py_20190108

1 - #2

Trillon, JM.
1968

white Pc7b6e4g5g3e2g2c2 Se5 Ra1h1 Ke1 black Pb7b5e3e6e7g7 Kc8 Bh4 Sa3

#2(12+9)
b) ♙c2-->b2

Solution

{

} a) {Black pawns have made 3 captures, corresponding to the 4 white units units (the Bf1 having been captured at home). It follows that, the Bf8 having was also captured at home, the Bh4 comes from a promotion without change of file (so ...h2xg1=B impossible). The only possibility is that of the The only possibility is that of the a7-pawn in a1 (the Pb5 coming from c7), which has broken the grand-rook.

} 1.0-0 ! threat: 2.Rf1-f8 # {

} b) wPc2-->b2 {As in the twin, Bh4 can only come from a promotion without change of file. The only possibility is the promotion of Pc7 to c1 (the Pb5 coming from a7), since a Ba1 would have been walled in. But this requires that the c2-pawn to have made two captures to get to c7. That is a total of 6 captures by white pawns. Now, 7 black units are missing, including the Bf8 which has never played. The participation of the Ph7 is thus necessary: it had to promote in h1, breaking the small-rook.

} 1.0-0-0 ! threat: 2.Rd1-d8 #

2 - 2# rook-lions bishop-lions and neutral pawn

Rook/Bishop: piece moving with the march of the rook or bishop using a sautoir

neutral pawn : The appellation of "neutral pawns" is questionable: these pawns are not "neutral", they are not "neither white nor black", they are alternatively white and black, in other words they flip-flop, like French politicians and others do. They are not neutral pawns, but pawns-swirlers. But don't let that stop us from enjoying this beautiful problem.

Caillaud, M.
2018

neutral Pg7 black BBc8e5 Ph7h6b4 Ke8 BRe1 white WBa2 Ph5 Kb3 Ra7 WRh8

C; #2(5+7+1)
🨶a2, 🨼c8, e5: Bishop-Lion
🨵h8, 🨻e1: Rook-Lion
🨅g7: neutral Pawn

Solution

{

tries :
} 1.Ra7-b7 ? BBc8-a6! {
} 1.g7-g8=nNB+ ? nNBg8*a2 {
} 1.g7-g8=nNR+ ? nNRg8-d8 {
And the last try : } 1.WBa2-g8+ BBe5*h8 2.g7*h8=nNR+ nNRh8-f8 !{

} 1.Kb3-c4 ! {Blocus (there is no threat by} 2.g7-g8=nNR {because of} 2...nNRg8-d8 ! {neither by } 2.g7-g8=nNB {because of} 2...nNBg8-b3 ! {.)
The key is "ampliative", making the Pb4 play.

} 1...Ke8-d8 2.nPg7-g8=nNR # {
} 1...BRe1-e6 2.nPg7-g8=nB # {
} 1...BRe1-e7 2.nPg7-g8=nS # {
} 1...BBe5*h8 2.nPg7*h8=nR # {
} 1...b4-b3 2.nPg7-g8=nNB # {
} 1...nPg7-g5 2.WBa2-g8 # {
} 1...nPg7-g6 2.WBa2-g8 # {
fairy AUW without the Queen (5 promotions).}

3 - 2# Eiffel

Gockel, H.
2017

white Ba3d3 Kh3 Qc2 Pe6b5a4b3c3d2 Sc8e1 Ra7g5 black Bh1c1 Kc5 Qd7 Pd5e4a2a5 Sb1 Rb6

#4(14+10)

: In Eiffel (invented by Petkov), the units paralyze according to the P-N-B-R-Q-P cycle.
The bK is not in check, the Ba3 being paralysed by the Nb1.
. The Qd7 is paralysed by the Ra7

Solution

{

Tries :
} 1.Bd3-f1 ? {threatening} 2.d2-d4 # {because Pe4, paralyzed by Qc2, cannot capture "en passant"} 1...Rb6-b7 ! 2.d2-d4 + Qd7-g7 {!!
} 1.Bd3-c4 ? {threatening} 2.Rg5*d5 # {(but not} 2.d2-d4 + ? d5*c4 {!! paralysing the Pd4 by bQ, while Rg5 is paralyzed by Bc1)} 1...Sb1*a3 ! {paralysing Bc4.( But definitely not} 1...Sb1*d2 ? {which is illegal as it frees the Ba3)} 1...Rb6-d6 2.d2-d4 # {by intercepting the bQ making it inoperative} 2...d5*c4 {.
} 1.Qc2*b1 + ? Bc1*a3 ! {

