Download David Van Brackle

David Van Brackle
Chief Judge, Southeast USA Region, ICPC
Take any four positive integers: a, b, c, d. Form four more, like this:
|a-b| |b-c| |c-d| |d-a|
That is, take the absolute value of the differences of a with b, b with c, c with d,
and d with a. (Note that a zero could crop up, but they’ll all still be non-negative.)
Then, do it again with these four new numbers. And then again. And again.
Eventually, all four integers will be the same. For example, start with 1,3,5,9:
1 3 5 9
2 2 4 8 (1)
0 2 4 6 (2)
2 2 2 6 (3)
0 0 4 4 (4)
0 4 0 4 (5)
4 4 4 4 (6)
In this case, the sequence converged in 6 steps. It turns out that in all cases, the
sequence converges very quickly. In fact, it can be shown that if all four
integers are less than 2n, then it will take no more than 3*n steps to converge!
If k is an integer, then [a+k, b+k, c+k, d+k] converges
in exactly the same number of steps as [a, b, c, d]
Proof:
[|(b+k)-(a+k)|,|(c+k)-(b+k)|, |(d+k)-(c+k)|, |(a+k)-(d+k)|]
= [|b-a|,|c-b|,|d-c|,|a-d|]
Corollary: [a-k, b-k, c-k, d-k] converges in the same number
of steps as [a, b, c, d]
If k is an integer, then [a*k, b*k, c*k, d*k] converges
in exactly the same number of steps as [a, b, c, d]
Proof:
[|(b*k)-(a*k)|,|(c*k)-(b*k)|, |(d*k)-(c*k)|, |(a*k)-(d*k)|]
= [k*|b-a|,k*|c-b|,k*|d-c|,k*|a-d|]
Corollary: [a/k, b/k, c/k, d/k] converges in the same number
of steps as [a, b, c, d] provided that a/k, b/k, c/k and d/k are
all integers.

Rotations all converge
in the same number of
steps:

Let E denote an Even
number, D an Odd
number. Then
 [a, b, c, d]
 |E-E| is E
 [b, c, d, a]
 |E-D| is D
 [c, d, a, b]
 |D-E| is D
 [d, a, b, c]
 |D-D| is E

Prove: if a, b, c and d are all non-negative
integers less than 2n, then [a, b, c, d]
converges in 3*n or fewer steps.

Proof: By Induction.
n=1, 2n = 2, so it’s all 0’s and 1’s
4 1’s: [1, 1, 1, 1] has already converged
3 1’s: [1, 1, 1, 0]  [0, 0, 1, 1]  [0, 1, 0, 1]  [1, 1, 1, 1]
2 1’s: either [0, 0, 1, 1] or [0, 1, 0, 1] Either way, it’s a subset of above
1 1: [1, 0, 0, 0]  [1, 0, 0, 1]  [1, 0, 1, 0]  [1, 1, 1, 1]
0 1’s: [0, 0, 0, 0] has already converged
So, it takes a maximum of 3 steps to converge.
Assume true for n-1. Assume [a, b, c, d] are all <2n
All Even: [E, E, E, E] converges in the same number of steps as
[E/2, E/2, E/2, E/2], all of which are less than 2n-1
All Odd: [D, D, D, D] converges in the same number of steps as
[D-1,D-1,D-1,D-1], which is All Even
3 E’s: [E, E, E, D]  [E, E, D, D]  [E, D, E, D]  [D, D, D, D]
2 E’s: either [E, E, D, D] or [E, D, E, D] Either way, it’s a subset
1 E: [E, D, D, D]  [D, E, E, D]  [D, E, D, E]  [D, D, D, D]
So, it takes a maximum of 3 steps to reduce to something that
takes 3*(n-1) steps. That’s a max of 3+3*(n-1) or 3*n