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MatPlus.Net Forum General Fairy Condition Question: Haan

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Fairy Condition Question: Haan

Is there a type of Haan where the hole left behind only lasts the (one) following ply (such that, following the first move, there is only one hole in the diagram, except after castling moves)?

If this is an original idea, I hereby propose the condition (and variant): Haan Solo. :)

note: this might give rise to an interesting retro stipulation: find the hole (not shown) in the diagram.

It would seem to play like duck chess, but without the need to make an extra duck move at each turn, and it's less confining than Haan.
So, I suspect this might be quite interesting both for problem chess and as a variant.

note: you can't castle in a game of Haan (in a problem, yes, but never in a Haan proofgame), but you can here.
Thus, it is worth noting that there is one special case, which is covered by Haan.
Whereas I would prefer that castling be considered a King move (read: after castling, I'd prefer a single hole be placed on the square vacated by the King, but not the Rook), it seems Haan has defined castling to leave two holes.
Thus, I think it best to defer to that condition's definition -- so, you can get two holes (temporarily) in a game of "Haan Solo" (who knows, maybe this unusual special case makes the aforementioned retro question more interesting).

If there are other special cases I may not have considered, please note them here.

[edit: I have revised the rules of Haan Solo -- there can be only one hole after any move; castling is a King move, which leaves a hole on the square vacated by the King.]

I think I saw something like 'Can't follow' which prohibits Umnov-like moves, but that's slightly different than a hole (can't follow moves can pass through the just departed square, whereas a hole can't be passed through).

Maybe "Transient Haan" is better?

>>> such that, following the first move, there is only one hole in the diagram, except after castling moves
After an en passant capture, I suppose there will be two holes?

@Shankar,

I don't believe there are two holes after en passant in Haan -- the holes go where the moving unit vacates, so it arguably makes some sense in castling, but it doesn't make sense to have two holes after a capture by en passant.

And just when I say that... sure enough, it seems this does happen in Haan, in the second (of three phases) of a problem by Zvonimir Hernitz Hernitz, feenschach 1972.

(= 2+2 )

h=3 Haan

b) 1.é5 d×é6 e.p. 2.R×é6 Rb5 3.Rd6 Rb6=

If there's not two holes after ep capture, I don't get why this is stalemate (bK should be able to move to e5 otherwise).
Sigh. That ep rule is clearly motivated only by a desire to make en passant more likely to yield weird results (read: let's add a goofy rule for en passant, so a vallado will be easier to achieve). Castling at least made sense. This en passant rule is destructive to idea of this condition.

I get that the pawn is 'vacated' from the board, but the capture should not create a hole (if this were the case, most capturing units -- locusts would be an exception -- would fall into the hole it created).
Do locusts create two holes in Haan?

OK, if that's the rule in Haan, it's also the rule in Haan Solo.
I am hoping this is not the rule -- that this is merely a joke problem (note: Win Chloe solver does not agree with this rule).
As for me, I'm going to look into this condition Joost mentioned. Hopefully, you are not prevented from following where an en passant capture occurred. :)

ps: no, Haan Solo is the better name (even despite the fact that there is an exceptional case, or two). This is obvious. Even ChatGPT should know why this is true.

@Joost,

"Can't Follow" is actually pretty close, but yeah, a hole is slightly different.
Are there problems with that condition in any databases?

In a way, the double hole for ep is explainable: A hole appears when a piece leaves a square. When being captured, a piece leaves the square.
You probably have the same with Locust captures: 2 holes (where piece leaves, and where piece is captured).

How do zeromoves work with Haan holes?

@Joost,

That's an interesting question!
Let me try a rose, and see what happens in Win Chloe...

Well, the only sound problem I find with a Rose and Haan (it also has an Ubi Ubi) is an INCREDIBLE problem by ... wait for it... Joost de Heer (from way back in 2022).
That's a beautiful problem, Joost!! Well done.
It's quite enjoyable to see how the chess board shrinks to nothing.

OK, let's see what happens if I make a zero-move...

According to Win Chloe, zero-moves (a unit returns to its starting square) creates no holes (as expected -- though, part of me was hoping for a Battle Chess animation showing a the zero-mover deteriorate into the hole it created).

OK, but seriously, zero-moves should not be possible in this condition. Right?
You should make a joke problem based on this -- it's a smart idea (and this would be a much better joke, because we're inclined to agree this should not be possible).

Another idea to illustrate this rule is to make an unusually long game (read: longer than 63 ply) with some zero-mover units.

