1 - #2 with Nightrider and Grasshopper
white Ph7h4g3 Kg8 Rh1 Gg4e6d5b8 Na4c6c8f5
black Bf1d4 Kh3 Qb2 Ph2 Sf3d2 Re1e2
#2(13+9)🨟g4, e6, d5, b8: Grasshoppers🨢a4, c6, c8, f5: Nightriders
{}
1.h7-h8=G + ?
but
1...Bd4*h8 !{}
1.Nf5-d1 + ?
but
1...Kh3*g3 !{}
1.Nf5-e3 + ?
but
1...Kh3*g3 !{}
1.Nf5-e7 + ?
but
1...Re2*e6 !{}
1.Na4*e2 ?
but
1...Re1*e2 !{}
1.Gd5-d3 ! threat:
2.Nf5-d1 #{}
1...Re2*e6
2.Rh1*h2 #{}
1...Re2-e5
2.h7-h8=G #{}
1...Re2-e3
2.Nc8-f2 #{}
1...Re2-f2
2.Na4-g1 #{}
1...Bd4-f2
2.Rh1*h2 #{}
1...Bd4-e3
2.Nf5-e7 #{}
1...Bd4-e5
2.Nf5-h6 #
very nice imbroglio of lines and diagonals
A small diversion to Chinese pieces
2 - h#3 with chinese pieces
white Pe7 Kg8
black VAa4f4 LEa1 MAc6e6e4 Kd5 Pb6d6c4e3g3 PAc7b4
h#3(2+14)4 solutions
🨻c7, b4: Pao
🨼a4, f4: Vao
🨓c6, e6, e4: Mao
🨺a1: Leo
{}
1.LEa1-e5 e7-e8=VA 2.MAc6-d4 VAe8-h5 3.PAc7-c5 VAh5-f3 #{}
1.PAb4-b5 e7-e8=LE 2.PAb5-c5 LEe8*a4 3.LEa1-d4 LEa4-b5 #{}
1.MAe4-c5 e7-e8=MA 2.MAe6-d4 Kg8-f7 3.VAf4-e5 MAe8-f6 #{}
1.MAc6-d4 e7-e8=PA 2.VAa4-c6 PAe8-a8 3.b6-b5 PAa8-a5 #
Very pleasant to search with a coffin construction made of different materials
Now here is a magnificent triplet.
3 - #2 with chinese piece
white VAg2 WNe2 Kd8 Pf4 Sc1 WPc7h6
black Bf1 BNh2 Kc5 Pa7 Se3 BPa1
h#2(7+6)
🨻c7, h6, a1: Pao
🨼g2: Vao
🨷e2, h2: Nao
b) ♘c1-->c2c) ♘c1-->c3
{}
a) {}
1.BNh2-f3 WNe2-b8 2.Kc5-d4 WPh6-c6 #{}
b) wSc1-->c2{}
1.BNh2-f6 VAg2-b7 2.Kc5-d5 WNe2-c6 #{}
c) wSc1-->c3{}
1.BNh2-d4 WPh6-a6 2.Kc5-d6 VAg2-c6 #
A superb work with a beautiful purity of mat
To expand the bestiary, we will take a tour of the Chameleons.
4 - h#2 Andernach
white Pa4e4h4 Kb7 Rg4
black Pd2f6 Ka5
black Chameleon Qg5 Rh1 Sf5 Be1
h#2 Andernach(5+7)4 solutions
{}
1.cBe1*h4=cR=w cRh4*h1=cQ=b 2.cQh1*e4=cS=w cSe4-c3=cB #{}
1.cRh1*h4=cQ=w cQh4*g5=cS=b 2.cSg5*e4=cB=w cBe4-d5=cR #{}
1.cSf5*h4=cB=w cBh4*e1=cR=b 2.cRe1*e4=cQ=w cQe4-c6=cS #{}
1.cQg5*h4=cS=w cSh4*f5=cB=b 2.cBf5*e4=cR=w cRe4-b4=cQ #
There are 4 black "chameleon" pieces and 4 solutions, it's up to you !
