Cross flips

 Published on Sunday, 3rd April 2011, 08:00 am; Solved by 371;
Difficulty rating: 100%

Problem 331

N×N disks are placed on a square game board. Each disk has a black side and white side.

At each turn, you may choose a disk and flip all the disks in the same row and the same column as this disk: thus 2×N-1 disks are flipped. The game ends when all disks show their white side. The following example shows a game on a 5×5 board.


It can be proven that 3 is the minimal number of turns to finish this game.

The bottom left disk on the N×N board has coordinates (0,0);
the bottom right disk has coordinates (N-1,0) and the top left disk has coordinates (0,N-1).

Let CN be the following configuration of a board with N×N disks:
A disk at (x,y) satisfying $N - 1 \le \sqrt{x^2 + y^2} \lt N$, shows its black side; otherwise, it shows its white side. C5 is shown above.

Let T(N) be the minimal number of turns to finish a game starting from configuration CN or 0 if configuration CN is unsolvable.
We have shown that T(5)=3. You are also given that T(10)=29 and T(1 000)=395253.

Find $\sum \limits_{i = 3}^{31} {T(2^i - i)}$.