py_20190528

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1 - hs#3 breton anti-andernach with rose

Breton : after each capture, a unit of the same nature must disappear (if possible) from the capturing side
Anti-Andernach : a unit changes colour if it does not capture

Petkov, P.
2018

white Pd6b6b5f5g6 Sd7 Rb4 Ke7 Bc2 black Qg8 Pa7b7a5d5e5f3f2 Ka8 ROe8h4

hs#3 breton(9+11)
anti-Andernach
e8, h4: Rose
b) d6 --> a6

Solution

{

} a) {
} 1.Bc2-a4[+bBa4] f2-f1=R[+wRf1] 2.Rf1-e1[+bRe1] {(unpins Rb4 on the line h4-g2-e1-c2-b4-c6-e7)} a7*b6[-a5] 3.Rb4*a4 + ROe8*a4[-e1] # {(via f6-e4-c3) unmasks battery.

} b) wPd6-->a6 {d6 --> a6
} 1.Rb4-e4[+bRe4] f2-f1=B[+wBf1] 2.Bf1-g2[+bBg2] {(unpins Bc2 on the line h4-g2-e1-c2-b4-c6-e7)} b7*a6[-d5] 3.Bc2*e4 + ROe8*e4[-g2] # {(Unmasks the battery) Specific Breton batteries.}

2 - h#2 3 sol take & make, zebu, night zebra

take & make: after a Take, the capturing piece is required to Make a move with the moving way of the captured piece
(a sort of posthumous tribute, as Alain ... Delon would say)
. Zebra: leaper (2,3)
Night Zebra : extended Zebra
. Zebu: conjunction of a Camel and a Giraffe, in other words a 1-3 and 1-4 Leaper at the same time
Leapers (2,4): To better familiarize ourselves with the 2-4 Leapers, we can observe that they are a Diminished Nightrider
(or a double Knight)
. neutral piece: piece that can be played indifferently by white or black

Barth, M & Pachl F
2014

neutral CHh6 ZEh3 Se2 Rb4 ZRd5 black Qh5 Pa7a5 Ka6 white Ke6

h#2 3sol(1+4+5)
h3: neutral Zebu
b4: neutre Rook
e2: neutral Knight
h6: neutral Leaper 2,4
d5: neutral Night Zèbra

Solution

{

} 1.Qh5*h3-g7 + {the Queen who captured a Zebu made a Zebu (in this case Giraffe) move
} ZRd5*g7-g2 + {(makes a Queen's move)} 2.CHh6-d4 Se2*d4-b8 # {by double check.

} 1.Qh5*h6-f2 {(the Queen who captured a 2-4 Leaper has made a 2-4 Leaper move)
} ZRd5*f2-g2 + 2.Se2-d4 + ZEh3*d4-b3 # {: the Zebu has played like a Giraffe, captured a Knight of which he has taken a step, made finally check since it is a Camel!!!;
checkmate also by double check

} 1.Qh5*e2-g3 {(the Queen who captured a knight has made a knight move)} ZRd5*g3-g2 + 2.ZEh3-d4 CHh6*d4-e8 # {(a Giraffe step); mate also by double check)

. Cyclic captures of 3 neutral pieces in d4. Zilahi also cyclic. The "Noctamzebra" comes to checkmate on g2 in all three solutions, accompanied three solutions, accompanied successively by a nNb8, a nZub3 & a nBone8.}

