journal of Mathematics and Mathematical Sciences
7
by Lemma 2.7, it follows that, for n > 1,
sn <
√
√ √
π
1 √
2 1 4n n1
4
2 n−2
√
n
√
π π √2
√
π
π √
4
1 √
.
2
n
21
4
2
2
n−1
2.23
References
1 H. T. Croft, K. J. Falconer, and R. K. Guy, Unsolved Problems in Geometry, Springer, New York, NY, USA,
1991.
2 E. Friedman, “Packing unit squares in squares: a survey and new results,” The Electronics Journal of
Combinatorics # DS7, 2009.
3 F. Göbel, “Geometrical packing and covering problems,” in Packing and Covering in Combinatorics, A.
Schrijver, Ed., vol. 106 of Mathematical Centre tracts, pp. 179–199, 1979.
4 H. Groemer, “Covering and packing properties of bounded sequences of convex sets,” Mathematika,
vol. 29, no. 1, pp. 18–31, 1982.
5 J. Januszewski, “Translative packing two squares in the unit square,” Demonstratio Mathematica, vol.
36, no. 3, pp. 757–762, 2003.
6 G. Fejes Tóth, “Packing and covering,” in Handbook of Discrete and Computational Geometry, J. E. Goodman and J. O’Rourke, Eds., pp. 19–41, CRC, Boca Raton, Fla, USA, 1997.
7 http://www.packomania.com/, updated by E. Sprecht.
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