A little break with a work by the great Michel which is much more affordable at the vulgum pecus, and which provides more aesthetic sensations than some other compositions of the dypterian sodomizers.
5 - #2 Circe
white Ba3 Kh4 Qf8 Pd3g5e5 Sc6d6 Rg3d7
black Pg7h3 Sf7 Rh2 Kf4
#2 Circe(10+5)
{
}
1.Qf8*g7 ? threat:
2.Qg7-f6 #
but
1...Sf7-h6 !{
}
1.Qf8*f7[+bSg8] + ?
but
1...Sg8-f6 !{
}
1.Rd7*f7[+bSg8] + ?
but
1...Sg8-f6 !{
}
1.Sc6-e7 ? threat:
2.Se7-g6 #
but
1...Kf4*e5[+wPe2] !{
}
1.Rg3-f3 + ?
but
1...Kf4*f3[+wRh1] !{
}
1.Ba3-c1 + ?
but
1...Rh2-d2 !{
}
1.Ba3-c5 ? threat:
2.Bc5-e3 #
but
1...Rh2-e2 !{
}
1.Qf8-g8 ! threat:
2.Rd7*f7 #{
}
1...Sf7-h6 {ou} 1...Sf7-h8 {ou} 1...Sf7-d8 2.Qg8-c4 #{
}
1...Sf7*d6[+wSg1] ! {1st correction}
2.Rg3-f3 #{
}
1...Sf7*g5[+wPg2] !! {2nd correction}
2.Ba3-c1 #{
}
1...Sf7*e5[+wPe2] !!! {3rd correction}
2.e2-e3 #
3 black corrections, sorry for the lack ! What a talent !
Another compositional genius approaches the Zebra and the equistopper.
Here's the thing :
6 - #2 Transmuted Kings with Zèbra and Equistoppers
white QFh6f4 Pd7g7a2 Kc8 Ze4 Sb3
black Pe5b7f3e2d2c3b4 Sa8 Bh2 Ke6
#2 Transmuted Kings(8+10)🨷e4: Zebra🩈h6, f4: Equistopper
{}
1.d7-d8=S + ? {(Double-check by Nd8 and EQf4 via d8)}
but
1...Ke6-c4 !{(via a2) or} 1...Ke6*e4 {(via e2)}
1.Ze4-b6 + ? {(check by EQh6 via b6)}
but
1...Ke6-c4 !{(via a2)}
1.Sb3-c5 + ?
but
1...Ke6*c5 !{}
1.a2-a4 ! threat:
2.Ze4-b6 #{(a2 having disappeared)}
1...Bh2*f4
2.d7-d8=S #{(g5 square is controlled by EQh6 and f4) }
1...Ke6-f6
2.d7-d8=B #{}
1...Ke6-e7
2.d7-d8=Q #{}
1...Ke6-d6
2.g7-g8=QF #{(via a4. Note that f6 is also controlled by EQg8 via e4)}
1...Ke6-f7
2.d7-d8=QF #{(via h6. Note that d5 is controlled by EQd8 via d2}
1...Ke6-f5
2.d7-d8=Z #{}
1...Ke6-d5
2.d7-d8=R #
{Super-AUW}
Here below is the definition of the Equistopper from an English site
"
The Equistopper (invented by Peter Harris and named by Brian Stephenson at Pitlochry, see The Problemist May 2004, p. 361) is the opposite of the Equihopper; it moves to any square that is exactly halfway towards any other unit of either colour, which must (naturally) be an even number of rows (ranks and files) away orthogonally. In J1, similarly to F1, we see how a Transmuting K, stripped of all powers except that of the attacking ES, is unable to escape the attack, and the backstop ESd2, separated from the Ks by an odd number of rows, cannot relieve it either. In J2 a Fersk”onig is chosen to be stalemated; e2 cannot be wK on account of many cooks, and surprisingly, a3 is the only square for the ES – all the units are placed with precision. J3 shows a neatly-stalemated Royal ES on an Anchor Ring, on which we note that if the ES can go to one square, it can also go to the squares 4 rows distant either way, corresponding to the choice of hurdles for an Equihopper going to the same square."