3 - 2# avec locustes et rose-locuste

locust : A grasshopper that swallows its sautoir

. Grasshopper : Piece moving like a Queen, but with the help of a sautoir behind which it lands
Note that the square just behind the sautoir must be free or occupied by an enemy unit

rose-locuste : Rose needing a sautoir to move, which sautoir she swallows

Cheylan, Y.
2015

white Bf1 Kf6 Lc1e8g8h8 Pg4 Se6 LSh1 black Ph5d4b3 Kd3 Bd1e2

#2(9+6)
c1, e8, g8, h8: Locusts
h1: Rose Locust

Solution

{

Tries :
} 1.Se6-c5 + ? Kd3-c4 ! {
} 1.Se6-f4 + ? Kd3-e3 ! {
} 1.Kf6-f7 ? b3-b2 ! 2.Se6-c5 + Kd3-c4 ! { the g8 locust was intercepted on the diagonal
} 1.Kf6-e7 ? {(and there, the defence} 1...Bd1-c2 {doesn't work because of} 2.Lh8*d4-c3 # {car} Kd3-e4 {is impossible because of} 3.LSh1*e4-f6 {through f2, as well as} 2...Kd3*c3 {because of} 3.LSh1*c3-b1 {passing by g3 and e4.) but } 1...Be2*f1 ! 2.Se6-f4 + Kd3-e3 ! {Locust e8 is intercepted on the column e ;
} 1.Kf6-g5 ? {ou} 1.Kf6-g6 ? h5*g4 ! {as the Locust g8 is intercepted on the g-file, we do not have checkmate by } 2.Lg8*g4-g3 {

} 1.Kf6-f5 ! {
} 1...Bd1-c2 2.Lh8*d4-c3 # {(car} Kd3*c3 {est impossible à cause de} 3.LSh1*c3-b1 {passing through g3 and e4 as already seen.
Note that this was not a threat because of 2...Kc2! slipping between the two Locusts.)
} 1...b3-b2 2.Se6-c5 # {
} 1...Be2*f1 2.Se6-f4 # {
} 1...h5*g4 2.Lg8*g4-g3 # {
} 1...h5-h4 2.Lh8*h4-h3 # {

Three test rebuttals become three variants. The judges (Dupin & Kerhuel) point out that a Nereid (Sea Bishop) could have been placed in d1 instead of a second Bishop of white squares. "A matter of taste", they say.
We obviously prefer the author's choice.
This problem is also the 5th of a competition in which... the first four seemed undrinkable to us.
It is therefore the best that our modest constitution can offer.}

4 - h#3 2 solutions back-to-back with Nightrider Mao

Nightrider Mao : The Mao is a creeping Knight who makes a rook step and then a bishop step,
but he needs the square for his rook step to be free.
. And "at night", the cats may not all be grey but the animals extend their march.
. So NMh1 can play f2, d3 (because e2 is free), b4 and g3 but not f5 because g4 is occupied
back-to-back: if two pieces of opposite colour are "back-to-back",
for example a white one on e5 and a black one on e4, they exchange moving way the time of the opposite.

Foster, G.
2018

white NAh1 Ke3 black Pb3 Kg4

h#3 back-to-back(2+2)
h1: Nightrider Mao

Solution

{

} 1.Kg4-h3 NAh1-b4 {making the modest b3 pawn a formidable Nightrider Mao... which remains a pawn all the same, able to promote itself !
} 2.b3-f1=R ! NAb4-d5+ 3.Kh3-h2 NAd5-h3 # ! {
NM, having acquired the King's moving way, attacks the bK which, having become a NM, cannot go anywhere: to f1, f3 and even less to g4 as h3 is occupied.

} 1.Kg4-g5 NAh1-e7 {(g4 square being freed)} 2.b3-b2 NAe7-g6 + {(NM having the moving way of a King, checks)
} 3.Kg5-a2 {(bK having the march of a NM)} NAg6-a3 # ! {
the bK is attacked by the NM who has become King; he himself, now NM, is completely immobilized because his rook steps are forbidden}

5 - h#4 circe with neutral pawns b) bKa8

Circe : a captured unit is reborn on its native square

Feather, Ch
2005

neutral Pg4f3 black Ka1 white Kd3

h#4 Circe(1+1+2)
g4, f3: neutral Pawns
b) Ka1-->a8

Solution

{

} a) {
} 1.nPf3-f2 Kd3-c2 2.nPf2-f1=nB nBf1-h3 3.nPg4*h3[+nBf1] nBf1*h3[+nPh7] 4.nBh3-e6 nPh7-h8=nB # {

