april 15 2014

Daniel's word


Due to the repeated and regrettable absence of the Chief Recorder, I was forced to replace him.

Unfortunately, as I am far from being as efficient as he is, it captures too much of my time to be able to carry out this task in a reasonable time.
Nevertheless, here is a brief account of the session, hoping for a rapid end to the difficulties that are keeping Rémy prevented and away from the course.

Attendance was reduced as only Marc was present. He therefore prevented the course from turning into a private lesson and allowed it to proceed more or less normally thanks to the invaluable support he provided for the analyses.

I prefer to develop the study solutions by adding comments that could be considered as chatter, but I hope that this will help some possible readers to better understand some ideas in order to be able to present these studies in their rooks in the framework of initiations that they could carry out in their club.

Correction of the exercises.
Exercise 1
. The position given was supposed to correct an erroneous analysis of Kramnik that I had erroneously attributed to Kasparov (but one only lends to the rich).




Master's words

NeukommThree helpmates 2# with one solution, which hardly exists any more. Only the first one will make you think. As well as another one in 3.5 moves (White to move).

Two 3# from a world champion. A tip for the second: 2 X 3 X 2! An old strategic 5#: immobilisation of a trick. A very nice moremover where I left the solution lying around. Before reading it, see if you think of the 5th move. I also leave the solution in the last miracle of the selfmate wizard: Quadruple promotion coupled with the 4 moves of a single pawn.

We know about Kasparov's sloppy analysis. It is much rarer to capture Kramnik at fault, but it can happen!

Conjugated squares in the Bishop and Pawns endgame, that changes us a bit... A recent prize, but a judge like we find less and less. 

Marshall rubinstein

A master of defence (and strategy) in... attack, against a master attacker. It is not sad. But curiously, the brilliant move that was praised, the opponent giving up 6 moves later, was not at all the best. Worse, it was one of the few not to win!

Watch out! See you next week, on Tuesday 22 April. May God keep you.

Have a good time.

1:...Rb5+ 2:Kc4! Ra5 3:Rg6 Ke7! seems to ensure the draw because if it is useful for the B to capture f4, it is even more urgent to bring back the K to participate in the defence. The bK confined on the "d" file cut by the wR on the "e" file would be hopeless against the wK accompanying his P "g". 4:Kb4! Rd5 5:Kc3! and I did not note the continuation of this very fine fight between the two camps leading the N to the draw because Marc proposed a very relevant winning plan.

I leave the reader to look for (after 3:...Ke7!) the W's plan to win the Pf5 while keeping the Pf4.

