june 10 2014

a word from the boss

An ample harvest of helpmates to last four months. The theme of the first h#3 is very explicit. Only the last h#3, one of the twins of the h#4 and one of the solutions of the h#5 seemed difficult to us..
Three #2 spread over nearly a century and a half. The #3 are more recent. The last one will probably be a surprise for many. We end with a little Lepuschütz festival (#6, #7 & #12).
A part of the fairy course concerning Andrade's trick (which was more successful than expected), then the growing men (2 of the 5 problems have already been presented in December). Obviously, some moves don't pass and must be in notation...
It only remains to wish you an exciting summer, hoping that you will bathe in beauty in all forms...

Daniel's word

Sorry for the very late reply, but between the football World Cup and the "household chores".

First of all, and although the Master has already shown this position in his report, I would like to review the problem of B again. Andrade (1958) which made us suffer so much.

1 - how many white moves to allow Black to escape #1

Cours2014061001 Hoping to provide an unambiguous statement :
How many white moves allow blacks to escape mate in 1 move? ?
For example 1:Qa5 ? is not such a move, because on all Black moves, whites will have a checkmate possibility in 1.
The answer is 6. it's up to you to find them.
I see that the Master has provided in electronic format all the positions that he was able to enter..
I didn't write them down (busy as I was looking for them, sometimes even understanding them) and I don't plan to look for them to provide them because I'm just as incapable of reinventing them as I am of remembering them (on this last point, I'm ashamed)..
I tested on Abdurahmanovic/Mihajloski's h#6 and it took me almost 45 minutes, even though I had seen the solution during the course and had appreciated it, which is to say where I am now.
Strong indication: the BK is being mated in c4.
Let's move on to the continuation of the course.
As you will see, and indeed as you would expect, there is only very big.
And I repeat, you don't have to be an expert to recognise beauty when you come across it..

2 - h#2 4 solutions

Click in the right part of the chessboard to move forward, left to move backward...
or directly on the move in the solution

Jubilé H. Fougiaxis-40
1st prize

white Kh6 Pc4 Rb8 WLa5f3 Bh7 black BLa8f5 Kc3 Qb7 Pa3d2e2 Sa6f2 Rd8d4 Bc1h5
h#2 4 solutions
🨴a5, f3, 🨺a8, f5: Leo

{Everything is normal, but the pieces that I have represented as Q to left oriented are Leos.
The Leo moves like a Queen but needs a sautoir to capture. He lands, and so captures, at any distance behind the sautoir, provided that no square is busy on the way. For example, A8 has three legal moves:
a simple movee } 1.BLa8-a7 {and two catches } 1.BLa8*a5 {et} 1.BLa8*f3 {The first solution: I remind you that in an helpmate, the B begins
. Hence this ridiculous notation consisting of writing 1: followed by a Black move :
} 1.Qb7-d5 WLa5*f5 2.Kc3-c2 WLf5-c8 # {
with double check. A battery case where both pieces check what, in classical chess, imperatively requires a Knight.Obviously, as the author is brilliant,
the solutions together form a perfectly harmonious whole. The second solution :
} 1.BLf5-d5 WLf3*b7 2.Kc3-b2 LEb7-g7 # {double check.
I leave the reader to enjoy the other two solutions, which begin respectively with } 1.Qb7-c7 {and} 1.BLf5-c5

Now a simple and charming thing from the great Heinonen.

