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| | | (1) Posted by Juraj Lörinc [Sunday, Feb 7, 2010 00:09] | Promoted pieces in problems with fairy pieces
Once upon a time it was required that fairy pieces in a problem should be considered as promoted from pawns. As a consequence, rule max. 8 fairy pieces + pawns per side applied.
Nowadays (almost) nobody cares about number of fairy pieces on the board, at least with regard to this kind of legality. (Almost) everybody just tries to have as little (especially non-thematical) fairy pieces on the board as possible.
But... what do you think about problems with fairy pieces as regards use of promoted orthodox pieces. To be very specific, imagine #2 with grasshoppers where White has 3 pawns, 3 grasshoppers and 3 queens. While such position would be legal in the strict sense (as above), would it be also acceptable? If promoted orthodox pieces are used, are they judged less strictly than fairy ones, more more strictly or about the same?
| | | (2) Posted by Joost de Heer [Sunday, Feb 7, 2010 00:34] |
Where a fairy piece came from is, for me, only interesting in compositions with retro-content. If there is no retro-content, the position may be illegal (not just with regard to promoted fairy-pieces, also things like 'crossed pawns' in Anticirce).
| | | (3) Posted by seetharaman kalyan [Monday, May 3, 2010 16:59]; edited by seetharaman kalyan [10-05-03] |
this doesnt answer Mr.Lorinc's question reg. judgement of the problem
| | | (4) Posted by Guy Sobrecases [Monday, May 3, 2010 18:05]; edited by Guy Sobrecases [10-05-03] |
I understand the answer of Joost. He does not pay attention to legality in non-retro fairy problems. I believe this answers all other specific questions, including the one about orthodox promoted units.
I know by experience that some other judges pay attention to legality, even when that's not mandatory to show a legal position. In my opinion, that's often interesting to look at the legal aspects, as it shows the technical ability of the author, especially when achieving a difficult "task". In this very case, orthodox or fairy units may be an important parameter.
| | | (5) Posted by Kevin Begley [Tuesday, May 4, 2010 06:28] |
@Juraj,
Good post w/ interesting questions raised!
Frankly, I think we'd all be more interested to hear your views, but, for the sake of discussion, I'll toss out a few of my own...
"Once upon a time it was required that fairy pieces in a problem should be considered as promoted from pawns.
As a consequence, rule max. 8 fairy pieces + pawns per side applied."
Though the constraint's downward push on economy was beneficial, the constraint itself was poor (and far to punishing -- rendering the problem "illegal," when really it may only be excessive).
This was akin to Arizona law.
If I can show 10 white Nightriders, using what is considered a "proper fairy condition" (Cavalier), why should 10 white Grasshoppers be dogmatically shunned (as if an illegal diagram)?
The obvious loophole: the author invents a "Grasshoppalier" condition.
On the whole, I am glad such artificial constraints were discarded.
"Nowadays (almost) nobody cares about number of fairy pieces on the board, at least with regard to this kind of legality. (Almost) everybody just tries to have as little (especially non-thematical) fairy pieces on the board as possible."
And, as raised in a later post, nor do they care about retrograde validity (unless published as a Retro).
There is some inherent error here -- in my view -- because solving then may require proper genre specification (unacceptable).
Clearly, "retro-illegal" circe exchange problems (read: anything with fewer than 32 units) should be accepted.
And, I am aware that a path has been contrived to achieve retro-legality in this condition -- and others --, though I'm still unclear exactly what that is (regardless, it seems a highly inelegant solution).
I feel, instead, the author should be obliged to state when Retro-legality has been surrendered.
And, I would not judge too harshly if this occurs for good reason (if it could have been simply avoided, judge accordingly).
"But... what do you think about problems with fairy pieces as regards use of promoted orthodox pieces.
To be very specific, imagine #2 with grasshoppers where White has 3 pawns, 3 grasshoppers and 3 queens.
While such position would be legal in the strict sense (as above), would it be also acceptable?
If promoted orthodox pieces are used, are they judged less strictly than fairy ones, more more strictly or about the same?"
I consider these constraints to be further examples of leftover dogma (originally erected to limit the directmate frontier -- now, only serving to fence us all in).