Also according to Win Chloe, locusts do not create two holes.

White : Kf1 Pe3 Locust f7
Black : Kg3 Sh8 Pg6

If there were two holes after a capture, this would be =1, but after 1.LO×g6-h5 ...Sg6 is playable (there is no second hole where the captured unit was "vacated", because holes are only created when a moving unit vacates a square).
I'm increasingly confident that the en passant problem which creates two holes was intended as a joke problem (it was not listed as C+ by Win Chloe, the solver doesn't agree with it, and it shows as incorrect in redaction mode).

This is yet another reason we need some official document for fairy condition rules (note: sometimes the original document which introduced a condition will clear this up, but sometimes we would need to know any rule modifications which have been adopted later).

And, we need joke problems -- like this one -- in such a document (they provide the best illustrations of what is, and is not, legal -- when accompanied by some authoritative decision).
If the joke makes sense, it should either be adopted (or, it may be listed as a sub-type for the condition).

In that respect, I applaud this Haan joke problem (if it is a joke problem).
That rule makes no sense to me, but it's still good somebody seeks to clarify what might be perceived differently.
The only trouble is: we now wonder whether it's the official rule, or just a joke. We have no idea whether it was officially adopted.

All we know for sure is that the latest version of Win Chloe doesn't agree with the two holes interpretation (except when it comes to castling).

I have confirmed two holes by en passant in Haan is that author's interpretation (the only way to officially create two holes in one move in Haan is by castling).

If you want to shrink the board to zero squares as fast as possible, maybe Haan + Rookagogo is the way to go. :)

Popeye implements this double hole for ep as well.

Bloody hell. This is real???!! This rule makes zero sense!

Does popeye similarly leave two holes after a locust capture?

This rule has encased Haan in a carbonite deep freeze.

Note that in Popeye, this is =1.

(= 3+2 )

=1
Haan
= White Locust at f7

Popeye gives: 1.Lf7*g6-h5 = !

Did the capturing locust create one or two holes? Let's add a bS on h8, and ask popeye:

(= 3+3 )

=1?
Haan
= White Locust at f7

No solution, because there no is hole on g6 (black can play 1...Sg6).

So, this proves Haan has a really, really, really dumb rule for en passant.
If the justification for the en passant capture rule were valid, it should apply here, too. The capturing locust should have analogously left two holes (on f7 and g7), and this =1 problem should have a solution.
This bad special case rule makes the condition (Haan) a programming nightmare, frankly. And the bad castling rule wasn't doing programmers any favors, either.

It is now evident that I need to thaw out Haan Solo.
And since I need to alter the bad en passant rule of Haan, I'm also going to remedy the castling rule.

In Haan Solo, there can be only ONE hole after any move.
Castling is a King move, and the hole is created on the square the King vacates.
Captures by en passant leave only a hole on the square that the moving unit has vacated.
But, in Haan Solo, this hole lasts only 1 ply (because there can be only one hole in Haan Solo -- never more than one, never fewer than one).

Also, no zero-moves are possible in Haan Solo. A unit that vacates its square may not return (as the hole will be there before their return).
There must always be one (and exactly one) hole following any move.

Note: Christian Poisson has confirmed that this is the official rule in Haan (aka Haaner Chess), as stated by the condition's inventor (Heinz Winterberg) in issue #5 of Feenschach (1971) -- the issue where this new condition was introduced.

The initial definition from that issue is reprinted here (on Die Schwalbe's page): https://www.dieschwalbe.de/lexikon.htm#sectionH

Roughly translated to English (perhaps somebody can provide a better translation -- as I'm only struggling to make sense of the Google Translator output):
Haaner Chess: A unit that vacates a square (by moving, by capturing, or as a victim of capture when the capturing unit does not occupy the square of the captured unit, e.g. in an ep capture) leaves a hole on the vacated square (which cannot be stepped on or crossed).

Notably, Winterberg also published a problem in feenschach 5 which clearly was intended to demonstrate that en passant capture leaves two holes; tragically, the problem is cooked.
So, however misguided, this was the intent of the the inventor (and this is official).

Heinz Winterberg
feenschach #5, September 1971.
(= 3+3 )

h#2 Haan C-

Intent:
1.d5 é×d6 e.p. 2.R×d6 Fb4‡

Cook: 1.d5 é6 2.Rd6 Fb4‡

This is a tragic cook, because the cook should have been the only solution, and the intent could have been ruled out with a more elegant definition of the condition.
Note that the black Pawn at e7 is necessary (otherwise the black King escapes to e7, and check is blocked by the hole on d6).