3 - h=8 Circe

Tehtäväniekka, 2011
2nd Prize

white Pg5e4f3g2c5 Ke8 black Pb2d2f2h2d4e6g6 Ke5
h=8 Circe (the captured pieces are reborn on
their original square if it is free. Otherwise, the piece disappears.
{Obviously, for an experienced solutionist, the solution is almost obvious.
For the neophyte, it's a miracle.
For the regular reader of the reports, the truth is somewhere in between. Let's take a closer look.
** The B's have a lot of pawns that will have to be stopped ... while we are in circe.
** The W's are very poorly equipped.
** We therefore deduce that the BPs will promote themselves, in order to move more easily,
. and that they will then disappear
in spite of the Circe rule which would naturally tend to revive them.
** The solution becomes more accessible (I remind you that the B starts). :
} 1.f2-f1=Q Ke8-d8 2.Qf1*f3 [+wPf2] g2*f3 {no rebirth cause King is on d8 square !
} 3.b2-b1=B Kd8-c8 4.Bb1*e4[+wPe2] f3*e4 {no rebirth
and I leave the reader to finish alone, because if he has arrived so far he has obviously understood the next move B and therefore the immediate move N. Etc.
I assumed that the Pg6 was preventing a demolition.
There is a solution not foreseen by the author :
} 1.b2-b1=Q Ke8-d7 2.Qb1*e4[+wPe2] Kd7-d8 3.f2-f1=Q g2-g3 4.Qf1*f3 [+wPf2] e2*f3 5.h2-h1=Q f3*e4 6.Qh1*e4 [+wPe2] f2-f3 7.d2-d1=Q f3*e4 8.Qd1-d3 e2*d3 {pat}

To continue in Circe :

4 - #10 Circe

Jubilé K.WENDA-70
feenschach, 2012 - 4th Prize

white Pb6 Ka7 Bc4f4 Re3 black Bc8g7 Kd4 Qh8 Pb7c5c3f3f5f6h4 Sf8 Rd7
#10 Circe
{A direct mat with a very nice manoeuvre presenting an echo between threat and achievement.
** Because of Circe, the Bc4 is not to be taken by the K - just as the Bf4 will not be engaged later.
** } 1.Bf4-e5 + {does not mate on the one hand because of} f6*e5[+wBc1] {and on the other hand
} 1...Kd4*e3 [+wRa1] {. The first two moves will thus remedy Kxe3 before attacking the bishop waltz.
} 1.Re3-e4+! Kd4*e4 [+wRh1] 2.Rh1-e1 + Ke4-d4 {
Now there is a mating maneuver that fails if the BQ is in h8 because in h8 it has control over e5.
. And, admirably, to deflect the Qh8 the Ws create a threat of exactly the same type but whose key square is d5 instead of e5.
. Obviously all this is obscure, but the clarity comes straight away :
} 3.Ka7-a8 ! {Where did this move come from?
The W's want to capture the black R with their white B, but being in Circe, this move would be illegal,
because the black R taken from a white square would be reborn in a8 ... checking. Hence this apparently surprising K move. The threat is now :
} threat: 4.Re1-d1 + Kd4-e4 5.Bc4-d5 + {to block the d5 square to play Bd3 without releasing the King
} 5...Rd7*d5 [+wBf1] {mat en 4} 6.Bf1-d3+ Rd5*d3 [+wBf1] 7.Bf1*d3+ {no rebirth !
} Ke4-d5 8.Bd3-c4+ Kd5-e4 9.Bc4-d5 # 3...Qh8-g8 { parrying the threat by adding a control to square d5 ... but removing a control from square e5 !
} 4.Bf4-e5 + {to obstruct e5 in order to play Be3.
} 4...f6*e5 [+wBc1] 5.Bc1-e3 + Kd4-e4 6.Be3-f4+ Ke4-d4 7.Bf4*e5[+bPe7] + Bg7*e5[+wBc1] {3 moves mat
} 8.Bc1-e3+ Kd4-e4 9.Be3-f4 + Ke4-d4 10.Bf4*e5# {no rebirth}
Continuons dans la diversité.

5 - h#2 3 solutions Patrol chess

Gerard SMITS
Probleemblad, 2010 - 5th Prize

white Pf6d3c4 Kg8 Rb2 Sf3 LId6 black Bh1a1 Kh2 LId1 Qf2 Ph3f4 Sd7g6 Rc3c6
h#2. 3 Solutions
Patrol Chess
The Lying Q's are Lions
{Lions: move like Queens, but they need a sautoir, both to move and to capture.
They can land as far as they want behind the sautoir as long as there are no pieces in the way.
Patrol: The pieces move normally, but to make a capture piece, and thus a check,
a piece must be observed by a piece of its camp.
Principle: 3 W pieces potentially check: b2, f3 and d6,
. but none of them fail because none of them are "patrolled" at the moment. In each of the solutions, one of these pieces will go and observe the other two simultaneously, triggering a double check. The piece playing this role obviously abandons a square close to the BK. The BQ will occupy this square so that the checkmate takes place. First solution, it is a helpmate, Black begins:
} 1.Qf2-g1 Sf3-d2 2.Rc6*c4 {Rc6 is activated by the Bh1, so it can make a capture.e
} 2... Sd2*c4 # {by double check by Ld6 and Rb2. The first solution having been given and the principle stated,
which for each solution allows you to know which piece W is playing and the square of the BQ, it will be very easy for the reader to find the other two solutions.
We have a new example of "exotic" double check.}
For a complete change, here is a #2 in Eiffel Chess.

