py_20151215

1 - h==5 toric chessboard

Hans Peter REHM
1983

white Pa7b7c7e7b3f2 Kd7 Bd1 black Pb4c2d2e2f4h5g2 Bb5 Rc6 Kf3 Na8

h==5 toric chessboard(8+11)
🨨a8 : Nightrider

: to simplify, we can put 8 chessboards around the main chessboard with the same position
. : a piece that can perform several successive jumps of the Knight in the same direction

Solution

{

Null Moves are forbiden (otherwise a stalemate would make no sense, for example in the position of the final ...Rc6-c6 or ...Rd1-d1)
} 1... c2-c1 [+bBc1] 2.b7*a8 [+wNa8] Bc1-d8 3.Na8-d6 {via b2} e2*d1 [+bQd1] 4.a7-a8 [+wBa8] Qd1-f7 5.Ba8-b7+ d2-d1 [+bRd1] 6.c7*d8 [+wSd8] {double stalemate ! A Nightrider would check, and therefore mate, via b1-h2.
AUW is now with its 5th promotion (in N).
Bb7 pins the Rd1, which pins the Nightrider d6, which pins the Bb5.
. Which Nightrider also pins the BQ (h8, b1, d2), which pins the Bb7.
Hence the cycle wB-bRd1-wN-bQ-wB. To these cyclic pinnings are added 3 pinnings, that of the Rc6 by the Bb7, that of the Bb5 by the Nd6 and that of the Sd8 by the Rd1, in total 7 figures and 3 pawns. The wNd6 controls g4 via b2-a8-h6 !}

2 - h#2 Madrasi 2 solutions

Lev GROLMAN
2010

neutral LIb6 Nb2 black Bc2h4 Kh3 Qh2 Pc4a4 Sc5 Rg7g4 white Kf3 LIb8 Na7

h#2 Madrasi 2sol.(3+9+2)
🨢a7=Nightrider
🨮b2=neutral Nightrider
🨴b8: Lion
🩀b6: neutral Lion

: two opposing pieces of the same nature that are "observing" each other are paralysed.

Solution

{

} 1.nNNb2-d3 Na7-b5 {
} 2.nNLb6-b3 nNNd3-f7 # {
The double check cannot be avoided by the return of the neutral Nightrider f7 to d3 because it is paralysed by the white Nb5. At least, as a black piece. It is not as a white piece (how would a white piece paralyse another white one ?) and therefore the mat cannot be trimmed by 3...axLib3: miracle of the neutral pieces ! Neither by Linb3xf7 because paralysed -- as a black piece -- by Lib8 !
} 1.nNLb6-f2 LIb8-b1 {
} 2.nNNb2-d1 nNLf2-f5 # {
the double check cannot be parried by the return of the neutral Lion in f2 because it is paralysed by Lib1; nor by Nnd1xf5 because it is paralysed by Na7 !
. Theme: reciprocal neutral batteries.}

3 - h=3 Circe

Attila BENEDEK
2010

white Pd5e5g4g3f2 Sc4e6 Bh3 Kf1 black Pd6f6h7 Sg7 Bf8 Kf3

h=3 Circe(9+6)

: a captured piece is reborn on its original square if it is free, otherwise it disappears

Solution

{

} 1.Sg7-h5 Sc4-e3 2.Bf8-h6 Se6-g7 3.Bh6-f4 g4-g5 4.h7-h6 Bh3-f5 {=
None of the 5 units e3, e5, g3, g5 & g7 are takable.}

4 - h#2 AntiCirce 2 solutions

Pierre TRITTEN
2010

white Sd4d5 Bf5g1 Ke5 black Qg4 Pe6 Ke8 Ba7

h#2 AntiCirce 2sol.(5+4)
2 solutions

: the capturing piece is reborn on its original square if this one is free, otherwise the move is illegal
. A distinction is made, however, in the case where the square of capture and rebirth are identical: in AntiCirce Calvet the move is legal, and in AntiCirce Cheylan it is not

Solution

{

} 1.Qg4*d4[bQd4->d8] Bg1-c5 {
} 2.e6*d5[bPd5->d7] Bf5-g6 # {

} 1.e6*f5[bPf5->f7] Sd4-c6 {
} 2.Ba7*g1[bBg1->f8] Sd5-f6 # {
Mat with 2B then matt with 2N}

5 - s#2 AntiCirce Cheylan

Ronald TURNBULL
2000

white Pg7h7d7 Sc1h1c7 Kf6 black Pc2d2f7 Sf3f5 Kd8

s#2 AntiCirce(7+6)

: the capturing piece is reborn on its original square if this one is free, otherwise the move is illegal
. A distinction is made, however, in the case where the square of capture and rebirth are identical: in AntiCirce Calvet the move is legal, and in AntiCirce Cheylan it is not

We notice that the Black King cannot capture either of the 2 white units, otherwise he would be reborn in e8 and would be captured by the remaining unit.

