june 14 2005

The night writer's report.

As tradition requires, and none of the participants complained, the last lesson of the season was entirely devoted to fairy chess.
The regulars were there, of course, reinforced by the presence of the Grandmaster and 2 new ones, David and Sam, whose fairy chess was the deflowering, Sam had a foretaste at the last lesson during the 3rd half.

The electronic version written by the Master appears at the end of this report, thanks to him.

To heat the neuron, some Proof Games in totally different genres

1 - in 28,5 moves
P.G. n moves: only sequence of moves from the starting position of the normal chess game to reach the position of the statement
Losing Chess: the goal is to lose the game, the capture is mandatory, you can be checked

The Master's words.

Five "Daniel" helpmates (it requires a bit of work for summer) and two three-movers, one of which (Warton) is quite spectacular.
In addition to our grandmaster, former world champion and former French champion (but one of the three best composers in the world, judging by the elo,
(i.e. the number of problems selected in the FIDE album), who honours us with his loyalty to the St-Lazare course, we had a (world?) champion this evening
of rollerblading and downhill skiing, the discreet (and no less famous) Shorbock (alias David) of France EChecs, who was sometimes the first to solve fairy problems, even though he was neophyte in this field.
Only a few problems of the fairy session will be in the electronic report. But not MC's PG (the aforementioned GM) in "who loses wins", which the master clerk will take care of. Neither the one in circe Rex inclusiv, but I'm giving the "problem that drives you crazy" of the same kind again, in case some people still don't. But you will find on the other hand the three "transmuted kings" the "functionnary chess" and the isardam. During dinner, in addition to a Laborczi with Chinese pieces, Maslar's three 7# helpmates made digestion easier. And a circe from Wong, author of a brilliant 3-solution PG
(attached).
Have a good time.
AV

Master's diagrams

W.: Kc5, Qc4, Ra1 and h1, Bb1 and e3, Nd2 and g1, pa3, b2, b5, d3, d5 and e4
B.: Ke8, Qd8, Ra8, Bc8 and f8, Nb8 and g6, pa7, b7, c7, d7, e5 and g7

This PG was composed by the Grand Master, hang on.

Let's continue the warm-up with the next PG in the Circe rules

2 - in 9,5 moves
Circe : the captured piece is reborn on its original square except if this one is occupied in which case it disappears
Kings included : Kings are also concerned by the condition Circe

W.: Kg1, Qd1, Ra1 and f1, Bc1 et c6, Ne3 and f3, pa2, b2, c2, d3, f2 and g2
B.: Kd7, Qd8, Ra8, g3 and h8, Bc8 and f8, Nb8 and g8, pa7, b7, c7, e5, f7, g7 and h7

Note that Black King is "en prise" but not in check, since he can reborn in e8!

Still in the Circe Rex inclusiv rules, a reflex problem.

3 - r#2 Circe Rex inclusiv
Reflex : when a checkmate in 1 move is presented to one of the 2 sides, he must play it

Uri AVNER & Dolf WISSMANN
Thematic Tourney Nunspeet, 2005
1st Place

white Ka4 Rc8 Sa5 Be8 black Bc3a6 Ka2 Qb7 Pb2c4d4g3g2 Sa1 Rb1e5

r#2(4+12)
Circé Rex inclusiv

Solution

{

} 1.Ka4-b3 + !{

} 1...Sa1*b3[+wKe1] + 2.Ke1-e2 2...Bc3-e1 #{

} 1...c4*b3[+wKe1] + 2.Ke1-d1 2...Re5-e1 # {display-departure-file} {

} 1...Qb7*b3[+wKe1] + 2.Ke1-d2 2...Rb1-e1 # {display-departure-file}

A totally destabilising problem.

