py_20200623

1 - hs=7 4 solutions with sparrow and eagles

Kotesovec, V.
2019

white Kd3 SWf1 black Ke6 EAg6f3c5

hs=7 4sol.(2+4)
🨢f1: Sparrow
🨺g6, f3, c5: Eagle

kind of Grashopper which, instead of arriving just after the sautoir, arises by going back 45 °
kind of Grashopper too but with a 90 ° angle on the side of the sautoir

Solution

{

} 1.SWf1-d2 {(f1-d3-d2 = 45° ; d2-d3-c4 = 135°)} EAg6-e2 + 2.Kd3-c2 Ke6-d5 3.Kc2-c1 EAc5-b1 4.Kc1-b2 Kd5-c4 {
} 5.Kb2-a1 Kc4-b3 6.SWd2-d1 {(d1-e2-f2 = 135°)} EAf3-b4 7.SWd1-b2 + Kb3-a4 {or} 7...Kb3-c4 {pat.
Black has no oher move, EAb1-a2 ou EAb1-c2 ou EAb4-a3 ou EAb4-c3 ou Ka3 ou Kc3 ou Kc2 would be a self-check.
for example, with Eagle c2, Sparrow b2 attacks King b3 since d2-c2-b3 = 135 degrees.
After 7...Ka4 or 7...Kc4 it's really pat : Sparrow is pinned by the Eagle b4, Ka2 is doubly prohibited, Kxb1 is prohibited by the Eagle e2.

} 1.Kd3-c4 Ke6-d6 2.Kc4-b5 EAc5-b6 3.Kb5-a6 EAg6-d7 4.Ka6-b7 EAf3-a6 + {(via b6)
} 5.Kb7-a8 Kd6-c7 6.SWf1-a5 {(imagine Sparrow going left from a7, then make an angle of 135 degrees with a6 and a5)
} EAa6-b8+ {(""discover check" by EAb6)} 7.SWa5-b7 + {
(access in b7 because a9-a8-b7 = 135 degrees, then fails because b5-b6-c7 = 135 degrees like also b9-b8-c7!)} Kc7-d8 {ou} 7...Kc7-d6 {pat

} 1.Kd3-e4 Ke6-e7 2.Ke4-f5 EAg6-g4 + {(check by EAf3)} 3.Kf5-g6 EAc5-d8 4.Kg6-h7 Ke7-f7 {
} 5.Kh7-h8 EAf3-e7 6.SWf1-g6 EAd8-h7 7.SWg6-g7 {(e8-f7-g7 = 135 degrees)} Kf7-e8 {or} 7...Kf7-e6 {pat
(EAh7-g8 or EAh7-g6 or EAe7-f8 or EAe7-f6 or Kf8 or Kf6 or Kg6 would be a self-check, this last move because g6-h7-i7 = 135 degrees)

} 1.Kd3-e3 EAc5-d2 2.Ke3-f2 Ke6-f5 3.Kf2-g2 EAf3-g1 {(via f1)} 4.SWf1-g4 + Kf5-f4 {
} 5.Kg2-h1 Kf4-g3 6.SWg4-h5 EAg6-g4 7.SWh5-g2 {(because g2-h1-h0 = 135 degrees)} Kg3-h4 {or} 7...Kg3-f4 {pat

Pat in the 4 corners.
Three identical mats, another slightly different.}

2 - 2# anti-circe

Gockel, H.
1998

white Bb6h5 Kb7 Qh3 Pe5f6c4 Sd8 Rd3a1 black Ph2d7f3d2b2c5 Kd6 Bb8 Rb1

#2 anti-circe(10+9)

capturing unit reborn on its start game square if free, else the move is not legal

Solution

{

Essais :
} 1.Ra1-a2 + ? Rb1-a1 ! {(occupy the rook rebirth square)
} 1.Ra1*b1[wRb1->h1] + ? Kd6-d5 ! {(no Rxd5 because rebirth square is not empty)

