december 10 2013

A word from Daniel

for this fairy session, it was Daniel who did it.


Growing men.


A piece cannot play a move shorter than the one it has played previously (i.e. the last time it played).
The following problems are originals of Paul Bissicks and are only there as an initiation.


1 - #1 Growing men

Paul BISSICKS

white Qd5 Kc7 black Pb7 Ka8
#1
Growing Men
{

} 1.Qd5*b7? {Which is a move of length 2.28 does not even check the Kh8 located at 1.4.} 1.Qd5-a5?! {A long shot of length 3 checks well against Ka8 ... who counteracts this one with
} Ka8-a7!! {shortening the distance and thus getting out of reach.
Henceù} 1.Qd5-a2# {because this move has a length of 4.24 and therefore makes it possible to reach squares at a distance of 5 or 6 on the next move..}


 

As you know, the fairy tale does not lend itself to the usual means of transmission for players. I will therefore content myself with providing you with a harvest of problems... orthodox or heterodox, but affordable by the playerlambda.
 
The 4 helpmates in this session are a good training for the Christmas period. When we understand the idea of the h#3, two solutions come, but the 3rd? The first h#6 has a trivial "apparent play": White moving would checkmate in one move; needless to say that the solution has nothing to do with it. And the last one is also quite ingenious.
 
The first #2, like the h#2, celebrates the 80th birthday of a nice and talented young man, although from across the Channel. In the next #2, can you tell me what the f3 pawn is for, since the BR is not allowed to play anyway? Well, forget this remark and solve the problem: everything will become clear. The last #2 could not be given in a "solving-show", because all the competitors would give the key in less than 3 seconds. !
 
The first #3 is surprisingly simple, when you know the usual productions of this author. The second one realizes an idea... 4 times. The great Vladimir's #4 is deceptive: apparently there are 2 or 3 ways of mate in 4. But no, that's not it at all! The #5 seemed difficult to me, but perhaps I had my head in the bag. For those who know the author for his batteries where the unmasking piece plays a 2nd move, that's not the idea at all.
 
The long problem that follows is... the easiest. But don't forget the second variant. I end with a nice selfmate; knowing the infatuation of most players for this compartment of art, I prefer, once again, to "forget" to remove the solution.
 
And all the same, a fairy problem, or rather an April's Fish from more than a century ago..
 
Appointment for the first course of the year 2014 on tuesday 7th january. Good food, not only in diagrams and games, but also good company, even good meal.
 
May God bless you however.
 
AV
Master's diagrams

2 - #2 Growing men

Paul BISSICKS

white Qe2 Pg3 Kf2 black Kh1
#2
Growing Men
{

We understand more quickly now that} 1.Qe2-f3+ ? Kh1-h2 2.Qf3-g2 {does not even check and that} 1.Qe2-e4 + Kh1-h2 2.Qe4-h7+ {is not mate because of} 2...Kh2-h3! {
We also note that} 1.Qe2-h5+ Kh1-h2 {would be mate after Qh10 ... on a checkerboard (except that on a checkerboard, the h-column does not allow checkmate from the couloir).

So} 1.Qe2-e5 ! Kh1-h2 2.Qe5-h8 #

3 - s#4 Growing men

Paul BISSICKS

white Qb1 Pd7d6e6e5h2 Kh1 Bd2 black Ph3 Kd8 Ra1
s#4
Growing Men
{

We understand that we will have to guide the R to a checkmate in the couloir. The mode lends itself quite well to remote control.. } 1.Bd2-a5 + Ra1*a5 { Now all you have to do is remove the Q, but how. The idea that comes is} 2.Qb1-b5 Ra5-a1 + 3.Qb5-f1 Ra1-a8 4.Qf1-a1 Ra8*a1# { This is twice untrue: on the one hand} 4.Qf1-a1 {is illegal because it is "too short" and on the other hand the Ws would have} 5.Kh1-g1 ! 2.Qb1-a1 Ra5*e5 3.Qa1-a2! {and 2 possibilities : } 3...Re5-a5 4.Qa2-d5! Ra5-a1# {or} 3...Re5-e1 {no check} 4.Qa2-e2! Re1-a1#