Solution

{

} 1.h7-h8=R + !{

    } 1...d2*c1=B[bBc1->f8]{

    } 2.g7*f8=Q[wQf8->d1] +{

    } 2...c2*d1=S[bSd1->g8] # {
AUW}

6 - serial s=10 AntiCircé

Ronald TURNBULL
2000

white Rc1a1 Sc7c6 Bb5 Kc2 black Pf2f3g3 Ke1

ser-s=10 AntiCircé(6+4)

Solution

{

} 1.Sc6-e7 2.Bb5-e8 {
} 3.Kc2-d1 {
does not check because of the Cheylan type and not "en prise" because square e8 is occupied
} 4.Ra1-a8 {
} 5.Ra8-d8 6.Sc7-a8 {
} 7.Rc1-c8 8.Kd1-e2 {no ...fxe2 because e7 is busy
} 9.Ke2-f1 10.Se7-g8 g3-g2 {only legal move of the blacks but it's stalemate !!
In fact, the White King is not "en prise", none of the promotions at the time of his capture being legal
. He can't move either
Moreover, none of the white pieces can move, which would make one of the captured pieces + promotions legal}

7 - r#2 with Pao, Vao et Mao

Brian D. STEPHENSON
2014

white Pf4e2 Bf5 PAh7g5 Ke4 black VAa2 Bg1h3 MAc6 Kh1 Qh2 Pe7

r#2(6+7)
🨵h7, g5: Paos
🨼a2: Vao
🨓c6: Mao

: When a mate in 1 move is presented to one of the 2 camps, he is obliged to play it
. : moves like a Rook but needs a sautoir to capture
: same thing but for the Bishop
: Crawling Knight performing first a Rook step, then a Bishop step.
The square of the rook step must absolutely be free

Solution

{

} 1.Ke4-f3 ! threat: 2.PAh7-f7 2...Qh2-g2 #{
} 1...Bg1-e3 2.PAh7-g7 2...Qh2*f4 #{
} 1...VAa2-f7 2.PAh7*e7 2...Qh2-g2 #{
} 1...MAc6-e5 + 2.e2-e4 2...VAa2-d5 #{
} 1...MAc6-d4 + 2.Bf5-e4 2...VAa2-d5 #

8 - h#5,5 with Andernach Grasshopper

Roberto CASSANO
2015

white Kf1 white HurdleColourChanging gb1 black Kh1 Bd1

h#5,5(2+2)
🨟b1: Andernach Grasshopper

: moves like a Grasshopper by changing the colour of the sautoir piece (unless it is a King)

Solution

{

} 1...Kf1-e1 {
} 2.Bd1-b3 hGb1-b4[b3=w] {
} 3.Kh1-g2 hGb4-b2[b3=b] {
} 4.Bb3-d5 hGb2-h2 {
} 5.Kg2-h1 Ke1-f1 {
} 6.Bd5-g2 + hGh2-f2[g2=w] #

9 - h#5 with Grasshopper et Equihopper

N.GUTTMAN & S.EMMERSON
2012

white Ke3 WEa7 black Ka2 BEa6

h#5(2+2)
🩈a7, 🩉a6: Equihopper
b) Grasshopper a6
c) -a6 +♞h3

: uses another piece as a sautoir and lands on the square symmetrical to the sautoir

Solution

{

} a) {
} 1.Ka2-a1 Ke3-d4 2.BEa6-g2 Kd4-e3 {retour} 3.BEg2-c4 WEa7-e1 4.BEc4-g2 {retour} Ke3-d2 5.BEg2-a2 Kd2-c1 #{
} b) +bGa6 {Grasshopper a6
} 1.Ka2-b2 Ke3-d3 2.BGa6-e2 Kd3-c4 3.BGe2-b5 WEa7-c3 4.BGb5-b1 WEc3-c5 5.Kb2-a1 Kc4-b3 #{
} c) -bEQa6 +bSh3 {-a6 +black Sh3
} 1.Sh3-f2 Ke3-e2 {tempo} 2.Sf2-d3 Ke2-d2 3.Sd3-b4 WEa7-c1 4.Ka2-a1 WEc1-e3 5.Sb4-a2 Kd2-c2 #

10 - serial =8 Equihopper 90°

N.GUTTMAN
2015

white Pc5 Ke3 EQe2 Sd7 black Pe6f5a2b3c4a6h3h7 Kd5 Sb1

ser-=8(4+10)
🩈e2: Equihopper 90°

: like the Equihopper of the previous problem, but with a right angle whose top is the sautoir.

