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MatPlus.Net Forum General Last move in a self-mate problem.
 
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(1) Posted by Dupont Nicolas [Monday, May 7, 2012 21:34]

Last move in a self-mate problem.


I don’t understand why this is the black side which is playing the last move in a self-mate problem. For me it would be much more logical if it was the white side, here are some arguments going in that way:

- With few exceptions (e.g. proof games) this is always the white side which is playing the last move of a problem's solution.
- The goal of a self-mate problem (forcing the black side to deliver checkmate) is already achieved with the last white move.
- It is superfluous to my mind to “verify” that each black answer to the last white move must indeed be a checkmate move. When the goal is, e.g., to force the black side to capture, we don’t “verify” that each black answer to the last white move must indeed be a capture.
- Letting black playing, as far as we already know that it has no other choice than delivering checkmate, is in contradiction with the “Dead Reckoning” principle, asserting that if the status of a game is already known (whatever the forthcoming moves), this game is over. Following this DR convention, black's ending move in a self-mate problem is even illegal!

So I would be glad to learn why a self-mate problem ends with a black move, and what is the underlying philosophy of such a choice.
 
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(2) Posted by Juraj Lörinc [Monday, May 7, 2012 21:43]

Dead Reckoning is much younger principle than selfmate idea. Thus it makes no sense to bring it into consideration if the question is about the reason to play the last move of selfmate.
 
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(3) Posted by Olaf Jenkner [Monday, May 7, 2012 21:55]

<< So I would be glad to learn why a self-mate problem ends with a black move, and what is the underlying philosophy of such a choice.

There is no philosophy needed: A problem should end with mate!
 
 
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(4) Posted by Dupont Nicolas [Monday, May 7, 2012 22:33]

"A problem should end with mate!"

This is correct if the aim of the problem is to checkmate! The aim of a self-mate problem is not to checkmate, but to force the opponent to checkmate... And this aim is already achieved after the last white move, because black has no other possibility than to checkmate.

In a # problem, this is logical to end the solution with a checkmating move, because white generally has other choices than delivering checkmate. But I think that, in a s# problem, listing the black's final checkmating moves is redundant, as black has no other choice than to checkmate.
 
   
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(5) Posted by Olaf Jenkner [Monday, May 7, 2012 23:12]

Yes, listing until checkmate is redundant, but it looks better!
The other issue is: When does a selfmate end? My opinion: With mate.
 
   
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(6) Posted by Jacques Rotenberg [Monday, May 7, 2012 23:44]

@ Nicolas
You asked "...So I would be glad to learn why a self-mate problem ends with a black move"
If I understand your question, you mean : why do we write the solution till the mate ? or why not to stop writing at the last white move ?
If it is the question, the answer is, I think : because it is ordinary practice (en Français : c'est l'usage)
 
   
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(7) Posted by Dupont Nicolas [Tuesday, May 8, 2012 00:22]

@ Jacques,

Yes, this is ordinary practice nowadays. But someone told me (I haven't verify though) that, at the beginning of the self-mate practice, the list of final black checkmates was not provided (hence the last move of a s# solution was done by the white side).
 
   
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(8) Posted by Kevin Begley [Tuesday, May 8, 2012 02:56]

If DR rules say the game is over, prior to black playing the move, then the question is moot (at least in orthodox s#).
It doesn't matter how "young" DR rules are -- if these are now accepted as the orthodox rules, and these rules clearly invalidate black's last move, then there is no last black move (and thus, no duals, when black has multiple moves which all checkmate white).

The alternative, in my opinion, is to allow the composer to DEFINE where there problem ends; and, their definition of the final aim (where the solver can stop), will inherently define what should be considered a dual (you simply conider anything that is part of the solution, and nothing else).

However, if you allow for this, the orthodox line (which now seems to have swung anyway, by changes to the rulebook), becomes considerably blurred.

Personally, I don't see how it helps to have such an orthodox division.
It would be better to have a universal standard by which the composer can provide a complete set of expectations (including where the solution ends, and how the solution should be formatted).
 
   
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(9) Posted by Kevin Begley [Tuesday, May 8, 2012 03:44]; edited by Kevin Begley [12-05-08]

@Nicolas,

You say that capture problems end the game after white's move -- that's not exactly true; certainly not the best analogy...
In self-capture problems (e.g., sx2), most (if not all) solving tools (and databases, and problem journals) would report black's as the final move.