4 - #4 Growing men

Paul BISSICKS

white Pb5 Kc7 Rd5 black Pb6 Ka7 Bh8
#4
Growing Men
{

We can imagine that we are going to checkmate from the couloir on column a.} 1.Rd5-d2 {and} 1.Rd5-d3 {must therefore be considered, but after } 1.Rd5-d2? Bh8-b2! {The Ws don't have a really threatening move, } 2.Rd2-f2 {facing} Ka7-a8 3.Rf2*b2 Ka8-a7 {et} 4.Rb2-a2 {is illegal! Hence:
} 1.Rd5-d3 Bh8-c3 2.Rd3-d1! {threatening }threat: 3.Rd1-a1# {
} 2...Bc3-h8 {to control a1} 3.Rd1-d4!! {and }threat: 4.Rd4-a4 {is unstoppable and mate.}


5 - #2 Growing men with twin

Paul BISSICKS

white Qf1 Kf3 black Kh2
#2 b) Rf3->h4.
Growing Men
{

A simple problem that allows you to check that you have understood. We will check that there is no dual !
} a){

} 1.Kf3-e3 Kh2-g3 2.Qf1-f2#{

} b) wKf3-->h4{

} 1.Kh4-g5 Kh2-g3 2.Qf1-g1#

Now, after the initiation, we arrive at Turnbull who gave back his honour to the first name.

. Luckily, in his great anticipation, the Master had chosen enlarging chess, an original but simple genre, so that the simplest works of RT remain accessible there. It was all the better as I was alone. I hope that the great Michel had only a simple impediment.

6 - #2 Growing men

Ronald TURNBULL

white Pe5f2 Kh4 Rh6g3 black Qb5 Pd4e6h5 Kh1
#2
Growing Men
{

We immediately think of capturing in h5 (with what?) then pushing the K to mate with the R.
} 1.Rh6*h5? Qb5-e8! {(easy).
Donc } 1.Kh4*h5! {Blacks's have 3 variants for intercepting the h-column, resulting in 3 echo variants.
} 1...Qb5-e2+{
} 1...Qb5*e5+{
} 1...Qb5-e8+{
} {How could anyone not like this problem ?}

7 - #5 Growing men

Ronald TURNBULL

white Se2 Kf5 black Ph3h4g3g5h6 Kh5
#5
Growing Men
{

Obviously, we couldn't expect anything else from Turnbull: only one white piece, and it's the only one that has no interest in the chosen genre, all the Knight moves inevitably having the same length! Question : how to avoid mate in 3 by Nf6 or Ng7 ? Knowing that the answer } 1...h3-h2 {followed by } 2...h4-h3 {is not the right one. Genius to serve the imagination is beautiful. ! if Black plays }1...h3-h2{, is that it's a legal move. THEREFORE the h3-pawn has never made a step of more than 1 in length, THEREFORE it comes in direct line from h7 and THEREFORE the h4-pawn comes from another file, has already made a move of 1,414 and THEREFORE it cannot play } 2...h4-h3 { This removes one defence from the Black, but at the same time offers them a new one: lift the stalemate. After a move of Knight, the Black answers }1...g5-g4! {and then, it is the proof that Pg4 comes from g7, so that h6 comes from h7, so that Pg3, h3 and h4 have already played diagonal moves (holds) and are therefore forbidden to advance. But then, what to do? The answer already appears in the above: lift the pat, and since the Knight has no interest in this kind of move ....} 1.Se2-d4! g5-g4! 2.Sd4-f3!! g4*f3 {all the black P's are now static !
}3.Kf5-f6! Kh5-g4 {BK is now condemned to diagonal steps !
}4.Kf6-g6! h6-h5 5.Kg6-g5# !
And now that we have understood, a little more difficult, but in a mate in 2.

8 - #2 Growing men

Ronald TURNBULL

white Pg6f5 Kh4 Rf7 black Pf4c7 Kh6
#2
Growing Men
{

Before going on to the explanations, let's clarify the idea: how do you #2? The only possible scheme is } 1.Kh4-h5# {with a WR controlling g7 and h7. But for the moment, this Kh5 move is not legal. For it to be, it would be necessary that the last Black move is 1...Kg7-h6.
And nothing allows to prove it because the position of the diagram can be obtained with for last move of the rook the move Rf4-f7. And in this hypothesis, the last White move can be g5-g6+ and the last Black move Kh7-h6, the Rf7 not checking Kh7.
If we have understood the above, the solution becomes simple :
} 1.Rf7*c7 { Here, one has no right to conclude on the stalemate, because nothing strictly prohibits the f3 move. So the B's have to play }f4-f3 {But then, the pawn comes directly from f7 and played f7-f6-f5-f4 and so the WR which was in f7 could come at best only from f5 (but not f4) and so it could not make a move longer than 2. But then, the last Black move, which is a move of K, is necessarily Kg7-h6, because in h7 the K would have been in double check. And so from now on, the BK is condemned to diagonal steps.
} 1.Rf7*c7 f4-f3 2.Kh4-h5# {And that's it. }

