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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

Copyrightq2012 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 narbitrary-orientedunit squares admits a translative packing into any square

of side length√2.n.

1. Introduction

Letibe 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 calledx-axis. Denote byia square in the plane with sides of unit length and with the angle between thex-axis and a side ofSαi equal toαi. Furthermore, byIsdenote a square with side lengthsand with sides parallel

to the coordinate axes.

We say that a collection ofnunit squares1, . . . , Sαnadmits a packing into a setC if there are rigid motionsσ1, . . . , σn such that the squaresσiSαiare subsets ofCand 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 intoI2, but they cannot be packed intoI2−for any >0. Three and four unit squares can be packed intoI2as wellsee Figure1a. Obviously, two, three, or four squaresS0can be translatively packed intoI2. If eitherα1/0 orα2/0, then two squares1and2cannot be translatively packed into

I2. The reason is that for everyα /0,the interior of any squaretranslatively packed into

I2covers the center ofI2 see Figure1b.

The problem of packing of unit squares into squareswith possibility of rigid motions is a well-known probleme.g., see1–3. The best packings are known for several values of

n. Furthermore, for many values ofn, there are good packings that seem to be optimal. In this paper, we propose the problem of translative packing of squares. Denote bysn

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a b Figure 1

a

Q3

Q2

Q1

λ1

λ2

5

b

Figure 2

a translative packing intoIs. The problem is to findsnforn1,2,3, . . .. Obviously,sn>n.

By4, Theorem 7, we deduce that limn→ ∞sn/n1. We show that

sn≤√2.n. 1.1

2. Packing into Squares

Example 2.1. We haves1

2. Each unit square can be translatively packed intoI√2, but it is impossible to translatively packSπ/4intoI√2−for any >0.

Example 2.2. We haves2

5see5. Here, we only recall that two squares:Sarctan 1/2 andSarctan 2cannot be translatively packed intoI√5−for any >0see Figure2a.

Example 2.3. We haves42

2. Four squaresSπ/4admit a translative packing intoI2√2

see Figure3, where√2/2 ≤ λ ≤ 3√2/2. In Figure3band Figure4a, we illustrate the cases whenλ√2 andλ√2/2, respectively. By these three pictures, we conclude that four squaresSπ/4cannot be translatively packed intoI2√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 Figure4b. Since any squareSαican be translatively packed into a circle of radius√2/2,

it follows thats4≤2 √

2.

Lemma 2.4see5. Every unit square can be translatively packed into any isosceles right triangle

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λ

a b

Figure 3

a b

Figure 4

Theorem 2.5. Ifn3, thensn≤√10√5/2√3·√n.

Proof. LetSα1, Sα2,and3be unit squares and put

λ1

1 2

10√5. 2.1

Three congruent quadrangles Q1, Q2,andQ3, presented in Figure 2b, of side lengths

λ1,

5,√2.5, andλ2 λ1− √

5, are contained in 1. Since the length of the diagonal of

Sαiis smaller than√2.5, by Lemma2.4we deduce thatSαican be translatively packed

intoQi fori 1,2,3. Consequently, the squares1, Sα2,and3can be translatively

packed intoIsand

s3≤λ1 √

10√5 2√3 ·

3. 2.2

Now assume that 4≤n≤16.

Denote bymnthe smallest numberssuch thatncircles of unit radius can be packed

intoIs. The problem of minimizing the side of a square into whichncongruent circles can be packed is a well-known question. The values ofmnare known, among others, forn≤16 see Table 2.2.1 in6or7. We know that

m44, m5<4.83, m6 <5.33, m7<5.74, m8 < m96,

m10<6.75, m11 <7.03, m12< m13 <7.47, m14< m15< m168.

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a b Figure 5

Since each unit square is contained in a circle of radius√2/2, it follows thatnunit squares can be translatively packed intoI√2mn/2. It is easy to verify that

sn≤ √

2 2 mn<

10√5 2√3 ·

n, 2.4

forn4,5, . . . ,16.

Finally, assume thatn >16. We use two lattice arrangements of circles. There exists an integerm≥4 such that either

m2< nm2m or m2m < nm12. 2.5

Obviously,m2 circles of radius2/2 can be packed intoI2m see Figure 5a.

Moreover,m2mcircles of radius√2/2 can be packed intoI√2m√2/2 see Figure5b; it is easy to check that

3m

√ 2 2 < √ 2m √ 2

2 , 2.6

providedm≥4.

Ifm21nm2m, then

sn≤ √ 2m √ 2 2 ≤ √

2·√n−1

2 2 <

10√5 2√3 ·

n. 2.7

Ifm2m1nm12, thens

n≤√2m1. Sincem2m1≤n, it follows that

m≤1/2√4n−3−1/2. Thus,

sn≤ √ 2 1 2 √

4n−31 2

<

10√5 2√3 ·

n. 2.8

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sn

1 π 2k

21k n2k2−4k2. 2.9

Proof. Assume that 1, . . . , Sαn is a collection of n unit squares. Letk be the greatest

integer not over√4n, letηπ/2k, and put

ζ 1η√21k n2k24k2. 2.10

For eachi∈ {1, . . . , n},there existsj ∈ {1,2, . . . , k}such that

j−1ηαi < jη. 2.11

Putϕαij−1η. We have 0≤ϕ < η. Moreover, letλ3andλ4denote the lengths of

the segments presented in Figure6a. Since

λ3λ4 cosϕsinϕ≤1ϕ <1η, 2.12

it follows thatSαi is contained in a squarePi with side length 1η and with the angle

between thex-axis and a side ofPiequal toj−1η. We say thatPiis aj-square.

To prove Lemma2.7, it suffices to show thatP1, . . . , Pncan be translatively packed into

. Denote byAjthe total area of thej-squares, forj 1, . . . , k. Obviously,

k

j1

Ajn 1η2. 2.13

Put

hj Aj

ζ−2√2 1η2

2 1η, 2.14

forj1, . . . , k. Observe that

k

j1

hj A1· · ·Ak

ζ−2√2 1η 2k

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Figure 6

Thus,

k

j1

hj n 1η

2

ζ−2√2 1η 2k

2 1η. 2.16

The equation

n 1η2

x−2√2 1η2k

2 1ηx 2.17

is equivalent to

x2x·22 1ηk1 8kn 1η20. 2.18

It is easy to verify thatis a solution of this equation. Consequently,

k

j1

hj n 1η

2

ζ−2√2 1η2k

2 1ηζ. 2.19

We divideintok rectanglesR1, . . . , Rk, whereRj is a rectangle of side lengthsζ

andhj. Since the diagonal of eachPiequals√21ηand

Aj

ζ−2√2 1ηhj−2√2 1η, 2.20

it follows that all j-squares admit a translative packing into Rj for j 1, . . . , k see

Figure6b. Hence,P1, . . . , Pn,and consequently1, . . . , Sαncan be translatively packed

into. This implies thatsnζ.

Theorem 2.8. Letnbe a positive integer. Then

sn≤√n

2π 2

4

nO1, 2.21

asn → ∞.

Proof. Letkbe the greatest integer not over√4n. Since

n2k2−4k2<n2k2≤

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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. G ¨obel, “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 T ´oth, “Packing and covering,” in Handbook of Discrete and Computational Geometry, J. E.

Good-man and J. O’Rourke, Eds., pp. 19–41, CRC, Boca Raton, Fla, USA, 1997.

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