Square packing in a square

5 unit squares in a square of side length ${\displaystyle 2+1/{\sqrt {2}}\approx 2.707}$

10 unit squares in a square of side length ${\displaystyle 3+1/{\sqrt {2}}\approx 3.707}$

The smallest value of ${\displaystyle a}$ that allows the packing of ${\displaystyle n}$ unit squares is known when ${\displaystyle n}$ is a perfect square (in which case it is ${\displaystyle {\sqrt {n}}}$), as well as for ${\displaystyle n={}}$2, 3, 5, 6, 7, 8, 10, 13, 14, 15, 24, 34, 35, 46, 47, and 48. For most of these numbers (with the exceptions only of 5 and 10), the packing is the natural one with axis-aligned squares, and ${\displaystyle a}$ is ${\displaystyle \lceil {\sqrt {n}}\rceil }$.[2][3] The figure shows the optimal packings for 5 and 10 squares, the two smallest numbers of squares for which the optimal packing involves tilted squares.[4][5]

The smallest unresolved case involves packing 11 unit squares into a larger square. 11 unit squares cannot be packed in a square of side less than ${\displaystyle \textstyle 2+2{\sqrt {4/5}}\approx 3.789}$. By contrast, the tightest known packing of 11 squares is inside a square of side length approximately 3.877084, slightly improving a similar packing found earlier by Walter Trump.[6]

For larger values of the side length ${\displaystyle a}$, the exact number of unit squares that can pack an ${\displaystyle a\times a}$ square remains unknown. It is always possible to pack a ${\displaystyle \lfloor a\rfloor \times \lfloor a\rfloor }$ grid of axis-aligned unit squares, but this may leave a large area, approximately ${\displaystyle 2a(a-\lfloor a\rfloor )}$, uncovered and wasted.[4] Instead, Paul Erdős and Ronald Graham showed that for a different packing by tilted unit squares, the wasted space could be significantly reduced to ${\displaystyle o(a^{7/11})}$ (here written in little o notation).[7] Later, Graham and Fan Chung further reduced the wasted space, to ${\displaystyle O(a^{3/5})}$.[8] However, as Klaus Roth and Bob Vaughan proved, all solutions must waste space at least ${\displaystyle \Omega {\bigl (}a^{1/2}(a-\lfloor a\rfloor ){\bigr )}}$. In particular, when ${\displaystyle a}$ is a half-integer, the wasted space is at least proportional to its square root.[9] The precise asymptotic growth rate of the wasted space, even for half-integer side lengths, remains an open problem.[1]

Some numbers of unit squares are never the optimal number in a packing. In particular, if a square of size ${\displaystyle a\times a}$ allows the packing of ${\displaystyle n^{2}-2}$ unit squares, then it must be the case that ${\displaystyle a\geq n}$ and that a packing of ${\displaystyle n^{2}}$ unit squares is also possible.[2]

A related problem is that of packing n unit squares into a circle with radius as small as possible. For this problem, good solutions are known for n up to 35. Here are minimum solutions for n up to 12:[10]

• Circle packing in a square
• Squaring the square
• Rectangle packing

• Squares in Squares, Erich's Packing Center, Erich Friedman, Stetson University