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No.1349 |
Original Fairy problems |
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No.1349 Arno Tüngler |
Solution: (click to show/hide) |
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white Ke8 Pa7d3d4f3h2h5
black Kd5 Qg7 Ba1b1 Rb2e3 Se6 Pc4c6f4g3h4h6
ser-!=66 (7+13) C? |
|
No.1349 |
Original Fairy problems |
|
No.1349 Arno Tüngler |
Solution: (click to show/hide) |
|
white Ke8 Pa7d3d4f3h2h5
black Kd5 Qg7 Ba1b1 Rb2e3 Se6 Pc4c6f4g3h4h6
ser-!=66 (7+13) C? |
what is the cook when the a-pawn is lower than a6 ?
For example: 1.Ke8 2.hxg3 3.gxf4 7.f8=B 8.Bb4 13.f8=B 14.Bfe7 18.Ka6 20.Bxh6 21.Bf4 24.h8=R 26.Rxc6 27.Rxc4 29.Ra8 30.Bb8 32.Ba5-c7 33.Ka7 35.a6 !=
So, with wPa6 that should already not be possible.
Is it so important to say « without capture « ?
Is there already a longer position « with captures » ?
Yes, the longest has 173 moves with and 119 moves without promoted force. See PDB P1333509 and P1273550. In long series-movers captures are usually quite important to get small changes in the position that add length.
cook 1.hxg3 2.gxf4 3.f5 4.f4 5.f6 6.f5 7.f7 8.f6 9.fxg7 10.a8=R 11.Ra4 12.Rxc4 13.Rxc6 14.Rxe6 15.Kf8 16.Rxh6 17.Rh8 18.h6 19.h7 20.Rg8 21.h8=B!=
Thanks to Václav for the cook that I should have seen…
Would like to correct this with wBd7 instead of wPa7 and one move shorter – a ser-!=65. The solution remains unchanged besides after 1.Bc8 2. Ba6 everything is one move less. Maybe you can test again?
A full test of this version is not possible by either Alybadix or Gustav. I tested this sd-auto=65 with Gustav 4.2c with the additional condition that pawn promotions can only be on the bishop, and the problem passed this test! Even this partial test required over 60 hours of computer time.
In his award to this informal tournament Kjell Widlert wrote: “1349 (Tüngler) This was cooked, and the composer’s correction was never officially published with a 1349.1 diagram. But the problem probably would not have made the award anyway, as it is a new variation on a very well-known matrix. I’m still impressed by the new record, however.” Would it still be possible to publish this 1349.1 version here? The only changes are: +wBd7, -wPa7, stipulation ser-!=65, solution: 1.Bc8 2.Ba6 3.Bb5 4.Ba4 5.Bd1 6.Be2 7.Bf1 8.Bh3 9.Bf5 10.Bg6 11.Bf7 12.Kd7 13.Kc8 14.Bg6 15.Bf5 16.Bh3 17.Bf1 18.Be2 19.Bd1 20.Ba4 21.Bb5 22.Ba6 23.Bb7 24.Kb8 25.Ka7 26.Ka6 27.Ka5 28.Ba6 29.Bb5 30.Ba4 31.Bb3 32.Kb4 33.Kc3 34.Bc2 35.Kd2 36.Kc1 37.Bd1 38.Be2 39.Kd1 40.Ke1 41.Kf1 42.Kg2 43.Kh3 44.Bd1 45.Ba4 46.Bb5 47.Ba6 48.Bc8 49.Bd7 50.Be8 51.Bg6 52.Kg4 53.Kf5 54.Be8 55.Bd7 56.Bc8 57.Ba6 58.Bb5 59.Ba4 60.Bd1 61.Be2 62.Bf1 63.Bh3 64.Bg4 65.h3 !=