} 1.Bb6-a5 ! {(threat} 2.Ba5*d2[wBd2->c1] # {)
} 1...Rb1-c1 {(2 Bxd2 is now forbidden)} 2.Ra1-b1 # {
} 1...Rb1-d1 2.Ra1-c1 # {but not} 2.Ba5*d2[wBd2->c1] + ? Rd1-d2 ! {
nor} 2.Ra1-b1 + ? Kd6-e6! {(b1 & d1 not free)
nor} 2.Ra1*d1[wRd1->h1] + ? Kd6-d5 ! {
} 1...Rb1-e1 2.Ra1-d1 # {but not} 2.Ba5*d2[wBd2->c1] + ? Re1*e5[bRe5->h8] ! {or} 2...Kd6-c7 ! {(e1 not free) nor} 2.Ra1-b1 + ? Kd6-c6 ! {
(b1 & e1 not free) nor} 2.Ra1-c1 + ? Kd6-c7 ! {(c1 & e1 not free) ; and} 2.Ra1*e1[wRe1->a1] {don't even check.
} 1...Rb1-f1 2.Ra1-e1 # {mais non} 2.Ba5*d2[wBd2->c1] + ? Kd6*e5[bKe5->e8] ! {
(3 Bxe8 is forbidden by occupying White Bishop rebirth square) while now Rxe8(Rh1).

Thème Loschinsky : la TB colle la TN à 4 reprises.
No "distancing".}

3 - h#4 circe with neutral units

Gandev, K.
2003

neutral Sg3 Be2h2 black Pb5h6e5d6 Ke6 white Kf2

h#4 circe(1+5+3)
b) ♟b5 --> g2
🨄g3, 🨃e2, 🨃h2: neutral pieces

a piece taken reborn on its original square if it is free
Can be played by both side

Solution

{

} a) {
} 1.nSg3-f1 nBe2*b5[+bPb7] 2.nBh2-f4 nBb5*f1[+nSg8] 3.e5*f4[+nBc1] nBc1*f4[+bPf7] 4.d6-d5 Kf2*f1[+nBc8] # {

} b) bPb5-->g2 {b5-->g2
} 1.d6-d5 Kf2*g2[+bPg7] 2.nBe2-c4 nBh2*g3[+nSb8] 3.d5*c4[+nBf1] + nBf1*c4[+bPc7] + 4.Ke6-d6 Kg2*g3[+nBf8] # {

Echo-chameleon.}

4 - h#2 3 solutions neutral marine units

Gandev, K.
2009

neutral TRh3e7 NDg1 MSf1 black Pg7f7f6f5f4d7 Ke3 Be4 Se6e5 white Pd5d4d3 Ka6

h#2 3sol.(4+10+4)
🩃f1: Neutral Marine Knight
🩂g1: Neutral Nereide
🩁h3, e7: Neutral Triton

marine units moves are standard but takes as a Locust.
Triton is a marine Rook (Trnh3 prevent ...Kxd3 because Trnxd3-c3 and ...Kf3 with Trnxf3-e3).
Nereide is a marine Bishop (Neng1 prevent ...Kxd4 because Nénxd4-c5 and ...Kf2 with Nénxf2-e3).
neutral marine Knight f1 prevent Kd2 with Cmnxd2-b3.

Solution

{

} 1.Be4*d5 {self-pin of the Bishop : if he play, Nmnxe3-d5.} d4*e5 + {: check with the threat Néng1xe3-d4.} 2.Se6-d4 e5*f6 # {
Knight d4 is pinned by neutral Nereide g1, Bishop d5 by neutral marine Knight f1.}

} 1.Se6*d4 d3*e4 + { : threat Trnh3xe3-d3} 2.Se5-d3 e4*f5 # {: Knight d3 is pinned by neutral Triton h3

} 1.Se5*d3 d5*e6 + { : threat Cmnf1xe3-d5} 2.Be4-d5 e6*f7 # {

AB-BC-CA cycle of black moves.}

5 - h#4 2 solutions face to face

Frankiss, Ch
2019

white Royal Kg8 white Pg3 Sb8g1 black Royal Ka2 black Pf3d3 Sb6e1

h#4 face to face 2sol.(4+5)

face to face if two units of opposit coulor are "face-to-face", as a white one on e4 and a black on e5,
they exchange the way they move.

Solution

{

} 1.rKa2-b3 Sg1-e2 2.rKb3-c4 Se2-c3=K[+brSc4] + {
the Knight moves like a King, so check the King, black King moves like a Knight.
} 3.rSc4-e5=K[+wSc3] Kg8-g7 4.Sb6-c4 Sc3-e4=K[+brSe5] # ! {
The Knight turned King dominates the King turned Knight.
Note that the escape in g4 is prohibited, the pawn g3 becoming King !
Moving :} 5.rSe5-g4=rP[+wKg3][+wSe4] {

} 1.d3-d2 Sg1*f3 2.d2-d1=Q Sf3-g5 3.Qd1-g4=P[+wQg3] Qg3*e1=P[+bQg4] 4.Qg4-e2=P[+wQe1] Qe1-a1=K[+brPa2][+bQe2] # {
the black King has become a pawn, the white Pawn a King !