So, if my translation of that definition is correct, it would seem Popeye's implementation is incorrect when it comes to the locust =1 example I provided (that problem should have a solution -- the black Knight on h8 should have no legal move).

I can hardly blame the Popeye team; this condition presents a major challenge for programmers to correctly implement the inventor's vision.

Note: I am not aware whether the inventor considered the special case of a zero-moving unit (zero-movers barely existed in 1971 -- the Rose was probably the first zero-mover seen, and that shows up in problems only in ~1969; I had somehow thought Dawson invented the Rose, which would have been decades prior, but maybe it was Boyer or Monreal, or both?! -- I do know they collaborated on the invention of the Lama).

Note: Anthony Stewart Mackay Dickins invented the Zero unit, but I'm not sure what year (as I'm not aware what this unit is called in Win Chloe, I can't search for it).
I suspect the Rose was the first zero-mover, unless such a unit existed in Shogi.

Note that it's also unclear in Haan (aka Haaner Chess) whether a kamikaze unit leaves a hole.

Is a kamikaze unit also considered the "victim of capture" by a unit (itself) which no longer occupies the square upon which it was annihilated?
If so, it should be possible to achieve three holes in one ply.

(= 2+2 )

h=2 Haan + Kamikaze

Intent: 1.Ke6 [-e7] e5 [-e4] 2.d5 [-d7] exd6 e.p. (d6 self annihilates) [-e5, -d5, -d6] =

note: neither popeye nor Win Chloe treat this combination of conditions. Is there any solving program which treats this combination of conditions?
I doubt it -- the Haan condition doesn't exactly play well with others (nor even with itself), because the inventor introduced some weird special case rules.
That's unfortunate, because the inventor's idea is a very good one (and there are some beautiful problems worthy of many replays in this condition, where you needn't even solve the problem to admire how the board shrinks to nothing).
Such a combination of conditions might have been far easier to program (definitely far easier to understand) had the inventor of Haan simplified the invention to one additional hole per ply.

Hopefully, with Haan Solo, there is never a doubt about the rules.
There is only one hole in a diagram, and it is always found on the square last vacated by the moving unit (castling moves are King moves), no zero-moves are possible (a unit may not move back to the square upon which it stands, because the hole will be there before it has time to return), and there is no question what happens with Kamikaze units (victims of special case captures do not leave holes -- holes are only the product of square vacations by moving units).

All that said, I invite anyone (and everyone) to have a go at this condition.
I would be particularly pleased to see any retros which ask the question: where is the hole?
May the force be with you!

Last point. It would have been interesting if there was some variation of Haan which created a new square (somewhere on the board -- possibly expanding the board), for every hole.

For example, imagine that by some algorithm, according to the location of the hole, we can determine a specific direction (up, down, left, or right -- from location of the hole created), and by this algorithm, we move from that hole to either fill in the first hole encountered in this given direction (or, if no hole is encountered, we expand the board in that direction).

This would be quite interesting. There would always be 64 squares, but the shape of the board would be altered by movements.

If Haan were more elegantly defined, such a condition would be spectacularly interesting.
Imagine achieving a contiguous board where a file (or a rank, or both) is gone, and a new file (or a new rank, or both) has been created on the other side of the board.
Imagine achieving a board where the board is split into 4 contiguous boards of 16 squares (with a missing rank and file running through the center).
Imagine achieving a board which is a 64x1 rectangle.

To do this, you need a condition with very elegant rules (and you need a very elegant algorithm to determine the direction of the new square).

I would suggest using an equipollent algorithm -- based movement and direction (the vector) of the last move (where castling is considered a King move!), and you either fill a hole in that direction (from the location of the hole), or you expand the board in that direction.

For example:

(= 1+0 )

Suppose the white Knight moves from Se4 to d6.
A new hole is created on e4. Where does the new square go?

From the hole, at e4, we move to d6. We know that's not a hole (or the Knight could not have jumped there).
Next, we consider c8. If there was a hole previously on c8, it is filled.
If not, we continue to b10 (note that it doesn't matter if b9 exists). If there is no b10 square, there is one now.
If the b10 square previously existed, we continue to a12, then to z14, then to y16, x18, etc.

Thus, the Knight can wander off infinitely in one direction, but getting back may soon become quite impossible.
If Chess imitates life, this condition would multiply that ten fold (keep moving without noticing the bridges burning behind you, you'll find yourself alone, with no hope to return).