It's also worth noting that some s#n problems may have been composed with a thematic intent to include black's final move (I know this is rare, but it's possible).
If DR destroys such works, it begs the question: how are we going to provide for these compositions (presuming a number of them, which go to construction extremes, can not be repaired for DR's impact)?
I have a suggestion here... keep reading!

Furthermore, consider the reflex-mate stipulation.
Some claim -- I think with some legitimacy -- that reflex is fairy condition; and, the real stipulation is s#n (under the condition that a player with #1, is compelled to play the #1).
Would you apply DR rules, along with the fairy condition, and end the problem prior to black's playing the #1?

The real issue here is duals -- it is something I became aware of when CapZug was invented (something which completely alters the old precedent for problem termination).
I had some disagreements with the FORM of CapZug, but I really have no disagreement with this kind of problem.

I think a preferable FORM (one that should be standardized) would be to allow the composer to define finality in his/her problem (where duals would be evaluated according to anything that is intended to appear in the solution).

However, this does not dodge entirely the dogmatic ideology surrounding duals!
It remains an honest question to ask why, in a s#2, black may have multiple 1st replies (which are considered variations), but if black has multiple 2nd replies, they are considered (by some) to be duals.

If a s#n is truly a problem for the white player, and the black player represents only a virtual opposition to ultimate aim (whether that be to [# the wK], or to [achieve a position where black has no moves other than mate]), it would seem illogical to allow the opposing player to undermine correctness, by simply possessing a variety of failing moves.
[note: in fact, I consider it absurd that some databases mark some s# problems (showing multiple mating finales) -- including some really spectacular accomplishments in the field! -- as C- -- cooked! In light of the Dead Reckoning impact, this dogmatic position begins to strain credulity.]

Furthermore, it is plain to see that duals depend upon the notation used (generally, we use threat-notation).
But, if you define "series-mover" (or "parry-mover") to be conditions, and stipulate #26 (rather than ser.#26, or pser-#26), it becomes clear that threat-notation no longer provides its intended function.

Without threat notation, the solution (for a series mover) may be written as:
1.A 2.B 3.C ... 25.Y 26.Z#

But, with threat notation, the solution becomes:
1.A {> 2.B {> 3.C ... {>25.Y {> 26.Z# }} ... }}}

The notation becomes only slightly more complex with parry-movers, because of variations.
But, ultimately, I think there must be a mechanism by which the composer determines what form the solution should take.

In summary, I believe the dogma surrounding "Orthodox" / "Fairy" (terms have never been credibly defined -- and I doubt they can ever be!) -- particularly with respect to duals -- is a serious impediment to progress.
I think composers -- whether using CapZug, s#n, r#n, or anything else -- must be allowed to stipulate both termination, and the solution's notation (preferably in an economical, universal, and standardized fashion).
 
   
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(10) Posted by Olaf Jenkner [Tuesday, May 8, 2012 07:41]

<< It's also worth noting that some s#n problems may have been composed with a thematic intent to include black's final move

Here is an example:

http://www.softdecc.com/pdb/search.pdb?expression=probid=%27P1081755%27
 
   
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(11) Posted by Frank Richter [Tuesday, May 8, 2012 08:39]

Well, we should better compose good selfmates than discuss such - in my opinion, very far from reality - questions.
Just my 2 cents ...
 
   
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(12) Posted by Dupont Nicolas [Tuesday, May 8, 2012 13:30]

Yes, it may appear that allowing a s# to end with black checkmating move(s) is a real bonus. If I remember there is such a problem with 4 solutions, each of them beginning with a Q-R-B-S move, and ending with a Q-R-B-S checkmating move, respectively.

This is also true that, in the WinChloe database, final black capturing move(s) is marked in a sx problem. But, curiously enough, this is not marked in a cap-zug problem, although this is very near from a sx problem. Why this lack of homogeneity?

Duals (and their incidence on the soundness of a problem) are another question. As far as I know, the situation is clear for any helped problem: no dual allowed to be C+. In a direct problem, the “French school” is distinguishing between “dual mineur” (still C+) and “dual majeur” (cooked). As an example, if the final checkmating move in a s# is a1=Q,R#, this is considered as a minor dual.
 
   
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(13) Posted by Frank Richter [Tuesday, May 8, 2012 13:36]

Black moves in s# are variants, not duals. No matter, whether they checkmate or not. Of course you may prefer to have unique black mating moves, but this is a matter of taste.
A white move that forces a checkmate like a8=Q,B+ can be considered as a minor dual.

Looking at the homogenity - what about the idea, that Winchloe is simply a product of human work?
 
