Klub Pongracz 10 JT C 13.8.2002 - SAT, 2nd group


The preliminary award was published in the issue No. 28, 29 of the bulletin Umenie 64. The tourney was announced for any problems using SAT, Sharp SAT (definitions see in updated older file) and some other SAT forms. It was judged by SAT inventor Ladislav Salai sr. There was 76 submitted problems by 10 authors from 5 countries. He decided to split the award to 2 groups, in the second group (whose award is shown below, the award of the first group is here) there were 33 problems in sharp SAT and in SAT combined with othe fairy conditions, with possible use of fairy units. The comments below marked LS are by judge.
Manfred Rittirsch
1st-2nd Prize e.a. Klub Pongracz 10 JT C 13.8.2002

1.Ba3+? Qxf6(f2) 2.f4+ Qxf4(f2) 3.Sc2+ Qxh2(Qd1)+
1...Qe3!

1.Sc3+! Sxc3(Sg1) 2.Ba3+ Qxa3(Bc1) 3.Se2+ Sxe2(Sb1) 4.Bxa3(Qd8)+ Qxf6(f2) 5.f4+ Qxf4(f2) 6.Sc2+ Qxh2(Qd1)#

LS: "Elegant binding of SAT and Circe. Circe moves in the try and the solution fully correspond with characteristics of the basic SAT. There is an inpiration for the future development of the fairy chess in the motivation of moves."









s#6 (9+5)
SAT, Circe

Lubos Kekely
1nd-2nd Prize e.a. Klub Pongracz 10 JT C 13.8.2002
NOT CORRECT!


1.Sg4+??

1.Bh7+! Rxh7(Bf1) 2.Sg4+ h5 3.Ba3+ bxa3(Bc1) 4.Bb2+ axb2(Bc1) 5.Bxb2(b7)+ cxb2(Bc1) 6.Ba6+ bxa6(Bf1) 7.Bb5+ axb5(Bf1) 8.Bc4+ bxc4(Bf1) 9.Bd3+ cxd3(Bf1) 10.Be2+ dxe2(Bf1) 11.f4+ Qxg4(Sb1) 12.Bxe2(e7)+ Qh3 13.Bf1+ Qg4 14.h3+ Qg3#

LS: "In the first phase it is necessary to shut off the black queen. The immediate 1.Sg4?? mates, that's why 1.Bh7+! The second phase prepares play of the light-squared bishop, while in the third white prepares his 11th move. Then there is the last act. Logical selfmate with the task of switchbacks of white bishops."

Unfortunately, Bojan Basic 3.8.2010 reported a cook:
1.Se4+ h5 2.Sd2+ Qg5 3.Bd3+ cxd2(Sg1) 4.f4+ Qg4 5.h3+ Qg3 6.Se2+ Qxd3(Bf1) 7.Sd4+ Qg3 8.Bc4+ cxd4(Sg1) 9.Se2+ Rxc4(Bf1)#









s#14 (6+8)
SAT, Circe

Ladislav Salai jr.
Special Prize Klub Pongracz 10 JT C 13.8.2002
NOT CORRECT!

1.CHRf7? CHRe5 2.g7~ ??

1.CHRf3!+ Se5 2.CHRd8+ Sdc6 3.CHRb3+ Sed4 4.CHRg5+ Sce6 5.CHRb7+ S4c5 6.CHRd2+ Sde4 7.CHRf7+ CHRd6 8.CHRa5+ Sc4 9.CHRf3+ S6e5 10.CHRd8+ Sdc6 11.CHRb3+ Sed4 12.CHRg5+ Sce6 13.CHRb7+ S4c5 14.CHRd2+ CHRe4 15.CHRf7+ Sd6 16.CHRa5+ Sec4 17.CHRf3+ S6e5 18.CHRd8+ Sdc6 19.CHRb3+ Sed4 20.CHRg5+ Sce6 21.CHRb7+ CHRc5 22.CHRd2+ Se4 23.CHRf7+ Scd6 24.CHRa5+ Sec4 25.CHRf3+ S6e5 26.CHRd8+ Sdc6 27.CHRb3+ Sed4 28.CHRg5+ CHRe6 29.CHRb7+ Sc5 30.CHRd2+ Sde4 31.CHRf7+ Scd6 32.CHRa5+ Sec4 33.CHRf3+ S6e5 34.CHRd8+ Sdc6 35.CHRb3+ CHRd4 36.CHRg5+ Se6 37.CHRb7+ S4c5 38.CHRd2+ Sde4 39.CHRf7+ Scd6 40.CHRa5+ Sec4 41.CHRf3+ S6e5 42.CHRd8+ CHRc6 43.CHRf3+ CHRd4#

LS: "The simple try doesn't work because of the white's stalemate. Almost textbook position for the Sharp SAT popularization, but also for that of fascinating chinese rose moves. We see 6-cycle round trip with the aim to move the white friend to the only suitable square c6! 6(!) white knights assist."

Unfortunately, Bojan Basic 3.8.2010 reported a cook:
1.CHRa1+ Se5 2.Sh6 Sc2 3.Sg4 Sed3 4.Se3+ CHRe5#









h#43 (8+5)
Sharp SAT
1+1 chinese rose

Lubos Kekely
1st-2nd HM e.a. Klub Pongracz 10 JT C 13.8.2002

1.Bb8+! Kxb8 2.Bc8+ Kxc8 3.Sd8+ Kxd8 4.f7+ Kc8 5.Rb7+ Kd8 6.Rc7+ Ke8 7.Rd7+ Kf8 8.Rd4+ Kxg8 9.Kh7=

LS: "Stalemate problem in the Sarp SAT is combined here with Madrasi elemenents. This synthesis allows unusual and forced walk of the bK."