If I had my way, I'd say anything goes (and this goes for all genres -- because anyway, I'd reduce all problems to a single genre)!
Directmates with promoted force (or even 9 pawns, if stated as retro-illegal) should be welcomed -- and, judged according to the overall impression, given their economy of form.
Self-imposed constraints become like outdated religious law.
Sanitary rules for meat consumption, for example, developed prior to refrigeration, were written as if commandments from God.
Later, it becomes very difficult to determine whether strict adherence is still required.
And big debates emerge as to what core principles constitute the essence of a religion.
There should be only one commandment for chess composition:
Judge thy adjacent problem as if you (personally) traveled the path of its creation.
| | | (6) Posted by Hauke Reddmann [Tuesday, May 4, 2010 13:49] |
Since a standing phrase from the SCHWALBE is
"Kronkorkenorgie" (crown cap orgy - referring obviously
to pre-computer era solving), I hereby suggest
the new legality check: number of fairy pieces may
not exceed number of bottles of beer on the wall. :-)
Hauke
| | | (7) Posted by Kevin Begley [Saturday, May 8, 2010 06:20]; edited by Kevin Begley [10-05-08] |
@Juraj,
I think I can make a good specific case...
I published problem D1 (see below) in the context of an article aiming to define three new fairy conditions.
Psychos = combo of Rook + Bishop + Knight, but always temporarily surrendering the movement last employed.
Thus, pS = psycho-Knight, moved last as Knight, and now has only powers of Rook + Bishop. etc.
As you can see, in D1, black has 3 fairy units + 7 pawns = illegal, by old standards.
I knew this, but I also knew the old standard was long abandoned (clear back in the Cyclasic Period). :-)
At the time I published this, I had already composed several other versions, including diagram D2.
For fun, I even tested a version showing 6 changes (18 thematic variations) on an 11x11 board, just to computer verify my idea to extend this out indefinitely)...
The second diagram has no such legality concern -- both sides have room to spare.
In fact, the second diagram actually extends the theme (to four changes) while saving 5 units!
But, which is really more economical?
Which one might a reader (or a solver -- our forgotten audience!) actually care to study?
In the context of my article, the choice was clear -- I must eliminate the non-essential fairy elements (only psychos are needed!).
But, in ANY context, I should prefer the economy of D1.
Economy of fairy elements, and overall aesthetics, are the critical questions...
Leave the piece counting for the Sentinells + Koko + Maximummer type problems. :-)
D1
K.Begley
StrateGems, 2009
 (= 9+11 )
#3
 Psycho-Bishops 0+3
1.Sb6! (> zz)
1…pBed4(pS) 2.Sd5+ A pSxd5(pR) 3.Be5# B
1…pBfe4(pS) 2.Be5+ B pSxe5(pR) 3.Rf5# C
1…pBdf3(pS) 2.Rf5+ C pSxf5(pR) 3.Sd5# A
1…pBfd3(pS) 2.Sd5+ A pSxd5(pR) 3.Rf5# C
1…pBef4(pS) 2.Rf5+ C pSxf5(pR) 3.Be5# B
1…pBde4(pS) 2.Be5+ B pSxe5(pR) 3.Sd5# A
1…pBxd7(pS) 2.Sd5+ Ke6 3.Sd8#
1.Qxh7?, 1.Qf8? [2.Qh8#]
1…pBf5(pR) 2.Rxf5+ gxf5 3.Qh6#
but 1…pBxd7(pR)!
1.Sc7? [2.Re5 [3.Se8#] 2...pBxd7(pR) 3.Sd5#
2…pBd6(pR) 3.Sd5#
2.Se8+ pBxe8(pR) 3.Be5#]
1…pBxd7(pR) 2.Sd5+ Ke6 3.Sd8#
But 1…pBde4(pS)!
D2
K.Begley
(version, improved?)