6 - #2 Eiffel chess


white Bh6c8 Ka1 Qd6 Pd3c4g6f3 Se5c1 Rh3e8 black Bc6 Kd4 Qh1 Ph5a2b4f6 Sc3d5 Rg3c5
Eiffel chess
{** In Eiffel chess, all pieces, except Kings, can paralyse.
And all, except the Kings, can be paralyzed.
** A piece having the power to paralyse paralyses the opponent pieces it observes.
When a piece is paralysed, it can neither play without capturing nor play by capturing.
It does not give check, which is a consequence of its inability to capture.
If she herself has the power to paralyse, she retains it when she is paralysed.
** A piece paralyses one or more pieces of the opposing camp which it observes according to the following associations :
. - a Pawn paralyses a Knight,
- a Knight paralyses a Bishop,
- a Bishop paralyses a Rook,
- a Rook paralyses a Queen,
- a Queen paralyses a Pawn.
Kings are excluded from this cycle of paralysis.
For example, the Nc1 can play because the Qh1 is paralysed by the Rh3 and would therefore not check.
But then, isn't there } 1.Sc1-b3 + { mat ?
The answer is "no" because Nb3 checks but the Blacks are not mated, for a reason that I let the reader discover).
} 1.Bh6-e3 {does not ckeckmate, not that it is taken by Nd5,
because the latter is paralysed by the Pc4, but because the Be3 would be paralysed by the Nd5 and would therefore not check.
} 1.Qd6*f6 ! {if you have answered the above question, you can see that the Ws are now threatening
} threat: 2.Sc1-b3 # {since Ne5 is no longer paralysed by Pf6 !} 1...Sc3-b1 2.Sc1-e2# {
} 1...Rg3-g1 {pins the Nc1 but
} 2.Se5-g4# ! {preventing Rxg6 paralysing the Q.} 1...Rg3*f3 2.Se5*f3# {
} 1...Rg3*g6 2.Se5*g6# {
} 1...Rg3-g5 {freeing the e3 square because the Re8 is paralysed but} 2.Se5-d7# {
} {and finally} 1...Rg3*h3 2.Se5-f7# ! { only square of Knight not releasing the Rh3 paralysed by the Bc8,
} 2.Se5*c6 ??{ would allow } Rc5*c6 {parrying check !
Beautiful work !}
To be complete, it should be pointed out that there was also a problem with Franz Pachl who is a first prize winner and who was selected by the Master, so there is no doubt that it was a very nice problem..
but it is a Take&Make triplet with Zebrariders, Camelriders and neutral pieces.
It knocked me out and I lost it. So I refuse to try to explain something that I couldn't get interested in..
Sorry for the author, who would have deserved the presence of the "Greffier en Chef", holding the post and a true amateur of all these things even when they are difficult.
Note from the virtual greffier: from the description given, Pachl's problem seems to be corresponding to it

7 - h#2 Take&Make b)♟e5->c4 c)♙b5->f5

Harmonie, 2011 - 1st Prize

neutral Rd1 CAg3 NZf1 Nh2 black Be4f8 Kd5 Pb7c7a5e5c3g6g4h4 Sb8a6 Rh7 BZd4 white Pb5 Kg8
h‡2 (2+15+4) C+
b) ♟é5->ç4
c) ♙b5->f5
Take & Make
🨙f1: neutral Zebra (bonder (2,3))
🩃g3: neutral Camel (bonder (1,3))
🨂d1: neutral Rook
🨮h2: neutral Nightrider
🨓d4: Zebra