   
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(14) Posted by Jacques Rotenberg [Tuesday, May 8, 2012 13:56]; edited by Jacques Rotenberg [12-05-08]

"if the final checkmating move in a s# is a1=Q,R#, this is considered as a minor dual." ???
It is not a dual at all!
 
   
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(15) Posted by Kevin Begley [Tuesday, May 8, 2012 15:34]; edited by Kevin Begley [12-05-08]

There is deliberate dual-suppression in CapZug.
This stipulation is actually a "sxn" stipulation, plus some fairy condition, plus some dual suppression mechanism.
I think it is unfortunate (if not short-sighted) for us to allow all of its fundamental components to be obscured by a faux-stipulation.

I would urge, instead, an explicit dual-suppression mechanism.
I think the best way to accomplish this is to elaborate the final-aim (where the solver may successfully terminate what is essentially a recursive solution).

You have 3 fundamental components in any stipulation:
1) the type of play (black helps/opposes the goal),
2) some number of moves,
3) the solver's goal -- this may be an aim (#, =, x, etc), or it may be another layer of stipulation (like s#0.5 is the first layer goal in a hs#4, though you *may* want the solver to follow that goal down, recursively, to a final mate).

If the solver's goal is a stipulation, we simply need a formal mechanism to dictate how far the solver must iterate.
Maybe they need to go down to finality of #, or down to capture of Kings, or perhaps they need only realize a s#0.5 position...

aside:
There are a variety of dual interpretations (and a lot of dogma) surrounding s#n -- there's no right answer here!
All I am suggesting is this: why not allow the composer a universal mechanism to inform solvers where the problem should end?

This way, not only could CapZug be presented using an existing set of established, fundamental, universal criteria; but, you could also avoid much of the argument about duals in s#n problems (since you would provide the composer alternatives).

I see no reason not to consider both interpretations as viable.
And, I see no distinction as to which set of dogma should be blessed as "Orthodox."
There is, as yet, no valid criteria established which separates (or even defines) the fairy/orthodox division.

If r#n can be considered "Orthodox," and hs#n is fairy, then we have much explaining to do!
-Why can't we allow an Orthodox s#n to have an alternative dual interpretation (from whatever is the norm)?

ps: I think it's easily possible to not only economically describe this dual suppression mechanism in a standard way, but it would also be possible to express a number of cutting-edge stipulations (which have been little explored).

For example (one I've used before):

Michel Caillaud
help-direct-mate TT, 2005
(= 3+6 )

h4#2

Personally, I don't like the stipulation form used, but it is intended to convey: black starts, and plays 4 helping moves (white gets 3), to reach a #2 diagram... and then, this #2 is intended to be solved.
This well illustrates one simple possibility of a recursive stipulation (with a goal that is a stipulation, rather than an aim)...

The intent is: 1.d1=S! g5 2.e1=R! g6 3.Rg1 g7 4.f1=B! & 1.g8=Q! [> 2.Qg3#] (AUW).

If instead, you used a standard recursive stipulation form, you could easily describe this in terms of a stipulation-goal (#2); and, the composer would have options as to how far down this must be iterated, for a successful solution.
That is to say, the composer need only employ a binary character, to say either: keep iterating -or- you can stop here (e.g., is the goal: >#2< or is the goal <#2> ?).
The very same treatment would help explicitly describe the dual-suppression in CapZug, and it would help provide for alternative dual-interpretations in s#n problems.
 
   
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(16) Posted by Dupont Nicolas [Tuesday, May 8, 2012 15:55]

Coming from StrateGems, chess problem terminology:

"dual: a defect that allows more than one defense or continuation in a thematic line of play"

a1=Q,R# clearly shows more than one continuation after the last white move, thus it is a dual (according to this website).

The fact that such a fundamental concept (dual or, subsequently, sound problem) has no neat definition, accepted and known to each problemist, is a serious flaw, thus Kevin's intention (more or less putting some order in the jungle) is of real interest.
 
   
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(17) Posted by Frank Richter [Tuesday, May 8, 2012 16:22]

Well, you should refer to the codex:
(http://www.saunalahti.fi/~stniekat/pccc/codex.htm)
"A dual is said to occur if, after the first move, there is more than one method of satisfying the stipulation."

In a selfmate, WHITE is satifying the stipulation.
No jungle here.
 
   
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(18) Posted by Dupont Nicolas [Tuesday, May 8, 2012 17:12]; edited by Dupont Nicolas [12-05-08]

Ok, but now be go back to my very first question: why adding superfluous black checkmating move(s) at the end of the solution, as far as the stipulation is already satisfied?
 