 (= 6+9 )
#3
= Psycho-Bishops 0+3
 = Psycho-Knights 0+1
 = Camel 0+1
 = Giraffe 1+0
 = (1,5)Leaper 1+0
1.b8=CH! (> zz)
1…pSd3(pB) 2.Sd5+ A pBxd5(pR) 3.CHc5# B
1…pBbd3(pS) 2.Sd5+ A pSxd5(pR) 3.GIb5# C
1…pBdc4(pS) 2.CHc5+ B pSxc5(pR) 3.Sd5# A
1…pBbc4(pS) 2.CHc5+ B pSxc5(pR) 3.GIb5# C
1…pBc3(pS) 2.CHc5+ B pSxc5(pR) 3.(1,5)a5# D
1…pBdb3(pS) 2.GIb5+ C pSxb5(pR) 3.Sd5# A
1…pSb3(pB) 2.GIb5+ C pBxb5(pR) 3.CHc5# B
1…pBab4(pS) 2.GIb5+ C pSxb5(pR) 3.(1,5)a5# D
1…pSa4(pB) 2.(1,5)a5+ D pBxa5(pR) 3.CHc5# B
1…pBba4(pS) 2.(1,5)a5+ D pSxa5(pR) 3.GIb5# C
1…CHe2 2.GIxe2+ pBxe2(pR) 3.Sd5#
.
| | | (8) Posted by Kevin Begley [Saturday, May 8, 2010 08:28]; edited by Kevin Begley [10-05-08] |
Err, after some digging in my archives...
I discovered D2 is not the version I meant to select...
This one, D3 (a much improved version, of the non-improved version :-)) is what I meant to post...
Closer study reveals an interesting threat concealment mechanism in this version (along with better threat in better economy).
I wondered to where this idea had vanished!?
The interesting cases are added into the solution as 2nd move tries.
K.Begley (version)
 (= 7+7 )
#3
= Psycho-Bishops 0+3
= Psycho-Knights 0+1
 = Camel 1+0
= Giraffe 1+0
= (1,5)Leaper 1+0
1.a8=GI! [> 2.Re7 [> 3.CHe8#]
2…pBd7(pR) 3.Sd4#
2…pBd6(pR) 3.Sd4#
2…pSc7(pR) 3.CHc4#
2…pSc6(pR) 3.CHc4#
2…pBxb7(pR) 3.GIb4#
2…pBb5(pR) 3.GIb4#
2…pBa4(pR) 3.(1,5)xa4#]
1…pSd2(pB) 2.Sd4+ A …pBxd4(pR) 3.CHc4# B
1…pBbd2(pS) 2.Sd4+ A …pSxd4(pR) 3.GIb4# C
. 2.Re7? [> 3.CHe8#]
.. 2…pSd7(pR) 3.GIb4‡
.. 2…pSc7(pR) 3.CHc4‡
.. 2…pSc6(pR) 3.CHc4‡
.. 2…pBa4(pR) 3.(1,5)xa4‡
. but 2…pSd6(pR)!
1…pBdc3(pS) 2.CHc4+ B …pSxc4(pR) 3.Sd4# A
1…pBbc3(pS) 2.CHc4+ B …pSxc4(pR) 3.GIb4# C
1…pBc2(pS) 2.CHc4+ B …pSxc4(pR) 3.(1,5)a4# D
. 2.Re7? [> 3.CHe8#]
.. 2…pBd7(pR) 3.Sd4#
.. 2…pBd6(pR) 3.Sd4#
.. 2…pSc7(pR) 3.(1,5)a4#
.. 2…pBxb7(pR) 3.GIb4#
.. 2…pBb5(pR) 3.GIb4#
. but 2…pSc6(pR)!
1…pBdb2(pS) 2.GIb4+ C …pSxb4(pR) 3.Sd4# A
1…pSb2(pB) 2.GIb4+ C …pBxb4(pR) 3.CHc4# B
1…pBab3(pS) 2.GIb4+ C …pSxb4(pR) 3.(1,5)a4# D
1…pSa3(pB) 2.(1,5)a4+ D …pBxa4(pR) 3.CHc4# B
1…pBba3(pS) 2.(1,5)a4+ D …pSxa4(pR) 3.GIb4# C
. 2.Re7? [> 3.CHe8#]
.. 2…pBd7(pR) 3.Cd4‡
.. 2…pBd6(pR) 3.Cd4‡
.. 2…pBxb7(pR) 3.GIb4‡
.. 2…pBbb5(pR) 3.GIb4‡
.. 2…pBa4(pR) 3.CHc4‡
. but 2…pBab5(pS)!
.
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