} 1.Bd3*e4 ! {threatening } 2.Se1-d3 # {but neither} 2.d2-d4 + ? d5*e4 ! {paralysing the pawn with Qd7, nor} 2.Rg5*d5 + ? Bh1*e4 ! {paralysing the Rook
} 1...Bh1*e4 2.d2-d4 # {
} 1...Bh1-g2 {(parries the threat because the Knight, by moving away, would no longer paralyse the Bg2)} 2.Rg5*d5 # {(because precisely the bB is immobilised)} 2.d2-d4 # {
} 1...Bc1*d2 2.Qc2*b1 # {
} 1...Rb6*b5 2.Ra7-c7 # {
"Tertiary threat correction".}

4 - 8# avec équistopper, leo & paos

Rehm, HP.
2017

white LEd5 Ka8 NEf1 Ph5g4h3c4b4 Sd3 WPd7 black Pd2d4b3b7 Kf3 BPb6

#8(10+6)
🨊d5: Leo
🨋d7, 🨑b6: Pao
🩈f1: Equistopper

: controls the middle of a segment of which one end is itself and the other an occupied square.
Thus Eq f1 controls e3 (thanks to Leo d5), g3 (thanks to Ph5), e4 (thanks to Pao d7), e2 & g2 (thanks to Cd3 & Ph3).
The only condition is that the said middle is itself... the centre of a square!
: Leo is a Chinese Queen, Pao a Chinese rook (need for a sautoir to capture)

Solution

{

Essai :
} 1.LEd5*b3 + ? BPb6*b3 ! {
The whole preamble will therefore consist in moving the Pao away from the b6 square

} 1.LEd5-f5 + {Leo f5 allows Eq f1 to check, but now the latter leaves the e3 square.
} 1...Kf3-e3 2.WPd7-d5 + {Eq retakes this square but leaves e4.
} 2...Ke3-e4 3.LEf5-d7 + Ke4-f3 4.Ka8*b7 {
(threatening} 5.LEd7-c6 # {, and here we can't parry by } 4...BPb6-e6 ? 5.LEd7-c6 + BPe6-e4 6.WPd5-f5+ Kf3-e3 7.LEc6-d5 # {)
} 4...BPb6-c6 5.LEd7-f5 + {(retour)} Kf3-e4 6.WPd5-d7 + Ke4-e3 7.LEf5-d5+ Ke3-f3 {
(diagram with the Pao b6 in c6 and incidentally the wK in b7)
} 8.LEd5*b3 # ! {
The e3 square is still controlled, not by the Eq but by the Leo.
Chinese dance on the triangle d5-f5-d7 to keep the black Pao away.}

5 - 2# Transmuted Kings

Murarasu, I.
2001

white Pa7d7e7f7h7g5 Sa6f3 Kg8 Ge4g2 Na1 black Pd6d4b5a4 Kd5 Bd3

#2 Transmuted Kings(12+6)
🨟e4, g2: Grasshoppers
🨢a1: Nightrider

: When a King is in check, it catches, for the time of the check, the moving way of the piece that threatens it
: a piece moving along the Queen's lines with the help of a sautoir behind which it lands
: extended knight

Solution

{

} 1.Sa6-c7 + ! {

} 1...Kd5*c7 2.d7-d8=B # {
} 1...Kd5-b4 2.d7-d8=N # {(all squares are controlled, including a2, d8 & h7)
} 1...Kd5-b6 2.a7-a8=S # {(c4 est contrôlée par la Se4, d7 par le Na1, et pas} 2.a7-a8=N + ? {le Roi pouvant s'échapper :} 2...Kb6-f4 {ou} 2...Kb6-h3 {)
} 1...Kd5-f4 2.f7-f8=R # {(e4 is controlled by Gg2, g4 by NIa1)
} 1...Kd5-c3 2.h7-h8=G # {(e3 is controlled by NIa1, e5 by Nf3, c8 by Pd7)
} 1...Kd5*e7 2.f7-f8=Q # {(} 2.Sc7-e6 + ? Bd3*e4 ! {)} {

