Tuesday evening the traditional end-of-season fairy festival was held.
The assembly was thus enriched by 2 elements: one which is no longer presented, and to which one can attribute the qualification of best French solutionist and composer, and the other which was at the beginning of the course a total beginner in the field of fairy.
The 2 extremes then!
As could be expected, the aforementioned beginner, carried away by inspiration, made rapid progress.
The appetizer is a retro one (rather rare in the Master's bag, for lack of amateurs).
1 – first move of a white Bishop ? John BEASLEY The Problemist 1972 (v)
t4tr1/1ppp1ppp/p4cf1/8/8/1P6/2PPPPPP/2F3R1 W : Kg1 Bç1 Pb3ç2d2é2f2g2h2 B : Kg8 Ra8f8 Bg6 Nf6 Pb7ç7d7f7g7h7a6 (9+12)
We must nevertheless stop a little on this position and answer the question: where does Fg6 come from?
The answer is simple but leads to a certain complexity in the sequence of moves.
A provision of problems for the summer.
Helpmate for Daniel, back from Bulgaria: in Boyer's problem, the 3rd triplet does not bring anything to the masterful unity of the first two, but makes one think in a tournament of solutions. The first h#3 presents a theme of direct mats. It is VERY EASY. This 1967 problem was signed almost 30 years later by a German author: but this is probably a coincidence, because the anticipated author did not "capture the precaution" of presenting it in mirror-position!
The others h#3 also have only one solution each. The h#4 has a repeated characteristic in all three solutions and yet it did not seem easy to me. The h#5 is humorous. h#12 (composed by computer) has already been presented in an old report. Just to see if you recognise it... A little #2 with double pins. If you didn't know the Wurzburg-Plachutta theme, here is the cheapest (and most pedagogical) version to my knowledge. Three other #3, two spectacular ones by the Serbian magician, the third one also but deeper, by a world champion in the specialty. A triple homage to another magician, Teuton this one, grandmaster of logic. Finally, a long problem by another Teuton logician with a Polish patronymic. I have left the solution of the four multimovers, hoping that you will have the courage to press the "training" button.
To close this selection, a curious s#2 with a surprise effect.
Only a few Fairy problems are compatible with the chessbasiste presentation. Rare exceptions all the same: a traditional retro (there were very few at the various fairy tuition). A fairy PG. A serial selfmate. And a novelty from Ronald Turnbull, the idol of the Saint-Lazarian listeners: Shrinking Chess! Rest well, play and analyse a lot. Rendezvous beginning of October, if God wills it.
: a captured piece can be put back in the box by the capturing player or put back on the board with the following restrictions :
- the square must be free
- the position must be legal after replacement
- a pawn cannot be placed in the 1st or 8th row .
- a Bishop must keep his colour at replacement
The position is simple but requires finesse in the 2nd and 3rd white moves .
the next is a cyclical problem whose mechanics need to be discovered.
3 - #15 with Equistopper and chinese pieces
Hans Peter REHM& Klaus WENDADie Schwalbe 2008
white VAe1e5 Bc3 Ke7 NEa7 Sc8 PAe3
black LEg1 Kd5 Pd3e4c7b6 Sa6 Ba1 Rb4a3
: attacks the central point of symmetry in relation to an occupied square of the chessboard provided that it falls on a full square
example: Ea7 control c6 because of e5, c5 because of e3, c4 because of e1, etc...
: Bishop needing a sautoir to capture
: Queen needing a sautoir to capture
: Rook needing a sautoir to capture
It is necessary to make a forcing maneuver without capturing gloves.
A very nice problem, the association of the 2 authors is certainly not foreign to it !
To follow, a very nice Gockel with a small menagerie
4 - #2 AntiCirce with Bishop-Lion and Rook-Lion
Hubert GOCKELProbleemblad, 20092nd Prize
white Be6 WBe5 Kg2 Qe1 Pb7d7d6g7g4f2d2d3 Sh1f3 Rg6 WRa4
black Bf8 BBb4h5 Kf4 Qf5 Pa5 Sb8e8 Ra8e7 BRg8a7
{}
1.WBe5-c7 ! {the f8-square being occupied, this move does not check} threat: 2.Qe1-e3 #{}
1... Qf5-c5 2.g7*f8=Q [wQf8->d1] #{}
1... Qf5-e5 2.g7*f8=R [wRf8->a1] #{}
1... Qf5-e4 2.g7*f8=B [wBf8->c1] #{}
1... Qf5*g4-d8 2.g7*f8=S [wSf8->g1] #{}
1... Qf5*e6-d8 2.Qe1-e4 #
: The capturing piece is reborn on its original square (which is a promotion square for fairy pieces) if the square is free, otherwise the capture is not legal.
: Bishop needing a sautoir to move and eventually capture
: Rook needing a sautoir to move and possibly capture
The first question to be answered is: why Qe3 don't checkmate ?
Afterwards, it becomes a little clearer. Although ...
A small remark : The Black King is not in check by the Lion-Rook a4 because of the occupation of the renaissance square f8
Still in the bestiary, a 2-moves
5 - #2 with Lion and Rose-Lion
James QUAHThe Problemist Supplément, 2011
white Pd5e2 RNb3e7 Rb5f5 Kd8
black Pf6f4 Sg8 Kd6 LIg3
: Queen needing a sautoir to move and possibly capture
: Rose needing a sautoir to move and possibly capture
The Rose is a rounded Extended Knight, if the definition can be imaged.
