The Chess960 Numbers Decoded
Convert to and from Chess960 numbers mathematically, without lookup tables
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0 NNxxx Qxxxxx Bxxx Bxxx 1 NNxxx Qxxxxx Bxxx xBxx 2 NNxxx Qxxxxx Bxxx xxBx 3 NNxxx Qxxxxx Bxxx xxxB 4 NNxxx Qxxxxx xBxx Bxxx 5 NNxxx Qxxxxx xBxx xBxx 6 NNxxx Qxxxxx xBxx xxBx 7 NNxxx Qxxxxx xBxx xxxB 8 NNxxx Qxxxxx xxBx Bxxx 9 NNxxx Qxxxxx xxBx xBxx 10 NNxxx Qxxxxx xxBx xxBx 11 NNxxx Qxxxxx xxBx xxxB 12 NNxxx Qxxxxx xxxB Bxxx 13 NNxxx Qxxxxx xxxB xBxx 14 NNxxx Qxxxxx xxxB xxBx 15 NNxxx Qxxxxx xxxB xxxB 16 NNxxx xQxxxx Bxxx Bxxx 17 NNxxx xQxxxx Bxxx xBxx 18 NNxxx xQxxxx Bxxx xxBx 19 NNxxx xQxxxx Bxxx xxxB 96 total 96 NxNxx Qxxxxx Bxxx Bxxx ... 192 NxxNx Qxxxxx Bxxx Bxxx ... 288 NxxxN Qxxxxx Bxxx Bxxx ... 384 xNNxx Qxxxxx Bxxx Bxxx ... 480 xNxNx Qxxxxx Bxxx Bxxx ... 576 xNxxN Qxxxxx Bxxx Bxxx ... 672 xxNNx Qxxxxx Bxxx Bxxx ... 768 xxNxN Qxxxxx Bxxx Bxxx ... 864 xxxNN Qxxxxx Bxxx Bxxx ...
The Chess960 numbering scheme devised by Reinhard Scharnagl begins with 10, 96-length blocks of each knight (N) position. The N positions include only the knights, rooks, and king. The rooks and king always fill the last three vacant squares in the order RKR, so they are indicated by x’s for simplicity and play no individual role in the determination of the Chess960 numbering.
Each N block is made of 6, 16-length blocks of each queen (Q) position; they consist of the pieces of the N position with the queen added, or the empty board squares after the bishops have been added.
Each Q block is made of 4, 4-length blocks of each dark-square bishop (Bb) position.
Each Bb block is made of 4, 1-length blocks of each light-square bishop (Bw) position. The B positions consist of the four dark or light squares in the entire row of eight pieces.
We will begin with this board position, and convert it to the Chess960 number mathematically.
First, we look at the N position. (Note: the location of each piece in a string of characters such as this is zero-indexed; it begins with zero on the left and increases to the right: 01234.)
NxNxx N1=0, N2=2
Here is the list of all the N positions. The number associated with the N position is multiplied by 96 as part of the position’s Chess960 number. Is there a way to determine the N number of any position mathematically, without using the list?
0 NNxxx 1 NxNxx 2 NxxNx 3 NxxxN 4 xNNxx 5 xNxNx 6 xNxxN 7 xxNNx 8 xxNxN 9 xxxNN
The N1 positions look like they belong to the series of triangular numbers:
Looking at the series as if they were indexed from the right side, converting N1 to the triangular number based on zero in the next-to-last position, and adding the index of N2 based on zero in the last position, gives us a series that descends from 9 to 0.
9 = 6 Nxxxx + 3 xNxxx 8 = 6 Nxxxx + 2 xxNxx 7 = 6 Nxxxx + 1 xxxNx 6 = 6 Nxxxx + 0 xxxxN 5 = 3 xNxxx + 2 xxNxx 4 = 3 xNxxx + 1 xxxNx 3 = 3 xNxxx + 0 xxxxN 2 = 1 xxNxx + 1 xxxNx 1 = 1 xxNxx + 0 xxxxN 0 = 0 xxxNx + 0 xxxxN
So, subtracting the N1 index from 3, and the N2 index from 4, then subtracting the sum from 9, gives us an ascending series from 0 to 9:
In our case, inserting the above N1 and N2 values reveals the N number:
The Q and B indices are much easier. Inserting the Q from our original setup into the N position gives a Q index of 1:
Including the B’s, and separating out the dark and light squares, gives the B indices:
x B x x Bb=1 xxBxxBxx x x B x Bw=2
Putting it all together, and simplifying the N equation a little, this is how we find the Chess960 number directly from the N, Q, and B indices, without resorting to lookup tables:
In our case, n = 118.
Starting with 118 as our Chess960 number, how do we mathematically determine the position that it represents, again without using lookup tables?
The Bw indices repeat in the sequence 012301230123 . . . . So we start by finding the remainder when we divide n by 4, in terms of how many units are left over; this is called the modulo function, and in this case it tells us what the Bw index is.
Then we are going to divide n by 4, and throw away the remainder; this is called the floor function. It lets us treat the Bb indices as if they repeat in the same 0123 sequence, and we can find the Bb index by the same modulo function.
Bw = n mod 4 = 2 x x B x n = floor(n/4) = 29 Bb = n mod 4 = 1 x B x x n = floor(n/4) = 7
Since the Bb indices also repeat in blocks of 4, we again divide n by 4 and throw away the remainder. Now the Q indices repeat in blocks of 6: 012345012345012345 . . . , and n modulo 6 gives us the Q index.
Q = n mod 6 = 1 xQxxxx n = floor(n/6) = 1
Now when we divide n by 6 and throw away the remainder, we are left with our N position number. How do we determine the N1 and N2 indices, again without looking at the list? Rearranging the N formula from above, we can find N2 as a function of N1.
Each possible value of N1 has a corresponding N2 value. How do we know which combination we want? We know that N1 is by definition less than N2, and that the possible N1 indices range from 0 to 3, and N2 from 1 to 4. The correct combination will uniquely satisfy the following conditions:
So all we have to do is try N1 = 0, 1, 2, and 3 until we find the right one. In our case,
So we know, without using the list, that the N position is
Now we can construct the position by starting with the B’s:
x x B x Bw=2 xxBxxBxx x B x x Bb=1
Adding Q = 1 to the empty squares:
Then filling the empty squares with the N position gives:
The remaining empty squares are filled with RKR:
And we are back to our original board position.
NQBRNBKR N xN xx N1, N2 indices (knights, rooks, and king) xQ xx xx Q index (queen added) x B x x Bb index (dark squares) x x B x Bw index (light squares)
To convert a position to its Chess960 number:
To determine a position from its Chess960 number:
Bw = n mod 4 n = floor(n/4) Bb = n mod 4 n = floor(n/4) Q = n mod 6 n = floor(n/6)
The entire list, with Shredder-FEN string, and mirror position:
p.s. The above method of trying all possible N1’s has something of a “brute force” quality to it. Another way to find the correct combination is to begin with the following:
if n<4 then N1=0
else if n<7 then N1=1
else if n<9 then N1=2
and then use the equation to determine N2. But this method has a “lookup table” quality to it, and seems less “mathematical.”
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