No.638 
Oliver Sick (Germany)

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Original Problems, Julia’s Fairies – 2014 (III): September – December

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No.638 by Oliver Sick – What happens if there’re more than one Imitator on the board? 

For Imitator I’ve used the definition from Fairings No.38, by Chris Feather. But it doesn’t explain what happens in case of more than one Imitator. Author’s additional explanation to his Imitators is: “A set of (one or more) imitators moves as a fixed structure, i.e. all at the same time. So the moves Ia5->b5 + Ib5->c5 is the same as Ia5,Ib5 –> Ib5,Ic5.”  I have to add, that this problem is correct by Popeye 4.69, has cooks by Popeye 4.65 and older, and has no solutions by WinChloe. So, I guess there’re many different interpretations. I’d appreciate your comments! (JV)


Definitions:

Sentinels Pion neutre: When a piece (Pawn excluded) leaves a square outside the first and last rows, a white piece leaves a wP, a black piece leaves a bP and a neutral piece leaves a nP unless 8 Pawns of that color are already on the board.

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)


No.638 Oliver Sick
Germany

original – 01.11.2014

Solutions: (click to show/hide)

neutral Ka6 Ia5 Ib5

ser-r#9      b) Ka6→b6    (0+0+1n)
Imitator a5, b5
Sentinels Pion Neutre


 

 

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Nikola Predrag
Nikola Predrag
November 2, 2014 23:12

I obviously didn’t get the rules, what about the flight to f4?

About the Imitator, it is probably not a piece if it does not occupy any square (can be “passed through”) and if it can pass through an occupied square.

oliversick
November 3, 2014 00:01
Reply to  Nikola Predrag

Hello Nikola,

The king cannot move to f4 (as a white move) because the evolving nBg5 gives check (as black) and could capture the nK (ok, I know neutral kings are kind of weird….)

An Imitator is a border case. In one way it can be seen as a kind of sensor, checking if the squares are occupied and it cannot move autonomous, so definitely not a piece… On the other hand side there are problems with Imitator promotion, so in this case it is a piece ?!?

As I said, imitators are a border case….

Oliver

Kenneth Solja
Kenneth Solja
November 3, 2014 09:06

King can’t move to f4 because then there would 9 pawns on the board, but I don’t understand the first solution how that is a mate. c4 is not possible, because then pawn in b5 would check in white’s move, but capture for ex. to b6 ??

Kenneth Solja
Kenneth Solja
November 3, 2014 09:07
Reply to  Kenneth Solja

Silly me of course the imitators 🙂 prevent that

Georgy Evseev
Georgy Evseev
November 3, 2014 11:54

It still looks like f4 is a flight, as there should be no 9th pion. Or am I wrong?

Nikola Predrag
Nikola Predrag
November 3, 2014 13:28

The rule is ambiguously formulated. In b):
I. 10.nKg5-f4 is legal because nK would not leave nPg5?
II. 10.nKg5-f4 is illegal because nK can not leave nPg5?

dupont
dupont
November 3, 2014 14:44

Dear Julia, this is a “rule-examination” period in JF for now! After the ep-capture subtleties and the not-so-clear distinction between leapers and riders, well-come to Imitators!

Maybe you should open a kind of sub-forum “discussion about fairy chess rules”?!

dupont
dupont
November 3, 2014 16:46
Reply to  Julia

To me, it is of course ok to discuss the condition in the same thread than the related published problem. I just hope that it doesn’t bother nor you neither the author… For example 68 comments in Sébastien’s 627 and it is not over!

Nikola Predrag
Nikola Predrag
November 3, 2014 16:32

Dear Julia, I wrote “IF it can pass…”, so “IF it can not pass…” might suggest that it is a piece, after all 🙂

dupont
dupont
November 3, 2014 19:10

I checked solution a) with WinChloe – every move is legal for this program, but doesn’t lead to a checkmate because of the “defense” nKxa5(Ib5c5). It means that priority is given to Imitators rather than to Sentinels.

