Put four numbers on the corners of a square. I’ll pick 1, 3, 14 and 39, in that order:

 1---

- 3
 |    |
 |    |
 |    |
39---

-14

Between each pair, write their unsigned difference:

( 1)- 2-( 3)
  |      |
 38      11
  |      |
(39)-15-(14)

Now you have a new square: (2, 11, 15, 38). I’m not going to keep drawing them; it’s a pain to typeset. But we repeat the process, getting differences of:

  • (9,4,23,36)
  • (5,19,13,27)
  • (14,6,14,22)
  • (8,8,8,8)
  • (0,0,0,0) – and we’re done.

Including the first and last boxes, that took (fairly impressively) seven boxes to converge to zero.

There are all sorts of questions you can ask:

  • Does every set of four integers eventually converge to zero?
  • Can you come up with a set that takes eight boxes to converge? Ten? A hundred?
  • What if you extend it to real numbers?

I should say: I heard about this game from Rob Eastaway, who talks about it for MathsWorldUK on this video.

Below the line are some mild spoilers; you may wish to play around before you read on.

One way to get fairly persistent Diffy boxes is to use the tribonacci sequence – it’s like the Fibonacci sequence, but you add three terms at a time instead of two. It starts 0, 1, 1, 2, 4, 7, 13, 24, and grows quite quickly.

For example, the box (24,13,7,4) becomes (11,6,3,20), then (5,3,17,9), (2,14,8,4), (12,6,4,2), (6,2,2,10), (4,0,8,4), (4,4,4,4), (0,0,0,0), taking nine steps to converge.

In fact, the tribonacci constant, $q \approx 1.839$, can be used to generate (effectively) the only Diffy box that doesn’t converge: (0, 1, $q(q-1)$, $q$). You may like to verify this!

There’s a write-up of the reasoning behind it here.

Have fun!