Super AUW.
note that} 1.Sf3-e5 + ? {is illegal, as for} 1.d7-d8=G + ? {, King becoming a Grasshopper at the moment he is checked}

6 - h#4 with Grasshoppers & kangourous

Foster, G.
2018

white Kd3 WGa7f2 black Kd1 BGa6a3

h#4(5+9)
🨟a7, f2, 🨥a6, a3: Grasshoppers
b) Grasshoppers --> Kangourous

: a piece moving along the Queen's lines with the help of a sautoir behind which it lands
: Grasshopper with double sautoir. It lands just behind the second sautoir

Solution

{

} a) {
} 1.Kd1-e1 WGa7-a5 2.BGa6-e2 Kd3-c2 3.BGa3-a6 Kc2-c1 4.BGa6-f1 WGf2-d2 # {

} b) +wKAa7 +wKAf2 +bKAa6 +bKAa3 {Grasshoppers --> Kangourous
} 1.Kd1-c1 WKa7-a2 2.Kc1-b2 WKa2-g2 3.Kb2-a2 Kd3-c2 4.BKa6-a1 WKg2-b2 #

7 - 1# duplex complete chess shrinking

Ravishankar, S.
2018

white Kc7 Rb6 black Ka8

#1 duplex(2+1)
complete chess
Shrinking Men

: The statement is valid with the move to White or Black
. : one can play one of its own pieces or one of the opponent's pieces
shrinking: a piece cannot play a move geometrically longer than its previous move
Note that a move is considered legal in the absence of a demonstration that it is not.
Thus, the two Kings are assumed to be "unshrunken" because each is on the colour of its initial square, thus assumed to have played only diagonal moves.

Solution

{

White to move, plays the black King :
} 1...Ka8-b8 # {checkmate the same black King, as the return} 2...Kb8-a8 {is illegal and } 2...Kb8-a7 {(as } 2...Kb8*c7 {) would be expanded. checkmate by double check, Black can't reply} 2.Rb6-c6 {because they are still in check by the white King.

Black to move, plays the white King :
} 1.Kc7-b7 {checkmate the same white king, as the return} 2.Kb7-c7 {is illegal and} 2.Kb7-a6 {or} 2.Kb7-c6 {or} 2.Kb7-c8 {(as Kxa8) would be expanded.}

8 - h#2,5 super-circé with camel, pao

Crisan, V.
2008

white VAd8 Sc8 PAb7 Ka8 black Qc7 CAe5 Kd5

h#2.5 super-circe(4+3)
b) 🨽e5 --> d4
🨵b7: Pao
🨶d8: Vao
🨽e5: Camel

: a captured piece is reborn at the whim of the side making the capture
: leaper (3,1) (Knight is a leaper (2,1))

Solution

{

} a) {
} 1...PAb7-b6 ! 2.CAe5*b6[+wPAa6] VAd8*b6[+bCAe4] 3.Qc7*b6[+wVAa7] Sc8*b6[+bQe5] # {

} b) bCAe5-->d4 {e5 --> d4
} 1...VAd8-e7 ! 2.CAd4*e7[+wVAf8] PAb7*e7[+bCAc4] 3.Qc7*e7[+wPAe8] Sc8*e7[+bQd4] # {

Echo checkmates after 4 successive captures on the same square.}

9 - h#2 madrasi

Madrasi : 2 opposing pieces of the same kind that observe each other are paralysed

Zidek, A.
1996

white Qg8 Pd7f7h7 Ke3 Bc6 Rc2 black Be8e7 Kh1 Qh8 Pf5g2e2 Sd6d3 Rc8c7

h#2 madrasi(7+11)
b) -Qg8 -Qh8

Solution

{

} a) {
} 1.e2-e1=B d7*c8=R 2.g2-g1=Q f7*e8=S # {(paralyse Nd6 to prevent } 3.Sd6-e4 {)
Sur } 1.e2-e1=S ? d7*c8=R 2.g2-g1=Q f7*e8=S + {doesn't checkmate :} 3.Se1-g2 {or} 3.Se1-f3 {

} b) -wQg8 -bQh8 {without Queens g8 and h8
} 1.e2-e1=S d7-d8=B 2.g2-g1=R h7-h8=Q # {
note that} 2...h7-h8=R + ? {doesn't checkmate because of } 3.Rg1-g8 ! {

Double AUW.}

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