But according to Christian Poisson, the mix “Sentinel + Imitator” has not been specifically programmed, so WinChloe is mainly working by itself and is wrong on that particular point.

dupont
dupont
November 3, 2014 19:34

Concerning solution b) WinChloe is considering that Rnf4{Id4,Ié4} is a defense against the check. It means that, once there are 8 pawns on the board, no more sentinel may appear, and thus this condition vanishes. As Popeye is claiming that solution b) is valid, it means that more than 8 pawns are allowed on the board for this program.

seetharaman
seetharaman
November 4, 2014 18:50

The imitators do not actually permit move of pieces through it. What happens is with Be4 and If3, the bishop can move to f3 and g2 because the imitator simultaneously moves to g2 and f1! But the Bf4 cannot move to h1 because the imitator is stopped from the edge of the board. Since it does this sort of travel with every piece on board, it is in effect a condition even though it has a physical “presence” on the board in the form of a symbol. So Chris Feather is correct – Imitator is restrictive condition.

seetharaman
seetharaman
November 4, 2014 18:56
Reply to  seetharaman

Sorry for the typos: Read ‘h1’ for ‘f1’ and ‘Be4’ for Bf4.

dupont
dupont
November 4, 2014 19:22

Imitators may be crossed by pieces, at least both for Popeye and WinChloe:

Stipulation ~1
Pieces white Pa2
cond Imitator a3

1.a2-a4[Ia5] !
1.a2-a3[Ia4] !

Strangely enough, the definition reported by Julia doesn’t fit this possibility – for Chris Be4-g2 is illegal with If3, though this move is legal both for WinChloe and Popeye:

1.Be4-g2[Ih1] !

seetharaman
seetharaman
November 4, 2014 19:59
Reply to  dupont

I visualise Imitators as being attached to any piece that move on the board (like a conjoint twin), so the question of crossing/overtaking them will never arise. ( I too dont understand why Chris Feather rules out the move Bg2 (with Be4 & I-f3)

dupont
dupont
November 4, 2014 21:01

There is another strange fact:

Stip ~1
Pieces white Ba2 black Bc4 Pb5d5
cond circe Take&Make
cond imitator b1

1.Ba2*c4[Id3][+bBe2] !
1.Ba2*c4[Id3][+bBb3] !
1.Ba2*c4[Id3][+bBa2] !
1.Ba2-b3[Ic2] !

Why is the move 1.Ba2*c4[Id3][+bBf1] considered illegal, although the other move where the Bishop is crossing the Imitator – 1.Ba2*c4[Id3][+bBe2], is allowed?

Geoff Foster
Geoff Foster
November 4, 2014 23:01
Reply to  dupont

The move 1.Ba2*c4[Id3][+bBe2] is interesting, because if the Imitator is visualised as being attached to the moving piece then it would arrive on d3 with the capture, thus blocking the “make” to e2! In this move I think the Imitator should finish on f1. The move 1.Ba2*c4[Id3][+bBf1] is considered illegal because the “make” part is not a legal move (with Bc4 and Id3, Bc4-f1 is not legal).

Nikola Predrag
Nikola Predrag
November 4, 2014 23:44

The definition from Fairings is written carelessly and shouldn’t be taken seriously. Reprinting the errors is a terrifying custom of modern media.

The question whether Imitator is a piece requires in the first place a general definition of what is a piece.
I would say that a piece occupies a square, if it may not be captured it’s a “colorless” or “immune” piece. If it may not move on its own accord it’s a “stone-piece” or dummy.
If it may be “moved” all over the board only by imitating a “real move” of some other piece it’s an Imitating-piece.

Imitator additionally gets the “characteristics” of a particular piece which starts to move, but remains colorless.
So, when wB is about to move, the Imitator is “awaken” transiently as a “(colorless) Bishop-Imitator”.
At the same time the piece which starts to move, additionally gets the characteristics of the Imitator, but retains own color.
Thus, wB transiently becomes a “white Bishop-Imitator”.
Then wBI+BI behave as one combo-piece, there’s no independent movement of either part of that combo-piece.

Actually, when an Imitator is on the board, no piece ever “may move of its own accord”, at least no more than the Imitator.

That is just one speculation, the question is whether it is consistent.
The given “official” definition doesn’t look consistent.

