Improved packings of n(n − 1) unit squares in a square

Let s(n) be the side of the smallest square into which we can pack n unit squares. The purpose of this paper is to prove that s(n2−n) < n for all n > 12. Besides, we show that s(182 − 17) < 18, s(172 − 16) < 17, and s(162 − 15) < 16. Mathematics Subject Classifications: 05B40, 52C15

2 Some squeezable packing of rectangles Let a packing of m unit squares in a rectangle R = (R x , R y ) be given. We assume that (R x − 1)(R y = 1) < m < R x R y and we can't pack a unit square in the waste area. This packing is called squeezable if both sides of a rectangle can be reduced, i.e., for some δ > 0 there exists a packing of m unit squares in a rectangle (R x − δ, R y − δ). The maximum of such δ > 0 is called the value of squeezing and is denoted by δ(R, m). We write δ(R, m) = 0 if the packing is not squeezable. The property of squeezability of a packing for small parameters can be proved rather simply. However proving this property for large parameters is a non-trivial mathematical problem. The following obvious formula connects δ(R, m) and s(n): s(n) = s(n) − δ(( s(n) , s(n) ), n).
If δ((R x , R y ), m) < 1 then the fact that for integer R x , R y δ((R x , R y ), m) δ((R x + 1, R y ), m + R y − 1) (2) can be proved by adding R y − 1 unit squares to the x-side of a rectangle (R x , R y ). Figure  1 shows the basic idea for efficiently packing unit squares in a square S, where rectangles C and D are integer and the waste is in rectangles A and B. It is easy to see that if the packing of unit squares in rectangles A, B is squeezable, then the packing of unit squares in S is squeezable and δ(S, ·) min(δ(A, ·), δ(B, ·)).
This bound can be increased if we note that after squeezing there is a little space between rectangles A, B. We can give this space to a rectangle with minimal squeezing value in order to increase that value and thus to increase the evaluation of δ(S, ·).
Let us consider a packing of 26 unit squares in a rectangle (4, 8) (see Figure 2). This packing is centrally symmetric and the waste is equal to 6.
In Figure 2 we see one of the main ideas for packing unit squares: using of stacks (4, 1) tilted by an angle α = arcsin(8/17). The main idea for squeezing a packing follows from it: tilting stacks (4, 1) by an angle α + ε so that the stack (4, 1) is located in a vertical strip of width 4 − δ, where ε and δ are sufficiently small. Hereinafter we determine the orientation of a unit square by a unit vector (x, y) with x > 0, y 0, x 2 + y 2 = 1 directed along the side of this unit square. If the bottom vertex of the unit square is at the origin then the three other vertices have coordinates (x, y), (x − y, y + x), (−y, x) . Note that if two points P t , P b are taken on the top side and the bottom side of this unit square then the scalar product P t − P b , (x, y) is equal to 1.
Continuing with the example in Figure 2, after increasing the tilt the stack (4, 1) in a vertical strip of width 4 − δ has orientation (x 1 , y 1 ), x 1 > 0, y 1 0 satisfying the system of equations 4x 1 + y 1 = 4 − δ, x 2 1 + y 2 1 = 1. To evaluate the squeezing value δ((4, 8), 26), we use the bisection method. The packing remains centrally symmetric. The distance between the point P = (P x , P y ) = (1 − δ/2, 2 − δ/2) and the upper side of the square S 2 intersecting the line For δ = 0.01 we have x 1 = .877695..., y 1 = .479219..., P y − P 1y = 0.021604 > 0. For δ = 0.02 x 1 = .87312663..., y ! = .48749347..., P y − P 1y = −0.0061309... < 0. The bisection method gives evaluation δ((4, 8), 26) > 0.0177702. Figure 3 shows a more complex example, a centrally symmetric squeezable packing of In this packing the left vertex of S 2 is on a side of S 1 . The square S 3 is placed so that the right vertices of squares S 2 , S 5 , and the top vertex of S 4 are on the sides of S 3 . Vertices of the squares S 8 , S 7 , S 9 are on sides of S 6 . Calculations show that there is a small distance between S 3 and S 6 , which guarantees squeezability of the given packing.
