may 27 2014

Daniel's word

Once again, we could only deplore the absences of the regulars, and once again the presence of Antoni (in great form, moreover) saved me from finding myself face to face with the Master.
So thanks be to Antoni and, since we are on devotions, we pray for an end to the difficulties that keep the Registrar away, firstly for his benefit, secondly for the benefit his friends will derive from his presence and finally for the obvious benefit of the eventual readers of the minutes!

As is now the rule, the course began with the correction of the exercises.

Exo 1
White to play and draw

A quite astonishing, I would even say mystifying, study by the great Timman.
I had looked for this exercise, but I must confess that I had found nothing, or more exactly, I had the impression that I had demonstrated that it had no solution (we will come back to this point later).

So, how to attack this study?
1) Firstly, by having a precise idea of the outcome of the fight R et B against two Bs. One is quickly convinced that it is a draw (except of course, for a very specific position) as the B cannot even afford to be exchanged.
2) Then by having a clear idea of the outcome of RB and P against a pair of Bs and possibly P (although the Pa6 looks lost). It is easy to convince oneself that this is usually won by the R camp because now the B can afford to exert much stronger pressure since he no longer fears exchanges (due to the presence of the P).
3) Except very particular case, it is thus enough for the B to preserve their "c" Pawn to win in a technical way and for the W to capture this P to make draw.
We see that, thanks to these reference marks and although we have not yet made any calculation, the analysis becomes much easier.

Let's move on to the analysis.
** Bc4 is in the grip and must therefore play. If he plays anywhere (e.g. 1:Bg4) the B's do not immediately play Rxa6 because of 2:Bg3+ winning c7 but 1:...c6! and the move 2:...Rxa6 with win cannot be avoided.
** I therefore considered 1:Bd3 with the idea Rxa6 2:Kb3 and the hope of persecuting the R which would be riveted to the defence of its Ba3 does not last long after 2:...Ra8 3:Bg3+ Kd4 winning a time on the Bd3.
** It remained to look at 1:Bd7 with the idea of Rxa6? 2:Bg3+ Ke4 (or Kf6) 3:Bxc7! and after 3:...Ra7 the W's have a check to save their bishops and draw.
But of course, the B's are not obliged to play the immediate Rxa6 and on 1:Bd7 Kd6! 2:Bb5 c6! the B's save their P win a6 and the game.

I said I felt I could demonstrate that there was no solution.
This is sometimes the case with apparently simple positions that seem to lend themselves to exhaustive analysis.
On a more complex study, you can't find it and you think you just haven't found the idea or the move. But when you feel you can demonstrate everything, it's more troubling.
Of course, I've encountered this phenomenon often enough to know that the study does have a solution and that my reasoning is wrong because I've jumped to a conclusion somewhere.

And this is of course the case here, and I leave the reader to find out for himself the gain in the 1:Bd7! Kd6! 2:Bb5: c6!

Master's words



Van dijk nSavchenko

Four helpmates, one of which (in 5) is in my opinion difficult and another (in 6) very easy. In the first one, these are really two solutions: one gives no indication of the other!

Three 2#'s spread over almost a century. The four 3# are spread over almost a century and a half. Then an easy 4# and a (not so easy) 5# that you probably don't know.


A f mackenzie

To close this selection, a long problem and a superb selfmate for which the solution deletion process broke down.

A little joke from the former 3rd world player. A trick to find, but what a trick... ! Let us recall for the understanding of this study that the pair R+B may win in 200 moves (!) against 2N, but it is powerless against 2B!

A trial whose refutation 10 moves later, or even more, we understand. But the motive is unusual: not to play, at the end of this delay, a move too strong. A small reminder in the pawn endgame about the conjugation rectangle and the overflow.

Mandler 1

An Alekhinian game, except for one detail: the strategic reconversion from Queen to King wing is done in the opposite direction! See you, God willing, on 10 June for the fairytale course. Have a good time.



Very lively commentary, especially of the Timman study which should be enjoyed without delay. There was also a pawn endgame but it was probably too banal for Daniel.