Example: if e7 was a Rose, it could move to g6, h4, g2, e1, c2, b4 and c6 or c8 and a6or capture g8 .
As it is a Rose-Lion, only squares e3, f1, e1, b4 and h4 are accessible because of the sautoirs d5 and f5. The sautoir of RLb3 is the Rd8
{}
1.g7-g8=R 2.Rg8-g5 3.Bf8-g7 4.Ke8-f8 5.e7-e8=Q 6.Qe8-b8 7.c7-c8=B 8.d7-d8=S + Qb6*d8 #
{AUW with a critical blow from the WQ passing on the other side of c8}
: White plays n moves in a row to force black to check them in 1 move .
Let's now move on to shrinking chess.
The rule is simple: a piece cannot play a move longer geometrically than its previous move .
Who better than Ronald Turnbull could highlight the rule ?
It starts with the simple (seemingly) following position
{h8 is not of the same colour as e8, so the Black King has taken at least 1 orthogonal move.
and is no longer entitled to the diagonal pitch which measures the root of 2..
As for the White King, e1 is of the same colour as h6, so it is not, a priori, "shrunk"..
In the position, there is a checkmate in 1 move by Kg7#; the first move, therefore, is not } 1.Kh6-g7
{if you want to make stalematethe solution : }
1.Kh6-g5 {} 1...Kh8-h7 ? 2.Kg5-g6 ! Kh7-h8 3.Kg6-g7 {stalemate}
1...Kh8-g8 2.Kg5-f6 {}
2...Kg8-f8 ? 3.Kf6-e7 ! {and stalemate at the 5th move}
2...Kg8-h8 3.Kf6-e5 !! {a certainly original way of capturing the opposition.
Kg7 would matt as seen above} 3...Kh8-g8 {on Kh7, 4.Kf5}
4.Ke5-e6 ! {the necessary degradation} 4...Kg8-g7 5.Ke6-f6 Kg7-g8 6.Kf6-f7 Kg8-h8 7.Kf7-g7 ! {stalemate}
To help you a little: we can prove that the Black King can only play orthogonal moves
the next one in the genre is a bit more loaded in material!
{From the colour of their squares, none of the 2 Kings is "shrunk".
As the Rh1 has probably not moved, it has all the latitude it needs to move.
Let's try } 1.Rh1-h6 {length 5} Kg8-f8 {The Black King is now "shrunk". }
2.Rh6-h8 + {length 2} Kf8-e8 ! {King escapes.
Indeed, the Rook can't make any more moves of a length > 2, and it would take 3 to capture the King.
The position being a little rough, we still have to find the 2 solutions}
Now a good brainstorm signed, as it should be, by Peter Harris.
9 - #3 Isardam Super-Circe
Peter HARRISThe Problemist, 2005Comm.
white Qb5 Rf2d5 Bc2 Sd4f5
black Qh8 Pe4 Sd2 Re7g8 Ke5
{ }
a) { }
1.Sf5*e7[+bRd1] ! Rg8-g5{ }
2.Sd4-c6+ Qh8-e8{ }
3.Qb5-b7 #{The only possible defensive moves } 3...Sd2-c4 {et} 3...Sd2-f3
{have the major disadvantage of putting Rooks d1 and d5 in mutual "observation". }
b) shift a2 ==> a1{1 row down }
1.Sf4*e6[+bRc8] ! Rg7-g4{ }
2.Se6-g5 + Sd1-c3{ }
3.Qb4*c3 [+bSc7] #{The capture of Ng5 } 3...Rg4*g5
{has the disadvantage of allowing the Black King to be taken by the Rd4 }
: any move putting 2 opposite pieces of the same kind in mutual control is illegal (the Black King is not in check because of the e7 rook)
: a captured piece is replaced at the discretion of the taker provided the position is legal (King of the taker not engaged or other case) .
And now a double stalemate: after the last move, neither side can move an ear.
10 - h==4 with Rose and Locusts
Vaclav KOTESOVECThe Problemist, 20054th HM
white ROb2 Pa2 Kd1 Ra4c4 WLd3h6
black Pa6d2 Kf5 Re4 BLb4
8/2pB1p2/k1P2N2/P1b1rp2/2K2n2/1p3r2/2Pq4/RQ5b W. : Kc4 Qb1 Ra1 Bd7 Nf6 Pc6a5c2 B. : Ka6 Qd2 Re5f3 Be5h1 Nf4 Pc7f7f5b3 ss=10 (8+11) Anticirce
Michel, as a dessert, presented us one of his creations made during his ablutions with a little-used genre: the "pressurised" chess, but as the judgement has not yet been made, you will not be entitled to it. You had to be there !
It remains for me to wish you, by joining the Master, very good studious holidays.
Yours sincerely
Le greffier
greffier's duvet
Excellent report.
Just two remarks: I don't understand the sentence "only squares e3, f1, e1, b4 and h4 are accessible".
The Rose-lion e7 can go to h4 indeed ( sautoir f5). And also in e3, c2 and a3. Thanks to the d5 sautoir, it can go to e3 and g2. And also to b4.
But in f1, I can't see. A Rose-Lion h6 could ( sautoir d6).
In e1, I can't see either. It would need a sautoir in c6 (or b4, or c2, or on the other side).
Isardam: one word is missing: any move putting two opposing pieces of the same nature under observation is illegal.
I haven't seen any other weirdness.
Let us specify that for the April fish... of June, the author had obtained a mention of dishonour of which he was immensely proud!
"Dr" seems to me more precise for the Dragon, although it is obvious that it is not a Queen in b8!
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