The attempt is as follows : 4:Rb6!
The idea is to prepare the R move on the 5th to win f5 while using the R beforehand to avoid losing f4.
So we do : Rb6-b4 then Kd4 then Rb4-c4-c5 (possibly preceded by a check on the 6th if the bK is there).
This plan occupied us a good moment and I leave it to the interested reader to look for the Black defence, which the Master will not fail to indicate in his electronic report.
Exercice 2.
White to play and draw.
I searched for this study, found some basic ideas including a position in which the B's lose because of a zugzwang but the deeper idea of the solution had escaped me.
1:e3 (obviously, if B plays e3 himself the win will be easy). Let us note in passing that a typical solutionist reasoning (this is not pejorative at all) is to say to oneself that if one plays e3 in a row, Bh1 will have to come back to defend via g2 or f3 and that there is no visible difference between these two moves, and thus that there would be a dual, and thus that Pe2 must first force the B to play the unique move Bg2...
It is necessary to recognize that to reason thus in a resolution of study is a sign of experiment and cannot be condemned, but in this case e3 is indeed the only move, the two possible returns of the B by g2 or f3 form a quite minor dual which one will despise .... and all is well thus!
1:...Bg2(or Be3) 2:f5 (why?) Bh3 3:f6!! and here is the idea so difficult to imagine: W's fight without the trump of their Pf4 but use the fact that the b1-h7 diagonal is shortened by the Pe4 to simply threaten to push h7 (without ever doing it) and mobilize K and the opposing B to oppose it because the Bf5 is persecuted and has no other support than its K.
Let us note to appreciate this idea at its true value that the idea would quickly prove insufficient if the B were not afflicted by the presence of the Pf6 which they cannot get rid of and without which they would have a clear gain! (Hence the two!! at the f6 move).
3:...ef 4:Kh7 Bf5+ 5:Kg8 and W's will draw through several zugzwang (reciprocal) positions. The moves are quite easy to find if you have the right cues: W must obviously always threaten to push h7 and prevent B from advancing by Bg6 and Kf5.
5:Kg8 (5:Kg7? Bf5!) Kd6 6:Kf8!! a move that seems difficult but is quite logical: 6:Kg7? Ke6! allowing Bg6 or 6:Kh8 Ke5 7:Kg7 Ke6. Here, one can afford to let Bg6 play because Kd6 is not in a situation to support this B in the next move.
Next session exercices.
Exo 1.
White to move. Position taken from an analysis of a Degraeve Bologan game played in Belfort.
Exo 2.
Black to move. Position from a variation of a Golutchabov - Tregubov game.
Among the 5 candidate moves a4, axb, Kf3, Kg4 and Rf3, only one is winning. Which one, and how
Game of the day.
Marshall - Rubinstein.
1.d4 d5 2.Bf4 Nf6 3.Nf3 e6 4.e3 c5 5.c3 Nc6 6.Bd3 Qb6 7.Qc1 Bd7 8.0–0 Rc8 9.Nbd2 Be7 [Tartakover played 9...Nh5; and Kmoch reports 9...cxd4 10.exd4 Nb4 11.Be2 Bb5 but 12.c4 and it's complicated] 10.Rb1 10:Qb1 is interesting to play b3, a3, followed by b4 or Ra2 or Re1 followed by Ne5 with a flexible and pleasant play. Kmoch proposes h3 but from now on, the continuation cxd; exd Nb4 is good because c4 is not possible after Bb5, exchanges and c4 and Nd3 at the end of the variation. 10...0–0 11.Qd1 Rfd8 12.Ne5 Nxe5 13.dxe5 Ne8 A brave move against Marshall. Ne4 was easier because of the face-off d1 d8. 14.Qh5! f5 15.Rfe1 [15.exf6 Nxf6 16.Qe2 Be8 17.e4? Bh5 18.Qe1 Bg6 gave nothing to W] 15...Bb5 16.Bc2 Qa6 17.Ra1 Bd3 18.Bd1?! [18.Bxd3 Qxd3 19.Nf3 was good for white]. 18...Qb6! now N catches the advantage. 19:b3 is no good because of Qa5! but what to play? 19.Nb3?! a5!? The Bs want to chase the N without giving d4 [19...c4 20.Nd4 Qxb2 21.Bf3 Nc7 left the advantage to B] 20.Bf3 Be4 21.Re2 Nc7 22.Rd2 a4 23.Nc1 g6 24.Qh3 Kh8 Kmoch says it's to play g5, but g5 can be played straight away if you want and besides 24:..d4! was stronger with the idea 25:cxd4 Qb4! 25.Qh6?! [It was necessary to play 25.Bh6 g5 26.Qh5; or 25.Ne2 g5 26.Bh5! Rd7 27.f3 and they escape] 25...Bxf3 26.gxf3 g5 27.Bg3 d4! 28.exd4 f4? [It was better to capture on d4 : 28...cxd4 29.cxd4 Rxd4 30.Nd3 Nd5] 29.Ne2 fxg3 30.hxg3 Qc6 31.Qh5 Rf8 32.Kg2 Qe8 33.Qg4 Qg6 34.Rh1 c4 with idea Nd5 followed by b5, b4. 35.Qe4 Kg7! 36.Qxb7 Nd5 threatens g4 37.g4 Rxf3? In this position, the brilliant move is the only one that does not win among the 3 candidate moves Qf7, Rb8 and Rxf3 38.Qxc8? [38.Kxf3! had to be played 38...Rf8+ 39.Kg2! Qe4+ 40.Kg1 Qxg4+ 41.Ng3 Rf3 threatening a3! 42.a3! and the Bs had to catch the draw by 42..h5 43:Rxh5 Rxg3+] 38...Qe4 without fear 39:Ng3 because of Rxg3 30:Kxg3 Qf4+ 39.Kg1 Ne3!–+ 40.Ng3 Rxg3+ 41.fxg3 Qb1+ 42.Kf2 Nxg4+ 43.Ke2 Qe4+ 0–1 and resigns, (because Marshall will soon end up in prison, a shame !)
Finally, after all these ordeals, I found myself alone with the Master for the meal and the fairy tale part, Marc having to get a minimum of rest in anticipation of his taking up his duties well before dawn! (it's the horror).
Fairy problèm.
h#2  3 solutions.
Rois KoBul
Anti Andernach
Piece on b5 is a Nightrider.

I insist that if this position seems inaccessible to the uninformed amateur, it is a misconception and the reader who would like to be led there is sure to come away enchanted.
** Helpmate : The Ws and Bs cooperate to achieve the required goal. Here, the Bs start and help the second W move to make mate.
** Kobul Kings: When a piece is captured, the king of his colour takes the march on that piece until another piece is captured.
** Take&Make: Pieces normally make moves without a capture. However, when a piece captures it must make a move according to the captured piece's movement. A capture is only possible if the associated move is possible. This will be illustrated in the explanation of the first solution.
** Anti Andernach: When a piece (except a K) plays, if it does not make a capture it changes colour.
** Nightrider (noted NI): This is a piece that plays as many moves of N as it wants, but all the steps must be in exactly the same direction. For example, the NIb5 can play c7, d6, f7, h8, d4, f3, h2, c3, d1 and a3.

