[Retros] Massacre SPGs

[Francois Labelle]

Gerd Wilts wrote:


> Does this list correspond to your list? What is the "missing" PG?


Yes, the list corresponds. The missing PG is:

wKg1, bKf8, bBd7

The solution is:

1. d4 e5 2. Bf4 exd4 3. Bxc7 Qxc7 4. Qxd4 Qxc2 5. Qxg7 Qxb2 6. Qxh7 Qxa2
7. Qxg8 Qxe2+ 8. Kxe2 Rxh2 9. Rxa7 Rxg2 10. Rxb7 Rxg1 11. Rxb8 Rxb8 12.
Rh7 Rxb1 13. Rxf7 Rbxf1 14. Rxd7 Rxf2+ 15. Kxf2 Bxd7 16. Qxf8+ Kxf8 17.
Kxg1

It'd be really interesting to know what your program thought for this SPG:

- If it saw 0 solutions, then your program had a bug.
- If it saw >= 2 solutions, then *my* program has a bug and I'd very much
like to know about it! (Alternatively people can try to find a cook by
hand if they're so inclined).

The PG does have the unusual property that the surviving piece is black,
i.e. it's a (1+2), all the 17 others are (2+1).


> x=5 has not yet been done, so there is still hope that a dualfree PG

> with two Kings only exists!


I wonder, would that be dismissed as a simple curiosity, or is it the holy
grail of retrograde analysis?

I started the computation. It's all a matter of having enough disk space
to list the distinct positions as the analysis progresses.

For x=3 the peak occurs at ply 21 with    160,061 distinct positions.
For x=4 the peak occurs at ply 23 with 30,590,355 distinct positions.

I can do x=5 if its peak is below about 6 billion (6e+09) positions.

	Francois