As I remember, there was severe insisting on the acceptance of Make-part in T&M as a non-capturing MOVE.
So the Imitator should imitate that part too.
As well as imitating the Make-part of the captured piece in Anti-T&M.

dupont
dupont
November 5, 2014 01:28

Geoff,

In your interesting viewpoint, you are considering that when a move is 2-fold (e.g. in T&M), the Imitator should also move twice. Popeye agrees but not WinChloe:

W: Pa2
B: Pb3
Imitator c1 T&M

WinChloe’s capture is 1.a×b3(b2){Id2} while Popeye’s capture is 1.a2*b3-b2[Id1].

If we put an obstacle for the Imitator in the first part of the T&M move (a Pawn in d2), the above move is still valid, which shows that Popeye doesn’t consider that the Imitator is playing twice, as the moving piece which he imitates!

In anti-T&M the problem is different, as the second part of the move is played by another unit, and I’m not sure the Imitator should also imitate this new guy! And indeed, always following the above setting but in circe T&M, Popeye’s move is 1.a2*b3[Id2][+bPb2]

Note that in T&M the Imitator’s move occurs at the end of the notation, while it occurs at the middle in circe T&M. I think this is correct.

dupont
dupont
November 5, 2014 01:59

Also interesting is to look at the same scheme (wPa2 bPb3 Ic1) in Anticirce. Popeye’s way of capturing is:

1.a2*b3[Id2][wPb3->b2]

To my mind this is fully correct as wPb3->b2, although a transfer of the moving unit Pa2, is not a move! So it is logical that the Imitator is not going from c1 to d1, despite the fact that the imitated Pawn is going from a2 to b2…

Geoff Foster
Geoff Foster
November 5, 2014 02:42

Sorry, I didn’t realise that Nicolas’s first example used anti-Take&Make. I thought it used Take&Make, in which case Popeye behaves as I expected:

1.Ba2*c4-a2[Ib1]
1.Ba2*c4-b3[Ic2]
1.Ba2*c4-e2[If1]
1.Ba2*c4-d3[Ie2]
1.Ba2-b3[Ic2]

I agree that for anti-Take&Make the captured piece’s “make” should not cause the Imitator to move.

dupont
dupont
November 6, 2014 14:26

It is also interesting to see how is working the Imitator inside an ep-sequence:

W: Pa5
B: Pb7
Imitator b8

The ep-sequence 1.b7-b5[Ib6] a5*b6 ep.[Ic7] is legal for Popeye, and the same 1.b5{Ib6} a×b6 e.p.{Iç7} is legal too for WinChloe. It seems to be a convincing logic.

It implies that if we keep as a fundamental rule that 2 pieces can’t stand at the same time on a given square (except for a few very special conditions, where it is explicitly allowed), then the Imitator is definitely not a piece.

Nikola Predrag
Nikola Predrag
November 7, 2014 05:27

Nicolas, the convincing logic is:
1.b7-b5[Ib6] a5*b6 ep.[Ic8]
or more precisely:
1.b7-b5/b6[Ib6/b7] a5*b6 ep.[Ic8]
since ep.capture determines that bP was occupying b6 and Imitator was occupying b7 as though 1.b7-b6[Ib7] was played.

peter harris
peter harris
November 7, 2014 12:38

A FUNDAMENTAL point is:

An Imitator MOVES [in unison with a piece].

If this is born in mind at all times many things would be better understood. So for example, the question of whether a piece can or cannot pass through simply cannot arise.

To make it clear that an Imitator is a Condition [which was the intention from the start] and not a Piece, a Definition could be as follows:

DEFINITION.
An Imitator is [only] a geometric location. The location changes in an identical manner to that of a piece being moved. The location cannot be off the board and the path of its change cannot be through, or end at, the location of any piece.

[Entertaining the eccentric notion that an Imitator is a piece brings in train a host of questions and difficulties. It introduces the possibility that an Imitator could for example, act as a hurdle or anchor point].

NOTE: In most cases it makes no difference at all whether an Imitator is deemed a piece or not.

Other than the absence of scotching the possibility of an interpretation that the Imitator can be a Piece, Chris Feather’s definition is 100%.

