No.1465, 1466 
Cornel Pacurar
(Canada)

Original Fairy problems
JF-2019/II:
July – December’2019


Definitions: (click to show/hide)


No.1465 Cornel Pacurar
Canada

original – 28.12.2019

Solutions: (click to show/hide)

White Ka8 Sg3 Pb1 Black Qg6 Pd4 Pb3 Pc3 Pe3 Pf2 Kg1

phser-=11       b) Pc3→c4           (3+7)
Circe Power Transfer


No.1466 Cornel Pacurar
Canada

dedicated to François Labelle
original – 28.12.2019

Solutions: (click to show/hide)

White Qe8 Ka4 Black Kc7 Pd4

ahser-s#7       b) Kc7→b7           (2+2)
Circe Power Transfer
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Vlaicu Crisan
Vlaicu Crisan
8 months ago

In phser the side to move is White. After the first n-1 moves, during the last move White fulfils the stipulation: stalemate in 1 (in 1465, which should be phser-=11), respectively selfmate in 1 (in 1466, which should be phser-s#7).
If during the series the white King is checked, due to parry, Black is obliged to avoid the check. The ph means that Black cooperates with White when removing the check.
Please play the solution of 1466. The white Queen occupies the rebirth square of the black King, so due to Circe Power Transfer the black King moves like a Queen. During the two phases the white King finds itself checked several times. In a), first two checks are given by the black King, then by the black Pawn. After the black Pawn promotes on white King’s rebirth square, the white King moves like a Bishop.
Something similar happens in the twin: the white King is checked three times by the black King, then two times by the black Pawn. After the black Pawn promotes, the white King moves like a Knight.
In conclusion: we have a superb Wenigsteiner, convincingly exploiting all the involved fairy elements. Hats off, Cornel!

François Labelle
François Labelle
8 months ago

In 1466, the stipulation ahser-s#7 is correct. What’s missing is a definition of Anti-parry series-mover (ahser-). Also, ideally, the auto-checks in the solution should be notated “!+” and not “+!”.

dupont
dupont
8 months ago

For those who are not familiar with anti-parry series, here is an explanation of 1466 a):

1.Ka5!+ is an auto-check as black is able to play Kc7xKa5 in Circe power transfert rex inclusive. Black is undoing this auto-check with 1…Kc2 (note than black is helping white to reach the goal – a normal s#1 after a 6-moves long anti-parry series). The second move 2.Ka4!+ Kf2 goes the same way than the first one.

3.Qd7!+ is an auto-check too as black is now able to play d4xKa4 in Circe power transfert. Black is undoing this auto-check with 3…b4. The fourth move 4.Qb7!+ e1=B goes the same way than the third one.

Now the wK is playing like a bishop and the next 2 moves 5.Kc6 6.Kh1 are preparing the goal: 7.Qf3+ It is obviously a check, and the only way for black to parry it is to play 7…Kg1# (for example KxQf3 is illegal as Kh1 is playing like a bishop). It is not difficult to see that this last move is legal and is a checkmate.

dupont
dupont
8 months ago

François said: ideally, the auto-checks in the solution should be notated “!+” and not “+!”.

The notation !+ is certainly better than +! but is still a bit ambiguous, as it might be condidered as a good (!) checking move (+).

“Ideally”, I would prefer a single-sign new notation for an auto-check, maybe a circle around the plus (\oplus in TeX).

Kjell Widlert
Kjell Widlert
8 months ago

Quite a good wenigsteiner, once you get used to the unusual kind of play, with analogous strategy to bring the bP to e1 so it can transform the wK into something the bK can mate.