py_20190205

{

} a) {
} 1.Rg1-h1 g2*h1=B 2.Rb1-b7 Bh1*f3 {(paralysing the Rook)} 3.e2-e3 + d2*c1=Q #{(Pawn is paralysed by the Queen)

} b) +wBb1 {Bb1
} 1.Rg1-f1 g2*f1=S 2.Bb1-f5 Sf1-e3 {(paralysing the Bishop)} 3.Qe7-c7 + d2*c1=R # {(Queen is paralysed by the Rook)

AUW.}

{

} 1...Ka6-b7 ? {seems forced, but no, it is only the thematic try, since it will be played on the 4th move

} 1...h7-h8=K {("parry" the check because now a6 is a non-royal King, like h8 itself !)
} 2.Kf7-e8 {(is cooking up a big surprise)
} Bh1*e4 3.h2-h1=K {(now it is e8 who is a non-royal King)
} Be4*h1 {(new consecration of the Ke8 -- thanks to the bishop! -- who, reset, can castle !)
} 4.0-0-0 Ka6-b7# ! {the impostor king can afford this privity thanks to his colleague h8.

Let us specify that the favourable prejudice applies concerning the right to castling of the rook,
which is not forced to have played the move preceding the diagram (...Ta8+):
there may have been a white pseudo-king capture like ...Nxe4.}

{

} 1.c2-c4 2.c4-c5 3.c5-c6 4.c6*b7=S 5.Sb7*d8=B 6.Bd8*c7=R 7.Rc7*c8=Q #

{

Tries :
} 1.kQd5-c4 ? {(threatening} 2.kROd1-d5 # {)} 1...kBa1-c3 ! {
} 1.kBf5*g4[-g4] ? {(threatening} 2.kBe3*f2[-f2] # {)} 1...kQc6-g6 + ! {
} 1.kBa3*b4[-b4] ? {(threatening} 2.kSb3*d4[-d4] # {)} 1...kBa1-b2 ! {
} 1.kRb1*a1[-a1] ? kRb4*b3[-b3] ! {
} 1.kQd5*c6[-c6] + ? Kb6-b5 ! {
} 1.kBe3*f2[-f2] ? kSd4*f5[-f5] ! {ou} 1...kSd4-e6=w 2.kBf5*g4[-g4] Qc6-d7 + ! {
} 1.kBf5-d7 ? {(threatening } 2.kBd7*c6[-c6] # {or} 2.kQd5*c6[-c6] # {)} 1...kQc6-g6 + !{

} 1.kBa3-b2 ! {blocus
} 1...kQc6-e6 2.kBf5*e6[-e6] # {
} 1...kQc6-g6 + 2.kBf5*g6[-g6] # {
} 1...kQc6*h6[-h6] 2.kRh2-h6 # {
} 1...kRb4*b3[-b3] 2.kBb2*d4[-d4] # {by double check
} 1...kRb4-c4=w 2.kRc4*c6[-c6] # {
} 1...kSa4-c3=w 2.kQd5*c6[-c6] # {b5 is controlled; (and not} 2.kBb2*a1 + ? Rb4-a4 ! {)
} 1...kSa4-c5=w 2.kSc5-d7 # {(the Knight remains white, because after a colour change it stabilises !)
} 1...kBa1*b2[-b2] 2.kSb3*d4[-d4] # {the rook being pinned like Na4
} 1...kPg4-g3=w 2.kBe3*f2[-f2] # {
} 1...kPf2-f1=kQ=w 2.kBf5*g4[-g4] # {
Note that} 1...kSa4*b2[-b2] {would be illegal, it is a self-check by the Rose d1 via b2 and a4
Note also} 1...kRb4-b5=w {which is also illegal

10 variations.}

{

} 1...Re4-a4=B 2.Ka5-b4 Ba4-c6=S + 3.Kb4-a4 Ba1-c3=S # {

} 1...Re4-c4=B 2.b3-b2 Ba1*b2=R 3.a3-a2 Bc4*a2=R # {

Mat du couloir and mat de Musset in the same problem !}

{

} 1.Rh2-e2 Rf3-b3 2.Sf5-h4 Bc4-d5 # {

} 1.Sf5-d4 Bc4-a6 2.Rh2-h5 Rf3-f8 # {

Reciprocal unpins, in both possible forms.}