   
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(19) Posted by Kevin Begley [Tuesday, May 8, 2012 17:18]; edited by Kevin Begley [12-05-08]

@Frank,

But, that definition is seldom employed...
In the Win Chloe database, there are some s#n problems marked C- (which I interpret as: cooked), due to multiple mating moves for black (some others are marked C+).
There are also problems marked C- due entirely to white "promotion-duals."
There are FIDE judges with differing opinions as to what constitutes a dual (in fact, even soundness) in s#n problems!
This can not be considered acceptable.

Finally, regarding my above suggestion (and more specifically, the example problem given), I'd like to note that I'm certainly not suggesting that problems with "fringe" stipulations are likely to abound by formalizing stipulations to provide for them (I'm somewhat skeptical of this).
I am simply suggesting that, while providing an explicit dual-suppression mechanism for composers, we would also cover a far wider variety of stipulations (including many which have yet to be invented!) -- using a very simple, formal methodology.

If you look at any formal problem, it has 3 fundamental components:
1) who will help/oppose in reaching the primary objective,
2) what is your deadline, and
3) what are your primary/ultimate goal(s)?

We generally deal with problems where the primary & ultimate goals are the same (#n).
However, there is the distinct possibility that our primary goal is to achieve another layer of formal problem -- we may want to push down to a new stipulation (one which *may or may not* need to be recursively resolved).

Therefore, we need a mechanism (call it a dual-suppression device), which allows the problem creator to formally define (and separate out) their ultimate goal (so the solver knows when to successfully cease resolving).

For example, let's define the goal of an orthodox s#n to be: s#0.5 ...
Then, you'd have to define a deadline -- some number of white moves (let's say 3).
Finally, suppose we say that the primary objective is for black to oppose.

Generally, I've now described a s#3 to be O(s#0.5)3 (where O = oppose, or direct play).
But, I have not yet shown whether the s#0.5 must be recursively solved.
So, now suppose I have two options: s<#>0.5 -or- <s#0.5> (where the former calls for iterative resolution down to the ultimate-aim <#>, whereas the latter instructs the solver that the ultimate-aim is the realization itself (that is: they can cease w/o further resolution).

This gives composers a formal mechanism to express the two distinct options: O(s<#>0.5)3 -or- O(<s#0.5>)3

To describe a hs#n:
H( O(s<#>0.5)1 )n-1 -- black helps white to reach the primary goal -- which is a stipulation itself: O(s<#>0.5) --, but after we push down to the next iteration, black opposes the new goal (s<#>0.5)... and we iterate until the ultimate aim <#> is reached.

note: unfortunately, hs#n was poorly defined (it's a rare case where white -- not black -- begins help-play).
This is arguably the case in help-games (PGn, and A->B) as well.
So, it may be necessary to add an extra 0.5 (at the very end)... but you get the point.

Or, we may decide that we have no interest in black's final moves -- in which case, we can simply eliminate these from the solution (thereby rendering duals a moot issue), by adjusting our ultimate aim (to: <s#0.5> ).
 
 
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(20) Posted by Frank Richter [Tuesday, May 8, 2012 17:33]; edited by Frank Richter [12-05-09]

@Nicolas: May be simply because of a mate needs a mating move?

Daniel Papack
Schach-Aktiv 2010
(= 9+15 )

s#2 (position corrected, thx to Jaques)
1.- Q~/d5 2.Rg7+/Dg6+ Lg6/T:g6#
1.Qe4! (2.Q:f5+ B:f5#)
1.- Q~/d5 2.Qg6+/Rg7+ L:g6/Tg6#
Black moves are superfluous? Please look carefully.
By the way, this highly original concept was excluded from the award due to ununderstandable reasons ...

Zoltan Labai
Schach 2012
(= 8+8 )

s#2
1.Qf3! Sc3 2.Qc6 e5#
1.- e5+ 2.Ke4 Sc3#
Superfluous too?

Yes, of course not in every selfmate are the mating moves of interest. This is a field for further activities, but a difficult one.

@Kevin:
>> There are FIDE judges with differing opinions as to what constitutes a dual (in fact, even soundness) in s#n problems! This can not be considered acceptable.

Yes, I'm fully agree, really ;-)

Regarding the stipulation - a solver has in every case to evaluate whether the position after the last black move is really mate or not. So why complicate simple things?
 
   
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MatPlus.Net Forum General Last move in a self-mate problem.