Laco Packa
Laco Packa
November 7, 2014 15:10
Reply to  peter harris

I would not like to start a similar comprehensive debate as in the case of e.p. However imitator offers similar doubts about its capabilities.
For example, a capture in Anticirce consists of imitators one or two moves? Sometimes it does not matter, but sometimes it is a question of possible moves.
bRb6, Ib5
Is the move Rxb1(R->a8, Ia7) possible?
a) Yes, it is one move. A new place for the imitator is defined by the vector of the original and ending square of moving piece.
b) No, they are two moves and the first of them is impossible due to imitators illegal move.

dupont
dupont
November 7, 2014 17:35
Reply to  Laco Packa

Your question has already been discussed a bit in my post Nov5 01:59. According to Popeye, your move is illegal (the Imitator is out the board) because the transfer Rb1->a8 is not a move, and hence should not have to be imitated.

Btw Popeye’s notation is fully clear on that point: Rxb3[Ib2][Rb3->a8], i.e. the imitation occurs after the move, not after the transfer.

dupont
dupont
November 7, 2014 17:54
Reply to  peter harris

For sure considering an Imitator as a piece would lead to great theoretical difficulties – for plenty of conditions the inventor would have to stipulate how Imitators behave within it. And the fact that a Pawn may promote to an Imitator is not enough to make it a piece ipso facto.

Geometric location is maybe a bit vague, why not defining an Imitator as simply coordinates? It would not bother me if a white Pawn reaching a8 would become an Imitator (a8)!

Finally, you said that Chris’s definition is ok. Could you please explain why Be4-g2[Ih1] is illegal, although Be4-f3[Ig2] is legal?

dupont
dupont
November 7, 2014 14:14

Nikola, your move 1.b7-b5/b6[Ib6/b7] a5*b6 ep.[Ic8] is logic according to your own point of view, which is that b7-b5 has not really been played when followed by ep-capture. This is for sure consistent and would be fully convincing if this reasoning were followed by Popeye and WinChloe, which is not the case.

My own point of view is that b7-b5 has really been played, even when followed by ep-capture. In this case the ep-capturing side is transferring b5 to b6, as a preliminary part of its move. This transfer notion is well-known e.g. in Anticircé – the final part of an Anticirce capture is a transfer of the capturing piece, i.e. is not a move.

As a consequence the Imitator needs not to imitate a transfer, and hence Popeye’s and WinChloe’s move 1.b7-b5[Ib6] a5*b6 ep.[Ic7] perfectly fits the above definition of the ep-capture via a transfer of the captured Pawn. Obviously I don’t claim that my logic is better than yours, only that it is more pragmatic.

Nikola Predrag
Nikola Predrag
November 8, 2014 01:29
Reply to  dupont

Nicolas,
standard rules are one thing and Popeye or Winchloe could more or less convincingly obey them but they can not “prove” them right or wrong.
Should the programmers proclaim or obey the standard rules?

It seems that you’ve mixed the Popeye’s interpretation of EP with your own one.
Popeye thinks that bP is captured on b5 after b7-b5, so Imitator may still occupy b6 and move to c7 as wPa5 moves to b6.

Julia,
I don’t know more about the Imitator than I’ve read here.
My speculations are just the questions that should be answered by those who know more about the original inventor’s intentions or common practice.

As far as I’m able to understand at present, Mr.Harris gave a short and consistent definition. It seems that it offers the clear guidelines for many practical cases.
However, some ambiguities are still possible and the discussion is desirable.

It doesn’t seem actually important whether Imitator is a piece or a condition. At least, as long as we don’t have the definitions of what is a piece and what a condition.

I’ve noticed that Popeye doesn’t combine Imitator with Madrasi, Isardam etc., perhaps because that piece/condition question is not clearly defined.

The notion that an Imitator is a piece, might look eccentric to a “sage one” who definitely knows the meaning of a “piece”.
The others should first reach the central point known to that “sage one”.

peter harris
peter harris
November 8, 2014 09:24

Nicolas:

You wrote:

“Finally, you said that Chris’s definition is ok. Could you please explain why Be4-g2[Ih1] is illegal, although Be4-f3[Ig2] is legal?”

Chris Feather’s definition is OK. I leave it to you to figure out why he said a move to g2 is illegal.