=9 (10+3)
Sharp SAT, Madrasi

Lubos Kekely
1st-2nd HM e.a. Klub Pongracz 10 JT C 13.8.2002

1.Bf3+! Bxf3(Bf1) 2.Be2+ Bxe2(Bf1) 3.Bxe2(Bc8)+ dxe2(Bf1) 4.Bxe2(e7)+ Sxe2(Bf1) 5.Bxe2(Sg8)+ Rxe2(Bf1) 6.Bxe2(Ra8)+ Qxe2(Bf1) 7.Bxe2(Qd8)+ Sxh2(Sg1)#

LS: "Again a lot of switchbacks and 12 Circe rebirths. It seems SAT selfmate has potential for the creation as wide as orthodox helpmates. The solution is spiced by tries that may fool the solvers."









s#7 (4+11)
SAT, Circe

Lubos Kekely
Milan Ondrus

3rd HM Klub Pongracz 10 JT C 13.8.2002

1.Bb6 c5 2.Bxc5(c2) cxd3(d7)#

1.Bc7 d6 2.Bxd6(d2) dxe3(e7)#

1.Bd8 g5 (f6?) 2.Bxg5(g2) gxf3(f7)#

LS: "SAT+Circe analogy in 3 solutions. Pity, that only the third one hints the possible complication of the idea."









h#2 (8+9)
SAT, Circe
3.1.1.1

Lubos Kekely
4th HM Klub Pongracz 10 JT C 13.8.2002

a)

1.Rxb8? 1...Be4! th. 2...Bxc2#
1.Rd8? 1...Bc6! th. 2...Bxa4#
1.Rh8? 1...Bc6! th. 2...Bxa4#
1.Rg1? 1...Bd6 th. 2...Bxb4#

1.Rg7!
1...Be4 2.Rg2!
1...Be5 2.Rg3!
1...Bd5 2.Rc7!
1...Bd6 2.Rb7!
1...Bc6 2.Ra7!

b)

1.Rg1? 1...Bd6! th. 2...Bxb4#

1.Rh8!
1...Be4 2.Rh2!
1...Be5 2.Rh3!
1...Bd5 2.Rc8!
1...Bd6 2.Rb8!
1...Bc6 2.Ra8!

LS: "Strange link between the less known condition of the "matverijdeling" and SAT. In the less attractive zeroposition one can be interested in the colour shift of the second defence moves after the same attacks by Black. Modern conception is underlined by the tries. Change of 5 white defence moves!"









White defends mate in 2 (9+6)
SAT
a) d7 -» f4
b) b8 -» f4

Lubos Kekely
Special HM Klub Pongracz 10 JT C 13.8.2002
NOT CORRECT

1.Bf1! zz
1...B~, GIe~ 2.Ba6+ ~ 3.Bxa6+ ~ 4.Bxa6#
1...c5, Sb5, GIf~ 2.Bb5+ ~ 3.Bxb5+ ~ 4.Bxb5#
1...d4, Sb~, Zb~ 2.Bc4+ ~ 3.Bxc4+ ~ 4.Bxc4#
1...e3, R~2, Zc~ 2.Bd3+ ~ 3.Bxd3+ ~ 4.Bxd3#
1...f2, Rd4, Rd1, GId~ 2.Be2+ ~ 3.Bxe2+ ~ 4.Bxe2#

LS: "Thematic squares on f1-a6 diagonal are guarded 3 times - once by pawn, once by orthodox piece and once by giraffe. Five-variation zugzwang problem with 10 switchbacks of white bishop to rebirth square. 18 black units make the construction heavy, giraffes with the long neck make the content more variable."
This intention is however refuted by 1...GIe~! as then 2.Ba6+? bxa6(Bf1) 3.Bxa6+ Bxa6(Bf1) 4.Bxa6(Bc8)+ Bxa6(Bf1)! (uncovered by Bojan Basic)









#4 (3+18)
SAT, Circe
0+5 giraffe

Milan Ondrus
1st Comm Klub Pongracz 10 JT C 13.8.2002

1.Rxh2(B) Rh8 2.Bxh8(B) Bf6#
1.Bxe2(B) Ba6 2.Rxa6(B) Rd6#

LS: "Synthesis of SAT and Andernach chess. The solutions are analogical, the mates SAT ones."









h#2 (7+14)
SAT, Andernach
2.1.1.1

Lubos Kekely
2nd Comm Klub Pongracz 10 JT C 13.8.2002

1.b8Q? zz
1...Bxb8(Qd1)!

1.b8R? zz
1...Bxb8(Ra1)!

1.b8B? zz
1...Bxb8(Bc1)!

1.b8S! zz
1...Bxb8(Sg1) 2.Sxe2(e7)#

LS: "AUW in zugzwang position."









#2 (4+9)
SAT, Circe

Lubos Kekely
3rd Comm Klub Pongracz 10 JT C 13.8.2002

1.f8==!! (no promotion)

LS: "Ecce joke double-stalemate in halfmove in the miniature!"









==1/2 (3+4)
SAT, joke problem

Comments to Juraj Lörinc.
Back to main page of Chess Composition Microweb.