py_20210629

2 - 3# circe b) -♙f2

Trillon, JM.
1971

white Ra2a1 Pa3d5f6h6g4f2 Kf5 Bg5h3 Sh2d6 black Bd1 Ke1 Pe5 Rh5h4

#3 Circe(13+5)
b) -♙f2

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

Solution

{

} a) {
The last black move could be neither ...BxNd1 (no Nb1), nor ...BxBd1 (no Bf1), nor ...BxQd1
(would follow Qxd1 but there is nothing on the 8th row), nor ...Bd1 (would follow Bc1xg5, idem),
nor ...Ke2-e1 (would follow Bd2xg5, idem). So it was ...e7-e5.
} 1.d5*e6 ep. [+bPe7] {(threatens} 2.Ra1*d1[+bBc8] # {, rebirth of B on c8 is no more a self-check.)
} 1...Rh5*g5[+wBc1] 2.Bc1*g5[+bRh8] Rh4*h3[+wBf1] {(so that after 3 Rxd1+? Kxd1, the Rh1 revival no longer checks)
} 3.Ra2-e2 # 1...Rh5*g5[+wBc1] + 2.Bc1*g5[+bRh8] e7*d6[+wSg1] {(so that after 3 Rxd1+? Kxd1, the Rh1 revival no longer checks)} 3.Ra2-e2 # {

} b) -wPf2 {-f2
the "prise en passant" is illegal, last move may have been ...Kf2xRe1 (Ra1) following Ra1-a2+
} 1.Sd6-c8 ! {(threatens} 2.Ra1*d1 # {)
} 1...Rh5*g5[+wBc1] + 2.Bc1*g5[+bRh8] {with 2 threats : } 3.Ra1*d1 # {et} 3.Bg5*h4 # {, this last one indispensable on} 2...Rh8*c8[+wSb1] {
} 2...Rh4*h3[+wBf1] 3.Ra2-e2 # {
} 1.Ra1*d1[+bBc8] {est illégal
} 1.Sh2-f3 + ? Bd1*f3[+wSb1] ! {
} 1.Ra1-b1 ? {(threatens} 2.Sh2-f3 # {)} 1...e5-e4 ! {
}

3 - 3# kamikaze b) ♕g5/♛a5

Trillon, JM.
1977

white Kamikaze ra1 bg5 sc3 white Pf5e2g2b5 Ke1 black Kamikaze ba5 sf6 black Rc8e8 Kd7 Pc7e7h7 Sg8

#3 Kamikaze(8+9)
b) Bishops become Queens

a unit (except the King) that captures disappears ipso facto.
So if a side is more populated, it must have made a capture by the King, and therefore can no longer castle.
Here, Black is 9, White 8. White castling is therefore a priori possible.
Pawns cannot change file.

Solution

{

} a) {
Castle is legal, the Queen captured the Ff1, having the good taste to disappear immediately.
} 1.kRa1-d1 + ? Sf6-d5 ! {(white Knight is pinned)} 2.kRd1*d5[-d5] {does not even check
} 1.0-0-0 + ! kSf6-d5 2.kSc3*d5[-d5] + kBa5-d2 + 3.kBg5*d2[-d2] # ! {

} b) +bkQa5 +wkQg5 {B --> Q
the only ways to remove Bf1 are: -- a promotion of Pd7 to d1, with passage through d2, -- a promotion of Pa7 to a1,
-- the capture of f1 by the black king, in each case, uncastling the white king.
} 1.kQg5-d2 + kSf6-d5 2.kSc3*d5[-d5] + kQa5*d2[-d2] 3.kRa1-d1 # !