Rest and aesthetics with the next problem

4 - h#4 3 solutions with and
helpmate : Black plays and help the White to mate them
3 solutions : there are three distinct solutions (which therefore start with different moves)
Grasshopper : Piece that moves on Queen's lines and needs a hurdle to move. It lands just behind the hurdle if the square is free or
occupied by an opponent's piece, otherwise the move is not possible.
Wazir : piece moving as a Rook but only one step

Daniel NOVOMESKY
The Problemisgt, 2002
6th ?omm.

white Kh1 WEa8 Gd5 black Ka5 Rb6 Sc5

h#4 3 solutions(3+3)
🨟d5: Grasshopper
🨠a8: Wazir

Solution

{

} 1.Sc5-b3 Gd5-a2 2.Rb6-b4 Ga2-c4 3.Sb3-d4 WEa8-a7 4.Sd4-b5 WEa7-a6 # {

} 1.Sc5-d7 Gd5-d8 + 2.Ka5-a6 Gd8-a5 3.Rb6-b5 Ga5-c5 4.Sd7-b6 WEa8-a7 # {

} 1.Sc5-b7 WEa8-b8 2.Ka5-b5 Gd5-a8 3.Kb5-a6 Ga8-c6 + 4.Ka6-a7 WEb8-a8 # {

mate on all floors !}

Very clean composition

Now, the Transmuted Kings

5 - h#2
Transmuted Kings : a checked King takes the step of the piece which puts it in check for the time it takes to make its move. Of course, in the case of a double check, he chooses which piece he will take the step from.

Peter HARRIS
The Problemist, 2003

white Kd1 Ba8 Sd5 black Qh3 Kh1 Rd7g6

h#2 Transmuted Kings(3+4)

Solution

{

} 1.Qh3-f1 + Kd1-h5 2.Qf1-f7 Sd5-c7 #

Great art

Again with the same rules

6 - h#2,5
h#2,5 : White begins

Peter HARRIS
The Problemist 2003

white Rb5 Kh8 Bg4 black Qa1 Kd4 Bg5

h#2 Transmuted Kings(3+3)

Solution

{

} 1...Rb5-d5 + 2.Kd4-a4 + Kh8-a8 3.Bg5-c1 Bg4-d1 #

Also a very nice composition, a true pleasure.

To keep the fun running

7 - h#3 with b) ♕h6-->a6
Nightrider : Extended Knight

Peter HARRIS
The Problemist, 2003-11
Comm.

white Qh6 Kd5 black Pf5g5 Kf4 Nd1

h#3 transmuted Kings(2+4)
🨨d1: Nightrider
b) ♕h6-->a6

Solution

{

} a) {
} 1.Nd1-h3 Qh6*g5 + 2.Kf4-h2 + Kd5-f1 3.Pf5-f4 Qg5-g2 # {

} b) wQh6-->a6 {Dh6-->a6
} 1.Pg5-g4 Kd5-e6 2.Kf4-g5 Ke6-f7 3.Nd1-h3 Qa6-h6 #

For those who are not familiar with the Nightrider, NRd1 can jump to b2, c3, b5, a7, e3, f2 and h3

For the continuation, now that the neuron is warmed up, we begin to work on Chinese pieces.

8 - h#2 2 solutions with
Pao : moves on the Rook lines but needs a hurdle to capture

????

white Ke1 Bh6 PAh5 Sf6 black Qd7 Kg6

h#2 Circe(4+2)
2 solutions
🨵h5: Pao

Solution

{

} 1.Qd7-g4 Sf6-h7 2.Kg6*h5[+wPAh8] Sh7-f8 # {

} 1.Qd7-g7 PAh5-g5 2.Kg6*h6[+wBc1] PAg5-g6 #

A little clarification : when a fairy piece is taken, it is reborn on the capture column and on the promotion row of its side.

Now a small look at Functionnary Chess

9 - #2
Functionnary Chess : a piece can only move and a fortiori capture if it is under "observation" of an enemy piece

Hubert GOCKEL
The Problemist, 1999
6th HM

white Sf3e8 Re4b7 Kc8 Bc5 black Qa1 Pg2f5d5 Sb1e5 Bc1c6 Ke6

#2(6+9)
Functionary Chess

Solution

{

} 1.Rb7-g7 ! {

} 1...g2-g1=Q 2.Rg7-e7 #{

} 1...g2-g1=S 2.Re4*e5 #{

} 1...g2-g1=R 2.Sf3-d4 #{

} 1...g2-g1=B 2.Sf3-g5 #

Good luck

Help yourself with a drink, if you haven't already done so for the next one.

