Award of 13th TT Spisska Borovicka C 3.8.2001

This is the award by Peter Gvozdják of TT that was announced on CCM too.

When looking back into the new FIDE Album 1992-94, I'm quite pleased that three Spisska Borovicka tourney problems from that three-year period have been selected into it...

In the last years, I tried to take the level of this tourney even higher by asking for more complicated theme, and, on the other hand, giving more time by announcing it on Internet some weeks earlier. And it really came true: this year?s problems are of the best ones ever. I've received 15 entries, some of them unsound, the other non-thematic or not original enough. But 8 problems left, all of them very good.

Special Honourable Mention ex aequo 10poi (Poisson): something to laugh about: At first don't look at the chessboard and try to imagine how to combine the Babson task with the 4x4 complete Lacny cycle. Then have a look at the position and learn just two pawns on 20 squares are enough.

Special Honourable Mention ex aequo 5rit (Rittirsch): Another good joke - the promotions just as refutations! But otherwise, the 3x3 Shedey cycle is shown in very elegant and most economical way.

2nd Honourable Mention 15kle (Klemanic): Shedey cycle, I only think also the third promotion should be orthodox, too.

1st Honourable Mention 11ret (Retter): This is one of the non-awarded WCCT problems, very suitable one for my tourney: three white orthodox promotions plus three black "random" promotions.

4th Prize 14asc (Aschwanden, Maeder): 4-fold Shedey cycle with w AUW in mating move. Very ambitious combination, could you imagine what would be higher?

3rd Prize 2qua (Quah): 3x3 Lacny cycle with three fairy promotions to the same-family-units. Very well done. (The author had sent almost the same position with the 3x3 Shedey, too, and left it up to me, to choose one of them. I like the awarded one a bit more, because the numeral "three" appears in all the elements, also in the b King's moves.)

2nd Prize 7asc (Aschwanden): 4-fold Rice cycle with black correction. Also very good orthogonal-diagonal analogy, and cycle of captures. Pioneer such example.

1st Prize 8asc (Aschwanden): Another pioneer example (of course!). 4-fold Lacny plus Allumwandlung. There are also fairy units and a big board here, but this one is much more sophisticated than the 4th Prize. For the complete understanding I add also the author's notices to the solution to see the whole motivation.

Peter Gvozdják, Wageningen, 3rd August 2001.
Christian Poisson
Special HM e.a. 13th TT Spisska Borovicka C 3.8.2001

a) 1.d4! 1...b1Q,R,B,S 2.d5Q,R,B,S etc.

b) 1.d4! 1...b1Q,R,B,S 2.d5R,B,S,Q etc.

c) 1.d4! 1...b1Q,R,B,S 2.d5B,S,Q,R etc.

d) 1.d4! 1...b1Q,R,B,S 2.d5S,Q,R,B etc.






r=2 (1+1)
4x5 board
Madrasi
b) Eiffel chess S-Q-R-B
c) Eiffel chess (S-R) (B-Q)
d) Eiffel chess S-B-R-Q
Manfred Rittirsch
Special HM e.a. 13th TT Spisska Borovicka C 3.8.2001

1.DGe8? (2.Qb6#)
1...e6, e5 2.Qb5, Qb7#
1...c1DG!

1.DGe3? (2.Qb5#)
1...e6, e5 2.Qb7, Qb6#
1...d1DG!

1.DGch5? (2.Qb7#)
1...e6, e5 2.Qb6, Qb5#









#2 (6+8)
2+0 double grasshopper
Emil Klemanic
2nd HM 13th TT Spisska Borovicka C 3.8.2001

(1.Qg5? (2.b8S#) 1...Rf5!)

1.Qg4? (2.c8R#)
1...Rf5, f5 2.a8BH, b8S#
1...Rxg4!

1.g7! (2.a8BH#)
1...Rf5, f5 2.b8S, c8R#









#2 (15+11)
2+1 bishopper, 0+1 transmuting king
Josef Retter
1st HM 13th TT Spisska Borovicka C 3.8.2001

1...cxd1, dxc1(dxe1), exd1, (Kxa5)
2.d8B, a8S, b8R, (d8B)#

1.Qf5! zz
1...cxd1, dxc1(dxe1), exd1, (Kxa5)
2.a8S, b8R, d8B, (a8R)#









#2 (14+7)
0+1 transmuting king
Reto Aschwanden
Thomas Maeder
4th Prize 13th TT Spisska Borovicka C 3.8.2001

1.mRg2(wRb2, bSf5)? thr. 2.c8mB(wSf5)#
1...mBg3(wSf5), mPfxg6(wSf5, bRc4), mPd2(bRc4)
2.c8mS, c8mR(wRc4), c8mQ(wRc4, wSf5)#
1...Sd6!