} b) bKa1-->a8 {Ra1-->a8
} 1.nPg4*f3[+nPf2] Kd3-e4 2.nPf2-f1=nR nRf1*f3[+nPf7] 3.Ka8-a7 nPf7-f8=nR 4.nRf8-b8 Ke4*f3[+nRa8] # {

Quite convincing orthogonal-diagonal echo !}

6 - h#2 isardam with first a) vao then b) pao

Isardam : It is illegal for 2 opposing units of the same kind to observe each other
. Chinese piece (Pao or Vao) : moves normally, but needs a sautoir to make a capture

Loerinc, J.
2005

white VAe3 black VAg3 Kc7 PAc1

h#2 Isardam(1+3)
c1: Pao
e3, g3: Vao
b) Pao e3

Solution

{

} a) {
} 1.Kc7-d6 VAe3-a7 2.PAc1-c7 VAa7-b8 # {
Pao and bK moves (other than ...Ke5, remaining under the attack of the Vao) are prohibited according to the Isardam effect.

} b) -wVAe3 +wPAe3 {Pao e3
} 1.Kc7-c6 PAe3-e8 2.VAg3-c7 PAe8-c8 # {
Vao and bK moves (other than ...Kc5, remaining under the attack of the Pao) are prohibited according to the Isardam effect}

7 - serial h#17 2 sol with nightrider-hopper and Grasshopper

Nightrider hopper: needs a sautoir to move, behind which he lands at a Knight's step

Novomesky, D.
2009

white Pc6 Kb6 black NHb2 Kb1 Gh1

ser-h#17 2sol.(2+3)
h1: Sauterelle
b2: Nightrider hopper

Solution

{

} 1.Kb1-c2 2.Kc2-d3 3.Kd3-e4 4.Gh1-d5 5.Ke4-f4 6.NHb2-h5 {
} 7.Kf4-e5 8.Ke5-d6 9.Gd5-d7 10.Kd6-e7 11.Ke7-d8 12.Kd8-c8 {
} 13.Kc8-b8 14.Kb8-a8 15.NHh5-b8 16.Gd7-b5 17.Gb5-b7 c6*b7 # {

} 1.Gh1-a1 2.Kb1-a2 3.Ga1-a3 4.Ka2-b3 5.Kb3-c4 6.Kc4-d5 {
} 7.Kd5-d6 8.NHb2-e8 9.NHe8-c4 10.Kd6-e7 11.Ga3-f8 12.Ke7-d8 {
} 13.Kd8-c8 14.Kc8-b8 15.Gf8-a8 16.Ga8-c8 17.NHc4-a8 c6-c7 #

8 - h#3 3 solutions with Paos & Vao

Wong, P.
2000

white VAd5 Ke6 PAb1h6 black Bc6 Kh1 Qh8 Pg7b6b5 Sa3 Ra1

h#3 3sol.(4+8)
b1, h6: Paos
d5: Vao

Solution

{

} 1.Qh8-f8 VAd5-b3 2.Bc6-g2 VAb3-d1 + 3.Qf8-f1 VAd1-h5 # {by double check, and not} 3...VAd1-f3+ {which also double check, but} 4.Kh1-h2 ! {saves itself honourably

} 1.Bc6-e8 PAb1-b3 2.Ra1-g1 PAb3-h3 + 3.Be8-h5 PAh3-f3 # {still double check

} 1.Ra1-a2 PAh6-f6 2.Qh8-h2 PAf6-f3 + 3.Ra2-g2 PAf3-f1 # {by double check too, and not} 3...PAf3-h3 {because of } 4.Kh1-g1 {

Anti-battery cycle.}

9 - serial h#3 4 solutions anti-circe

Anti-circe : During a capture, the capturing piece returns to its native square...
if it is free, otherwise the capture is illegal

