Fairy chess composition

# No.511 (EH)

 No.511 Eric Huber (Romania) Warm welcome to Eric in Original Problems section of JF! Original Problems, Julia’s Fairies – 2014 (I): January – April   ?Previous ; ?Next ; ?List 2014(I) Please send your original fairy problems to: julia@juliasfairies.com

No.511 by Eric Huber – A nice dedication and complicated to solve three-man in typical Eric’s style! Not long ago I welcomed Eric as a judge of JF-2013/III, but this time I’m happy to welcome his first publication in Original Problems section of JF! (JV)

Definitions:

Maximummer (Black Maximummer): Black must play the geometrically longest move or may choose from among longest moves of equal length, distances being measured from the center of each square. Diagonal and oblique distances are measured from the orthogonal coordinates by using Pythagora’s theorem (take the square root of the sum of the squares of the orthogonal distances). All other orthodox chess rules apply.

Anti-Circe =Anti-Circe Calvet (the default type): After a capture the capturing piece (Ks included) must immediately be removed to its game array square (necessarily vacant, else the capture is illegal). Captures on the rebirth square are allowed. Game array squares are determined as in Circe.

Sentinels Pion Advers: When a piece (not a Pawn) moves, a Pawn of the colour of the opposite side appears on the vacated square if it is not on the first or the last rank, and if there are less than 8 Pawns of that colour on the board.

 No.511 Eric Huber Romania original – 10.03.2014 Dedicated to Peter Harris   hs#4,5                                           (1+2)b) bKc7->b6; c) bKc7->d6MaximummerAnti-CirceSentinels Pion Advers     Hint for solvers: (click to show/hide)   Solutions are not homogeneous.   Solutions: (click to show/hide)     a) 1…Kc7-b6[+wPc7] 2.c7-c8=Q Kb6-a5[+wPb6] 3.Qc8-e6 Ka5-a6[+wPa5] 4.Ka3-b4[+bPa3] a3-a2 5.Qe6-e1[+bPe6] Ka6-b5[+wPa6] # b) 1…Kb6-a5[+wPb6] 2.b6-b7 b3-b2 3.b7-b8=R b2-b1=R 4.Rb8-b3 Rb1-h1 5.Rb3-b5[+bPb3] + Rh1-a1# c) 1…Kd6-e5[+wPd6] 2.Ka3-a4[+bPa3] Ke5xd6[bKd6->e8][+wPe5] 3.e5-e6 a3-a2 4.Ka4-a3[+bPa4] a2-a1=S 5.e6-e7 Sa1-c2 # (C+ by Popeye 4.65/4.63/4.61/4.59 and WinChloe 3.24) The problem is dedicated to Peter because his amazing 491 brought me to compose it. The only common points between the two problems are the 3 fairy conditions Maximummer, AntiCirce and Sentinels Pion Advers. All mates are specific. The Maximummer condition forces Black to play the mating move. The Sentinels condition prevents 6.Kc3? in a) or 6.Kb2? in b) and c). The AntiCirce condition prevents the capture of b3/a4 pawns by the white King (wK rebirth would be illegal). Observations: In a) one can notice the “try” 3.Qc8-e8?/Qc8-c3? … 5.Qe1 Kb5[+wPa6]+ 6.Qe8! Leaving a black Sentinel on e6 to close the e-file is mandatory. In b) bPb3 leaves square b3, which is why the white Rook must leave another black Sentinel on b3. 2…b3-b2 is a move with AntiZielElement. In c) the “try” 1…Kd6-e6[+wPd6]? 2.Ka3-a4[+bPa3] Ke6*d6[bKd6->e8][+wPe6] fails because White has no available tempo by the wK on the 3rd move and 3.e6-e7 would abandon the guard of d7/f7 squares. The 5th white move is a tempo buy generic levitra from india online move. (Author)

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Kjell Widlert
March 11, 2014 14:17

I like the way Eric explains all the subtle points of the solutions, instead of just giving them “straight from the computer” – or almost all of them, for he doesn’t mention the clever move 5.Rb5! which stops 5. – Rh8.

March 11, 2014 20:50

In fact we have two different ways of considering the same events:

From Kjell’s point of view, the Maximummer condition offers Black two candidate moves 5…Ra1 and 5…Rh8. White forces one of them thanks to 5.Rb5.

From my point of view 5.Rb5 gives check and for this reason Black only has three legal moves 5… Ra1, 5…Kxb5[bKe8][+wPa5] and 5…Ka6[+wPa5]. The Maximummer condition reduces this choice to only one move, which I did mention in my comments (“The Maximummer condition forces Black to play the mating move.”)

I will stick to mine, because in Maximummer Black must play the longest of the legal moves. However in the present situation it doesn’t really matter.

Kjell Widlert
March 12, 2014 01:02

Eric states my point of view very clearly and correctly. So according to me, the move 5.Rb5+ is a smart way of forcing Ra1# rather than Rh8 (by an AntiCirce effect). According to the composer, Rb5+ is just a check that Black must parry (with his longest legal move). So in this case, I think more highly of that effect than the composer does! That is unusual…

Nikola Predrag
March 12, 2014 02:51

Any check to bKa5 by wR could be parried only by bRh1-a1, so 5.Rb5 is just one of potential checks (after 3.b8R and some random 4th move by wR). But 5…Ra1 is mate only if b3 is blocked and 6.wRh8 not possible.

The smart point is 4.Rb8-b3! (to create the block of b3 after 5th white move), which leaves only one check (on b5) possible, (avoiding dualistic checks). ZugZwang with closing h-file fails in principle, 5.Rh3? Ra1+! 6.Rh8.

The smart point is NOT how to provoke CHECK 5…Rh1-a1+, but to arrange it to be a CHECKMATE. 5.Rb5+ comes when everything is already smartly arranged.
Of course, the whole smart plan includes 5.Rb5+, so it is also a smart move 🙂

Kjell Widlert
March 12, 2014 10:46

Agreed 🙂

Nikola Predrag
March 11, 2014 20:18

5.Rb5+ is check, by far too strong effect to be considered as “stopping one particular move”.
5.Rh3(+bPb3) would STOP 5…Rh8 but since it fails for other reasons, White FORCES 5…Ra1#.

AntiZielElement is not about some move having ANY useful/harmful effects (a pretty usual feature in many problems). The effects of some move should be OPPOSITELY related to the SAME particular goal (Ziel) of that move.

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