The preliminary question is : why is d8=N doesn't mate ? (to help you a little : it's a double-check !)
Un petit détour par le Triton
7 - h#3 with Pao and Triton
white Kd1 PAd4 TRa4
black Pd2d3e3e5c5 Ke4 Rf3 Sd5 Bh3
h#3(3+9)b) 🨋a4-->h4🨋a4: Triton🨵d4: Pao
{}
a) {}
1.Ke4-f5 PAd4-f4 2.Kf5-e4 TRa4-c4 3.Bh3-f5 PAf4-h4 # {}
b) wTRa4-->h4 {}
1.Bh3-d7 TRh4-g4 2.c5-c4 TRg4-f4 3.Bd7-a4 + PAd4*a4 #
due to the absence of a strike, here is a problem in Functionary Chess
8 - #2 - Functionary Chess
white Bc8e7 Kh8 Qf3 Pe3 Sb5b4f7 Rc2h4
black Qb8 Pg3g5 Rb3g4 Ba2g1h7 Ke5
#2 Functionaries(10+9)
{}
1.Rc2-c5 ? threat:
2.e3-e4 #
but
1...Ba2-b1 !{}
1.e3-e4 ! threat:
2.Sb4-d3 #{}
1...Rg4*e4
2.Qf3-f6 #{}
1...Bh7*e4
2.Rc2-c5 #
Something more complicated
9 - series s#15 - AntiCirce - Madrasi
white Pc5e5a7g7d3 Ka5
black Kd4 Pb2b4b6f6 Bd1 Rh1 Sd6
ser-s#15 AntiCirce Madrasi(6+8)
{}
1.g7-g8=S 2.Ka5-b5 {} 3.a7-a8=S 4.Sa8-c7 {} 5.c5-c6 6.Sc7-a6 {}
7.c6-c7 8.c7-c8=S {} 9.Sc8-e7 10.e5-e6 {} 11.Se7-d5 12.e6-e7 {}
13.e7-e8=S 14.Sg8-e7 15.Se7-c6 + Rh1-g1 #
h#2 Mars Circe 5 (!) solutions
white Ba8 Kh4 Qc7 Pe7 Sg8 Rd5
black Bg3 Kh3 Qh2g1 Pf2a2b4g5h5 Se3e1 Rg2a1
h#2 Mars Circe 5sol.(6+13)5 solutions
{}
1.Qg1*c7 e7-e8=Q 2.Qc7-c3 Sg8*c3 #{}
1.Qg1*d5 e7-e8=R 2.Qd5-d2 Sg8*d2 #{}
1.Qg1*a8 e7-e8=B 2.Qa8-a3 Sg8*a3 #{}
1.Qg1*g8 e7-e8=S 2.Qg8-f8 Se8-g7 #{}
1.Qg1*e7 Qc7*a1 2.Qe7-g7 Sg8-h6 #
The Master managed to discreetly put a position
11 - s#11 - AntiCirce with Leo et Moa
white LEg7 Kf8 Pf7f3c5a2b2c2 Sc3 Bc4f2 Rb1 MOe2
black Pd2 Sh1 Ba1 Kd1
s#11 AntiCirce(13+4)🨷e2: Moa🨴g7: Leo
{}
1.MOe2-d4! Kd1-c1 {} 2.Sc3-e2 + Kc1-d1 {}
3.LEg7-g1 Kd1-c1 {} 4.LEg1-g5+ Kc1-d1 {}
5.Se2-f4 Kd1-c1 {} 6.MOd4-e2 + Kc1-d1 {}
7.MOe2-c3 Kd1-c1 {} 8.Sf4-e2+ Kc1-d1 {}
9.LEg5-g1 Kd1-c1 {} 10.LEg1-g7 + Kc1-d1 {}
11.Bf2-e1 d2*e1=Q[bQe1->d8] #
Hoping that you will not have abused strong drinks to try to understand the statements and solve the problems, it remains for me to wish you a good holiday.
The holiday homework is in the electronic version of the Mestre to which I wish a well-deserved rest.
Happy reading to all.
Yours sincerely
Le greffier
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