Chess960 starting position

A Chess960 starting position is one of 960 possible initial game positions in the chess variant Chess960. The special arrangement of pieces on the players' first ranks is selected randomly before play according to Chess960 rules, and can be generated either by a computer program, or using dice, coin, cards, etc.

Starting position requirements

White pawns are placed on the second rank as in standard chess. All remaining white pieces are placed randomly on the first rank, with two restrictions:

The bishops must be placed on opposite-color squares.
The king must be placed on a square between the rooks.

Black's pieces are placed equal-and-opposite to White's pieces. For example, if the white king is randomly determined to start on f1, then the black king is placed on f8. (The king never starts on the a - or h -files, since this would leave no space for a rook.)

Methods for creating starting positions

There are several procedures for generating random starting positions with equal probability.

Single die method

Ingo Althofer

	a	b	c	d	e	f	g	h
8									8
7									7
6									6
5									5
4									4
3									3
2									2
1									1
	a	b	c	d	e	f	g	h

Chess960 starting position generated by the following die rolls:

3 (bishop on e1)
5 (skip and reroll)
1 (bishop on b1)
4 (queen on f1)
6 (skip and reroll)
2 (knight on c1)
1 (knight on a1)
king on g1; rooks on d1 and h1

A common method for selecting a starting position is one proposed by Ingo Althofer in 1998, which requires only a single cube die.^[1]^[2]^[3] The position of White's pieces is determined as follows:

Roll the die and place a bishop on the black square indicated by the die, counting from the left, a through h.
Thus, 1 indicates the first black square from the left (a1), 2 indicates the second black square (c1), 3 indicates the third (e1), and 4 the fourth (g1). Since there are no fifth or sixth positions, reroll a 5 or 6 until another number shows.
Roll the die and place a bishop on the white square indicated.
1 indicates b1, 2 indicates d1, and so on. Reroll a 5 or 6.
Roll the die and place the queen on the first empty position indicated, always skipping filled positions.
Thus, 1 is the first (leftmost) empty square, while 6 is the sixth (rightmost) empty square.
Roll the die and place a knight on the empty position indicated. Reroll a 6.
Roll the die and place a knight on the empty position indicated. Reroll a 5 or 6.

This leaves three empty squares. Place the king on the middle empty square, and the rooks on the remaining two squares. Place the white and black pawns on their usual squares, and Black's first-row pieces to exactly mirror White's. (So, Black should have on a8 the same piece type White has on a1.)

The above procedure uses an average of 6.7 die rolls. Note that one of the random positions (rolled by 2–3–3–2–3 or 2–3–3–4–2) is the standard chess starting position, at which point a standard chess game ensues.

Optimization

On average, the single die procedure uses 6.7 die rolls. An optimal procedure would use on average log(960)/log(6) = 3.83 die rolls. It is straightforward to reduce the average number of dice rolls to 6.2. Instead of rerolling 5s and 6s until something else comes up, reroll once and (4/36) for each bishop use the following scheme:

5 followed by 5 subtract 4 → 1
5 followed by 6 subtract 4 → 2
6 followed by 5 subtract 2 → 3
6 followed by 6 subtract 2 → 4

For the queen one indication (1-6) is required.

For the first knight one indication (1-5) is required. If it is 6, then the roll is repeated.

For the second knight one indication (1-4) is required. If it is 5, then the relative position from the first knight is the next empty square, else if it is 6 then the next after. If the empty square does not exist, then the five empty squares are considered in circle.

Polyhedral dice method

With polyhedral dice shaped like each of the five Platonic solids, one never needs to reroll any dice.

A set of Platonic solids dice, from left: tetrahedron (d4), cube (d6), octahedron (d8), dodecahedron (d12), icosahedron (d20)

	a	b	c	d	e	f	g	h
8									8
7									7
6									6
5									5
4									4
3									3
2									2
1									1
	a	b	c	d	e	f	g	h

Starting position using four polyhedral dice:

d8=6 (bishop on f1)
d4=4 (bishop on g1)
d6=3 (queen on c1)
d20=20 (knights on h1 and e1)
king on b1; rooks on a1 and d1

	a	b	c	d	e	f	g	h
8									8
7									7
6									6
5									5
4									4
3									3
2									2
1									1
	a	b	c	d	e	f	g	h

