Imitator – definition, features, implementation

The discussion about Imitator has started in the comments to the problem No.638 – https://juliasfairies.com/problems/jf-2014-iii/no-638/#comments

The original source (Fairy Chess Review, April, 1939, 5-minute Paper 34) indicates another Dutchman as the inventor: Gerrit Jansen! The later publication referred to by Dickins (FCR Oct. 1954) doesn’t mention any inventor at all. So which is right, Kok or Jansen?

The original definition in 1939 is the following:
The Imitator (…) is a neutral man which imitates EVERY move like a shadow. In other words, when any White or Black man moves, the imitator moves simultaneously in a parallel direction for the same distance.
The imitator may not be captured and may never capture, so that any move of White or Black which would bring one or more imitators to occupied squares (or, of course, off the board) is inadmissible and may not be played. Incidentally this includes the fact that the imitator is a non-checking man.
It must be realised, also, that the ability of the imitator to prevent various moves makes it readily possible for the Kings to stand in what would normally be checks, because the capture of the King has ceased to be admissible.
(…) With the final remark that Pawns may promote to Imitators as well as normally, No. 3615 may be left to the reader.

End of quote. Judging from the language, I assume that T. R. Dawson himself was involved in the writing, although the paper is signed by Gerrit Jansen and A. Wijker.
At least two noteworthy points: the paper implicitly allows several imitators on the board, and it explicitly allows pawn promotions to imitators!

I should say that I’ve started to enjoy Imitator, its features with a different moves and conditions, implementation by Popeye and WinChloe.. and how much of it is based on the definition.

At first, I’ve tried to find more about the inventor’s definition, but all I have is:
I’ve got to know that on Chess Composers blog the information about Imitator’s inventor came from the Retro Corner – http://www.janko.at/Retros/Glossary/Imitator.htm (written by Philippe Schnoebelen) – where it is said:

The Imitator (invented by Th. Kok in 1939) is a fairy piece depicted as a big black spot, and denoted by “I”. It does not move by itself, but it imitates every move by normal pieces. Moves that cannot be imitated are illegal, so that the Imitator acts as a constraint on possible moves.
The Imitator does not belong to any side. It cannot be captured.

And one more interesting addition – I don’t know either it is a part of the definition, or it is based on the implementation is solving programs:

Castling is imitated by decomposing into a King move followed by a Rook move.

This addition made me think about the features Imitator can add to the castling move:

IMITATOR with Castling

(Imitator’s behavior is analogical in the last versions of the both programs – Popeye and WinChloe)

1. Castling short – the both moves are imitated in a sequence, the first is a King, the second is a Rook. But as a result Imitator returns on its initial square.
F.i., in the scheme below:
white ke1 qe5 rh1 | bla kf3 pg4 | imitator a4
after Kg1 Imitator goes to c4, but after Rf1 Imitator returns again on a4!

1.0-0{Ia4}‡!

Popeye (Imitator is a condition)

beg sti #1
pie whi ke1 qe5 rh1
bla kf3 pg4
con imitator a4 end

  #1                          3 + 2
             Imitator a4

   1.0-0[Ia4] # !

2. Castling long – again, the both moves are imitated in a sequence, the first is a King, the second is a Rook. But this time a place of return for Imitator is one square right from its initial square.
F.i., in the scheme (2) below:
white ke1 qe5 ra1 | bla kd3 pc4 | imitator g4
after Kc1 Imitator goes to e4, but after Rd1 Imitator returns to the right and comes on h4 (one square to the right from its initial position)!

1.0-0-0{Ih4}

Castling with Imitator gives some special features, like in the Scheme (3):

The only solution here is 1.0-0+! – where Imitator goes together with Kg1 (If1) and than with Rf1 (Id1) – returns to its initial position!
A check with 1. Rf1?? is impossible as together with this move Imitator goes to b1, and a Rook from f1 can’t give a check as a capture RxKe8 can’t be imitated as b8 is blocked!