[I suggested in my last comment, that an Imitator if a piece, could be used as a hurdle. This is not so, because the movement of the Imitator would prevent the hurdling piece say an Equihopper, from reaching its intended destination].

Nikola Predrag
Nikola Predrag
November 8, 2014 10:28

Hm,
Be4-f3 is legal because f3 is not occupied by a piece and then the I imitates the move, moving from f3 “now occupied” by wB?
Be4-g2 isn’t legal because then the I can’t imitate the move through g2 “now occupied” by a piece?
Or something more obvious or more hidden reasoning?

If the Feather-definition is indeed correct, it is at least not very clear to many of us, including the Popeye’s programmers.

peter harris
peter harris
November 8, 2014 12:26

The Be4-g2[lh1] is not illlegal.
This is the explanation.

peter harris
peter harris
November 8, 2014 13:52

To the two Ns:

As I was eating my lunch I thought it may help to say the following regarding the inter-action between a Piece and an Imitator;

It should not be viewed / imagined that the piece moves and then the Imitator moves.

They both move together simultaneously as if joined together.

So in the example Chris Feather gave the Bishop and the Imitator move as if joined – the B to g2 and the I to h1 – and not that the B moves to g2 and then the I moves to h1 – which would require that the I has to pass through the Bg2 to get to h1.

[Chris did say that the I is not blocked by the moving piece – but it would not be necessary to say this if the simultaneous movement was appreciated].

So with Ba8 and Id5, the B could move to b7, c6, d5, e4 with I moving to d4, d3, g2 h1.

[So Chris erred in including g2 in his example].

dupont
dupont
November 8, 2014 14:59

Convincing, thanks. Now, what about Imitators in T&M?

White Ba1
Black Bc3 Pe2f2
Cond Imitator e1
Cond take&make

Popeye (4.68 & 4.69) gives as legal white moves:

1.Ba1*c3-a1[Ie1] !
1.Ba1*c3-d2[Ih2] !
1.Ba1*c3-d4[Ih4] !
1.Ba1*c3-a5[Ie5] !
1.Ba1*c3-b4[If4] !

I can’t understand these results. Why is the Imitator not blocked by Pawns e2 and f2?

peter harris
peter harris
November 8, 2014 16:14

Nicolas:

I have found that whilst programs deal perfectly with conditions SEPARATELY they go awry when the conditions are COMBINED.

This is surely the case here with T&M + Imitator.

It is to do with the intricacies of programming.

If the bug [for it is a bug you have illustrated] is shown to a programmer he will probably be able to fix it.

But do not let the bug make you doubt your understanding of the 2 conditions or the ability of a program to handle them [correctly] separately.

What would be of concern is if when testing a Condition on its own, a program does run according to your understanding of its purported Definition.

Nikola Predrag
Nikola Predrag
November 9, 2014 00:50

So, it seems that we are where we’ve started but actually, after the contradictory and incorrect opinions, I know less than before.
It’s hard to detect was serious here and what was not, starting with the definition.

“…For sure considering an Imitator as a piece would lead to great theoretical difficulties – for plenty of conditions the inventor would have to stipulate how Imitators behave within it….”
Any examples?
Or should I simply believe on the basis of all those “clear” examples?

I should better seriously consider a concept with the Imitator as the only chess piece, while all other units are just the geometric locations or simply coordinates.
Perhaps even in the orthodox chess, the only piece is the Imitator 🙂

dupont
dupont
November 9, 2014 01:18

Ia1 Pb3c2 UU. Is the null-move from the ubi-ubi allowed?

And now something specific to the question of whether or not an Imitator is a piece: Condition Sentinels Imitator. Is it allowed to get more than 8 Imitators on the board?

Nikola Predrag
Nikola Predrag
November 9, 2014 03:15

What is the point in the question “a piece or not a piece”, if we don’t know what makes something a piece?
And guessing about various combinations of rather exotic fairy elements, while the more common situations are still unclear?
What about Madrasi or Isardam? Or a “simple” Locust which is considered as quite an orthodox concept in some interpretations of the standard rules?
What does Winchloe say about a simple ep. capture when the Imitator is Included?