A packing of 58 unit squares in a rectangle (6,11-2/35) can be obtained by removing one stack (6,1) in Figure 3 and lifting up by 37/35 all the squares that are below this Object Formulae or system of equations Numerical value δ 0.004 Orientation (x 1 , y 1 ) 1.00378910536129684 and top side of S 6 Table 1: Calculations with δ = 0.004. stack. Similar calculations give the evaluation of the squeezing value δ((6, 11), 58) > 0.01681735886.
Consider a more difficult problem of a squeezable packing of 43 unit squares in a rectangle (5,10). In Figure 4 six unit squares S 1 , S 4 , S 9 , S 10 , S 11 , S 12 have not the orientation ( 5 13 , 12 13 ) nor (1, 0). The square S 1 has a vertex on the side of the rectangle (5,10), one on a side of S 2 , and one on a side of S 3 . The right vertex of S 1 is on the bottom side of S 4 . S 4 is tilted so that the bottom right vertex of S 3 is on the left side of S 4 and the top vertex of the stack (3, 1) is on the right side of S 4 . The left vertex of S 5 is on the side of S 6 . The squares S 9 , S 10 are tilted by the same angle so that the vertex of S 8 is on the side of S 9 , the vertex of S 5 is on the bottom side of S 9 , and the vertex of S 7 is on the bottom side of S 10 . The squares S 11 , S 12 form a rectangle (2,1). The right vertex of S 12 is on the right side of a rectangle (5,10). The vertex of S 13 is on the top side of S 11 . The bottom sides of S 11 and S 12 are parallel to the line connecting the right vertices of S 9 and S 10 . The vertex of S 14 is on the bottom side of S 15 . Calculations show that there is a small distance 0.0055111... between the bottom side of the rectangle (2, 1) = S 11 ∪ S 12 and the line connecting the  To prove conjecture (1), we need the following lemma.
The proof is technically simple and can be understood from Figure 5, showing a centrally symmetric squeezable packing of 86 unit squares in a rectangle (8, 12). For an arbitrary k 3, the centrally symmetric packing in the upper half of a rectangle (2k, 2k + 4) consists of 2 staircases. A staircase with orientation (1,0) having k(k+1) 2 unit squares, and a staircase with orientation (x 1 , y 1 ) = ( 4k 2 −1 4k 2 +1 , 4k 4k 2 +1 ) that has (3k−1)(k+2) 2 unit squares. The top vertex of S k+1 has ordinate i.e., S k+1 is in rectangle (2k, 2k + 4). The top vertex of S 0 has ordinate i.e., S 0 does not intersect the staircase with orientation (1,0). Each square S j , 1 j k intersects the vertical line x = k − j in the point The ordinate of this point satisfies i.e., none of the S j , 1 j k intersects the staircase with orientation (1,0). We see that there is a positive distance between the two staircases. Therefore, this packing is squeezable. The lemma is proved.
3 Improved squeezable packing of some squares As mentioned in the introduction, in [3] Nagamochi mistakebly says that in [2] it is proved that s(n 2 − n) < n ∀n 17.
Thus he implicitly formulates the conjecture (4). For the proof of this conjecture we use lemma 1 as follows.
Thus the conjecture (4) is proved for n 13. For the proof of this conjecture for n = 12 see Figure 6. The packing in Figure 6 is obtained from the squeezable packing in rectangles (8,4), (5,10). In the packing in (5,10) we tilt the angular squares S 1 , S 2 by an angle arcsin(10/26) so that the bottom vertex of S 1 has an integer y-coordinate and S 2 has intruded space in the rectangle (8,4). From the packing in (8,4) we remove two right top squares and move to the left by 1/20 unit squares tilted by an angle arcsin(8/17) so that the bottom vertex of S 3 is on the side of S 4 . The small distance between S 2 and S 5 makes the packing in Figure 6 squeezable.
Thus we have proved that s(n 2 − n) < n ∀n 12.