————
Also, the en passant-discussion of the last weeks couldn’t leave me indifferent, so I’ve checked the

IMITATOR with En Passant
(Imitator’s behavior is analogical in the last versions of the both programs – Popeye and WinChloe)

Without an Imitator the mates would be:
1.b4! axb3 e.p.‡; 1.b3+! axb3‡.

Adding the Imitator in a Scheme (5) changes a situation cardinally! Now it is important, where should play the white Pawn – on b4 or b3. It is not very visible from the first glance why 1.b3+? Doesn’t work for (a) and 1.b4? doesn’t work for (b).

a) 1.b4{If6}! axb3 e.p.{Ig5}# (1.b3?(If5) axb3+(Ig4) but 2.Kb2(Ih4)!! and the capture cxb3 is impossible because the Imitator h4 is blocked and can’t come on g3!!)
b) 1.b3{If5}! axb3 {Ig4}# (1.b4(If6)? 2.axb3 e.p(Ig5)+ but 2.Sxc3(Ih7)!! and the capture bxa2 is impossible because the Imitator h7 is blocked and can’t come on g6!!)

————–

That’s all for today :). But further I’d like to discuss with you some combinations of Imitator with other conditions, especially differently implemented in Popeye and WinChloe.
And.. I’ve started to compose my first problem with Imitator!

Another non-trivial question is this:
When an imitator is imitating a move by a hopper (grasshopper, lion, etc) – does the imitator need a hurdle too? Or does the imitator move like a normal line-piece (a rider) even when imitating a hopper?

Some small tests indicate that Popeye 3.69 and WinChloe 3.30 agree that the imitator needs a hurdle too, at least as long as the hopper is a grasshopper.
When the hopper is a lion, Py doesn’t want to play at all (“Too much fairy for an imitator”), and WC still requires a hurdle – but it must be in the same relative position to the imitator as the other hurdle is relative to the lion! (So with wKd4 wLId1 – bKb1 – Imitator f4, LId1-a1(Ic4) is legal. But with wK moved to e4, LId1-a1 is no longer legal in WC, even though f4-c4 would still be a legal lion move.)
When the hopper is a leo, we have the funny situation that with wKd4 wLEd1 – bKb1 bSa1 – Imitator f4, White can play LEd1xa1(Ic4), so the imitator doesn’t need something to capture like the piece it imitates does … in fact, it is not even allowed to have something to capture. This is in WC; as expected, Py doesn’t want to play with the combination imitator + leo either.

An interesting fact – the name of the inventor!
About a week ago Thomas Maeder advised me to turn to Hans Gruber with a questions about Imitator‘s inventor and definition.
And this is what I’ve got today:

1). “it is true that Gerrit Jansen (a pseudonym used by Theodorus Kok) presented the Imitator in the Fairy Chess Review, which was at that time edited by Thomas Dawson.“
So, Gerrit Jansen and Theodorus Kok is just one and the same person!

2). Some more materials below are in German. Maybe German friends would comment or translate the main things:

Imitator [Gerrit Jansen (Pseudonym von Theodorus Kok), The Fairy Chess Review, Band 3, Nr. 17, IV/1939, S. 180‑181]

Der Imitator ist ein Schatten [Gerhard Wolfgang Jensch, Schachmatt 55, 16.11.1947], der die Bewegung jedes ziehenden Steins auf einer Parallelen in derselben Richtung mitmacht. Er kann selbst weder schlagen noch Schach bieten. Für den Imitator gelten die gleichen Zugregeln wie für den gerade ziehenden Stein: Die Transitfelder müssen (bei Langschrittlern) leer sein. Ein Zug, durch den der Imitator über besetzte Felder oder über den Brettrand hinaus bewegt würde, ist nicht zulässig. Kommen mehrere Imitatoren in einer Stellung vor, so imitieren alle in der beschriebenen Art.

