No.712, 713 
Stephen Emmerson (England)

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Two White and black retracts by Stephen Emmerson:

  • No.712 – White and Black retract 1 move, then helpmate in one. Author has called this problem “a twin and cousin for the famous Sunyer position” (wKh5,bKe8 in Dr. Julio Sunyer‘s position);
  • No.713 – White and Black retract 1 move, then helpmate in one, with Circe condition. (JV).

Definition:

Retractor: In Retractors, White and Black retract (take back) a given number of moves (-n moves), in order to reach a position in which a forward stipulation is met.

Circe: Captured units (not Ks) reappear on their game-array squares, of the same colour in the case of pieces, on the file of capture in the case of pawns, and on the promotion square of the file of capture in the case of fairy pieces. If the rebirth square is occupied the capture is normal.


No.712 Stephen Emmerson
England

original – 05.02.2015

Solutions: (click to show/hide)

white   kh6 black ke8

-1wb & h#1                               (1+1)
b) Kh6?g5    -1wb & h=1


No.713 Stephen Emmerson
England

original – 05.02.2015

Solution: (click to show/hide)

white   kc2 black ka8

-1wb & h#1                              (1+1)
Circe


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Juraj Lörinc
February 5, 2015 23:34

See page 138 in the bulletin, then also pages 54 and 60. Closely related to both problems, really. 🙂

Stephen Emmerson
Stephen Emmerson
February 6, 2015 00:27
Reply to  Juraj Lörinc

Your -1 & H&1 AntiCirce on p.60 – a nice problem! Only 8HM? Who was the judge?!

Juraj Lörinc
February 6, 2015 01:01

I do not remember. Strangely, my WinChloe database says S. Emmerson, but that can’t be true, can it? The TT was dedicated to problems with two pieces on the board and it was in fact very strong tourney for times 18 years ago and overall popularity of fairy chess (that was lower than today).

Juraj Lörinc
February 6, 2015 01:05

Actually, it seems it could be S. Emmerson as my page with 5 placed problem from the tourney says the same
… so you were joking, weren’t you? 🙂
Interestingly, at that time I was slightly disappointed by place of this two-king retro too. 🙂

Stephen Emmerson
Stephen Emmerson
February 6, 2015 10:26
Reply to  Juraj Lörinc

It was my first formal tourney, I think. I wonder now how I made any sense of such a wide variety of problems. The award was very long. I was also embarrassed to find I’d given 1 Pr to VC’s fairy editor – I’d convinced myself it was by someone else!
Of, course, had my #713 been in the tourney, I’d have been embarrassed to find I’d given it 7HM!! 🙂 (Yes, definitely joking! Unfortunately there’s no Circe in my forward play unlike your problem!)

Kjell Widlert
Kjell Widlert
February 6, 2015 01:18
Reply to  Juraj Lörinc

712 a-b: Sunyer’s problem is quite famous. Fascinating that there are two positions so close to it, with completely different solutions, even forming an “all-uncapture”!

(I know, it is not really “all”, as pawns can be uncaptured too, but the similarity with AUW is obvious. And I also know that there are two changes from a to b: position and stipulation.)

Stephen Emmerson
Stephen Emmerson
February 6, 2015 10:48
Reply to  Kjell Widlert

Yes, interesting that (a) involves the same uncaptures as Sunyer too.

Incidentally, there is an alternative setting for (b) where it is shifted so the Ks stand at e6 & h4. -1(w,b) & H=1, two solutions.

B/s as before: -1.Kf5xS -Kh5xB, 1.Sg5 Bxg5=
Also Q/p: -1.Kf5xP f7xQ, 1.Kh5 Qf6=.

I don’t think there’s any way of getting all these results together for a full uncapture set without ZP.

Cornel Pacurar
February 8, 2015 21:31

Stephen, the best you can have is probably:

a) Ke6 / Kb8
-1(w,b) & h#1
-1.Kd7xRe6 Ra6xQe6 & 1.Ra6-a8 Qe6-b6#

b) Kb8 -> Kh4
-1(w,b) & h=1
-1.Kf5xPe6 f7xQe6 & 1.Kh4-h5 Qe6-f6=
-1.Kf5xSe6 Kh5xBh4 & 1.Se6-g5 Bh4xg5=

adrian
adrian
March 27, 2015 09:09
Reply to  Cornel Pacurar

a) ain’t good, fully anticipated (rotated position) by PDB/P0002479 – Mrs. W. J. Baird, 1877. “Morning Post” 02/1910, Kf5 / Kh2 -1(w,b) & h#1…

adrian
adrian
March 27, 2015 09:41
Reply to  adrian

Which PDB/P0002479 in effect anticipates Stephen’s 712 part a) just as well…

Nikola Predrag
Nikola Predrag
February 6, 2015 18:00

No one mentions it so I’m obviously missing something.
What makes a difference between 1…Ra2xQ and 1…Rc7xQ, in No.713?

Kjell Widlert
Kjell Widlert
February 9, 2015 00:50
Reply to  Nikola Predrag

If you are missing something, then so am I: this must be a fatal dual!

Stephen Emmerson
Stephen Emmerson
February 9, 2015 10:00
Reply to  Nikola Predrag

I have to agree. What an annoyingly simple thing to miss.
No 7 HM for me, then.

S.N. Ravi Shankar
S.N. Ravi Shankar
February 24, 2015 19:15

713 is anticipated by my 4835, Die Schwalbe Aug, 1984,
Kb2/Kg8 : Circe : -2wb & h#1
-1. Kc1 Kh8 -2. Kd1xRc1 Rc8xQc1 and 1. Rg8 Qh6.
Which in turn was anticipated by another composer!
But I could not trace the name and source.