The Knight seems to wander off like a, uhh, umm, like a Skywalker! LOL.
And what do you get when a Skywalker is mixed with Haan?
I would call this variant "Kylo Ren".

Indeed, it may be an interesting problem just to organize a circuit in such a condition (yes, it is possible, but some long leapers may need considerable help).
That's a perfect analogy to tell a story with a chess problem. That's exactly what we want from a fairy condition: an opportunity to tell an interesting and relatable story.

Consider how very interesting it would be to see what happens to the board after a Knight's Tour (we know what happens in Haan, and that's quite interesting to witness; now imagine what a balanced form might do to the board).
In fact, it would be very interesting just to observe how a board breathes under such condition (anybody who has ever admired patterns from the Game of Life will know what I mean).
You needn't even solve a problem to appreciate the board patterns created by the condition and the composer -- it's enough to admire patterns on a balanced form (where the board is both living and dying).

To clarify how the algorithm works...
Consider a rose moving from e2 to g3 (along the route e2-c3-b5-c7-e8-g7-h5-g3).
note: it can not return to e2, as the hole is created there before it may return.
So, we know the hole appears on e2, and the patch vector goes from e2 to g3 (the patch vector doesn't consider the path -- you only consider the vacation square and the arrival square).
Thus, it doesn't patch the e2 square (as it might if you considered the path the Rose takes to reach g3). Instead, this creates a new square on i4 (the same as if the rose -- or a knight -- moved directly from e2 to g3).

I think this algorithm will always add one new hole behind the moving unit, and simultaneously add one new square (or patch up one hole) in front of the moving unit; unless I'm missing something, the hole and the patch will never meet (which is what we want).

But, nobody wants to program this with the artificial special case rules of Haan (such a condition is difficult enough as it is -- and Haan is already prohibitively challenging because the special case rules undercut our ability to even understand the condition -- which is why nobody seems to have it programmed correctly).

This is an important lesson for fairy condition inventors: the benefits of simplification are immeasurable.
Haan could have been one of the simplest (and I daresay one of the most interesting) fairy conditions imaginable (it could have been more elegantly defined than even than PWC), but needless complexity was tragically added.

Even a simplified form of Haan remains an unbalanced force. It is like the death star destroying squares.
By comparison, Kylo Ren is a balanced form (and simplified)-- it creates as many squares as it destroys (one of each, per ply, which is ideal).

At present, even a simplified Haan condition is a formidable programming challenge; adding new squares is a far more formidable challenge (there's no good way to extend our coordinate notation to infinity -- perhaps there's a way to further simplify this, by making a larger torus structure, such that the unit that wanders off too far in one direction gets looped back, but to do this, we would need to adjust the square creation algorithm).
Eventually, I predict, we will see a realization of simplified, balanced conditions. When we do, the challenge will be to look at any other condition again.

Royal Rose e1, zero moves allowed, Haan, s#0.5

@Hauke,

Obi Juan: You want to go home and rethink your life.

Royal Rose in Haan: What, you think you're some kind of Jedi, waving your hand around like that? I'm a Royal Rose in Haan. I'm not sure whether I can legally go home.

New idea by me: Quantum measurement pieces. Pieces that can take more than one route to their destination take all of them at the same time (as you can either measure movement or position).

Obviously that is only relevant with conditions like Haan.

SH, original
neutral quantum measurement double-move-rook a1
Haanover (was that the right one?)
stalemate in 1

1.Rh8 (this leaves holes on all squares on the 1st/8th rank and a/h-file except h8, as the rook traveled over all of them, i.e. took all routes at once)

@Siegfried,

LOL. I wondered where you were going with the Haantum Theory. Well played. :)

Actually, wait. The Rook makes a double move.
So, it can only go a1-h1-h8 or a1-a8-h8. Right? Thus, there's only three holes on the board (not 27 holes), and it's not stalemated.
It never occupied h7 or g8, thus there's no hole there.

Either I am missing something obvious here (and if so, it's because nobody understands quantum theory), or you need a triple-move rook (I think that works).

It leaves holes on all squares it moves over, not only where it lands.
Triple rook could be stalemated in any corner (except a1 obviously).

Better to have a triple-move Q-Rook in Haan than a double-move Q-Rook in Haanover.

You never defined Haanover.

Haanover is an established condition.

https://juliasfairies.com/fairy-terms/fairy-classification-project/fairy-classification-tabular/

QUOTE
Haanover
After each move, the departure square of the moving piece and all squares passed over by the move become ‘holes’ which no unit may occupy or pass through again.