Karpov kamsky

Checked, in the game of the day after 6 Nbd2, we play ...Bf5, but after castling. And on 6 Nc3, one must take into account ...dxc4 and ...Be6. On move 15, Karpov's idea is g4 and f5, but one move later: 15...Rfc8 16. Rc1 Bf8 17. g4 Bf8 17. g4 Rc4 18. f5!

Variation 23...Nd3! Black continues with 38...Rc2! accepting the loss of a pawn willingly. On 36...b5! I forgot during the lecture the amazing defence (if White continues as in the game) 37 Be3? g5! 38 hxg5 Ndxe5! after which the Knight is impregnable, the wQ being locked in the middle of the board.

The variation 52 Kg3? Qc3+! is a master greffier's trick, completely forgotten by your fake master. However, I will only give one "?" because White has the 53 Qf3! save.


Dinner: the "illegal cluster" that becomes legal on removal of any unit was so. In circe, at the position bPh3 & h5, add wR + wP + bR + 2bP to get an illegal bundle. Twin: bPh2 & h5. Have fun. 


White to play and win (from the no less great Pervakov).

I had also looked for this exercise.
As with the previous one, I didn't find it, but here the position is sufficiently complex that I didn't for a moment have the idea that I had demonstrated anything (except a certain weakness).
The B's threaten (among other things) to capture the Qd8 or to play a1=Q or to mate by 1:...g2+ 2:Kg1 Qh2#.

1) So I obviously considered 1:Qxc7 Rxc7 2:a8=Q (which has the advantage of warding off the main threats) 2:...g2+ 3:Kg1 Rc1 4:Nxf3! (not 4:Ne2? Bg3+! 5:Nxc1 f2#) 4:...Bg3+ 5:Ne1! and wins.
I thought to myself that this didn't look like a solution and that something (an B-resource) must be missing.
Indeed, the preceding variation is well forced until the 3rd move, but after that the B's have a better suite.

2) I looked (and I also advise the reader to look) if there were other candidate moves (in the first move).
Indeed, there is a very strong W-move which equals all the threats stated above and which moreover threatens strongly.
1:Qg5! certainly, this move page g2+ in a very effective way. Moreover, it threatens mate in 1 move.
I will now indicate what I had found at home. Obviously, if I give the following variation, although it is not the solution, it is because it is absolutely essential to see it to understand and appreciate the solution.
1:...Qf4! 2:Qxf4 g2+ 3:Kg1 Bf2+ 4:Kxf2 g1=Q+ 5:Kxg1 a1=Q+ 6:Kf2 Rc2+! 7:Ne2 and I leave it to the reader to find out how the B's draw in 3 moves in a forced way.

3) This is therefore not the solution since the B's draw. But the solution is a refinement of this variation.
1:Qg5! Qf4! 2:a8=Q! (why?) Rxa8 3:Qxf4 and play the variation already seen.
The reader will easily see how putting the Rc8 in a8 changes things. Finally, the reader will remain vigilant in the manoeuvre of his K at the end of the variation to win with Q against R and P.