1.mQb1(wRb2, wPa2)! thr. 2.c8mS#
1...mBg3(bRc4), mPfxg6(bSf5, bRc4), mPd2(bSf5)
2.c8mR(wRc4), c8mQ(wRc4, wSf5), c8mB(wSf5)#









#2 (11+14)
magic units f1, g4, b8, a7, a7 - f4, h3, g8, d3, f7, g5, h2
James Quah
3rd Prize 13th TT Spisska Borovicka C 3.8.2001

1.c3! zz
1...Kg4, Kg3, Kg2
2.g8M, g8EA, g8SP
2...Qb3#

1.c4! zz
1...Kg4, Kg3, Kg2
2.g8EA, g8SP, g8M
2...Qb4#

1.Sc5! zz
1...Kg4, Kg3, Kg2
2.g8SP, g8M, g8EA
2...Qb5#



+++ Composition In the Spotlight (CIS) No. 21 +++


Spotlight comment by Juraj Lörinc:

The 65th TT feenschach has deadline 30.11.2011, i.e. only 9 days are left. It is dedicated to #2, #3 and ser-h#n using fairy pieces eagles, sparrows or mooses and no other fairy elements. Its announcement in the magazine listed only few problems, thus the field is mostly unexplored. Present problem would not be thematical duie to use of grasshoppers too, as well as fairy conditions Beamtenschach (Functionary chess) and stipulation r#2.

Neverthless it is very instructive study in the movement possibilities of the selected set of direction-changing hoppers. In three phases white keys provide hurdles for black hoppers present on the f-file. Only the black king can move in the first black moves, providing hurdles for white hoppers newly promoted in the 2nd white moves. After 1.c3 it is Gf3 who has to be activated, after 1.c4 it is Mf1 or Gf4 and after 1.Sc5 it is Gf5 or Mf2. In all cases, 2nd white moves both control important hoppers on f-file over the bK and also open SPh7 to b8 via a7. Finally bQ mates using the guard from f-file.









r#2 (12+11)
Beamtenschach
0+3 grasshopper, 2+2 moose
sparrow h7, eagle c8
3 solutions
Reto Aschwanden
2nd Prize 13th TT Spisska Borovicka C 3.8.2001

1...kRLf4~, kRLxd4! 2.f8BL, b8RL#

1.kPdxc5? (2.kQc5#)
1...kRLf4~, kRLd4! 2.b8RL, f8Q#
1...kRLxc4!

1.kRxb5? (2.kQc3#)
1...kRLf4~, kRLxd4! 2.f8Q, b8Q#
1...kRLxf7!

1.kSxd4! (2.kQc3#)
1...kRLf4~, kRLd4! 2.b8Q, f8BL#









#2 (9+14)
Bicaptures
kamikaze units e3, c5, b5, d4 - b4, h6, f4, g5
rook lion f4, bishop lion d1
0+6 nightrider lion
Reto Aschwanden
1st Prize 13th TT Spisska Borovicka C 3.8.2001

1.Bf6? zz
Closes RHf10-f4 but opens RHf10-f5. The square c4 is not reachable for the black King. Preliminary opens d1-d6.
1...d1nB a 2.NHc8# A
Guards e2. d4 must be guarded by Bf6 because NHc8 intersects the patrolling Nd10-b6 therefore Bb6 is not guarding d4.
1...d1nS b 2.Ni2# B
Patrols RHf2 to guard f4. In this phase, c4 is not reachable for the black King.
1...d1nR c 2.Qe10# C
Patrols nSi1 to guard g2. f5 is guarded by RHf10.
1...d1nQ! d 2.Ni6# D
Corrects the promotions to Bishop and Rook. 2.NHc8+?? d5! and 2.Qe10+?? nBe6! Guards d3 and f3 because Nc9 abandons the control of f3 and the patrolling of Nb7 that guarded d3.
but 1...NHg7!
Gives a flight g4.

1.NHj1! zz,
The square g2 is not reachable for the black King and f3 is guarded by NHj1. Opens RHf2-f5 and closes RHf2-f4. Preliminary opens d1-i6.
1...d1nB a 2.Ni2# B
Patrols nBb3 that guards c4. f4 is guarded by RHf10.
1...d1nS b 2.Qe10# C
Patrols RHf2 that guards f5. In this phase, g2 is not reachable for the black King. 2.Qg2+?? nSxg2! because nSi1 is patrolled by NHe3.
1...d1nR c 2.Ni6# D
Guards d3 as Nc9 abandons the patrol of Nb7. f3 is guarded by NHj1.
1...d1nQ! d 2.NHc8# A
Corrects the promotions to Bishop and Rook. 2.Ni2+?? nSg2! intersecting RHf2-i2 and 2.Ni6+?? nQxi6! Guards d4 and e2, because NHa4 abandons the guard of e2 and because intersepting the patrol of Nd10-b6 for Bb6.
1...NHg7 2.Qg2#
RHf2 is patrolled by Bd4. 2...nSxg2 is not possible any more because of not being patrolled.











#2 (18+7+3)
No fairy promotions, Black Sting Chess
2+0 rookhopper, 4+2 nightrider hopper
4+1 nightrider, 3 neutral units
ultrapatrol units b6, f2, d10, b7, g6 - a7, c3, d6, j7 - b3, i1
10x10 board

Comments to Juraj Lörinc.
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