Problem of the Month (September 2007)

We say that two shapes in the plane can "see each other" if there is a horizontal or vertical line segment that intersects the interior of both shapes and no other shape. This month we are interested in the problem of placing squares so that each square sees the same number of other squares. In particular, let Sk(n) be the maximum number of non-overlapping integer sided squares that can be placed (on a square grid) in a square of side n so that each square sees exactly k other squares. And let Mk be the minimum value of n for which Sk(n)>0.

What are the small values of S3(n)? Is there a formula for S3(n)? How about Sk(n)? What are the small values of Mk? Is there a formula for Mk? What about placing squares in a rectangle? How do these values change if we allow the squares to be rotated or placed off the square grid?

Konstantin Knop found an error in one of my diagrams, and provided a figure proving S5(30)≥24.

David Bevan noticed Sk(n) ≤ n/ k/4  2, since a square can't see k other squares if it is perimeter is less than k.

David Bevan also showed S5(n) ≥ (n2–12n+27)/4 for n≥18, S6(n) ≥ (n2–30n+181)/4 for n≥29, and S8(n) ≥ (n2–103n+2687)/5 for n=40r+6 and r≥2.

David Bevan was interested in the limiting density of solutions L(k). It is fairly easy to see that L(0)=0, L(1)=0, L(2)=0, L(3)=0, L(4)=1, L(5)=1/4, and L(6)=1/4. He thinks L(7)=2/9, L(8)=1/5, L(9)=2/25, L(10)=1/14, L(11)=1/18, and L(12)=1/20, since these are the best on an infinite grid. Are these correct? Is there any solution at all for n≥9 ?

David Bevan also made the following graph-theoretic observation: if k is odd, then Sk(n) is even.

Here are the values of Sk(n) for k≤2:

 S0(n) = n     M0 = 1 S1(n) = 4n/3     M1 = 2 S2(n) = 2n    M2 = 2

Here are the best known values of Sk(n) for other small k:

 S3(n) ≥ 4n-14 S3(5)=6 S3(6)≥10 S3(7)≥14 S3(8)≥18 M3 = 5
 S4(2n) = 4n2-12n+4 S4(8)=20 S4(10)=44 S4(12)=76 M4 = 8 S5(13)≥6 S5(16)≥8 S5(19)≥54(David Bevan) S5(23)≥88(David Bevan) M5 ≤ 13 S6(20)≥8 S6(30)≥82(David Bevan) M6 ≤ 20 S7(31)≥8(Trevor Green) M7 ≤ 31 S8(86)≥245(David Bevan) M8 ≤ 86
Trevor Green was interested in this problem without the restriction of placing squares on the grid. Here are some of the best solutions in this case. S*3(2+ε)=4(Trevor Green) M*3 = 2 + ε S*4(3+ε)=9(David Bevan) S*4(4+ε)=16(David Bevan) S*4(6+ε)=36(David Bevan) M*4 = 3 + ε S*5(4+ε)=16(David Bevan) M*5 ≤ 4 + ε S*6(4+ε)=16(David Bevan) M*6 ≤ 4 + ε

If you can extend any of these results, please e-mail me. Click here to go back to Math Magic. Last updated 9/21/07.