Game of the day.
It opposes Karpov to Kamsky at the time when the latter was the rising young prodigy (with a father who pissed everyone off).
1.d4 Nf6 2.c4 g6 3.Nf3 Bg7 4.g3 c6 5.Bg2 d5 6.cxd5 Karpov is a player who is in no particular hurry to smooth out the position, so one wonders why this exchange which seems to eliminate tension.
The Master gave an answer during lesson but I have not managed to write down the reasons in enough detail to report them here. Basically, it comes out that the W's have no good move other than this one in this position; 6:Nbd2 running into Bf5 followed by who knows what else.
6...cxd5 7.Nc3 0-0 8.Ne5 to try to oppose Nd7 and drive B to play e6, conceding Ws some space advantage.
8...e6 9.0-0 Nfd7 10.f4 Nc6 11.Be3 Nb6 12.Bf2 Bd7 Previously, against the same Karpov, Kasparov had tried 12:...Ne7 with the idea of countering B's e4 by dxe4 and occupying the d5 square.
13.e4 Ne7 14.Nxd7 Qxd7 15.e5 Here is a good example of Karpov's contribution to the strategic conceptions of the time. The closure of the centre does not bother him despite his two Bs behind the pawns. 15...Rac8 deciding which R to put on c8 is a matter of taste. Depending on whether one prefers to keep a R on the "f" file or to clear f8 to play Bf8.
Note, for the attackers, that Karpov indicates an idea on Rf-c8 which impressed me a lot: 15...Rfc8 16.g4 Bf8 17.f5! gxf5 18.Be3! (why not Bh4, by the way) followed by 19:Bh3! the take on g4 leaving the Pf7 very exposed.
16.Rc1 a6 17.b3 Rc7 18.Qd2 Rfc8 19.g4 Bf8 [19...f5 was strongly to be considered.] 20.Qe3 Nc6 21.f5! a move that requires precise calculation on the kingside but especially on the queenside and in the centre as the B's next move continuations must have been correctly judged.
21...Ba3 22.Rcd1 and now the undefended Nc3 seems to allow the B's several combinations, but neither Nxe5 nor Nxd4 goes because of the undefended Nb6! 22...Nxd4? 23.Qxd4 Bc5 24.Qf4 Bxf2+ 25.Qxf2 Rxc3 26.Qxb6.
22...Nb4 23.Qh6! and Nc3 is not immediately capturable because Nb4 prevents the return of the B to f8 to parry the mate after f6
. If 23.Nb1 Nc2! 24.Qh6 Bf8.
23...Qe8 Karpov had planned 23...Nd3!? 24.Nxd5! Nxd5 25.Rxd3 Bf8 26.Qg5 which indeed seems dissuasive, but the Master associated with the computer showed that this move 23:...Nd3 was in fact excellent because this position is curiously at the B advantage when one pushes the analyses which I followed with attention but which, for this reason, I could not note.
24.Nb1! Bb2 25.Qd2 Nc2 Karpov had planned 25...a5 26.a3! (26.Qxb2? Rc2 27.Qa3 Rxa2–+) 26...Rc2 27.Qe1 Qb5 28.axb4 Re2 29.Qxe2 Qxe2 30.Rd2 which seems excellent for W's, but once again if we push the calculation we discover that after 30:...Qb5! 31:Rxb2 Qxb4 the White play will not be so easy.
26.Kh1 Qe7 27.Bg1 Nd7 28.Rf3 Qb4 29.Qh6 Qf8 30.Qg5 Qg7 31.Qd2 (31:f6 was seriously to be considered).
31...b6 32.Rdf1 a5 33.h4 Nb4 34.a3 Rc2 35.Qf4 Nc6 36.Bh3 Here again, we thought that 36:f6!? was interesting but the Master found the defence 36:...g5! 37:hxg Qg6
36...Nd8? 36:...b5! seemed necessary and perhaps allowed B to hold, based on the conviction that W has no winning attack... This debate has occupied us for some time but it seems to be true.
37.Be3! It is quite extraordinary that in this position, Karpov judges that the W win will be made on the queenside by exploiting the bad position of the black pieces!  To do this, the R's must be exchanged twice by Rf2, and to do this, Be3 must first be played to maintain the d4 defence while preventing a subsequent Qh6.
Faced with an alien player who sees this while everyone else is looking for mate in force on the kingside, it is certain that the young prodigy of the time has something to occupy himself
37...b5 [37...g5? 38.hxg5 Nxe5 39.Qxe5] 38.R3f2! b4 39.axb4 axb4 40.Rxc2 Rxc2 41.Rf2 Rxf2 42.Qxf2 Ba3 43.Qc2+- Nxe5 44.dxe5 Qxe5 45.Qc8! Qe4+ [45...Qxe3 46.Qxd8+ Kg7 47.f6+ Kh6 48.Qf8#]
46.Bg2 Qxb1+ 47.Kh2 Bb2 48.Qxd8+ Kg7 49.f6+ Bxf6 50.Bh6+ Kxh6 51.Qxf6 Qc2 it's completely over, but you have to stay focused because 52:Kg3?? threatening g5+ and Bf3# would allow 52:..Qc3+ while 52:g5+ Kh5 53:Kh3? would allow Qf5+
52.g5+ Kh5 53.Kg3! now, Qc3+ would be faced with Bf3 check! 53...Qc7+ 54.Kh3 1–0
Finally, there was the fairytale session at the dinner table.

