# [Retros] Maximum number of officiers in legal positions

peufe at tin.it peufe at tin.it
Tue May 22 07:02:19 EDT 2007

>From Franco Fedeli

Italy
Hi friends,
in order to have the maximum number of officiers
one must sacrifice 4 pawns to open 8 lines so
it is possible to promote 12 pawns.
In this way we can have positions with 28 officiers
on the board (Kings included).
If all pawns are promoted to Bishops, then some
interesting restrictions arise about the square's colours
of the Bishops ,which could be used in retro analysis.
To find them we must start from the simple rule:
every take open 2 near lines and makes possible
2 promotions for one colour and 1 promotion for the other.
The colour of the squares where the pawns will be promoted
is the same.
As example a4xb5 makes possible to promote 2 W pawns
on the B square b8 and 1 B pawn on the B square a1, so
we can have 2+1 Bishops on black squares.
In the following table I summarize all the possibilities:

B Pawns captured W Pawns captured Final Squares colours Final Squares colours W promoted Bishops B promoted Bishops
of double W promotions of double B promotions

4 0 (4,0) --- (8,0) (4,0)

4 0 (3,1) --- (6,2) (3,1)

4 0 (2,2) --- (4,4) (2,2)

3 1 (3,0) (1,0) (7,0) (5,0)

3 1 (3,0) (0,1) (6,1) (3,2)

3 1 (2,1) (1,0) (5,2) (4,1)

3 1 (2,1) (0,1) (4,3) (2,3)

2 2 (2,0) (2,0) (6,0) (6,0)

2 2 (2,0) (1,1) (5,1) (4,2)

2 2 (2,0) (0,2) (4,2) (2,4)

2 2 (1,1) (1,1) (3,3) (3,3)

+ Symmetric.

PROMOTED BISHOPS:

one colour (B or W) other colour (W or B) Squares Colours Total Promoted Bishops on opposite colours squares

8 4 (8,0) (4,0) 12 + 0

8 4 (6,2) (3,1) 9 + 3

8 4 (4,4) (2,2) 6 + 6

7 5 (7,0) (5,0) 12 + 0

7 5 (6,1) (3,2) 9 + 3

7 5 (5,2) (4,1) 9 + 3

7 5 (4,3) (2,3) 6 + 6

6 6 (6,0) (6,0) 12 + 0

6 6 (5,1) (4,2) 9 + 3

6 6 (4,2) (2,4) 6 + 6

6 6 (3,3) (3,3) 6 + 6

TOTAL BISHOPS:

one colour (B or W) other colour (W or B) Total Bishops on opposite colour squares

(9,1) (5,1) 14 + 2

(8,1) (6,1) 14 +2

(7,1) (7,1) 14 + 2

(7,3) (4,2) 11 + 5

(7,2) (4,3) 11 + 5

(6,3) (5,2) 11 + 5

(6,2) (5,3) 11 + 5

(5,5) (3,3) 8 + 8

(5,4) (3,4) 8 + 8

(5,3) (3,5) 8 + 8

(4,4) (4,4) 8 + 8

I hope that you can have some good idea for retro analysis problems
using this theme that, if you allow me, I call " Fedeli Theme".
Thanks
Franco Fedeli

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