# FIDE-Albums

Solutions

M. Caillaud
2nd Prize feenschach 1979

12+10. (Maos g1, e7) Both sides may castle.
The game was played with Maos instead of Knights. (a) Where have the Bishops been captured?
(b) Least number of moves by Mao b8?

Solution: 5 white captures are axb, b2xc3, e2xd3xb4 and g2xh3; 1 black is e7xf6.

In a game with Maos, the 1st move is by a Pawn. Here White could only start with a2-a3 or f2-f3. These moves could only free wKe1 and wRa1 but we are told that White may still castle. Thus Black had to quickly allow a White capture, so that the game started with 1. a3 (or f3) d6 2. f3 Bh3 3. g2xBh3 (slightly) freeing wBf1. After 1. a3, the wBc1 never left its SW corner.

Because of 1 ... d6, freeing square d6 for the wMe7 would require cross-captures d6xc5 and later c7xd6. But then all 3 missing Wh. men (discounting entombed wBc1) have been captured on black squares, which cannot account for the missing wBf1. Therefore the original bPd7 is still now on d6, and the wMe7 did reach e7 through the only other route, b6-c8-e7, requiring cross-captures bxc and c7xb6 (in that order), and precisely b7xBc6.

Let's go back to the beginning of the game. After the first moves we saw, the bQd8 had to go and get captured on d3, freeing wQd1, wMg1 & wRh1. The wQ or the wR could then get captured on f6, further freeing Black's game.

Observe that between e2xQd3 and e7xQ/Rf6, Black had to play Mb8-d7 because he had not enough waiting moves (h7-h6 is not enough). The bM was then stuck in d7, but the Ra8 was now free.

After e7xf6, the bMg8 can get captured on c4, fully freeing wKBf1 who can now get captured on c6, fully freeing the bQRa8. The wMb1 is only free after two captures: b2xc3 and a3xb4. So that it really is the wMg1 who went to b6-c8 (and now lies on e7) before the b6 door was closed by c7xb6.

bRa8 got out of the NW corner and only then the remaining wQ or wR can be captured on b6, freeing bMd7. Now bR and bM must somehow elimitate the wQB and get captured on c3 and b4. It turns out that we cannot have the bM captured on c3. The bMd7 could only reach this square after c6-c5 has been played, and this move must be delayed because it blocks the path of the wMb1 who still has to reach g1. Therefore the bR must get captured on c3 and let the bM capture the entombed wQB and then get captured on b4.

Starting from b8, the bl. Mao cannot capture the wQB on square b2 ! His shortest trip is b8-d7-b8-a6-b4-a2xBc1-e2-f4-d5-b4. Then a3xMb4 follows.

Now while wMb1 reaches g1, Black needs waiting moves. c6-c5 and h7-h6 are not enough. So that the unaccounted for bKB must have been retained till these last moments, and can only get captured near the very end, and thus by wMc8xBe7.

Finally, we have (a) wKB taken on c6, wQB on c1, bKB on e7, bQB on h3; (b) the bMb8 did play at least 10 moves.

(Many thanks to Henrik Juel who helped me spell out this fantastic analysis.)