Phénix 117 - April 2003

Evgeny Bourd
4224 Phénix 117 - April 2003

a) 1.Ka6 Bxd4 2.Ka7 Ra8#

b) 1.Kb4 Rg3 2.Kc3 Bd2#

Echo diagonal-orthogonal with reciprocal exchange of functions of Be3 and Rg8 (pinning and mating piece). What is a bit special, it is the fact that all black moves are made by bK.

h#2 (5+9)
b) h8 -» h6

Alexandr Pankratiev
4227 Phénix 117 - April 2003

1.Se3+ Qf5 2.Sg5 Qc8#

1.Sg5+ Qe4 2.Se3 Qa8#

Inversion of Black moves is motivated by need to open double masked line b1-h7 and simultanously not to attack wK too much. White queen is temporarily pinned and then mates on the 8th rank. Note the function of wSh7 - he only prevents bK from entering f8, but doesn't contribute to mating nets. Brave concept - too brave perhaps...

h#2 (4+13)

Jorma Pitkänen
4241 Phénix 117 - April 2003

1.Qe6! th. 2.Rxb2+ Kxb2#
1...Se5 2.Sxh4+ Sg6,Sd3 3.Rxb2+ Kxb2#
1...Be5 2.Sd4+ Rxh7 3.Rxb2+ Kxb2#
1...Qxb7 2.Sh6+ Qxh7 3.Qe4+ Qxe4#

Compare this problem to s#3 by the same author awarded in J. Havran 50 JT. They have the same basic scheme of wKh1, bKb1, bRa1, black linemover on a8 aiming to h1, wRb7, wRc6, active wQ, wBh7 - that's quite a lot of similarities. Nevertheless, the final result is different - while in Havran JT there was a hunt for bS, here we see main role of wS in White's second moves.

s#3 (6+12)

James Quah
4249 Phénix 117 - April 2003

1.LEf1? A th. 2.VAd3# B
1...NEc7 a 2.VAe2# C
1...NEc5 b 2.VAc4# D

1.VAc4! D th. 2.VAhe2# C
1...NEc7 a 2.LEf1# A
1...NEc5 b 2.VAhd3# B

Using AMU for creating new Cyclone schemes is James' patent. Here we see Djurasevic cycle 4-2, also known as double Jerochin, in the other James' AMU problem with chinese pieces there is Djurasevic 4-1. The idea is astonishing in its simplicity: moving any pair of chinese pieces to f1-a6 diagonal causes unstoppable check - it is just necessary to choose the one observed exactly once by black paos.

#2 (8+6)
vao h7, d5, h5, leo f7
nonstop equihopper g7, g5, pao a7, a5

André Davaine
4252 Phénix 117 - April 2003

1.Qg3! zz
1...h2, ROxf3 2.Kb3#
1...Bf8, Bxh8, Kb6 2.Bb3#
1...Bxh6, Kb8 2.Bb2#
1...Bxf6 2.Kb2#
1...ROxe2 2.Rxa1#

Less usual fairy condition Antikings means that a side is checked if its king is not attacked. The fight concentrates along b1-b8 line. E.g. the error of Kb6 defence is in the fact that bK no longer can reach eight rank and thus 2.Bb3+ cannot by parried by bK entering check on the horizontal line opened by mating wB. Strange condition... these conditions changing the rules for check (SAT, Vogtlaender chess and others) are sometimes hard to grasp...

#2 (12+7)
rose g1

Vlaicu Crisan
4255 Phénix 117 - April 2003 (v)

1...Bc8 2.Ke6#
1...Ba6 2.Kc4#
1...Ba8 2.Kxc6(Rd5)#

1...Sc5, Sd8 2.Ke6#
1...Sd6, Sa5 2.Kc4#
1...Ra6, Rb6, Rh6, Rg6, Rf6, Re6 2.Kd6#
1...Rc1, Rc2, Rc3, Rc4, Rc8, Rc7 2.Kc5#

Vogtlaender chess is really strange as is presented by this twomover. After keys White has lonely king and this is in fact precisly everything what he needs to mate Black! Mates 2.Ke6# and 2.Kc4# are transferred in the unusual manner. There are 2 things to regret: symmetry of the position and purely technical reasons for 3 out of 4 conditions: Glasgow, Andernach chess and PWC.