And for dessert, some positions :



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

9 - #3 with chinese pieces

Hans Peter REHM

white VAf2 LEd7 PAe3f3 NAf1 Kg6 Pa4b6c5d6g2 Sh3 Bf4 black VAh7 MAb7 Ke4 Pc6 PAg4 Re1
#3(13+6)
f1: Nao
e3, f3, g4: Pao
f2, h7: Vao
b7: Mao
d7: Leo
chinese piece : moves normally, but to capture, need a sautoir
the path must be clear between the start square and the sautoir, as well as between the sautoir and the finish square.
le Léo move on the Queen's lines, Pao on the Rook's lines and Vao on Bishop's lines
Le Mao is a special case, it is a Crawling Knight: 1 Rook step then 1 Bishop step.
The square of the rook step must be free. Le Nao is a chinese Nightrider

Solution

10 - s#10 Mars Circe

René J. MILLOUR

white Bb8c2 Ka5 Qb6 Pb4c4e4f4g2 Sa6b5 Rd4e2 black Qd1 Pd5d6 Kd3 Bh1
s#10 Mars Circe(13+5)
Mars Circe : the pieces normally move, but to capture, they virtually return to their original square.

Solution

11 - h#4 Functionnary Chess with Grasshoppers and Dragon

Stephan DIETRICH

white Kf5 DRd5 Gc5c4 black Kd4 Sf6 Gg5
h#4 Functionnary Chess(4+3)
b) turn 90° to the right
c5, c4, g5: Grasshoppers
d5: Dragon
Functionnary Chess : so that a piece can move, and a fortiori capture, it must be "observed" by at least one opposing piece
Grasshopper : piece moving like a Queen but needing a sautoir behind which it lands
the path must be clear between the starting square and the sautoir, and, behind the sautoir,
the square must be free or occupied by an opponent's piece
Dragon : a piece that moves like a knight as well as a pawn.

Solutions

12 - serial s#37

Unto HEINONEN

white Pa2b2e2f2h2 Ke4 black Ph3g7d6a4 Se3h1 Bg8 Kg1
ss#37(6+8)
serial selfmate : White plays n moves in a row to bring a position in which Black is forced to checkmate in one.

Solution

13 - h#2 AntiCirce Isardam with Equihoppers

Manfred RITTIRSCH
dédicate to F. PACHL for his 60 years
Die Schalbe, 2011

white Ph5 Kh3 EQh7 Se3c7 black Bg5 Kf6 Qe2 Pd2c3d6e5g4 Sg7h1 Rf8 EQd1
h#2 AntiCirce Isardam(5+12)
2 solutions
h7, d1: Equihoppers
AntiCirce : the capturing piece is reborn on its original square if it is free, otherwise the move is illegal.
Isardam : it is illegal to put two opposing pieces of the same nature in mutual "observation".
Equihopper : piece moving only with a sautoir which is the centre of symmetry between the start and target squares.

Solutions

14 - #3

Sam LOYD
American Chess Magazine, 1898

white Pb5b4f7c7 Ke4 Ba3c8 Sa8 Re5 black Rd8 Kd6 Bd7
#3(9+3)
Solutions

Thanks to Daniel for his pedagogical approach
next class on January 7th
Have a nice holiday
Yours faithfully
Le greffier virtuel

Gallic hair

Qd5xb7: I would have said "length 2.8".

Nao : I don't really like the name "Mao of the night" because the Mao crawls, but not the Nao! The Mao b7 can capture the Pd6 but not the Pc5.

The Nao is to the Nightrider what the Leo is to the Queen, the Pao to the Rook and the Vao to the Bishop: he only needs the sautoir to capture.

Dietrich : clearer than this 270° without any precise meaning, to say that in the twin, the WK comes in d6 and the BK in e4.

 
Last pb (Rittirsch) :  2 solutions.
 

Merry Christmas.

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