py_20240430

1 - 2# two-coloured chameleon

Lazard, F.
1927

white Kh4 Pd5c5 Se4c3 Bb2b3 Qf2 black Ke5

#2 two-coloured Chameleon(8+1

: each unit (other than King) change nature after every move according to the natural cycle P-N-B-R-Q-P.
In two-coloured Chameleon, it changes also color.

Solution

{

} 1.Se4-d6[+bBd6] {

} 1...Bd6-b8[+wRb8] {; or on c7, f8
} 2.Sc3-e4[+bBe4] # {

} 1...Bd6*c5[+wRc5] 2.d5-d6[+bSd6] # {

Note that} 1...Bd6-e7[+wRe7] {is illegal.}

2 - h#2 exclusive chess

Cheylan, Y.
1996

white Pf7g5 Rc7 Sb2 Kc3 Bg2 black Bh8b3 Sg8b4 Rc5d1 Pb5c4a2f4e6 Qa5 Ke5

h#2 exclusive chess(6+13)

: checkmate must be unique.

Solution

{

} 1.a2-a1=R ? f7-f8=Q 2.Ra1-b1 Sb2-d3 + 3.c4*d3+ ! {legal because do not mate, c5 being capturable : ...Bd5

} 1.a2-a1=Q f7-f8=Q 2.Qa1-a3 Sb2-d3 # {
It's really mate because...} 3.c4*d3 + {checkmate (so illegal) cause there is other checkmates} 3.Sb4*d3 # 3.Rd1*d3 # {
On the other hand} 3...Rc7*c5 + {is illegal because there is also Qxc5#
} 3...Qf8*c5 + {idem} 4.Sb4-d5 # {(} 4.Bb3-d5 # {)
Black would have 2 replies with checkmate, therefore cannot play any.}

3 - h#2# 4 sol take & make rose giraffe

Pachl, F & Barth M
2015

neutral rob2 GId1 CAd2 DQc1 white Kh5 black Pa2c5g2f3g5 Ba4 Re5 Sh1g1 Kh3

h#2 T&M 4sol.(1+10+4)
󼎞b2: neutral Rose
🨙d1: neutral giraffe
🩃d2: neutral Camel
🨮c1: neutral bouncer (2,4)

: after a capture (Take), you must pay tribute to captured unit
by doing a movement without capture with the movement of said unit
: unit moving according to its characteristics :
(1,2) for Knight, (1,3) for Camel, (1,4) Giraffe and (2,4) for bouncer (2,4)
: extended rounded Knight

Solution

{

} 1.Re5-e2 nDQc1-e5 2.Ba4-b3 nROb2*e5-c1 # {

} 1.Re5-d5 nROb2-e5 2.Ba4-d7 nCAd2*e5-g6 # {

} 1.Re5-e6 nCAd2-e5 2.Ba4-c6 nGId1*e5-d2 # {

} 1.Re5-f5 nGId1-e5 2.Ba4-b5 nDQc1*e5-d1 # {

No mate is by double check, which is rare with neutral pieces.
But the mating unit cannot refuse check, because it is each time incarcerated by black Rook and Bishop.

AB-BC-CD-DA cycle.}

4 - s#8 with elephants

Caillaud, M.
1997

white Pa3b4b5c5e2 Rd3 WEe8 Kf7 black BEa4 Pd5h6g5h4 Kh5

s#8(5+7)
🨍e8, 🨓a4: Elephant

: Combines the powers of the Queen and the

Solution

{

} 1.Kf7-f6 + Kh5-g4 2.Kf6-e6 + Kg4-f4 3.Rd3-f3 + Kf4-e4 4.Ke6-d6 + Ke4-d4 {

} 5.Rf3-d3 + Kd4-c4 6.Kd6-d7 + Kc4*b5 7.Kd7-c7 + Kb5-a6 8. Kc7-b8 + BEa4*e8 # !

5 - h#2 double circe

Ott, M.
1991

white Kh8 Qf3 Be3 black Pg7f4h4c6d2e2 Bg6a1 Kg5 Rg4 Qd6

h#2 double circe(3+11)
b) ♛d6 <--> ♟f4

double circe : (sometimes oddly called "Circe Kamikaze"), the capturing unit is reborn as well as the captured unit.

Solution

{

} a) {
} 1.Kg5-h5 Qf3*f4[wQf4->d1][+bPf7] 2.e2*d1=S[bSd1->g8][+wQd1] Qd1*g4[wQg4->d1][+bRa8] # {

} b) bQd6<-->bPf4 {d6 <--> f4
} 1.Rg4-g1 Be3*g1[wBg1->c1] 2.d2*c1=B[bBc1->f8][+wBc1] Bc1*f4[wBf4->c1][+bQd8] # {

AUW in black rebirth.}

6 - h#2 2 sol symmetry point

Ooms, A.
2019

white Kd4 Rb2 black Ke6 Sg7g5

h#2 symetry point(3+3)
2 solutions

symmetry point : (a Japanese invention), two units placed symmetrically about the center of the board
(common point of d4, e4, e5 & d5 squares) exchange their march.
So, bNg7 walks like a Rook, the wRb2 like a Knight.

Solution

{

} 1.Sg7-g6 Rb2-e2 + {
(the Rook has recovered its usual march)
} 2.Ke6-f5 Re2-e5 # {
It's really mate : Re5 walks as a King (therefore controls e6, f6, f5, f4 & e4),
white King as a Rook (so control g4). Knight g6 cannot capture this Rook,
because it would give the White King his march, so that the Black King would find himself in check.
Neither the Black King can capture the Rook: the two Kings would "exchange their steps", in this case would not change them !

} 1.Sg7-e7 Rb2-b5 2.Sg5-f7 Rb5-e5 # {
Nf7 cannot capture the Rook, for the White King would then walk like a Knight.
And ...Kd7 is impossible, the said white King moving as a Rook.
Which walks like a King and therefore controls d6, e6, f5 & f6.

Echo mate... symmetrical !}

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