#2 (2+1)
Glasgow, Vogtlaender chess
Andernach chess, PWC
2 solutions

Václav Kotesovec
4259 Phénix 117 - April 2003

1.d5 Kd1 2.Kd3 Ge3 3.d4 Ge5=

1.Kd5 Ge6 2.Kd4 Ge7 3.d5 Ge5=

1.Kc5 Kd3 2.Kd5 Ge6 3.d6 Ge7=

Once upon a time I have been pondering Circe problems with no captures, no rebirths... could they be interesting? Could they be Circe intensive? I would say Yes, but it requires quite creative thinking. Here the author uses Circe for guarding one of grasshoppers in 3 chameleon echo stalemates. Older h=3,5 by VK shows other, but very similar stalemate. Or see ser-h=24 by Ivan Skobe with 5 Circe-selfcheck illegal captures in final position after precise play.

h=3 (3+2)
2+0 grasshopper

Václav Kotesovec
4260 Phénix 117 - April 2003

1...Se3 2.Bh7 Sg4+ 3.Kf5 Kd5 4.Bg6 Sh6#

1...Kb4 2.Kd5 Kb3 3.Kd4 Be3+ 4.Kd3 Sf4#

It is possible to use Circe selfguarding in mating net as well. No capture here either.

h#3,5 (3+2)

Daniel Novomesky
4263 Phénix 117 - April 2003

1.LIg8 2.LIg5 3.LIh5 4.LIf5 5.LIc1 6.Kg1 7.LIh1 8.LIff1 Kg3#

1.Ke2 2.Kd3 3.LIb1 4.Kc3 5.LId2 6.Kb2 7.LIa2 8.Ka1 Ke5#

Daniel works in the field of longer series helpmates quite actively. As one can see also from his 1st Prize 9th TT CCM, more solutions are not impossible even with excellent economy.

ser-h#8 (2+3)
1+2 lion

Chris J. Feather
4265 Phénix 117 - April 2003


1.Gg4 6.Kc4 7.Gb4 8.Kb5 9.Ga5 10.Ka6 11.Ga7 12.Kb7 13.Gb8 14.Ka8 Bf3#

Chris Feather, well known helpmates expert, has recently produced a lot of set play seriesmovers. See e.g. his ser-h#20*. Here despite the same mating move we have mates in opposite corners.

ser-h#14* (2+3)
0+2 grasshopper

Daniel Novomesky
4266 Phénix 117 - April 2003

1.Gf8 2.Gf6 3.Gf5 4.Gf4 5.Gf3 6.Gf2 7.Gf1 8.Gc1 9.Ge1 10.Gg3 11.Gh4 12.Gh1 13.Gh3 14.Gf3 15.Ge3 16.Gd3 17.Gc3 18.Gb3 19.Ga3 20.Ga8 21.Ga6 22.Kb6 23.Kb5 24.Kc4 25.Gd3 26.Kd4 27.Ke3 28.Gf3 29.Kf2 30.Gg3 31.Ge1 31.Gf1 32.Gc1 33.Gce1 Kg1#

Again Daniel and again with seriesmover. It takes quite a lot of time to get the bK into action. A very Köko specific ideal mate (Kg1 Bd1 - Ka7 Ge1 Gg3) is attained after the complicated round trip of bG.

ser-h#33 (2+3)
0+2 grasshopper

Vlaicu Crisan
dedicated to Ion Murarasu
4268 Phénix 117 - April 2003

a) 1.Kxc3(nPAb8) nVAxb8(nPAb3) 2.LExb3(nPAe5) nVAxe5(nPAc8)#

b) 1.Kxf4(nVAc8) nPAxc8(nVAg5) 2.LExg5(nVAc4) nPAxc4(nVAb8)#

Compare this to Vlaicu's excellent 4th Prize 14th TT CCM... Another excellent diagonal-orthogonal correspondence of many-many elements. All moves are captures and all rebirths are analogous - they lead to double check model mates.

h#2 (1+6+2)
Super Circe
2 neutral units
leo e3, pao c3, vao f4
b) c4 -» e5

Bernard Rothmann
4269 Phénix 117 - April 2003

1...Bd2 2.Qd5 Bb4#

1...Ogc1 2.Be2 Oxd4#

1...Ogxg7 2.Qh5 Oxd4#

1...Oxg8+ 2.Sbc6 Oxc6#

1...Ohg6 2.Bf1 Oxe6#

Our 15th TT CCM was dedicated to helpmates with exactly 5 phases. It is hard to find any common link among all the solutions here - in fact I failed to do so. Fortunately the entries to our 15th TT were better.

h#1,5 (11+15)
5+4 orphan

Michael Grushko
4276 Phénix 117 - April 2003

1...Ka7 2.Kxg2(Gg8) Gc4 3.Gb3 Gf3+ 4.Kh2 Gg2 5.Gxg2(Gg8) Gxh1 6.Kxh1(Gh8) Ga2=

Asymmetry in this position is reached by the determination of Circe squares on the board. Note that Gh8 is not active in the final position.

h=5,5 (4+4)
3+3 grasshopper

Comments to Juraj Lörinc.
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