Solution

{

} 1.EQe2*e6{ sautoir c4
} 2.EQe6*b3{sautoir e3
} 3.EQb3*b1{sautoir a2
} 4.EQb1*h3!{fantastic sautoir d5
} 5.EQh3*h7{sautoir f5
} 6.EQh7*a6!!{fabulous sautoir e3
} 7.EQa6*a2{sautoir c4
} 8.EQa2-e2{again c4
stalemate !!!
back home. Model stalemate. Neither ...c3 nor ...f4 are playable; c6 (via f5) and e6 are controlled.}}

11 - h#2 Transmuted Kings

R.GANAPATHI
2012

white Pc5 Kg3 Rb3 Nd7 black Pg4 Kc2

h#2 Transmuted Kings(4+2)
2 solutions
🨢d7: Nightrider

: when a King is in check, he catches the march of the piece that puts him in check. If the King is not in check, he moves normally.

Solution

{

} 1.Kc2-d2 Nd7-f3 + 2.Kd2-f1 Rb3-b1 # {
} 1.Kc2-c1 Rb3-c3 + 2.Kc1-a1 c5-c6 #

12 - h#3 Circé Einstein 2 solutions

Peter HARRIS
2015

white Ph5 Bb6 black Qg3 Ke8

h#3 Circe Einstein 2sol.(2+2)

: after their movement, and according to whether they capture or not, the pieces evolve.
- with capture : ♙->♘->♗->♖->♕
- without capture : ♕->♖->♗->♘->♙

Solution

{

} 1.Qg3-c7=R Bb6*c7=R[+bRh8] {
} 2.0-0[+bBf8] h5-h6 {
} 3.Bf8-g7=S Rc7*g7=Q[+bSb8] #{

} 1.Qg3-g6=R h5*g6=S[+bRa8] {
} 2.Ra8-a7=B Bb6*a7=R[+bBf8] {
} 3.Bf8-e7=S Ra7*e7=Q[+bSb8] #

13 - h=3 Isardam Super-Circé

Peter HARRIS
2005

white Qb1 Sh2 black Pa3e3f5f7 Kh6

h=3 Isardam Super-Circe(2+5)

: it is illegal to put two opposing pieces of the same nature in mutual "observation"
: the rebirth square of the taken piece is left to the taker

Solution

{

} 1.Pe3-e2 Sh2-f1 {
}2.Pe2*f1 [+bRf1] [+wSa1] Qb1*f5 [+bPa2] {
}3.Rf1*a1 [+wSe4] Qf5*f7 [+bSb1] {

model stalemate. The wNe4 "paralyses" the bNb1.}

14 - hs#3 Anti-Andernach with Set play

Michel Caillaud
2008

white Qa1e1 Ph2c7h7 Kh1 black Qb3 Pe2e3 Kh3

hs#3 Anti-Andernach(6+4)

: whites play and blacks collaborate until their penultimate move,
after which the whites play and force the blacks to check them out in one move.
: when a piece plays without capturing, it changes colour
Solution

{

Apparent play
} 1...Qb3-b7=w 2.Qa1-g7=b Qg7-g1=wQ 3.c7-c8=bB Bc8*b7 # {
jeu réel
} 1.Kh1-g1 Qb3-g8=wQ {
} 2.Qa1-a8=b Qa8-h1=wQ {
} 3.h7-h8=bR Rh8*g8 #

15 - h#4 Mars Circe

René J. MILLOUR
1985

white Pg6 Ka6 black Ke4

h#4 Mars Circe(2+1)b) ♔a6-->g1

: the pieces move normally, but in order to capture, they virtually pass through their original square
. For example a White Bishop in f6 can capture an opponent's piece on the diagonals c1-a3 and c1-h6.

Solution

{
The Mars Circé was invented in 1980 by RJM precisely.
} a) {

} 1.Ke4-d5 g6-g7 2.Kd5-c6 g7-g8=R {
} 3.Kc6-b7 Rg8-b8 4.Kb7-a8 Ka6-b7 # {
The line a1-a8 is now free, so it is checkmated, the Black King being unable to escape from the a-file.