4 - h#2 4 sol with lion

Stepochkin, A.
2017

white LIf1 Ka1 Sb3d6 Bb5 Rf6 black Bc6 Rc7g6 Pf7g5h3e4 Qh7 Ke3

h#2 4sol.(6+9)
🨴f1: Lion

Grasshopper can reach further than the square behind the sautoir

Solution

{

} 1.Bc6*b5 LIf1-a6 2.Bb5-c6 {(back)} Sd6-c4 # {

} 1.Rg6*f6 LIf1*f7 2.Rf6-g6 {(back)} Sd6-f5 # {

} 1.Qh7-h5 Bb5-d3 2.Qh5-e2 {(activation of Lion)} Sd6-c4 # {

} 1.Qh7-h4 Rf6-f4 2.Qh4-f2 {(activation of Lion)} Sd6-f5 # {

Solutions in 2 distincts couples.}}

5 - #2 Breton opponent

Gockel, H.
2020

white Be5a2 Pb5d6e7 Rc3g4 Sc8e6 Kf6 Qg1 black Ba7 Sa6h8 Qc4 Kd5 Pd4e4f4h3g6b3

#2 breton adverse(11+11)

Breton opponent after each capture, a unit of the same nature must disappear (if possible)
from the side that suffered the capture. If there is more than one, the choice belongs to the capturing side.

Solution

{

Essais :
} 1.Rg4*f4[-b3] ? {(thematic try)} Sh8-f7 ! 2.Ba2*c4 + Sf7*e5[-c4] ! {
} 1...d4*c3[-f4] ? 2.Se6-f4 # {
} 1...Qc4*a2[-e5] ? 2.Qg1-g5 # {

} 1.Se6*f4[-d4] + ? Qc4*c8[-f4] ! 2.Qg1-d4 + Ba7*d4 {
} 1.Se6*f4[-g6] + ? Kd5-c5 ! {and not } 1...Qc4*c8[-f4] 2.Qg1*d4[-b3] # {double check ;
or even} 1.Se6*f4[-b3] + ? Kd5-c5 ! 2.Rc3*c4 + Kc5*b5 {

} 1.Be5*f4[-b3] ! {(threatens} 2.Ba2*c4 # {, but not non} 2.Rg4-g5 + ? d4*c3[-g5] ! {)
} 1...d4*c3[-g4] 2.Qg1-g5 # {
} 1...Qc4*a2[-f4] 2.Se6-f4 # {
} 1...Qc4-b3 2.Ba2*b3 # {
Changed mats (AB-BA).
Other variations
} 1...g6-g5 {(pour} 2.Ba2*c4 + ? g5*f4[-c4] {)
} 2.Rg4*g5[-d4] # {avoiding 2...dxc3 (-Tg5)
} 1...e4-e3 2.Qg1-h1 #

6 - pat aidé 1,5= patrouille super-circé madrasi einstein sentinels adv pawn a) andernach b) anti-andernach

Should be in the Guiness Book for the record number of conditions in a chess problem !

Harris, P.
2007

white Qc5f3 Kd4 Pe4 black Rf5c3 Kf4

h=1.5(4+3)

patrouille
Super-circé
madrasi
Einstein
Sentinelles en pion adverse
a) andernach
b) anti-andernach

Solution

{

} a) {Andernach
} 1...Qf3-g2=R[+bPf3] 2.Rf5*c5=wQ[+wPf5][+wQf2] {display-departure-file} Qc5-c4=R[+bPc5] {stalemate.
Note that} 3...f3*g2=wS {, would be a self-check, pawn becoming a white Knight.
} 3...Kf4-e5 {impossible, the wK being supported,} 3...Kf4-g5 {also, the Rook being also.

} b) {Anti-andernach
} 1...e4*f5=S[+bRg8] 2.Rc3-c4=wB[+wPc3] Bc4*g8=R[+bRg4][+bPc4] {pat.
} 3.Kf4-g5[+wPf4] {impossible because of the wPf4 that appears.
} 3.Kf4*f3 {is also impossible, the paralysed Rook not supporting its King
nor} 3.Kf4*f5[+wSd5] {
} 3.Kf4-g3 {impossible because Nf5 is supported.
Rg4 prevents 2...Kg4, Qf3 not being supported.

The 2 stalemate pictures are very dissimilar !}

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