py_20240430
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1 - 2# two-coloured chameleon
Lazard, F.1927
white Kh4 Pd5c5 Se4c3 Bb2b3 Qf2
black Ke5
#2 two-coloured Chameleon(8+1
{
} 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.}
} 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)
{}
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 M2015
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)
{}
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
{}
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
{
} 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.}
} 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
{}
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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