From: Colin Paul Adams <colin <at> COLINA <dot> DEMON <dot> CO <dot> UK>
Date: 13 feb 2000
Subject: CHU: more on A15

In fact, even after

1. +R - 2a,      K x 2a
2. Ph - 4c=,     K - 2b
3. Ph - 2a,      K x 2a

the published solution fails to mention the longest (though one might
query this as being better, as it merely inserts an extra (pointless)
move into the sequence) defence.

My program (with the promotion bug now fixed), finds:

4. Ln x 5c x 4b, R x 4b
5. DK x 2c+,     K - 3a
6. FK - 7a,      +FL - 5a
7. FK x 5a,      R - 4a
6. FK x 4a Mate

The defence of K - 3a puts up much longer resistance. After:

1. +R - 2a,      K x 2a
2. Ph - 4c=,     K - 3a
3. FK - 7a,      R - 6a

my program finds:

4. Ln x 5c x 4b, R x 4b
5. Ph x 6a+,     +FL - 5a
6. +Ph x 5a,     R - 4a
7. +Ph x 4a,     K - 2b
8. +Ph - 2a,     K - 1c
9. FK - 7i,      K - 1d
10. FK - 6i,     K - 1c
11. FK - 6h,     K - 1d
12. FK - 5h,     K - 1c
13. FK - 5g,     GB - 4f
14. FK x 4f,     P - 3e
15. FK x 3e,     C - 2d
16. DK x 2d+ Mate in 31, which is the same length as Mr. Hanazawa's
suggestion, and is much the same, except for a slightly different
finish (note that there is a typo in Mr. Hanazawa's suggestion - as
MSM says Ln x 4c x 4b - it should be Ln x 5c x 4b, as above).
--
Colin Paul Adams
Preston Lancashire