py_20230228

1 - 2# circe

Caillaud, M.
1983

white Sa2b2 Ra1h2 Pa6e5f6 Be1 Qg5 Kh6 black Pa5c2 Bb3 Ka3

#2 Circe(10+4)

a captured piece is reborn on its home square if it is free.

Solution

{

} 1.Qg5-e3 ! {(blocus)
} 1...c2-c1=Q {(Pinning wQ.)} 2.Sa2*c1[+bQd8] # {
} 1...c2-c1=R 2.Qe3*b3[+bBc8] # {(} 2.Sa2*c1[+bRh8] {is illegal)
} 1...c2-c1=B 2.Sa2-b4 # {(} 2.Sa2*c1[+bBf8] {is illegal)
} 1...c2-c1=S 2.Qe3-c5 # {(} 2.Sa2*c1[+bSb8] Sb8*a6[+wPa2] {
} 1...Ka3*b2[+wSg1] 2.Qe3-c1 # {
} 1...a5-a4 2.Sb2-c4 # {

AUW}

2 - serial helpmate consequent 9#

Dupin, J.
1981

white Pf5e4c2c4 Sa5 Ka8 black Sg2c6 Pf3g4f4d5d3b4b3 Kd4 Rb7

ser-h#9 consequent(6+11)

the legality of each position must be considered in its own (independently of the other moves).

Solution

{

} 1.Sg2-e3 ? {retrostalemate, i.e., no last white move

} 1.Sg2-e1 ! 2.Se1*c2 3.Sc2-e3 {
Here, the last White move was necessarily c2-c4, so...
} 4.b4-c3[-c4] 5.Se3-c4 {
here, the last white move was necessarily e2-e4, so...
} 6.f4-e3[-e4] 7.Rb7-e7 8.Re7-e4 9.Sc6-e5 {display-departure-rank} Sa5*b3 # !

3 - selfmate 9# with double grasshoppers

Wenda, K.
2012

white DBb3 Qg2 Ph2e5d4c3 Rh3f5 Bh7c1 Sg5 Ka3 black DKd8g3 Kg4 Pc2d5 Qe6

s#9(12+6)
🨌b3: Doublebishopper
🨑d8, g3: Doublerookhopper

piece that makes two successive grasshopper moves.
Here we have a bishop-grasshopper and two rooks-grasshopper.

Solution

{

} 1.Sg5-e4 ? d5*e4 2.Rf5-f3 + {(check by BGb3 via d1)} DKd8*b3 ! {(via d3
and not} 2...Qe6*b3 # ?{which checkmate because of the protection of RGd8 via d3)

} 1.Rf5-f4 + ! Kg4*g5 2.Rf4-f5 + Kg5-g4 {
(here, the move } 3.Qg2-f2 ? {is countered by } Kg4*h3 ! { )
} 3.Qg2-f1 ! DKg3*e5 {(via g5)} 4.Qf1-g2 + DKe5-g3 {(via g5)} 5.Rf5-f3 + {(by BGb3 via d1)} Qe6-e2 {
(here, the move } 6.Bc1-e3 ? {is countered by} c2-c1=Q + {)
} 6.Bh7*c2 Qe2-d1 {(only move, staying on the pinning line)} 7.Bc1-e3 Qd1-e2 {(idem)
} 8.Rf3-f5 + {(check by BGb3 via e6)} Qe2-c4 9.Bc2-b1 ! Qc4*b3 # {
It is easy to see that this is the only black move.
The Queen is protected by RGg3 (via d3).
Other moves of Bc2 :
} 9.Bc2-d3 ? Qc4*b3 + 10.Ka3*b3 {
} 9.Bc2-e4 ? d5*e4

4 - 2# quintessences-lions

Quah, J.
2022

white QNa8b8f8 WLd5g1 Ke2 Rc4 Sa4 black BLe7 Pc7a7c3 Ka3

#2(8+5)
🨴d5, g1, 🨺e7: Lions
🨷a8, b8, f8: Quintessence-lions

Moves along the Queen's lines with the help of a jumper
A Quintessence is a Nightrider bifurcating 90 degrees with each movement.
For example 🨷a1 can play b3, d2, e4, g3 & h5.
A Quintessence-Lion needs a sautoir.
The 🨷f8 can go to b5 via e6-c7, the 🨷a8 also via c7.

Solution

{

Try :
} 1.QNf8-b5 ? {display-departure-file} {(threatens} 2.LId5-a5 # {, the b3 square being then protected by the QLb8 via c6 sautoir a5)
} 1...c7-c5 2.LIg1*a7 # {
} 1...c7-c6 2.QNa8-d4 # {display-departure-file} { (it's mate via b5 and not} 2.LId5-a5 + ? Ka3-b3 ! {).
But} 1...LIe7-e1 ! {controlling a5.

} 1.QNa8-d4 ! {display-departure-file} {(via b5 sautoir c7; threatens} 2.QNf8-b5 # {display-departure-file} {via e6 sautoir c7, check "à la couverte")
} 1...c7-c5 2.LId5-a5 # {
} 1...c7-c6 2.LIg1*a7 # {

Djurasevic theme : ABCD-DABC.}

5 - h#2 breton

Crisan, V & Huber E
2019

white Ra4d2 Sb5f7 Pa3b4g6 Bg3h7 Kd7 black Bc4g5 Rc3f5 Ke3 Sd6f4 Pe4

h#2 breton(10+8)
b) ♞f4-->d3

breton When a piece is captured, another piece of the same kind must disappear into the capturing camp at his choice

a) On which square mate will certainly not be given?
The square d3, five times controlled, of course !
b) the mate will certainly not take place on the square f4, four times controlled, will it? ?

Solution

{

} a) {
} 1.Sf4*g6[-e4] Sb5*c3[-a4] 2.Bc4*f7[-g6] Rd2-d3 # {

} b) bSf4-->d3 {Nf4-->d3
} 1.Sd3*b4[-e4] Sf7*g5[-h7] 2.Rf5*b5[-b4] Bg3-f4 # {

In each twin, the black knight initially occupies the mating square of the other twin.}

6 - serial helpmate 15# with Pao

Emmerson, St
2014

white PAg2 Sg4 Ka1 Pf6 black Pd7f7g7 Kg1

ser-h#15(4+4)
🨵g2: Pao

moves like a Rook but needs a sautoir to capture

Solution

{

} 1.d7-d5 2.d5-d4 3.d4-d3 4.d3-d2 5.d2-d1=PA 6.PAd1-d8 7.PAd8-f8 8.PAf8*f6 {
} 9.PAf6-h6 10.f7-f5 {(note that Pah6-h1 immediately would check the White King via g1)
} 11.f5-f4 12.f4-f3 13.f3-f2 14.f2-f1=PA 15.PAh6-h1 PAg2*g7 # {

Epaulette mate}

7 - h#2,5 take & make

Tritten, P.
2014

white Bb4 Kb1 Ph4 Rd8 black Rc3 Bd1 Ke6

h#2.5 Take&Make(4+3)
b) -♙h4

After a capture (Take), the capturing piece must pay tribute to the captured one
by making a move without a capture (the Make) with the march of the opposing piece it has just captured

Solution

{

} a) {
} 1...Rd8-d5 2.Bd1-c2 + Kb1*c2-h7 3.Ke6*d5-h5 Bb4*c3-f3 # {
Black's king cannot take h4, as he should return to h5 due to the "Make".

} b) -wPh4 {-Ph4
} 1...Bb4-d6 2.Rc3-c2 Kb1*c2-f2 3.Ke6*d6-h2 Rd8*d1-h5 #

Add a comment

Name

E-mail

Website

Anti-spam