Der Imitator wird in der Originaldefinition als neutraler Stein beschrieben, in den sich ein Bauer umwandeln darf.

Thur Row sagt, dass die Figur besser Companion genannt worden wäre, was aber nun nicht mehr ginge [Chess Ultimates, Band 3, Nr. 2, VII/1973, S. 81]. In Rows Version darf man in Stellungen mit Imitator nicht rochieren; eine Bauernumwandlung in einen Imitator ist erlaubt [Chess Ultimates, Band 1, Nr. 7, X/1970, S. 25].

Ein schwarzer Imitator [E. W. Bennett, 7669. The Fairy Chess Review, Band 6, Nr. 18, VI/1948, S. 135] imitiert nur schwarze Züge, ein weißer Imitator nur weiße.

Ein Lone Imitator [Thur Row, Chess Ultimates, Band 1, Nr. 8, XI/1970, S. 32, sowie Nr. 10, I/1971, S. 38 (Definition)] ist ein einsamer Imitator Rowscher Version: Außer ihm darf es keinen weiteren Imitator auf dem Brett geben; eine Bauernumwandlung in einen Lone Imitator ist nicht erlaubt.

As with S, Imitator normally works with N, and I see no problems for combining it with other riders (haven’t tried it in Popeye yet). For now I can’t imagine a piece which is impossible to imitate… But I can’t tell it doesn’t exist ?
If even some combinations are not implemented for now, it should be possible in the future.

The inventor of a concrete fairy element can’t know what kinds of new pieces and conditions will be invented in the future. So, I’m not sure if he can describe all possible combinations. On another hand, the inventor of the next element might define how it should be combined with the invented earlier. If it is not done, then I believe that some other person(s) might extend the definitions and behavior of a fairy element, but with some justification, basing on the author’s definitions.

About Imitator with Take&Make. A very interesting combination! For now we have two different implementations of it.
1) WinChloe: only the 1st part (Take-) of the move is imitated, like in the example:

2) Popeye: the both – Take and Make – moves are imitated, the same example:

beg sti +1
pie whi bb1 bla ke5 sc2
con imitator d1
cond take&make  end

+---a---b---c---d---e---f---g---h---+
|                                   |
8   .   .   .   .   .   .   .   .   8
|                                   |
7   .   .   .   .   .   .   .   .   7
|                                   |
6   .   .   .   .   .   .   .   .   6
|                                   |
5   .   .   .   .  -K   .   .   .   5
|                                   |
4   .   .   .   .   .   .   .   .   4
|                                   |
3   .   .   .   .   .   .   .   .   3
|                                   |
2   .   .  -S   .   .   .   .   .   2
|                                   |
1   .   B   .   I   .   .   .   .   1
|                                   |
+---a---b---c---d---e---f---g---h---+
  +1                          1 + 2
             Imitator d1
           Take&MakeChess

   1.Bb1*c2-a1[Ic1] + !
   1.Bb1*c2-d4[If4] + !

Combination with Take&Make is not defined by Imitator’s inventor. So, we can either consider that the combination is undefined, or to add it to the definition. I’d prefer it to be added, as I like Take&Make and I’d be interested in the usage of it with Imitator.
My personal view is that Popeye’s implementation is more correct, as Take&Make can be compared with a castling. If the imitation of castling uses its decomposition into two moves – King’s and Rook’s – then it looks logical to decompose Take&Make moves into “Take” and “Make” move and to imitate it in the same sequence. Other views?

Quote from Julia on November 20, 2014, 12:13

My personal view is that Popeye’s implementation is more correct […]

I agree, but nevertheless there is something which I think is debatable: add an obstacle on e2 to your position, Julia. Then 1.Bb1*c2-a1[Ic1] + is still considered legal. It means the Imitator is not imitating the trip b1->c2->a1 but the direct road b1->a1.