For once I won't report the whole problems because the first problem presented a Take&Make with Nightriders, which has already been explained last time and therefore rather nice, but in a twin selfmate helped, and I don't like hs and if I liked one solution, the twin didn't seduce me.

Then, the second problem was more fun, but in my opinion much too difficult to find and I don't see what to explain to help find the solution. So I'll just give it, and show why it's a solution, which should be more interesting to curious readers.

Problem 1.
hs#2,5.    Twin : Na4 on b4.

NIb1 and NId5: Nightriders.

I remind what I have already explained in the report of the April 15 session.
** Take&Make: The pieces normally make the moves without capture. On the other hand, when a piece captures it must imperatively (to finish its move) make a move according to the movement of the captured piece. A capture is only possible if the associated move is possible.
** 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 NId5 can play f6, h7 but not g8 because the 2 steps to get there are not identical.
** I skip the self helpmate, which for me is almost always a sign of impotence: the author has a selfmate one move position, but he is unable to make it into a worthwhile selfmate n move, so he gives another starting position and says that one must first play as in a helpmate to reach this mysterious position which will be a selfmate one move. Obviously, this becomes much harder to find, especially in Take&Make where the possibilities in helped are multiplied.
To sum up: both sides cooperate (assisted) in a first step to reach a position which is a selfmate one move.



hs#2,5 Take&Make with Nightriders

white Qc7 Ke1 Rd7h5 Sa4 Nd5 black Qc8 Pb3 Kb5 Ra1h2 Nb1
b) Ca4-->b4
d5, b1: Noctambules
} a) {first twin : Here are the preparatory moves :
} 1...Qc8*d7-e7 +{(the Q capturing a R ends its movement with a move of R). } 2.Nd5*e7-e8 +{(the Nightrider ends its capture of the Q with a Q move) } Kb5-a6 {and we arrive at the selfmate one-move position intended by the author
selfmate 1 move.
That is to say that the W's play a move that FORCES the B's to checkmate them. (the reader will have understood that NIb1 must be forced to play).

} b) wSa4-->b4{twin a4-->b4
} 1...Qc8*c7-f4 2.Nd5*f4-f8 + Kb5-b6 { selfmate 1 move
I am not sensitive to this genre or to the twin, which is of no interest to me. One can say that the matrix is shifted one row to the right, but in a helper it's not difficult, especially since in the first one, if one adds a column to the right of column "a" there is still a mate and it's that of the twin.
Oh yes, I forgot: the two solutions.
selfmate 1: 3:Qe7+ forcing NIb1xe7-c7#
selfmate 2: 3:Re7+ forcing NIb1xe7-d7#
white Qc7 Ke1 Rd7h5 Sa4 Nd5 black Qc8 Pb3 Kb5 Ra1h2 Nb1
b) Ca4-->b4
d5, b1: Noctambules
} a) {
} 1...Qc8*d7-e7 + 2.Nd5*e7-e8 + Kb5-a6 3.Qc7-e7 + Nb1*e7-c7 # {

