|(1) Posted by Administrator [Saturday, Dec 25, 2010 04:06]|
7th International Solving Contest
As you all know this event will take place on January 23rd 2011. In order to provide a quick and complete (preliminary) information about the results a dedicated application will be lounched on this site. It is now in a testing phase and we hope that there's enough time to detect and fix possible errors.
ISC controllers who are members of MatPlus.Net are invited to try it (they can find the link above the main menu) and report in a personal note to Administrator any problem they may come upon.
|(2) Posted by Administrator [Monday, Jan 24, 2011 00:55]|
The problems and problems+solutions from 7th ISC have just been published on this site. Click the ISC 2011 link above the menu and then choose what you want to see from Problem submenu.
I estimate that about half of results have already arrived. Hopefully we'll have complete preliminary tables in day or two.
|(3) Posted by Neal Turner [Monday, Jan 24, 2011 11:45]|
Thanks Milan for this very easy and convenient system for entering the provisional results!
Just one little niggle - it doesn't seem to recognise the difference between entering '0' (an incorrect solution) and ' ' (no solution given) as when I go to the results all the non-scoring cells have a zero.
|(4) Posted by Administrator [Monday, Jan 24, 2011 12:17]|
Neal: Thanks Milan for this very easy and convenient system for entering the provisional results!
I'm glad to be of some use :-)
Neal: Just one little niggle - it doesn't seem to recognise the difference between entering '0' (an incorrect solution) and ' ' (no solution given) as when I go to the results all the non-scoring cells have a zero.
My omission, unfortunately too late to fix. Fortunately does not affect the results. Will be corrected for the next ISC :-)
|(5) Posted by Brian Stephenson [Monday, Jan 24, 2011 14:51]|
I have a 'niggle' too. In GB we had a 'foreign' solver, Kieran O'Driscoll of Ireland. I entered him in the middle of my list of solvers. Thereafter, when I entered Paul Valois, Paul Valois came out as Ireland, which was uneditable. So I had to delete Paul and enter him again. When I had finished entering everybody, I checked the overall results only to find that Kieran O'Driscoll was not amongst them. I tried entering him again, but failed because him name was no longer in the drop-down list of foreign solvers provided.
|(6) Posted by Administrator [Monday, Jan 24, 2011 17:37]|
Brian, I think you forgot to load the pre-created record for Paul Valois, but typed the name and the rest over the old record. I recovered the initial record for Kieran O'Driscoll. Will you please re-check results for him and for Paul?
|(7) Posted by Brian Stephenson [Monday, Jan 24, 2011 20:24]|
All now correct. Thanks.
I really don't understand what I did wrong, but no matter - all is now fixed. :-)
|(8) Posted by Miodrag Mladenović [Tuesday, Jan 25, 2011 20:30]|
I am wondering if some endgame experts can give their opinion about endgame #4 from the "Category 1" competition:
4) Martin Minski,
Original for Rochade Europa 2011
(= 4+4 )
1.f8Q d1Q [i] 2.Qg7+ (1) [ii] Kh2 3.Bf2! (2) [iii] Kh1 [iv] 4.Qg4 (1) [v] Qf1 5.Qh4+ Kg2 6.Qg3+ Kh1 7.Sg4 (1) Qg2 [vi] 8. Qh4+ Qh~ 9. QxQ#
[i] 1.... Rh2+ 2. Kg6 d1Q 3. Sxe4+ +-
[ii] 2.Qg8? Kh2 3. Bf2!? Rxf2! 4. Sg4+ Kg2 5. Se3+ Kh1 6. Sxd1 Rh2+ 7. Kg7 Rg2+ =
[iii] 3. Qh6+? Kg2 4. Qg5+ Kh1!; 3. Sg4+ Kh3!; 3.Bb8+ Kh1 4. Qh6+ Kg1 =
[iv] 3.... Rxf2 4. Sg4+ Kg2 5. Se3+ Kh1 6. Sxd1 Rh2+ 7. Kg8 Rg2 8. Sf2+! Kh2 9. Sg4+ +-
[v] 4. Sg4? Qc1 5. Qe5 Rxf2! =
[vi] 7.... Rxf2 8. Sxf2 Qxf2 9. Qxf2 +-
1) I am wondering why there is no variation:
[ii] 2...Kh3 3.Qg4+ Kh2 4.Bf2 Qf1 5.Qg3+ Kh1 6.Sg4 Qf1 7.Sg4 Qg2 8.Qh4+ Qh~ 9.QxQ#
I wrote this variation and not the main one. Beside's being the same lenght as the main variation it does ends up with identical position as the main variation. However it's not even mentioned in the solution. Should not this variation cary at least 1 point in solution?
