Translative Packing of Unit Squares into Squares

Abstract

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≤𝛼_𝑖<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 𝑆(𝛼_𝑖) a square in the plane with sides of unit length and with the angle between the 𝑥-axis and a side of 𝑆(𝛼_𝑖) equal to 𝛼_𝑖. Furthermore, by 𝐼(𝑠) denote a square with side length 𝑠 and with sides parallel to the coordinate axes.

We say that a collection of 𝑛 unit squares 𝑆(𝛼₁),…,𝑆(𝛼_𝑛) admits a packing into a set 𝐶 if there are rigid motions 𝜎₁,…,𝜎_𝑛 such that the squares 𝜎_𝑖𝑆(𝛼_𝑖) 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 𝐼(2), but they cannot be packed into 𝐼(2−𝜖) for any 𝜖>0. Three and four unit squares can be packed into 𝐼(2) as well (see Figure 1(a)). Obviously, two, three, or four squares 𝑆(0) can be translatively packed into 𝐼(2). If either 𝛼₁≠0 or 𝛼₂≠0, then two squares 𝑆(𝛼₁) and 𝑆(𝛼₂) cannot be translatively packed into 𝐼(2). The reason is that for every 𝛼≠0, the interior of any square 𝑆(𝛼) translatively packed into 𝐼(2) covers the center of 𝐼(2) (see Figure 1(b)).

(a)

(b)

Figure 1 

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 𝑠_𝑛 the smallest number 𝑠 such that any collection of 𝑛 unit squares 𝑆(𝛼₁),…,𝑆(𝛼_𝑛) admits a translative packing into 𝐼(𝑠). The problem is to find 𝑠_𝑛 for 𝑛=1,2,3,…. Obviously, 𝑠_𝑛>√𝑛. By [4, Theorem 7], we deduce that lim_𝑛→∞𝑠_𝑛/√𝑛=1. We show that 𝑠_𝑛≤√2.5⋅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 𝑆(𝜋/4) into 𝐼(√2−𝜖) for any 𝜖>0.

Example 2.2. We have 𝑠₂=√5 (see [5]). Here, we only recall that two squares: 𝑆(arctan1/2) and 𝑆(arctan2) cannot be translatively packed into 𝐼(√5−𝜖) for any 𝜖>0 (see Figure 2(a)).

(a)

(b)

Figure 2 

Example 2.3. We have 𝑠₄=2√2. Four squares 𝑆(𝜋/4) admit a translative packing into 𝐼(2√2) (see Figure 3, where √2/2≤𝜆≤3√2/2). In Figure 3(b) and Figure 4(a), we illustrate the cases when 𝜆=√2 and 𝜆=√2/2, respectively. By these three pictures, we conclude that four squares 𝑆(𝜋/4) cannot be translatively packed into 𝐼(2√2−𝜖), for any 𝜖>0. Consequently, 𝑠₄≥2√2. On the other hand, four circles of radius √2/2 can be packed into 𝐼(2√2) (see Figure 4(b)). Since any square 𝑆(𝛼_𝑖) can be translatively packed into a circle of radius √2/2, it follows that 𝑠₄≤2√2.

(a)

(b)

Figure 3 

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(b)

Figure 4 

Lemma 2.4 (see [5]). Every unit square can be translatively packed into any isosceles right triangle with legs of length √5.

Theorem 2.5. If 𝑛≥3, then 𝑠_𝑛≤((√10+√5)/2√3)⋅√𝑛.

Proof. Let 𝑆(𝛼₁),𝑆(𝛼₂),and𝑆(𝛼₃) be unit squares and put 𝜆₁=12√10+√5. (2.1) Three congruent quadrangles 𝑄₁,𝑄₂,and𝑄₃, presented in Figure 2(b), of side lengths 𝜆₁,√5,√2.5, and 𝜆₂=𝜆₁−√5, are contained in 𝐼(𝜆₁). Since the length of the diagonal of 𝑆(𝛼_𝑖) is smaller than √2.5, by Lemma 2.4 we deduce that 𝑆(𝛼_𝑖) can be translatively packed into 𝑄_𝑖 for 𝑖=1,2,3. Consequently, the squares 𝑆(𝛼₁),𝑆(𝛼₂),and𝑆(𝛼₃) can be translatively packed into 𝐼(𝑠) and 𝑠₃≤𝜆₁=√10+√52√3⋅√3. (2.2)
Now assume that 4≤𝑛≤16.
Denote by 𝑚_𝑛 the smallest number 𝑠 such that 𝑛 circles of unit radius can be packed into 𝐼(𝑠). 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 𝑚_𝑛 are known, among others, for 𝑛≤16 (see Table 2.2.1 in [6] or [7]). We know that 𝑚₄=4,𝑚₅<4.83,𝑚₆<5.33,𝑚₇<5.74,𝑚₈<𝑚₉𝑚=6,₁₀<6.75,𝑚₁₁<7.03,𝑚₁₂<𝑚₁₃<7.47,𝑚₁₄<𝑚₁₅<𝑚₁₆=8. (2.3)
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(√2𝑚_𝑛/2). It is easy to verify that 𝑠_𝑛≤√22𝑚_𝑛<√10+√52√3⋅√𝑛, (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≤𝑚²+𝑚or𝑚²+𝑚<𝑛≤(𝑚+1)². (2.5)
Obviously, 𝑚² circles of radius √2/2 can be packed into 𝐼(√2𝑚) (see Figure 5(a)). Moreover, 𝑚²+𝑚 circles of radius √2/2 can be packed into 𝐼(√2𝑚+√2/2) (see Figure 5(b)); it is easy to check that √⋅√3𝑚+222<√√2𝑚+22, (2.6) provided 𝑚≥4.
If 𝑚²+1≤𝑛≤𝑚²+𝑚, then 𝑠_𝑛≤√√2𝑚+22≤√2⋅√√𝑛−1+22<√10+√52√3⋅√𝑛. (2.7)
If 𝑚²+𝑚+1≤𝑛≤(𝑚+1)², then 𝑠_𝑛≤√2(𝑚+1). Since 𝑚²+𝑚+1≤𝑛, it follows that 𝑚≤(1/2)√4𝑛−3−1/2. Thus, 𝑠_𝑛≤√212√4𝑛−3+12<√10+√52√3⋅√𝑛. (2.8)