} b) wSa4-->b4{twin a4-->b4
} 1...Qc8*c7-f4 2.Nd5*f4-f8 + Kb5-b6 3.Rd7-e7 + Nb1*e7-d7 #
Problem 2.
Cours2014052704       +       Cours2014052705
The position has only 2 B pawns.
The problem consists in adding white K, R, P and black K, R, P, P (shown next to it) to obtain an illegal cluster in circe.
And with the same statement, there will be a twin by moving Ph3 to h2.
** placing the pieces, well that's classic.
** Circe: when a capture is made, the captured piece is reborn on the square it occupied before the start of the game. This square is perfectly defined for Q and B. For the rest, Knights and Rooks are reborn on the original square of the same colour as that of the capture (thus a wR captured on d6 is reborn on a1, whereas if it is captured on d5 it is reborn on h1, the initial square of R of the colour of the square where the capture is made). Pawns are reborn on the same file as the one on which they were captured.
Finally, if the rebirth square is occupied, the piece does not reborn (it disappears from the board, as in orthodox chess).
The reading of this explanation can be frightening, but really if one reads it without preconceived ideas one realizes that it is very simple.
I have indicated the bare essentials, but for the curious reader, circus chess is defined more exhaustively here.
** Illegal cluster.
The resulting position must be illegal, but it becomes legal as soon as any of the pieces are removed (excluding Kings, of course).
1st twin. Let's examine solution.
The position is obviously Black to move.
** First question: why is it illegal?
** The last W move can only be the move Rf4 (and not Rxf4 because the captured piece on f4 should be on the board because of the circe rule and all the potential resurrection squares are free).
** Now consider the position just before this last W move.
Now that wR is elsewhere on the f-file, (e.g. on f6) we have a position White to move.
What was the last B move ?
There is none:
-- It cannot be a capture made by a bP because the captured piece should be on the board, with all the potential rebirth squares free.
-- It cannot be Kg5-g4 because the bK would have been in check.
-- It cannot be Kg5xg4 because the g4 piece should be visible after rebirth.
The position is therefore illegal.
** Second and more interesting question: Why is it legal to remove any piece or pawn ?
** If you remove a bP, you no longer have the problem of finding a last black move before Rf4+.
If you remove h7, you have h7-h6. If you remove h5 you have Rh5-h4, etc.
** The question is how to legally obtain this position if you remove the Pf2.
Answer: The last W-move is Rf1-f4, which allows the previous B-move Kg5xFg4 (the Bg4 being prevented from reviving by the presence of the R on f1)
and the position comes legal.
This position does not pose a problem of legality and B can play
1:...Kxg4 (the B does not revive) 2:Rf4+ and we get the diagram position without the Pf2.
Second Twin (pawns h5 and h2).
Examine the solution.
This time, it is the wK who is in check (he is even mate, but it is not important here).
** Why is this position illegal ?
** The last B move can only be h3xg2 or f3xg2. But what was captured on g2?
** It cannot be a R for two reasons. It cannot come from f2 or h2 since these squares are occupied. And it cannot come from g1 when playing Rg1xg2 because nothing could be captured on g2 since all the black piece respawning squares are free!
** It is thus a pawn.
Cours2014052709  Position A.
B plays f3xg2.
But what is W's last move?
There is none.
** Second and more interesting question: Why is it legal to remove any piece or pawn ?
** If we remove the Rh6, the last W-move we were vainly looking for in position A becomes h7-h6.
** If we remove the Ph7 (or Ph5 or Rg6), the last W-move we were vainly looking for in the A position becomes Rh6.
** If we remove the Ph2 or Pf2, then the position of comes legal by making a R-catch on g2.

W plays Rf2-g2 (or Rh2-g2 if Ph2 is missing) and B answers fxg2.

The position thus becomes legal if one removes an element, whatever it is.

A last remark which is purely technical and which concerns only the drafting: in this type of problem, the English can say "... because the captured unit would have been reborn" but I did not find a simple solution to say that such a capture is impossible because the captured piece would be ? The verb reborn has no past participle (and consequently no compound past tense either).
So I had to do each time figures like "because nothing could be captured in g2 as all the black piece rebirth squares are free".

Last session of the season: Tuesday 10th. It will be dedicated to the fairy tale. We hope the great Michel will come.

light down

Very good maestro, what courage!

The initial comment from "For once" to "seduced" should be changed or deleted.

The last comment in my own entry should also be deleted.

Twin b of the cluster: without the Rh6, the last move is, not h7-h6 but h6-h7.

One could say "he is reborn" but this form is unusual, and it is not even a tribute to Olivier, the pronunciation being different!


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