py_20160412

h#4 Mirror Circe Take&Make

R.KUHN
2015

white Kg6 Bh1 black Pe6f5g2 Kb8 Ba8

h#4 Mirror Circé(2+5)
Take&Make
2 solutions

: A captured piece is reborn on the original square in the opponent's side.
example: a white Bishop from a black square, if captured, will be reborn on f8 if the square is free
. : A captured piece is required, just after the capture to make a move with the characteristics of the captured piece. the captured piece.
This move does not have to be a capture.

Solution

{
}1.g2-g1=R + Kg6*f5-f4[+bPf2] {
} 2.Rg1*h1-b7[+wBc8] Kf4-g3 {
} 3.f2-f1=B Bc8*b7-a7[+bRh1] + {
} 4.Kb8*a7-g1[+wBf8] Bf8-c5 # {

} {
}1.g2-g1=B Bh1-f3 {
} 2.Bg1-a7 Kg6*f5-f4[+bPf2] {
} 3.Ba8-e4 Kf4*e4-c6[+bBf1] {
} 4.Kb8-a8 Kc6-c7 #{
Mats in two opposite corners.}

r#2 Circe

Shlomo SEIDER
1976

white Pe5a7h4 Kh8 Bc1 black Pf6d7c7h5 Kh6 Bh2c4 Sg4c3 Re3a2

r#2 Circe(5+11)

: when a mate in 1 move is presented to 1 of the 2 camps, it must play it : a captured piece is reborn on its native square if it is free, otherwise it disappears
Solution

{

} 1.a7-a8=Q ? Ra2*a8[+wQd1] 2.Qd1*g4[+bSg8] {
} 1.a7-a8=R ? Ra2*a8[+wRh1] 2.Rh1*h2[+bBf8] {
} 1.a7-a8=B ? Ra2*a8[+wBf1] 2.Bf1*c4[+bBc8] {
} 1.a7-a8=S ? Ra2*a8[+wSb1] 2.Sb1*c3[+bSb8] {

} 1.e5*f6[+bPf7] ! {So there will be no ...Bg8. The threat is 2 Bb2, after which Black must play ...Re8#.} threat: 2.Bc1-b2 2...Re3-e8 #{
Black will therefore endeavour to obstruct the e column
} 1...Bh2-e5 2.a7-a8=R 2...Ra2*a8[+wRh1] #{
} 1...Sc3-e4 2.a7-a8=S 2...Ra2*a8[+wSb1] #{
} 1...Bc4-e6 2.a7-a8=B 2...Ra2*a8[+wBf1] #{
} 1...Sg4-e5 2.a7-a8=Q 2...Ra2*a8[+wQd1] #{

AUW.}

h#4 AntiAndernach White UltraMaximum

Peter HARRIS
2013

white Bf2 black Qh8 Pa2 Ka1 Ba8 Rf4

h#4 AntiAndernach(1+5)
White UltraMaximum
b) h=4

: when a piece plays without capturing, it changes colour
. : White is required to play the geometrically longest moves

Solution

{

} a) 1.Rf4-d4[+wRd4] Rd4-d8[+bRd8] 2.Ba8-e4[+wBe4] Bf2-a7[+bBa7] 3.Rd8-d5[+wRd5] { pour interdire 4 Fa8.
} Be4-b1[+bBb1] 4.Ba7-d4[+wBd4] Bd4*h8 # {

} b) {
} 1.Qh8-h1[+wQh1] Qh1*a8 2.Rf4*f2 Qa8-h1[+bQh1] 3.Qh1-h2[+wQh2] Qh2-b8[+bQb8] 4.Rf2-b2[+wRb2] Rb2*b8{
stalemate : a Q would not stalemate because on ...Kb1, it would be forced to play in h2}

h#6 with Locusts

bernd ellinghoven &
Petko PETKOV
2001

white Ka5 WLa1 black Pb5 Sa8 Kc4 BLb4

h#6(2+4)
🨊a1, 🨐b4: Locusts

: Grasshopper that grabs its sautoir

Solution

{

} 1.Sa8-b6 Ka5-a6 {
} 2.Sb6-c8 Ka6-b7 {
} 3.Sc8-a7 WLa1*a7-a8 {
} 4.Kc4-b3 Kb7-a6 {
why not Kb6? because Ka4 would become an eatable sautoir !
} 5.Kb3-a4 Ka6-a7 {
} 6.Ka4-a5 Ka7-b7 #

h#3 reciprocal

Theodor STEUDEL
2010

white Pf7d7a2 Ka1 black Pe2g2 Kh8

h#3 reciprocal(4+3)