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(b)

Figure 5 

By Theorem 2.5, Examples 2.1 and 2.2, we conclude the following result.

Corollary 2.6. Every collection of 𝑛 unit squares admits a translative packing into any square of area 2.5𝑛, that is, 𝑠_𝑛≤√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 √⁴𝑛. Then 𝑠_𝑛≤𝜋1+√2𝑘2(1+𝑘)+√𝑛+2𝑘²−4𝑘+2. (2.9)

Proof. Assume that 𝑆(𝛼₁),…,𝑆(𝛼_𝑛) is a collection of 𝑛 unit squares. Let 𝑘 be the greatest integer not over √⁴𝑛, let 𝜂=𝜋/2𝑘, and put 𝜁=1+𝜂√2(1+𝑘)+√𝑛+2𝑘²−4𝑘+2. (2.10)
For each 𝑖∈{1,…,𝑛}, there exists 𝑗∈{1,2,…,𝑘} such that 𝑗−1𝜂≤𝛼_𝑖<𝑗𝜂. (2.11)
Put 𝜑=𝛼_𝑖−(𝑗−1)𝜂. We have 0≤𝜑<𝜂. Moreover, let 𝜆₃ and 𝜆₄ denote the lengths of the segments presented in Figure 6(a). Since 𝜆₃+𝜆₄=cos𝜑+sin𝜑≤1+𝜑<1+𝜂, (2.12) it follows that 𝑆(𝛼_𝑖) is contained in a square 𝑃_𝑖 with side length 1+𝜂 and with the angle between the 𝑥-axis and a side of 𝑃_𝑖 equal to (𝑗−1)𝜂. We say that 𝑃_𝑖 is a 𝑗-square.
To prove Lemma 2.7, it suffices to show that 𝑃₁,…,𝑃_𝑛 can be translatively packed into 𝐼(𝜁). Denote by 𝐴_𝑗 the total area of the 𝑗-squares, for 𝑗=1,…,𝑘. Obviously, ^𝑘_𝑗=1𝐴_𝑗=𝑛1+𝜂². (2.13)
Put ℎ_𝑗=𝐴_𝑗𝜁−2√2√1+𝜂+221+𝜂, (2.14) for 𝑗=1,…,𝑘. Observe that ^𝑘_𝑗=1ℎ_𝑗=𝐴₁+⋯+𝐴_𝑘𝜁−2√2√1+𝜂+2𝑘21+𝜂. (2.15)
Thus, ^𝑘_𝑗=1ℎ_𝑗=𝑛1+𝜂²𝜁−2√2√1+𝜂+2𝑘21+𝜂. (2.16)
The equation 𝑛1+𝜂²𝑥−2√2√1+𝜂+2𝑘21+𝜂=𝑥 (2.17) is equivalent to 𝑥²−𝑥⋅2√21+𝜂(𝑘+1)+(8𝑘−𝑛)1+𝜂²=0. (2.18)
It is easy to verify that 𝑥=𝜁 is a solution of this equation. Consequently, ^𝑘_𝑗=1ℎ_𝑗=𝑛1+𝜂²𝜁−2√2√1+𝜂+2𝑘21+𝜂=𝜁. (2.19)
We divide 𝐼(𝜁) into 𝑘 rectangles 𝑅₁,…,𝑅_𝑘, where 𝑅_𝑗 is a rectangle of side lengths 𝜁 and ℎ_𝑗. Since the diagonal of each 𝑃_𝑖 equals √2(1+𝜂) and 𝐴_𝑗=𝜁−2√21+𝜂ℎ_𝑗−2√21+𝜂, (2.20) it follows that all 𝑗-squares admit a translative packing into 𝑅_𝑗 for 𝑗=1,…,𝑘 (see Figure 6(b)). Hence, 𝑃₁,…,𝑃_𝑛, and consequently 𝑆(𝛼₁),…,𝑆(𝛼_𝑛) can be translatively packed into 𝐼(𝜁). This implies that 𝑠_𝑛≤𝜁.

(a)

(b)

Figure 6 

Theorem 2.8. Let 𝑛 be a positive integer. Then 𝑠_𝑛≤√𝑛+√𝜋2+2⁴√𝑛+𝑂(1), (2.21) as 𝑛→∞.

Proof. Let 𝑘 be the greatest integer not over √⁴𝑛. Since √𝑛+2𝑘²−4𝑘+2<√𝑛+2𝑘²≤𝑛+2√𝑛<√𝑛+1, (2.22) by Lemma 2.7, it follows that, for 𝑛>1, 𝑠_𝑛<𝜋1+2√⁴√𝑛−221+√⁴𝑛+√=√𝑛+1𝑛+√2+𝜋2⁴√𝑛+√𝜋2+12+√+1𝜋+𝜋2₄√.𝑛−1 (2.23)