py_20071113

h#1,5 King in disguise 2 solutions

Kohey YAMADA
7° Saké exemple original

white Kg6 black Pe7h6 Rh8

h#1,5 King in disguise(1+3)
2 solutions

{

} 1...Kg6-g7 2.e7-e6 {proves that Rh8 is not a King in disguise } Kg7-f6 # {

} 1...Kg6-f7 2.Rh8-h7 + {now it's pe7's turn not to be a King in disguise} Kf7-g8 #

h#1,5 King in disguise

Tadashi WAKASHIMA
7° saké exemple original

white Pe5 Kg6 Bh5 black Pd6 Ke8 Ra8

h#1,5 King in disguise(3+3)

Solution

{

} 1...Kg6-f7 {
pe5 cannot be a disguised King and as the White King cannot checkmate, it is therefore Bh5 which is the disguised White King} 2.0-0-0 {If Ke8 had been a King, it could not have castled as it was in check at the start of the move, so it is the Black Rook which is a King in disguise, hence the mate : } Kf7-e8 #

h#2 King in disguise

Kohey YAMADA
7° Saké exemple original

white Kh8 black Pg7f7 Bh5h6

h#2 King in disguise(1+4)

Solution

{

In the diagram, both f7 and g7 cannot be kings in disguise because the white king must have passed through the 8th row to reach h8
} 1.g7-g5 Kh8-h7 2.g5-g4 {
now we realise that Bh5 is the Black King in disguise
} Kh7*h6 #

h#2 King in disguise

Michel CAILLAUD
7° Saké - 1° prix e.a.

white Pc2 Ke1 Rh1 Bb1 black Ph2 Sh7 Bg8g5 Ra3

h#2 King in disguise(4+5)

Solution

{

} 1.Bg5-d2 {the Bishop is therefore not the Black King in disguise
} 0-0 {that the rook is not the white king in disguise,
because it could not have passed through g1.
The last white move before the diagram position was therefore Ba2-b1, with or without a capture.
If he is king in disguise, the preceding black move was necessarily Rb3-a3 +
which prohibits the black rook from being king in disguise likewise for the Bg8,
it's the Nh7 that does it !
} 2.Bd2-g5 {switchblack capturing the escape square g5 at Nh7
} c2-c3 # {so Bb1 is the King !}

h#2 Take&Make with twin

Volker GULKE & Peter SCHMIDT
Feenschach, 2006
recommandé

white Sc3 Ke7 Rd4 Bb6e6 black Se8g3 Pd6d7f3g5 Ke5 Rf8 Bh2

h#2 Take&Make(5+9)
b) Pg5-->d2

Solution

{

} a) {

} 1.Sg3-f5 + Be6*f5-g3 + {
} 2.Rf8-f4 Rd4*f4-f5 #{

} b) bPg5-->d2{g5-->d2

} 1.Sg3-e4 Rd4*e4-g5 + {
} 2.Rf8-f5 Be6*f5-f4 #

h#2 AntiCirce

Klaus WENDA
The Problemist Supplement, 2007

white Kg7 Rd8 Ba8 Ga1c3 black Pc2c4d2 Kd4

h#2 AntiCircé(5+4)
2 solutions

Solution

{

} 1.c2-c1=G Gc3-c5 {
} 2.Gc1-e3 Ga1-e5 #{

} 1.d2-d1=R Kg7-f6 {
} 2.Rd1-d3 Rd8-h8 #

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