I’m curious to know if problemists agree with Popeye’s viewpoint here or consider it as a bug, like the one which was detected with Transmuted Kings. Indeed this is the same kind of “mistake”, the Imitator imitates moves as if there were played by leapers.

Quote from Julia on November 20, 2014, 12:13
Combination with Take&Make is not defined by Imitator’s inventor. So, we can either consider that the combination is undefined, or to add it to the definition. I’d prefer it to be added, as I like Take&Make and I’d be interested in the usage of it with Imitator.
My personal view is that Popeye’s implementation is more correct, as Take&Make can be compared with a castling. If the imitation of castling uses its decomposition into two moves – King’s and Rook’s – then it looks logical to decompose Take&Make moves into “Take” and “Make” move and to imitate it in the same sequence. Other views?

Generally, in my view, all approaches are equally possible on the starting line. I remember my probably first encounter with similar ambiguity: it was shortly after Bratislava PCCC meeting in 1993, where we met the Andernach chess – promoted that year for the first time in Andernach and immediately chosen for WCCT fairy section. Not in the WCCT framework, but rather general, I was considering what happens if neutral piece is captured under Andernach chess rules. My first idea was that capturing piece gets the colour of the captured piece – so it should become neutral, right? In fact, it turned out that capturing piece gets the colour of the opposite than the side that moved it. I.e. no piece can become neutral by Andernach capture.

We have been discussing this and later many other combinations and I try always to keep my mind open to various models. Bedrich Formanek said something along lines: “It is always wise to choose the interpretation that potentially yields the best quality problems. It is just difficult to foretell which one it would be before the problems are composed.”

That is why we have now two Anticirce types (both proved to be fruitful enough), both Popeye and WinChloe interpretation of Anticirce rebirth squares for locusts, etc. etc. I see no need to have everything defined single way.

Quote from Julia on November 7, 2014, 18:02
The discussion about Imitator has started in the comments to the problem No.638 – https://juliasfairies.com/problems/jf-2014-iii/no-638/#comments

Imitator. If in solution a piece move from (x) to (y), two or more times, can be considered the same move if this moves imitators on different squares
Example: ‘Bc5 (Ib6,Ig2)’ and ‘Bc5 (Ih8,Id2)’ is same move?

Model mates with Imitators … that’s really an interesting question!
I actually don’t think the concept of “model mate” has been defined clearly enough to cover all kinds of fairy conditions.

Remember that a model mate has to be pure (all K flights covered in just one way) and economical (all white units except possibly K and P’s take part).

Pure: What happens with K moves that are prohibited by a fairy condition? Should such moves be counted as not existing (and therefore irrelevant to the purity of the mate) or as potentially existing (and relevant to the purity of the mate)?
Example: wKc3 wRh1 bKa1 in UltraMaximummer. Black is mated, as he is in check from Rh1, and the condition allows only Ka1-b2 which is guarded by the wK. But is this a model? Ka1-b1 is not legal, but if Black played that move, he would also still be in check from Rh1. Does it make a difference if the wK is on c2 instead, so that two white pieces could capture if Black played that illegal move?
It seems to me most natural to consider moves that are illegal for some other reason than self-check as non-existing, and therefore as irrelevant to the concept of purity. That would include moves that are illegal because an Imitator cannot imitate. So from that point of view, Julia’s Scheme 1 is model.

Economical: Even in orthodox chess, a white piece can’t be active in just any way for the mate to be model. It must guard a flight or pin a black piece that could otherwise defend; it is not considered enough to close the line of a black piece that could otherwise defend (so wKa3 wRf1 wSg1 bRh1 is not a model). If you think this seems like an arbitrary restriction, I can sympathize; but that’s the way our Fathers wanted it to be.
So the question is if it is enough for a white piece to block an Imitator so that some other black piece cannot defend…
To me, it seems most natural to say that if it’s not enough to close a line, then it also isn’t enough to block an Imitator. From that point of view, Julia’s Scheme 3 is not a model.

I would like to hear other views on these questions!