10 - s#9
selfmate : Whites plays and forces Black to mate them.
Isardam : it is totally illegal to place 2 opposing pieces of the same nature in mutual control

Turnbull R.
1999

white Pb7h7 Ka5 black Pa2d6 Ka1

s#9 Isardam(3+3)

Solution

{

} 1.b7-b8=Q d6-d5 2.Qb8-b3 d5-d4 3.Ka5-b4 d4-d3 4.Qb3-d1 + Ka1-b2 {
} 5.h7-h8=B + a2-a1=B 6.Qd1-a4 d3-d2 7.Kb4-b3 + d2-d1=Q 8.Bh8-c3 ! Qd1-c2 {(Kxc3#)} {
} 9.Qa4-a3 + ! {(9 Qb4#? etc.)} Qc2*c3 #

To help you a little, this condition does not apply to Kings. Moreover, the capture of a King is illegal if it puts 2 opposing pieces in mutual control as indicated in the
Isardam rule

A short excursion in the Köko

11 - h#3 2 solutions
Köko : in order for a move to be legal, the piece must be in contact with another element (friend or opponent, no matter) at the end of the move.

Aleksandar POPOVSKI
The Problemist, 2004

white Ke2 Bc4 black Qc3 Pf5f4g4 Ke4 Ba6 Rd5

h#3 Köko(2+7)
2 solutions

Solution

{

} 1.Qc3-e1 Ke2-f1 2.Qe1-e3 Bc4-d3 + 3.Ke4-f3 Kf1-g2 # {

} 1.Rd5-b5 Bc4-b3 2.Qc3-f6 Ke2-d3 + 3.Ke4-e5 Kd3-d4 #

A little rest with another beautiful problem

12 - h#2 Circe with Grasshopper

Friedrich CHLUBNA
Die Schwalbe, 1979
1st Prize

white Pf3 Kb1 Gf4 black Bh6 Kh8 Qf5 Pe4g3 Sg8d5 Rh7g7

h#2 Circé(3+9)
🨟f4: Grasshopper
b) ♘g8=♗

Solution

{

} a) {
} 1.Sg8-f6 Pf3*e4[+bPe7] 2.Qf5*f4[+wGf8] Pe4*d5[+bSg8] # {

} b) +bBg8 {Bg8
} 1.Rg7-g6 Gf4-f6 2.Qf5*f3[+wPf2] Pf2*g3[+bPg7] # {

}

To finish off your neuron's connections, the Andernach rule

13 - h=2
Andernach : a piece changes colour after capturing (Kings are not affected)

Anatoly STYOPOCHKIN
The Problemist, 1999

white Pg5g7g4c7a6c4 Kf3 Bf7 Sf5 black Pc2a7 Ke5 Rb8

h=2 Andernach(9+4)
b) ♜b8-->h8

Solution

{

} a) {
} 1.c2-c1=R c7*b8=B=b 2.Rc1*c4=w Rc4-c7 [+wRc7] {stalemate

} b) bRb8-->h8 {
} 1.c2-c1=Q g7*h8=S=b 2.Qc1*g5=w Qg5-g6 [+wQg6] {stalemate

AUW}

Very nice

Rule is now reversed with the anti-Andernach

14 - h#2
anti-Andernach : when a piece move without capturing, it changes colour (Kings are not included)

John M. RICE
Die Schwalbe, 1998
1st Comm.

white Qb8 Pc5e7g7g5 Sh8f5 Rc7e1 Ke4 black Ph5 Ke6 Bb6 Rb3

#2(6+9)
AntiAndernach

Solution

{

} 1.e7-e8=B=b ! threat: 2.Qb8*e8 # {
} 1...Bb6*c5 2.Qb8*b3 # {
} 1...Be8-a4=w 2.Ba4*b3 # {
} 1...Be8-b5=w 2.Qb8*b6 # {
} 1...Be8-c6=w 2.Ke4-f4 # {
} 1...Be8-g6=w 2.Ke4-d4 #

Very nice problem too

A new incursion into Chinese pieces

15 - h#4,5 with and
4,5 moves : White begins
Pao : see 8
Vao : as Pao, but on Bishop lines