Of course, the two solutions "lack unity", says Mr Balasubramanian, but what fun !}

6 - h#2 symetry point

Argirakopoulos, T.
2019

white Royal Ke8 white Pe6 black Royal Kh6 black Pf5g4f4b2 Sd3

h#2 symetry point(2+6)
b) ♟b2-->c2

symatry pointtwo units placed symmetrically with respect to the center of the chessboard
(common point to squares d4, e4, e5 & d5) exchange the way they move.
Pe6 move as a Knight, Nd3 as a pawn.

Solution

{

} a) {
} 1.b2-b1=S e6-f8=R 2.Sb1-a3=K[+brSh6] Rf8-h8 # ! {, black King moves as a Knight.

} b) bPb2-->c2 {pb2-->c2
} 1.c2-c1=B e6-d8=Q 2.Bc1-a3=K[+brBh6] Qd8-f6 # {, black King moves as a Bishop.

AUW}

7 - h#4 symetry point

Foster, G.
2019

white Royal Kd2 white Se1 black Royal Ke8 black Rh8

h#4 symetry point(2+2)
b) ♜h8 --> a8

Solution

{

} a) {
} 1.0-0 rKd2-e3 2.Rf8-d8=S[+wRe1] Re1-b1=K[+bRd8][+brSg8] {
} 3.Rd8-d6=K[+wrRe3][+bKd6] rRe3-e8[+wrKe8][+bRd6] 4.Rd6-h6 rKe8-f7 # ! {

} b) bRh8-->a8 {Th8 --> a8
} 1.rKe8-c8[-a8][+bSd8][+wRe1] {(It's a real 0-0-0, but with some transformations : it does not check, the Rook moving as a Knight, but} Re1-h1=S[+bRd8] ? {would be a self-check !
} 1...Kd2-c1 2.rKc8-b8 Re1-h1=S[+bRd8] 3.Rd8-f8=K[+wrRc1] rRc1-c6=rK[+bRf8] 4.rKb8-a8=rS[+wKh1] rKc6-b7 # !

8 - h#2 3 solutions with Nightrider hopper and Grasshopper

Mueller, D & Pachl F
2007

white nhg7f8 Bg4 WGe3 Kh8 Pb3e5d5c5 Rf3c6 black Pf4 Sh6g2 Kd4 BGc1

h#2 3sol.(11+5)
🨟e3, 🨥c1: Grasshoppers
🨷g7, f8: Nightrider Hopper

Extended Knight needing a slot for his last jump

Solution

{

} 1.BGc1*c6 WGe3-g3 2.BGc6-e4 Bg4-e6 # {(Gg3 looks at e5)

} 1.Sh6*g4 Rc6-d6 2.Sg4*e5 WGe3-e6 # {(NHf8 attack the black King and NHg7 looks at c5)

} 1.Sg2*e3 Bg4-f5 2.Se3*d5 Rc6-e6 # {

3 mats par "anti-batterie" sur e6.
Squares e4, d5 & e5 are cyclically guarded by G, R and B or blocked by a black unit.}

9 - 2# with chinese pieces

Stephenson, B.
2020

white NAg1 Ka6 WPe7 Sc2 Pa2 black VAf2 Ka4 BPe1c4 Pc5

#2(5+5)
🨵e7, 🨻e1, c4: Paos
🨷g1: Nao
🨼f2: Vao

play normally but need a jumper to capture (no distance limit)
chinese Nightrider

Solution

{

} 1.WPe7-e3 ! {
} 1...BPe1-a1 {or} 1...BPe1-b1 {or} 1...BPe1-c1 {or} 1...BPe1-d1 {or} 1...BPe1-f1 2.WPe3-c3 # {
} 1...BPc4-b4 {or} 1...BPc4-h4 {or} 1...BPc4-g4 {or} 1...BPc4-f4 {or} 1...BPc4-e4 {or} 1...BPc4-d4 2.WPe3-e2 # {
: "covering" mate by Nao g1, the Pao no longer being able to intercept in c3
} 1...VAf2-h4 {or} 1...VAf2-g3 2.WPe3-e4 # {:} 2...Vd4 not possible and} 2...BPc4-c3 {activate Nao g1

}

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