: unlike the normal helpmate, black has the additional possibility to checkmate (or stalemate) White themselves on their last move

Solution

{

normal helpmate :
} 1.e2-e1=S f7-f8=B 2.g2-g1=R d7-d8=Q 3.Se1-d3 # {

the reciprocal :
} 3.Rg1-g8 Qd8-h4 #{

AUW}

h#2 Couscous AntiCirce Madrasi

Pierre TRITTEN
2015

white Kc4 Rf5 Sf8g3 Bh5 black Kc6 Rc7 Bb7

h#2 Couscous AntiCirce(5+3)
Madrasi
2 solutions

: a capturing piece is reborn on the native square of the captured piece. : Opposing pieces of the same kind that observe each other are paralysed

Solution

{

} 1.Bb7-c8 Bh5-f3+ {
} 2.Bc8*f5[bBf5->h1] Sg3*h1[wSh1->c8] # {

} 1.Rc7-h7 Rf5-f6 + {
} 2.Rh7*h5[bRh5->f1] Sg3*f1[wSf1->a8] #

h#2 couscous Circe

Pierre TRITTEN &
Chris FEATHER
2015

white Kc3 Rb3 Bd3 black Pe6a2c4c5 Kd5 Se5

h#2 couscous Circe(3+6)
2 solutions

: a captured piece is reborn on the native square of the capturing piece

Solution

{

} 1.c4*b3[+wRb7] Bd3-b1 {
} 2.a2*b1=R[+wBa8] Rb7-d7 #{

} 1.c4*d3[+wBd7] Rb3-b1 {
} 2.a2*b1=Q[+wRd8] Bd7-c6 # {
Bien sûr,} 3.Kd5*c6[+wBe8] {is not allowed due to the revival of the wB in e8.}

serial h#8 with Sparrow

Miodrag MLADENOVIC
2011

white Bd6a8 Kf4 WSc3 Pd2 Se5 Re8 black BSd3 Pd7 Kd4

ser-h#8(7+3)
🨊c3, 🨐d3: Sparrow

: grasshopper that lands at an angle of 135 degrees to the sautoir, i.e. turns back at 45 degrees, on a square obviously adjacent to the sautoir

Solution

{

} 1.BSd3*c3 {sautoir d2 or d4
} 2.BSc3-c4 {d4
} 3.BSc4-c5 {d4
} 4.BSc5-d5 {d4 or d6
} 5.BS5*e5 {d4 or d6
} 6.BSe5-e4 {d4 or f4
} 7.BSe4-e3 {d4 or f4
} 8.BSe3-d3 {d2 ord4
this is the starting position without the Sparrow c3 and Ne5
} Re8-e4 #{

Echo orthogonal-diagonal.}

serial h#9 with Sparrow

Friedrich HARIUC
2011

white Pf2g5 Kb3 Bg6 WSa7f7 black Pg2 Ka1

ser-h#9(6+2)
2 solutions
🨊a7, f7: Sparrow

Solution

{

} 1.g2-g1=SW 2.SWg1-f4 3.SWf4*g6 4.Ka1-b1 5.SWg6-b2 6.Kb1-c1 7.Kc1-d2 8.SWb2-c3 9.Kd2-d3 WSf7-e3 # {display-departure-file} {
Sparrow a7 checks and controls e4, Sparrow e3 controls e2 and will control d2 and d4 if bK plays it !
} 1.g2-g1=R 2.Rg1*g5 3.Rg5-f5 4.Ka1-b1 5.Kb1-c1 6.Kc1-d2 7.Kd2-d3 8.Kd3-d4 9.Rf5-d5 WSa7-c4 # {display-departure-file} {
The Sparrow c4 checks by relying on the bR, which is pinned by the Sparrow f7. Each White Sparrow uses his colleague as a sautoir for mate. }

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