Zoltan LABORCZI
Mémorial E.Holladay - StrateGems, 2004
1st Prize

white Pb4b2f3 Kg1 PAf4 VAb6 black Pb5e3 Sh1c4 Bd6e6 Ke7 Rf6

h#4,5(6+8)
b) ♙b2-->e2
🨵f4: Pao
🨶b6: Vao

Solution

{

} a) {
} 1...VAb6-d4 2.Bd6-e5 VAd4-c3 3.Ke7-d6 PAf4-d4 4.Kd6-d5 PAd4-d1 5.Kd5-d4 VAc3-d2 #{

} b)wPb2-->e2{pb2-->e2
} 1... PAf4-d4 2.Be6-d5 PAd4-d2 3.Ke7-e6 VAb6-d4 4.Ke6-e5 VAd4-a1 5.Ke5-d4 PAd2-b2 #

A series of helpmate in 7, but from remembering, so there will be mistakes from hand or head.

16 - h#7

Zdravko MASLAR
Diagrammes, 1981

white Ph2 Sh8 Kh4 black Kf3

h#7(3+1)

Solution

{

} 1.Kf3-e4 Kh4-g4 {
} 2.Ke4-e5 h2-h4 {
} 3.Ke5-f6 h4-h5 {
} 4.Kf6-g7 h5-h6 + {
} 5.Kg7*h8 h6-h7 {
} 6.Kh8-g7 Kg4-f5 {
} 7.Kg7-h6 h7-h8=Q #

17 - same énoncé

Zdravko MASLAR
Feenschach, 1982

white Pa2 Kf8 Ba4 black Kb1

h#7(3+1)

Solution

{

} 1.Kb1-c1 Ba4-e8 {
} 2.Kc1-d2 a2-a4 {
} 3.Kd2-e3 a4-a5 {
} 4.Ke3-f4 a5-a6 {
} 5.Kf4-g5 a6-a7 {
} 6.Kg5-h6 a7-a8=Q {
} 7.Kh6-h7 Qa8-h1 #

18 - s.é.

Francisco SALAZAR
T.T. The British Chess Magazine, 1973
1st HM

white Pb3b2 Kh2 black Ka7

h#7(3+1)

Solution

{

} 1.Ka7-b6 Kh2-h3 {
} 2.Kb6-c5 b3-b4 + {
} 3.Kc5-d4 b4-b5 {
} 4.Kd4-e3 b5-b6 {
} 5.Ke3-f2 b6-b7 {
} 6.Kf2-g1 b7-b8=Q {
} 7.Kg1-h1 Qb8-h2 #

19 - s.é.

Zdravko MASLAR
Feenschach, 1982

white Pd2 Kd3 Bc3 black Kh3

h#7(3+1)

Solution

{

} 1.Kh3-g4 Kd3-c2 {
} 2.Kg4-f5 d2-d4 {
} 3.Kf5-e6 d4-d5 + {
} 4.Ke6-d7 d5-d6 {
} 5.Kd7-c6 d6-d7 {
} 6.Kc6-b5 d7-d8=Q {
} 7.Kb5-a4 Qd8-a5 #

And to conclude, a splendid helpmate

20 - h#2

Peter WONG
Phenix, 1995
2nd Prize

white Pe3f6 Sd1 Be2 Kh7 black Qd3 Pe6e5f5 Ke4 Rh2 Sg2h3

h#2 Circe(5+8)
2 solutions

Solution

{

} 1.Rh2-h1 Be2*d3[+bQd8] + {
} 2.Qd8*d3[+wBf1] Bf1*g2[+bSg8] #{

} 1.Qd3*e2[+wBf1] Bf1*e2[+bQd8] {
} 2.Qd8*f6[+wPf2] Sd1-c3 #

Saving time by seeming to lose time and vice versa, this is the solutionist's guiding principle for this beautiful problem.

It remains for me to wish you good reading and good holidays, during which I am sure you will not fail to take the various reports
to read them on the beach.

Rendez-vous very soon for the corrections.

Kindly yours

Corrections.

The master-greffier does not want to underestimate the readers, but specifying that promotion to King is possible in the "who loses wins" can't hurt!
Pb 8: the pao is in h5 (Pa h5).
I did not notice any error, the master-greffier has in particular reconstituted perfectly the 4 h#7. But I don't have a double of some fairy positions.
Good " headache " on the beaches...

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