} b) wKa6-->g1{

} 1.Ke4-f4 {Kf3 would be illegal because of the white pawn controlling h3 and f3.} g6-g7 2.Kf4-g3 g7-g8=Q {
no check, the King is not on the diagonal d1-h5
} 3.Kg3-h2 Qg8-g2 4.Kh2-h1 Kg1-h2 # {! It is checkmate, the white King having opened the line d1-h1 and the black King not being able to escape from the 1st row. }

16 - #6 Maximum Circe

Kjell WIDLERT
2015

white Pb2c2a4 Kh8 Rh5 Ba1 black Pa5b6a7a2f3c3 Kc4 Ra3

#6 Maximum Circe(6+8)

: blacks are obliged to make the geometrically longest legal moves

Solution

{

} 1.b2-b4 ! 1...a5*b4[+wPb2]{
} 2.b2*a3 2...a7-a5{
} 3.a3*b4[+bPb7] 3...a5*b4[+wPb2]{
} 4.b2*c3[+bPc7] 4...c7-c5{
} 5.c3*b4 5...c5*b4[+wPb2]{
} 6.b2-b3 # {
Albino successive}

17 - #16 Growing Men

Ronald TURNBULL
2013

white Ke8 black Pa2a6b7d7d6d4d2 Kb8 Bc8e1

#16 Growing Men(1+10)

Growing Men : a piece is required to play a legal move of equal or greater length to that of his previous move

Solution

{

The bK is obviously "enlarged", as well as the a2, d2, d4 & d6 pawns, which are therefore currently paralysed.
} 1.Ke8-d8 Kb8-a7 {
1... Bf2 2 Kxc8+ Ka7 3 Kc7 and 4 Kxb7#
} 2.Kd8*c8 Ka7-b6 3. Kc8-b8 Be1-f2 4.Kb8*b7+ Kb6-c5 {
4... Ka5 5. Ka7 Kb4 6. Kxa6 (idem)
} 5. Kb7-a7 Bf2-g3 {
5... Bh4 6. Kxa6 Bf6 7. Ka5 Bd8+ 8 Kb5#
} 6.Ka7*a6 Kc5-b4 7.Ka6-b6 Bg3-h4 8.Kb6-b5+ Kb4-c3 9. Kb5-a5 Bh4-g5 10. Ka5-a4 {
threatening 11 Ka3 and 12 Kb3#.
} Kc3-b2 11. Ka4-b4 Kb2-c1 {
11... Bf6 (12 Kb3+)
} 12.Kb4-b3 Bg5-f6 13. Kb3-a3 Bf6-d8 14. Ka3*a2 Bd8-a5 15. Ka2-a1 Ba5-d8 16.Ka1-b1#!

18 - #3 Growing Men

Ronald TURNBULL
2013

white Qf5 Pf6 Kf3 Bc7 black Pf7c6c4 Kh4

#3 Growing Men(4+4)

Solution

{

} 1.Bc7-b6 {
1. Be5 c5 (now, Pc3 is "enlarged" and so immobile) 2.Bg3 (threatening 3 Be1# but it's stalemate !)
(1. Ba5 (threatening 2 Bc3 & 3 Be1#) c3 (now Pc5 is immobile) 2. Bxc3 (stalemate)
(1. Ke3 ? c3 !
} 1...c6-c5 {
the two bP are now immobile. 1... c3 2. Bd4 c2 3. Bf2#
} 2.Kf3-e3 {
2. Ke4 Kg3 3. Qf4+ (Kxf4!)
} 2... Kh4-g3 3. Qf5-f4+ {#, bK is now "enlarged".}

19 - #3 Growing Men partial retro analyse

Ronald TURNBULL
2013

white Qa7 Pc4d3e5f2h3a2 Ke2 Bb1a1 black Kc1

#3 Growing Men PRA(10+1)

In "partial retro-analysis", retro-game is not imposed as in Lapierre's famous problem, on the contrary, all the "components" are considered (Ceriani said) and therefore there can be several solutions WITHOUT DUAL, depending on the past position. A priori, the bK was able to access in c1 without being enlarged. Let us place the pawns c4 & d3 in d2 & e2, with bKc4 and the two wB a1 & b1 enlarged. The continuation could be d3+ Kc3 followed by ... Kc2-c1, dxc4 and exd3.}

Solution

{

} {

} 1.e5-e6 {A concession, however, to the "limitation of history" : this move affirms that the e5-pawn comes from e2, excluding gxf3xe4-e5. And thus the previous game with exd3 is excluded. Assuming the unenlarged bK, the last move must be ...Kc2-c1. He must have played ...Kc4-c3 in response to a pawn check in b3 or d3. However, there was at this moment a pawn in b2 or d2, hence impossible, or in c2 (for a later cxb3 or cxd3), but then the bB could not already be in b1, which also makes it impossible, because there can be no access only if it is not enlarged. The bK is therefore enlarged and comes from b2 or d2.
In the first case, the Ba1 is and there is no reason for the wK to be increased, hence the continuation} 1... Kc1-b2 2.Ke2-d2! Kb2*a1 3.Qa7-g7#! {3 Qd4+? Kb2!
In the second case, the wK is enlarged. (his last move cannot have been Re1-e2+) hence the continuation } 1... Kc1-d2 2.Ba1-b2! Kd2-e1 3.Bb2-c3#! {not 3 Qa5+? Kxf2).}

Add a comment