[Retros] HAP problems up to ply 7

[Francois Labelle]

First some statistics...

Number of possible HAP strings in plies 1..7:
2, 4, 15, 75, 468, 3152, 23500

Number of HAP strings that uniquely determine a game in plies 1..7:
0, 0, 0, 1, 5, 59, 341

Those 1+5+59+341 problems, classified by White's first move:

a3:  0 0  0   0
a4:  0 0  0   3
b3:  0 0  1   1
b4:  0 0  3   9
c3:  0 0  0   2
c4:  0 0  0  13
d3:  0 0  3  10
d4:  0 0 14  86
e3:  0 0  4  22
e4:  0 2 18 111
f3:  0 0  1   0
f4:  0 0  4   6
g3:  0 0  0   1
g4:  0 1  4  17
h3:  0 0  0   1
h4:  0 0  0   8
Na3: 0 0  1   6
Nc3: 1 1  5  25
Nf3: 0 1  1  18
Nh3: 0 0  0   2

An example of a shortest problem for each:

a3: (requires at least 8 plies)
a4: 1.P P 2.A H 3.A PxP 4.AxA
b3: 1.P P 2.H H 3.HxH HxA (posted earlier by me)
b4: 1.P A 2.H P 3.HxA HxP
c3: 1.P P 2.H H 3.HxP HxA 4.HxH+
c4: 1.P P 2.A P 3.AxP AxA 4.HxA
d3: 1.P P 2.A H 3.HxP HxA+
d4: 1.P A 2.H AxP 3.H AxH+
e3: 1.P P 2.A H 3.HxP HxA
e4: 1.P P 2.H PxP 3.HxH+
f3: 1.P P 2.H H+ 3.H H#
f4: 1.P P 2.A H+ 3.AxH PxP
g3: 1.P P 2.H H 3.HxH HxP 4.H+
g4: 1.P P 2.H HxP 3.H+
h3: 1.P P 2.A HxP 3.AxP HxA 4.PxH
h4: 1.P P 2.A H 3.A PxP 4.AxA+
Na3: 1.A P 2.P PxP 3.HxP HxA
Nc3: 1.A P 2.AxP HxA (posted by Joost)
Nf3: 1.A P 2.AxP A 3.AxA (posted by Joost)
Nh3: 1.A P 2.P H 3.PxP HxH+ 4.AxH

There are 10 problems in 7 plies with castling. One example:
1.A P 2.P H 3.HxP HxA 4.O

There are no problems with en passant in 7 plies or less.

Happy solving!

 	Francois