# [Retros] Colour Rules in 20 Bishops Chess Positions

peufe at tin.it peufe at tin.it
Wed May 23 12:34:31 EDT 2007

By Franco Fedeli
Italy

In order to have 10 + 10 Bishops on the board,
we must promote all 16 pawns.
To do this it's necessary to open all 8 lines sacrificing 4 W officiers and 4 B officiers,
so we may have at most another W officier (Q,N,R) and another B officier (Q,N,R) on the board.
Each sacrifice open a line on which 2 pawns of the same colour may promote.
If we assign a bit to each line, "0" standing for 2 W promotions on that line and "1"
standing for 2 B promotions on that line, we easily see that, in positions with
10 B + 10 b + 1 (Q,N,R) + 1 (q,n,r) + K + k
we can assign to each couple of lines (a,b), (c,d), (e,f), (g,h) the couple (0,1) or (1,0)
independently, so we have 16 choices.
Each byte among the 16 implies different numbers of blake and white final squares
of promotions, as in the table.
WPB= White Promoted Bishops
BPB= Black Promoted Bishops
WTB= White Total Bishops
BTB= Black Total Bishops

10 B + 10 b + 1 (Q,N,R) + 1 (q,n,r) + K + k

Squares of promotion Choices WPB BPB WTB BTB
(W,B)

(4,0) 1 (8,0) (8,0) (9,1) (9,1)

(3,1) 4 (6,2) (6,2) (7,3) (7,3)

(2,2) 6 (4,4) (4,4) (5,5) (5,5)

(1,3) 4 (2,6) (2,6) (3,7) (3,7)

(0,4) 1 (0,8) (0,8) (1,9) (1,9)

In positions of the kind:

10 B + 10 b + 1 (q,n,r) + K + k

the missing white officier could have been captured by a black pawn,
changing the line and the final square colour of promotion.
In that case all legal situations are:

10 B + 10 b + 1 (q,n,r) + K + k

WTB BTB

(9,1) (9,1) , (8,2)

(7,3) (7,3),(6,4),(8,2)

(5,5) (5,5),(6,4),(4,6)

and symmetric (switching all numbers inside ( ).

10 B + 10 b + 1(Q,N,R) + K + k

This situation is the specular one.

In positins of the kind:

10 B + 10 b + K + k

the 2 missing officiers could be captured by pawns, changing
the lines and the final square colour of promotions.
In this case all legal positions are:

10 B + 10 b + K + k

Bishops of one Colour Bishops of the other Colour

(9,1) (9,1)

(9,1) (8,2)

(8,2) (8,2)

(8,2) (7,3)

(8,2) (6,4)

(7,3) (7,3)

(7,3) (6,4)

(6,4) (6,4)

(6,4) (5,5)

(6,4) (4,6)

(6,4) (5,5)

(5,5) (5,5)

and symmetric (switching all numbers inside ( ).

In positions such that we can prove that last move was O-O
or O-O-O then new rules arise on promoted bishops.
As example:
White O-O-O

Black could'nt promote in a1,e1 (occupied) nor in d1, f1 (because check).
Black can promote no more than 7 pawns:
2>b1 2>b1
2>c1 OR 2>c1
1>g1 3>g1
2>h1

White O-O

Black could'nt promote in e1,h1 (occupied) nor in d1,f1 (because check).
Black can promote no more than 7 pawns:
2>a1 2>b1
2>c1 OR 2>c1
3>g1 3>g1
In order to do this Black captures Q,R,N,N.
Black e pawn could be captured by c white pawn.

An easy counting shows that the tables are:

Last move White O-O
10 B + R + 9 b + 1(q,n,r) + K +k

WTB= White Total Bishops
BTB= Black Total Bishops
Numbers in ( W,B ) means Bishops on White or Black squares.

WTB BTB

(3,7) (1,8)

(4,6) (1,8)

(5,5) (3,6)

(6,4) (3,6)

Last move White O-O
10 B + R + 9 b + K +k

WTB BTB

(2,8) (1,8)

(3,7) (1,8)

(4,6) (1,8)

(4,6) (3,6)

(5,5) (1,8)

(5,5) (3,6)

(6,4) (3,6)

(7,3) (3,6)

Last move White O-O-O
10 B + R + 9 b + 1(q,n,r) + K +k

WTB BTB

(8,2) (5,4)

(7,3) (5,4)

(6,4) (3,6)

(5,5) (3,6)

Last move White O-O-O
10 B + R + 9 b + K +k

WTB BTB

(9,1) (5,4)

(8,2) (5,4)

(7,3) (5,4)

(7,3) (3,6)

(6,4) (5,4)

(6,4) (3,6)

(4,6) (3,6)

(5,5) (3,6)

I hope that this could be useful for new ideas in retroanalysis colouring problems.
Bye,

Franco Fedeli

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