To checkmate, one or more B-pieces will appear thanks to the Anti Andernach rule.
And the bK will have to be locked in a suitable cage: the Take&Make rule will be used once to make it travel and the Kobul Kings rule will be used to give it a mode of movement adapted to its cage.
Let's look at this in more detail:



black Pa7b6a5g3g5f6g7 Kc2 Rc4 Bh1 Nb5
b5: Nightrider
right side of the board to advance and left to retreat or directly on the move

. First solution (I remind you that the B's start):
} 1.Bh1-c6=w {As it is not a catch, Anti Andernach makes the B become White (it is now W's turn to play) !).} Bc6*b5-f7[c2=rN] {(this is a W move). This move is a catch so:
1) The B stays W
. 2) Take&Make so the Bb5 has not finished his move, he has made the capture so he also has to make "in the process" a knight move of the night. He goes to f7.
3) Kobul Kings. So the bK will now move as a nightrider.

} 2.rNc2-a6 Bf7*c4-c8[a6=rR] # {This move is a capture so:
1) The B remains White.
The B is the only one that remains White. 2) Take&Make so the Bc4 has not finished his move, he has made the catch so he must also make "in the process" a move of R. He goes in...
3) Kobul Kings. So the bK will now move as a rook.
The bK is in a6, moves as a R, and it must be mate. To do this, the Ws do have a B on c4 but that B still has to play a R.
move. So the mate move was 2:Bf7xc4 - c8!.
I hope I have been understandable. Note how all the pieces were used in the solution. And, as the authors are extremely brilliant, the 3 solutions will have these same qualities, without ever repeating the diagrams.
Very high art !

Solution 2.

} 1.Rc4-h4=w {(becomes white)} Rh4*h1-d5[c2=rB] {(the wR goes to d5 and the bK moves as an B)} 2.rBc2-h7 Rd5*b5-h8[h7=rN] # {(the R has completed its move with a knight move of the night, and the bK moves as a Nightrider. He is therefore mate because he can only go to f8 where he would still be in check.
Perfect unity with the previous solution.
Examine the process carefully and the 3rd solution will not escape you.
Solution 3.

} 1.Nb5-d6=w Nd6*c4-b4[c2=rR][+wNb4] + {(the bK moves like a R)} 2.rRc2-h2 Nb4*h1-f3[h2=rB] # !! {(The bK h2 moves as a B and cannot escape the check of the NIf3).}


A perfect achievement. You don't need to be an expert to admire beauty when you encounter it.

spring fuzz

Kramnik's analysis was 54 Kd4? (instead of Kc4!) Rb4+ 55 Ke5 Re4+ 56 Rf6 and here the great Vladimir omits a saving move, which you can easily find.

Mark's plan seems to be countered by the bK's counter-attack via h5, but it seems that 54 Kc4! still wins: 54...Ra5 55 Rxg6 Ke7 56 Kb4! (instead of Rb6) Rd5 (threatening ...Rd4+) 57 Kc3! Ra5 58 Rf6! and, as a result, it is the wK who aims for h5.

Note that the "!" given to 54...Kd7 after 54 Kd4? is no longer justified. My apologies.

For the study of the Welsh, I give the conjugated square pairs, which will help the most Cartesian minds: g7/e6, f8/d6, f7/e5 & g8/d5.

Exercise 1: the diagram did occur in the game and one could ask "how to undo", but I am not that cruel. The question was simpler: in Ke2, Bb4, Pa2, b3 & c2 / Kf5, Pe3, f2 & f4, does White to move lose as one player writes at 2650?

Exercise 2: White's player (and author of the analysis we are criticising) is called Goloshchapov.

Concerning the game of the day, I cannot resist the pleasure of sharing with the receivers of the report the remark I made the night after the course: Knowing that I had prepared, in view of the course initially planned for April 8, but postponed to the 15th, a game beginning with 1 d4 d5 2 Bf4, an opening that is not currently common, what was the probability that, on this same April 15, this opening would be found in the Korobov-Edward game (in Dubai), the game ending with the same result (black win) and thereby authorising one of the most beautiful tournament victories ever achieved by a Frenchman?

In the 37th, it is of course the 3 candidate moves Qf7, Rb8 and a3!

Perfect presentation of the fairy problem.

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