No.557, 558 Chris Feather (England) Original Problems, Julia’s Fairies – 2014 (II): May – August   →Previous ; →Next ; →List 2014(II) Please send your original fairy problems to: julia@juliasfairies.com

Two problems by Chris Feather are nicely presented by the author: “They are both rather heavy (3 units :)) and both use Couscous Circe and a single neutral pawn, with an imitator (NB: not a piece, but a condition!)

No.557 – Imitator in a long and interesting series-helpmate;
No.558 – Surprising neutral promotions motivated by Imitator! (JV)

Definitions:

Imitator (I): All moves must be exactly imitated in length and direction by the I, else they are illegal. The I may be blocked by the board edge or by a unit of any colour. However it is not blocked by the moving piece. Thus with If3 and any unit on d1 on an otherwise empty board, a Be4 may play all its usual moves except to b1, c2, g2, h1 and h7 – these moves being blocked by the unit on d1 or by the board edge. Note that Be4-f3(Ig2) is perfectly legal. The imitator is a condition which, confusingly, looks like a piece. However it may not move of its own accord, it may not be captured and its presence does not allow pawns to be promoted to imitators (at least never in my compositions). Thus it has no real piece-like characteristics. (from Fairings No.38, by Chris Feather)

Couscous Circe: As Circe, but the captured piece reappears on the Circe rebirth square of the capturing unit. Pawns reappearing on promotion squares are promoted instantly, at the choice of their own side.

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.557 Chris Feather England original – 16.06.2014   ser-h#29                                 (1+1+1) Circe CouscousImitator e7     Solutions: (click to show/hide)   1.nPb7-b5[Ie5] 2.nPb5-b4[Ie4] 3.nPb4-b3[Ie3] 4.nPb3-b2[Ie2] 5.nPb2-b1[ Ie1]=nB 6.nBb1-e4[Ih4] 7.nBe4-a8[Id8] 8.Kb8xa8[Ic8][+nBe8] 9.Ka8-b7[Id7] 10.nBe8-b5[Ia4] 11.Kb7-b6[Ia3] 12.Kb6xb5[Ia2][+nBe8] 13.Kb5-c4[Ib1] 14.Kc4-d4[Ic1] 15.Kd4-e4[Id1] 16.Ke4-f5[Ie2] 17.Kf5-f6[Ie3] 18.nBe8-g6[Ig1] 19.Kf6xg6[Ih1][+nBe8] 20.Kg6-g7[Ih2] 21.Kg7-f8[Ig3] 22.nBe8-f7[Ih2] 23.Kf8xf7[Ih1][+nBe8] 24.Kf7-f8[Ih2] 25.nBe8-d7[Ig1] 26.nBd7-c8[If2] 27.Kf8-e8[Ie2] 28.Ke8-d8[Id2] 29.nBc8-d7[Ie1] Kd6xd7[Ie2][+nBe1] # (C+ by Popeye 4.67) No.558 Chris Feather England original – 16.06.2014   ser-h#8                                  (1+1+1) Circe Couscousb) shift b1→a1Imitator g2     Solutions: (click to show/hide)   a) 1.nPe2-e1[Ig1]=nR 2.nRe1-d1[If1] 3.nRd1-d6[If6] 4.Ke8-e7[If5] 5.Ke7xd6 [Ie4][+nRe8] 6.nRe8-f8[If4] 7.Kd6-e7[Ig5] 8.Ke7xf8[Ih6][+nRe8] Kg6-g7[Ih7] # b) 1.nPd2-d1[If1]=nQ 2.nQd1-c2[Ie2] 3.Kd8-d7[Ie1] 4.nQc2-c8[Ie7] 5.Kd7xc8[Id8][+nQe8] 6.nQe8-e2[Id2] 7.Kc8-d7[Ie1] 8.Kd7-e8[If2] Kf6-e7[Ie3] # (C+ by Popeye 4.67)

Subscribe
Notify of
Inline Feedbacks
Geoff Foster
June 17, 2014 07:53

In No.557, Black wants to reach a position with bK on d8, a neutral piece on d7, and the Imitator on e1, because after White plays Kxd7 the Imitator on e1 will go to e2 and the captured neutral piece will be reborn on e1 (the wK’s home square). Black will then be unable to capture the wKd7, because the piece on e1 will block the Imitator on e2. The neutral piece must be a nB or nS, because a nQ or nR on d7 would self-check the bKd8. It turns out that a nS won’t work, so the neutral piece must be a nB.

The bK/nB can’t go to d8/d7 without capturing, because the Imitator would finish on i7 (which is not even on the board!). Therefore the bK must capture the nB in such a way so that the Imitator finishes 4 squares to the left and 6 squares lower down the board (-4,-6). Each time the nB is captured by the bK it is reborn on e8 (the bK’s home square), so captures on a8 (-4,0), b5 (-3,-3), g6 (2,-2) and f7 (1,-1) move the Imitator the required (-4,-6) squares.

Some of the move ordering is cleverly determined by the Imitator (e.g. moves 9, 10 and 11).

Kjell Widlert
June 18, 2014 00:25
Reply to  Geoff Foster

Thanks for the analysis, Geoff! I thought there must be some mathematics behind the long sequence, and now I know.

Kjell Widlert
June 18, 2014 00:12

Could we perhaps agree that the Imitator actually has some characteristics of a piece and some characteristics of a condition – and it’s not necessary to decide it’s just one or the other.
This is like the electron: it’s like a wave and like a particle at the same time! Confusing, but we can live with it.

Geoff Foster
June 18, 2014 02:32
Reply to  Kjell Widlert

Now for No.558. In the mate in (a), the Ih7 stops the bK and nR from moving to the right, which is why the bK can’t capture the wKg7 and the nRe8 does not attack the bKf8. The bK can’t escape to e7 because it would then be attacked by the nR.

In (b) the position is moved to the left, so the previous solution will not work, because the Imitator will not finish on the edge of the board. But why doesn’t the (b) solution work in (a)? The answer is that Black could play 9.nQf2-a7[Ia8], where the Imitator finishes on the top rank and the wKf7 can’t capture the bKf8!

shankar ram
June 18, 2014 08:52
Reply to  Kjell Widlert

Well.. the discussion on whether the imitator is a piece or a condition is an old one..
But I’ve never heard it compared to an electron’s wave/particle duality before..!
There are some amazing babson tasks composed with it by Rene Millour..

Nikola Predrag
June 18, 2014 18:49

If there were any reasons for an “old discussion”, they are not mentioned here. The I works as piece, even if it is not visible on a diagram. Any piece probably could be defined as a condition, what makes the I so special?