py_20200310

1 - hs#3 5 solutions with locusts

Millour, RJ.
2019

white Pg7g6f5 Sc8c6 Be3a4 Kd3 black Ba1 Kd5 Lf3 Pd7h3g2 Se8f8 Re7h2

hs#3 5sol.(8+10)
🨐f3: Locust

piece moving on the Queen's lines with the help of a sautoir which she swallows
. (of course, the sautoir must be an opposing unit and the square just behind it must be free)

Solution

{

} 1.g7*f8=WL g2-g1=L 2.WLf8*e8-d8 Rh2-b2 3.WLd8*d7-d6 + Lg1*e3-d4 # {

} 1.g7*f8=Q g2-g1=Q 2.Qf8*e7 Qg1-g4 3.Qe7-e4 + Qg4*e4 # {

} 1.g7*f8=R g2-g1=R 2.Rf8-f6 Rg1-b1 3.Ba4-b3 + Rb1*b3 # {

} 1.g7*f8=B g2-g1=B 2.Bf8*e7 Bg1*e3 3.Sc8-b6 + Be3*b6 # {

} 1.g7*f8=S g2-g1=S 2.Sf8-e6 Sg1-e2 3.Se6-f4 + Se2*f4 # {

Super-Babson.}

2 - h#4 with andernach Mooses

Novomesky, D.
2013

black Kc4 black HurdleColourChanging Mb1e3f3f5

h#4(0+5)
🨐b1, e3, f3, f5: andernach mooses
b) 🨐b1-->f6
c) 🨐b1-->h7

playing like a Grasshopper following the path of a Queen, makes an angle of 45 degrees from the sautoir,
. in relation to the arrival box of a Grasshopper.
Moose b1 can go to g5 (sautoir f5) because the angle g6-f5-g5 is 45 degrees. The same applies to f6.
. makes the sautoir change colour (unless it is a King)
Andernach is a tribute to the congresses that popularised the colour change.
Thus, the move Eb1-g5 makes the Moose f5 white. In the same way, Moose f3 can move to e6 (or g6) because f6-f5-e6 = 45 degrees.
. And there again, Moose f5 becomes white ipso facto. For a very short ride, Moose's march is that of the Knight.

Solution

{

} a) 1.hMb1-g5[f5=w] hMf5-h4[g5=w] 2.Kc4-b5 hWMg5-a6 3.hMf3-d4[e3=w] hMa6-b4 4.Kb5-a4 hMh4-c3[d4=w] # {

} b) bhMb1-->f6 1.hMf6-e4[f5=w] hWMf5-e2[f3=w] 2.Kc4-d3 hWMe2-d4 3.Kd3-c2 hWMf3-e5[e4=w] 4.Kc2-d1 hWMe4-d2[e3=w] # {

} c) bhMb1-->h7 1.hMf3-g6[f5=w] hWMf5-h6[g6=w] 2.Kc4-b3 hWMh6-d3[e3=w] 3.Kb3-a2 hWMg6-d2[d3=b] 4.hMh7-c3[d3=w] hWMd3-b2[c3=w] # {

Triple écho.}

3 - h#2 breton b) ♝d8 --> c5

breton after each capture, a unit of the same nature must disappear (if possible) from the capturing side.

Bonavoglia, M.
2018

white Rd1 Ka8 Bd4g2 Sf3 Pb2 black Pe7 Kd5 Bh1d8 Sc4

h#2 breton(6+5)
b) ♝d8-->c5

Solution

{

} a) {
} 1.e7-e5 b2-b4 2.Bd8-g5 Sf3*g5[-d4] # {

} b) bBd8-->c5 {Fd8-->c5
} 1.e7-e6 b2-b3 2.Sc4-b2 Bd4*b2[-f3] # {

Typical Breton double chess.
(One can also note the aesthetic side of the first white move echoing the first black move)}

4 - 2# functionary chess

functionary chess one can only play (and a fortiori capture) on the condition of being observed by an enemy unit.

Ducak, J.
2009

white Pd7g7 Sd6 Ra8c8 Kg3 Bb5 black Bf6h1b6a3 Ke7 Qg6 Pe6a7b2 Se8 Ra1

#2 functionary chess(7+11)

Solution

{

Tries :
} 1.d7-d8=Q + ? Bb6*d8 ! {
} 1.d7-d8=R ? {blocus} 1...a7-a5 ! 2.Rd8-d7 + Ke7*d7 ! {
} 1...Se8-c7 2.Rd8-d7 # { Nc7 giving vigour to Bb5 ; on the other hand, Ba3 gives vigour to Nd6 and therefore forbids 2...Kf7 or Ke8 ;
} 1...Se8*d6 2.Rc8-c7 # {exploits the d6 blocking ;
} 1...Se8*g7 2.Ra8*a7 # {(...Kxd8 not possible, bK rejuvenates the Rc8) ;
} 1...a7-a6 2.Rd8-d7 # {, the Pa6 giving effect to the Bb5.
. The trial offers us 3 rook mates, the real game will offer us 3 pawn mates, and a 4th as a bonus !

} 1.Ra8*a7 ! {(threatening} 2.d7*e8=S # {but no move allows it to be realised) ;
} 1...Se8-c7 2.d7-d8=Q # { 2...Kxd6 impossible because of Bf6 and 2...Kf7 because of Ba3 ;
} 1...Se8*d6 2.d7-d8=S # {(} 2.d7-d8=R + Sd6-b7 ! {or} 2.d7-d8=B + Bb6*d8 ! {
: Bb5, activated by Nd6, prevents ...Ke8)
} 1...Se8*g7 2.d7-d8=R # {(to control f8, and} 2.d7-d8=Q + ? Bb6*d8 ! {)
} 1...Ba3*d6 2.d7-d8=B # {mate by double check :} 2...Bb6*d8 {is no longer sufficient because of the Ra1 ;
et} 2.d7-d8=Q ? {is illegal, the activation of Bd6 checks against the White King. ;
} 1...Ba3-b4 {or} 1...Ba3-c5 2.d7-d8=Q # {(a Bishop would allow ...Kxd6) by double check: Bxd8 does not go, the Ra1 activating its counterpart a7.

AUW with 3 changed mates.}

5 - 2# patrol

patrol it is necessary to be supported by a unit of one's side to capture or give check.
The b4-pawn does not check, being unsupported. And 1 Rxb5? is illegal.

Gockel, H.
2004

white Pb4b3a2d4e2c6 Rh3d5 Be3f7 Kb7 black Qg8 Pd2f6e6b5c7 Sf3a8 Bd1 Ka5

#2 patrol(11+10)

Solution

{

} 1.a2-a3 + ? Ka5-a4 ! {
} 1.Rd5-g5 ? {(supports this Rook by Be3 and therefore threatens} 2.Rg5*b5 # {supporting Pb4 but} 1...f6-f5! {parry the threat;)
} 1...e6-e5 2.a2-a3 # {(...Ka4 now impossible)
} 1...Sf3-e5 2.Be3*d2 # {(the e3 Bishop being supported, can capture, he in turn supports Pb4)
} 1...f6*g5 ? {is illegal, as} 1...Qg8*g5 ? {&} 1...Sf3*g5 ? {

} 1.Be3-g5 ! {(threatening} 2.Bg5*d2 # {)
} 1...e6-e5 2.Rd5*b5 # {
} 1...Sf3-e5 2.a2-a3 # {
} 1...f6-f5 2.Bg5-e7 # {
ABC-BCA (the hipsters call it "Dombro-Lacny").
A cycle based on 2 variants and the threat.}

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