Starting position using all five polyhedrals:

d8, d4, and d6 (same as with four polyhedrals)
d20=20 (knight on h1)
d12=11 (knight on d1)
king on b1; rooks on a1 and e1

Roll all the dice in one throw and place White's pieces as follows:

Place a bishop on one of the eight squares (counting from the left, a through h ) as indicated by the octahedron (d8).
Place the other bishop on one of the four squares of opposite color as indicated by the tetrahedron (d4).
Place the queen on one of the remaining six squares as indicated by the cube (d6).
Take the value of the icosahedron (d20), divide by four (round up), and let x = the quotient, and y = the remainder + 1. Place a knight on the xth empty square. Then place the other knight on the yth remaining empty square. In other words, see the d20 as a d5 for the first knight: 1-4, 5-8, 9-12, 13-16 and 17-20. Then for the second knight, look within the group to get a d4. For example, a 20 is in the fifth group and the fourth spot in that group, so place the knights on the fifth square and the fourth square. An 11 is in the third group and the third spot.
You can also use just a d10 since there are only ten unique placements of the knights once the bishops and queen has been placed. Hold one knight on the leftmost square and count one, two, three, four with the other knight on the empty square, then when it loops, move the leftmost knight one square to the right, five, six, seven, then it loops again, eight, nine, and finally with ten both knights are as far right as they go. For example, with a six the knight would be placed on the second of the five empty squares, then the second knight would be place on the second of the three squares that are empty to the right of the knight. Using a d10 in this way after two different colored d4:s and a d6 is a minimal one-roll way since 4×4×6×10 is exactly 960. (And, by subtracting one from each die and multipying with 1, 4, 16 and 96 respectively, then adding those numbers together, you find the number in the Chess960 numbering scheme.) The d8, d4, d6, d20 still give equal chance for all 960 positions, but with every position being represented in four different ways.
Or alternatively (using an additional die and different calculations): Place the first knight according to the value of the d20 die, by counting the five empty squares and looping back to the left whenever reaching the rightmost empty square. Then with four empty squares remaining, do the same for the other knight using the dodecahedron (d12) die. With this method, every position is represented in 48 different ways.
Place the king between the rooks on the remaining three squares.

Place the white pawns and mirror the position for Black.

Coins (binary) method

Two coins (small and large) are used to randomly generate numbers with equal probability. Tails on the smaller coin counts as 0, tails on the larger coin counts as 1, and heads on either coin counts as 2. To create numbers in the range 1 through 4, toss both coins and add their values together. To create numbers in the range 1 through 3, do the same but retoss whenever 4 is the result. To create numbers in the range 1 through 2, just toss the larger coin (tails is 1, heads is 2).

There is a way of using coins and making all starting positions equally likely. It uses a third coin for which tails counts as 0, and heads counts as 4. Tossing all three coins generates the values 1 through 8 with equal probability. The method follows the piece placements used for a die. Two coins are used for the bishops as before. Then six squares are available for the queen. All three coins are tossed and retossed until a number in the range 1–6 comes up. Then five squares are available for the first knight. Now the three coins should be tossed and retossed until a number in the range 1–5 shows up. For the other knight, only a four-way choice is needed, so a single toss of two coins suffices. The average number of tosses needed for this method is 5 ¹⁴⁄₁₅.

A similar coin-tossing method uses one coin to generate all starting positions with equal probability. Toss the coin four times and record the results. If the four coin tosses are all tails, start again. Otherwise toss the coin an additional six times and record the results. Then convert the sequence into a binary number counting heads as 0, tails as 1. The resulting number is a number between 0 and 959 that can then be converted into a starting position using the Chess960 numbering scheme. For example, if the tosses are T, H, T, T, H, H, H, H, T, T this converts to the binary number 1011000011, or 707, which in the Chess960 numbering scheme is the starting position BRKQNNRB.

References

↑ Althöfer, Ingo (1998). Computerschach und Spiele (CSS) (in German). Dieter Steinwender and Frederic Friedel. Missing or empty |title= (help)
↑ Althöfer, Ingo (March 1998). van den Herik, Jaap, ed. "LIST-3-HIRN vs. Grandmaster Yusupov". International Computer Chess Association (ICCA). Universiteit Maastricht. 21 (1): 52–60. ISSN 0920-234X.
↑ Hans Bodlaender (2002-05-10). "Fischer Random Chess: Manual Procedure for Generating Piece Placements". The Chess Variant Pages. Retrieved 2013-01-26.

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