s(1) = √1 = 1.000		s(2) = √2+√1 = 2.414+		s(3) = √3+√2 = 3.146+
s(4)=√4+√3=3.732+		s(5) = √5+√4 = 4.236+		s(6) = √4+√3+√2 = 5.146+

What are the best packings you can find for larger values of n? How many of them can you prove are optimal? Can you pack squares of area 1 through 11 inside a square of side 8.5? Are there values of n where the optimal packings involve tilted squares? Does s(n) ≈ n/√2 ?

David Cantrell improved my best packings for n=9, 10, 12, 13, 14, 15, 16, 17, 18, 20, and 21, and then did n=22, 23, and 24!

Here are the best known packings:

s(7) ≤ √6+√5+√(3/2) = 5.910+		s(8) ≤ √7+√5+√3 = 6.613+		s(9) ≤ 7.246+
s(10) ≤ 7.829+		s(11) ≤ √9+√8+√7 = 8.474+		s(12) ≤ √10+√9+√8 = 8.990+
s(13) ≤ √9+√7+√5+√4 = 9.881+		s(14) ≤ √12+√8+√5+√4 = 10.528+		s(15) ≤ √14+√11+√8+√2 = 11.300+
s(16) ≤ √13+√9+√8+√7 = 12.079+		s(17) ≤ √16+√14+√13+√2 = 12.761+		s(18) ≤ √15+√12+√11+√8 = 13.482+
s(19) ≤ √17+√16+√10+√8 = 14.113+		s(20) ≤ √12+√9+√7+√6+√3+√2 = 14.705+		s(21) ≤ √20+√15+√13+√12 = 15.414+
s(22) ≤ √22+√21+√14+√10 = 16.176+		s(23) ≤ √21+√20+√18+√13 = 16.902+		s(24) ≤ √15+√14+√12+√11+√10 = 17.557+

Sasha Ravsky notes that s(n) must be at least √n+√(n-1), and this is enough to prove the values of s(n) for n≤5. He also notes that a collection of squares with total area S and the side of the largest square x can always be packed inside a square of side x+√(S-x²) without tilting any squares. Thus √(n(n+1)/2) ≤ s(n) ≤ √n + √(n(n-1)/2).

Philippe Fondanaiche managed to pack squares of areas 1 through 100 inside a square of side 71.647+, with less than 2% wasted area. But then David Cantrell packed them inside a square of side 71.344+, with less than .8% wasted area!

If you can extend any of these results, please e-mail me. Click here to go back to Math Magic. Last updated 6/30/04.