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id, title, challengeType, forumTopicId, dashedName
id title challengeType forumTopicId dashedName
5900f4b71000cf542c50ffca Problem 331: Cross flips 5 301989 problem-331-cross-flips

--description--

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.

animation showing game on 5x5 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 C_N 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. C_5 is shown above.

Let T(N) be the minimal number of turns to finish a game starting from configuration C_N or 0 if configuration C_N is unsolvable. We have shown that T(5) = 3. You are also given that T(10) = 29 and T(1\\,000) = 395\\,253.

Find \displaystyle \sum_{i = 3}^{31} T(2^i - i).

--hints--

crossFlips() should return 467178235146843500.

assert.strictEqual(crossFlips(), 467178235146843500);

--seed--

--seed-contents--

function crossFlips() {

  return true;
}

crossFlips();

--solutions--

// solution required