2) My second question is about missing variaton:
[iv] 3... Rxf2 4. Sg4+ Kg2 5. Se3+ Kf3! (I think this is stronger variation because after 6. Sxd1 Rh2+ 7.Kg8 Rg2! is draw) 6.Qg4+ Kxe3 7.Qxd1+. I thought that this is the main variation even if it's shorter than other two variations. However it does not carry points at all.
I am wondering what is opinion of other solvers about this endgame?
|(9) Posted by Mihail Croitor [Wednesday, Jan 26, 2011 22:52]|
For first endgame i showed variation with 3... Rxf2 4.4.Ng4+ Kg2!? etc as main line. and i'm really impressed that this variation not was specified. Variation 1... Rh2+ not so easy to win also.
|(10) Posted by Miodrag Mladenović [Thursday, Jan 27, 2011 07:23]; edited by Miodrag Mladenović [11-01-27]|
This is second endgame from this competition:
10) Alexander Kazantsew, Shakmaty v SSSR 1949
(= 4+5 )
1.d6! Sb5! (i) 2. dxe7 (ii) (1) Ke5 3. e8S! Bh8 4. Kg8 (1) Kxe6 5. Kxh8 Kf7 6. h7! (1) a3 7. Sd6+ Kf8! 8. Sxb5 a2 9. Sd4! (iii) (1) a1R! 10. Se6+ Kf7 11. Sd8+ Kg6 12. Kg8 Ra8 13. h8S+! Kf6 14. Shf7 (1) =
(i) 1…. Sc4 (?) … 8. Sxc4 a2 9. Se5 a1R 10. Sd7+ Kf7 11. Se5+ Kf6 12. Sd7+ = points for variation max. 3 points
(ii) 2. d7? Sd6+ 3. Kg8 Sb7 -+
(iii) 9. Sc3? a1R! 10. Sb5 Kf7 11. Sc7 Ra7 12. Se8 Rd7 13. Sd6+ Kg6 -+
I am very confused with scoring of variation (i). What does mean max 3 points? How is 3 points spread over this variation? "points for variation max. 3 points" is very wide definition. It may mean that you get 3 points only if you wrote all moves from this variation. It would be nice if points were written explicitely so that there is no confusion. I can already see that different controlors may give different number of points for same moves written (if not all 12 moves are written).
|(11) Posted by Kevin Begley [Friday, Jan 28, 2011 07:49]; edited by Kevin Begley [11-01-28]|
>2...Kh3 3.Qg4+ Kh2 4.Bf2 Qf1 5.Qg3+ Kh1 6.Sg4 Qf1 7.Sg4 Qg2 8.Qh4+ Qh~ 9.QxQ#
There seems to be a typo in your line...
And, I'm no endgame expert, but I suspect that the author may have left this variation off, because it would present a dual...
Your 4.Bf2 certainly wins (either black directly transposes to the main line, or all hope falls with the loose Rook on f6).
But, in your line, 4.Sh5 and 4.Sxe4 also seem to win (in the latter case, if 4...Qd3 5.Bb8+ ...Kh1 6.Qxe2! etc).
I am not qualified to say what score your answer merits [not a political dodge, just being honest]
ps: Why doesn't the solution give the following line (it seems more interesting than the main line):
1.f8Q d1Q 2.Qg7+ Kh2 3.Bf2! Rxf2 4.Sg4+ Kg2 5.Se3+ Kf3!? 6.Qg4+! (6. Sxd1 Rh2=) Kxe3 7.Qxd1 +-
|(12) Posted by Joose Norri [Thursday, Feb 17, 2011 17:57]|
It has to be said that nr. 4 was not quite a successful choice for a solving competition. There is no reason to think that the given variation is the main one. It could be argued that after 3.-Kh1 it takes longer to get a winning material advantage than after 3.-Rxf2, but firstly, that has never been a criterium for defining the main variation and secondly, after 1.-Rh2+ it seems to take even longer. I think it's a mistake to state in that variation "3.Sxe4+ and wins"; material is even. Yes it's an obvious win, but to whom?
But the main point is that there is no study-like play. These are ordinary precise moves that could happen in any game; no surprise anywhere. So when the solver sees (...Qg3+ Kh1 Sg4...) he starts looking for some interesting play in other variations (e.g. is there a chance of black counterplay after 1.-Rh2+ 2.Kg6 d1Q 3.Sxe4+ by sacrificing his pieces for stalemate).
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