(a little reminder: the knight is a bonder (1,2))
{Indeed, there is enough to drop the pen, especially when one knows the richness of the Take&Make !
Little help: three black pieces can go to c5 and the Zebra to d4 can be captured in 4 different ways
by a neutral piece which will have to make a Zebra move afterwards, but BRxd4 would be a weakening}
Finally, as an introduction to the course, we had an extraordinary presentation involving different types of retrograde analysis.
When you explain these different types, it seems a bit artificial (even very artificial!) and the simple theoretical presentation makes you think about what life must have been like for flies in some English denominational orphanages....
The great Turnbull provided the material and the master explained it to us. Quite luminous presentation !
The Master wrote about it in his report, but the reader will have to hold on if he really wants to follow, proof that when the subject is difficult, the oral presentation is superior to the written explanation. For the same reason, I will not go into it either, except for the simplest example which I will try to develop a little more..
Growing Men. Whites play and mate in 5 moves.
** We have already seen the Growing Men.
A piece cannot play a move shorter than its previous move..
This means that as soon as a piece has played a move of geometrical length n, all its subsequent moves must have at least this length..
** The length of a move is determined by measuring the distance between the centres of the start and finish squares..
A K therefore makes moves of length 1 (when it makes an orthogonal move) or 1.4 (root of 2, when it makes a diagonal move).
The Knights are not directly impacted by this rule, as all their movements have the same length (root of 5, i.e. about 2.24)..
** When we are interested in the moves played previously (retrograde analysis) we will apply the following principle :
All moves that cannot be shown to be prohibited are allowed.
In the case of Growing Men (it would be more precise to say "non-shrinking"), a K or a P will have the right to play a move of length 1, unless the contrary can be demonstrated.

8 - #5 Growing Men


white Se2 Kf5 black Ph6g5h4h3g3 Kh5
Growing Men
{(Apparently) obvious solution :} 1.Se2-d4 ! { threatening } threat: 2.Sd4-e6 ! {et} threat: 3.Se6-g7 # {It seems that in such a simple position, the Bs cannot do much. But therein lies the genius of Turnbull. The B's really have the 3 moves g2, h2 and g4 because we cannot prove the impossibility of any of these moves. ** g2 is of no interest in the face of the aforementioned threat. ** h2. Here is indeed a good idea of defence: free h3 to play h4-h3 and give a square to K. Alas, after } 1...h3-h2 {which is legal !, we can affirm that the Ph3 has played ONLY moves of length 1 (since we are in enlarging chess and he has just played a move of length 1). -- So it comes from h7. -- But then h4 comes from another column. So h4 has already played a move of length 1,4. -- But then h4 is not allowed to go to h3. AND THEREFORE the W's always checkmate by Ne6 and Ng7#.. ** } 1...g5-g4! {This is a very nice move. It is indeed: -- We can't prove that g5-g4 is illegal, so the B's can play it. -- But then, g5 comes from g7. -- But then, h6 can only physically come from h7! -- But then, g3 and h3 come from other columns than g and h! -- So g3 and h3 are not allowed to play g2 and h2 respectively. And so on} 1.Se2-d4! g5-g4! 2.Sd4-e6?? {the B's are pats! Composing this would already please a beginner composer like me. Of course, for Turnbull, it's not enough....
} 1.Se2-d4! g5-g4! 2.Sd4-f3!! g4*f3 {only one move, but from now on none of the f3, g3 and h3 pieces can play
} 3.Kf5-f6!! {So far, there is no evidence that either of the two Kings has played a move longer than 1, so they are not constrained. The B's are now in zugzwang and will be forced to play a move that will strongly degrade their possibilities..} 3...Kh5-g4 4.Kf6-g6! { BK having now played a diagonal move, it no longer has a legal move..
} 4...h6-h5 {forced} 5.Kg6-g5# ! {note that} 5.Kg6-f5 {would be illegal because it would be checked !}
All this is perfectly explained by the Master in his report, I just wanted to go into a little more detail. The other examples would deserve the same work to make them accessible to the greatest number, and to pay all the more homage to Turnbull's genius, but it is also necessary to leave an advantage to the face-to-face course (not to say "live"). Have a good holiday.

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