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Rex's RK

My Works

My Account Names

ThisIsRex (2241 rated rk peak and two rk shield wins)

BlindRex (2124 rated blindfolded rk peak)

Tesset12 (2289 rated rk peak, 2587 first hour shield perf, previous daily rk record, 3rd place in TURKL S1, Lichess Swiss RK Swiss Winner)

TheUsualDumbKid (2312 rated rk peak, Board 3 winner of Racing Kings Team Olympiad)

james126 (2367 rated rk peak, one rk shield win)

TheUsualDumbKid (Silver Medalist in MSO Racing Kings 2023)

ThisIsRex (Current #1 on chess.com rk lb, qualified as lifetime chaturaji master)

Purpose of Page

The purpose of this page is to mainly give out the information that I have learned in Racing Kings and publicize my studies and the tournaments I host in the Racing Kings community.

While I did win two shields, once hold the 4th highest rated spot on the leaderboard, have a 2587 performance in the first hour of a Racing Kings shield, have three of the top five Daily Racing Kings highest scores, and pioneer theory in three different openings. That is basically all the base level accomplishments that I've held. To me, my biggest accomplishments in the Racing Kings community were: being the first to theorize 1. Nd4, hosting the first RK1440 tournament, and popularize the variant in a time where the amount of active players and enthusiasts at least tripled.

What Is Rk1440?

As many know, Fischer Random, or Chess 960, is the version of chess where the pieces are randomized on the back ranks. Racing Kings 1440 (or RK1440 for short) is the Fischer Random version of Racing Kings where all the pieces are randomized. There are two versions, symmetrical and asymmetrical. The RK1440 Open Championship will host near equal symmetrical positions.

Like Fischer Random there are some rules making it so that there aren't extensive positions. The kings can only be on the four outermost squares (or two outermost files). The second rule is that the bishops must start on opposite colors.

In mathematical terms: The king has 4 squares to start on. This results in one bishop having 3 squares and the other bishop having 4 squares. The queen then has 5 squares it can be placed on. There are 6 remaining combinations for the knight pair and rook pair to be placed.

In equation form it's: 4 x 3 x 4 x 5 x 6 = 1440

For ordering go here for the Direct Derivation algorithm: https://github.com/thomas-daniels/Chess.NET/wiki/Algorithm-for-RacingKings1440-positions

528 Lost Ordering

Position 365 (365 in lost only |1805 in RK 1440 + 528 | 1098 in RK Random) is used as the example here.

Divide the position (N) by 2 and get quotient N1 and remainder K. K is the position on the board the white king corresponds to: 0 is K on f2, 1 is K on f1.

365/2 is 182 with remainder 1 so N2 is 182 and position is:

Divide N2 by 3 to get quotient N3 and remainder B1. If king on f2, B1 corresponds to the dark bishop and f1 to the light. Starting at e1, go right, down then left. If 0, it corresponds to first available square, 1 to second, and 2 to third.

182/3 is 60 with remainder 2 so N2 is 60 and position is:

Divide N3 by 4 to get quotient N4 and remainder B2. The other colored bishop is represented by this number. Same order, start at e2 go clockwise: 0 corresponds to first available square, 1 to second, 2 to third, etc.

60/4 is 15 with remainder 0 meaning N4 is 15 and the position is:

Divide N4 by 5 to get quotient N5 and remainder Q. The remainder Q corresponds to the Queen's position. Same ordering and 0 is first position available, 4 is last.

15/5 is 3 with remainder 0 meaning N5 is 3 and the position is:

N5 corresponds to the knight and rook pair's position. Use this table below to discover the knight and rook placement.

The final position is to the right.

The 528 Lost Positions

Racing Kings 1440 has a rule that the King must be placed on one of the 4 outermost squares. Due to this 528 positions that are legally possible (by Racing Kings rules alone). Here's how that number is calculated.

The white King can be placed on f1 or f2. If placed on f1, the light-squared bishop has 3 squares and the dark-squared bishop has 4 squares. Vice versa for f2. The queen then has 5 squares to be placed on regardless. For the king placed on f1, a knight placed on d2 would be illegal. Therefore, since the position is mirrored, the knight cannot be placed on e2 for white. Therefore, the light-squared bishop can either cover the e2 square in 1 of its placements, or not in two of its placements. This is the opposite color and number if the king is placed on f2. For king on f1, if the light-squared bishop is not placed on e2, the queen has 1 position that covers the e2 square and 4 positions that do not. In the 4 positions that do not cover the square, only 3 knight-+rook pair combinations can legally be made (since the knight cannot be placed on d2). So for 1 position of the bishop, 6 rook-knight pair combinations can legally be made regardless of the queen's position. For the other 2 positions for the bishop, 1 queen position makes 6 rook-knight pair combinations legally while the other 4 come out to 3 rook-knight pair combinations.

Now to the equation. The f1 example works with opposite colors and reversed number with f2. The king then has 2 squares it can be on. One bishop has 4 squares. The other bishop has 1 square that results in 6 rook-knight combinations regardless of the 5 queen positions. The other 2 positions have 1 queen position resulting in 6 rook-knight pairs and the other 4 result in 3 rook-knight positions. The equation resulting is this:

2 x [ (4 x 1 x 5 x 6) + (4 x 2 x 1 x 6) + (4 x 2 x 4 x 3) ] = 528

For numbering these it's a similar algorithm to the Direct Derivation of the Chess 960 and RK 1440 positions. Look to the left for ordering.

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