Millour, RJ.
2005

white Bg8 Kh5 Qf8 Pe7f7 Sd7d4 Re3 black Pc3b2 Kc4 Re8 Bb4

ser-h#3 Anti-circe(8+5)
4 solutions

Solution

{

} 1.b2-b1=Q 2.Qb1-e4 3.Qe4*e7[bQe7->d8] f7*e8=Q[wQe8->d1] # {(it is indeed checkmate, because} 4.Kc4*d4-e8 {is prevented by the e3 Rook, the capture by Qf8 being illegal)
Rook promotion (} 3...f7*e8=R[wRe8->h1] {) would have allowed} 4.Kc4-d3 ! {,
Bishop promotion (} 3...f7*e8=B[wBe8->f1] {)} 4.Kc4-d5 ! {,
Knight promotion (} 3...f7*e8=S[wSe8->b1] {)} 4.Kc4-b5! {.

} 1.b2-b1=R 2.Rb1-d1 3.Rd1-d3 f7*e8=R[wRe8->h1] # {

} 1.b2-b1=B 2.Bb1-e4 3.Be4-d5 f7*e8=B[wBe8->f1] # {

} 1.b2-b1=S 2.Sb1-a3 3.Sa3-b5 f7*e8=S[wSe8->b1] # {

Babson Theme}

10 - h#2 2 solutions take & make

Tritten, P.
2010

white Sd7 Rc8 Bb8 Ke8 black Be5d3 Kd5 Qh4 Pb6a6d4 Sb7g6 Rc5f1

h#2 Take&Make(4+11)
2 solutions

Solution

{

} 1.Rc5-c7 Sd7*e5-f4 + 2.Kd5-d6 Rc8*c7-c8 # {

} 1.Be5-c7 Sd7*c5-c3 + 2.Kd5-c6 Bb8*c7-b8 # {

Another placement of the Bishop or Rook on the last move would allow a black capture immediately replacing a unit on c7.

Reciprocal batteries and returns.}}

11 - s#2 Fuddled Men

Fuddled Men : a unit must rest after each move to digest its alcohol

Bissicks, P.
2013

white Kf3 Ra1 black Pg6 Kh1 Bf1

s#2 Fuddled Men(2+3)

Solution

{

Tries :
} 1.Ra1-a4 ? Bf1-c4 ! 2.Kf3-e2 g6-g5 + 3.Ra4*c4 {
} 1.Ra1-a7 ? Bf1-a6 ! 2.Kf3-e2 g6-g5 + 3.Ra7*a6 {

} 1.Ra1-a6 ! {
} 1...g6-g5 2.Kf3-f4 Kh1-h2 # {(or K elsewhere, or B plays)
} 1...Bf1-e2 + 2.Kf3-g4 g6-g5 # {or Kh1 plays
} 1...Bf1-g2 + 2.Kf3-e4 g6-g5 # {or Kh1 plays
} 1...Bf1-h3 2.Kf3-g4 g6-g5 # {or Kh1 plays
} 1...Bf1*a6 {(or on b5, c4)} 2.Kf3-e2 g6-g5 # {or Kh1 plays
} 1...Kh1-g1 2.Kf3-f2 g6-g5 # {or Bf1 plays
} 1...Kh1-h2 2.Kf3-g3 g6-g5 # {or Bf1 plays

. The white King moves to 6 separate squares (of the 8 available).}

12 - #2 sat

SAT: a King is in check when... he threatens to play on a square where he is not in check in the traditional sense.
. It is checkmate if the King cannot deny himself this "free" square.

Gockel, H.
2008

white Bb8 Ka8 Qb4 Pa7g4 Sg3g7 Ra4 black Qg2 Pg6h6g5h3 Kh4 Bh1 Sf2

#2 sat(8+8)

Solution

{

} 1.Sg3-e4 ? {is illegal

} 1.Sg7-e6 ! {threatening} 2.Se6-f4 # {giving 2 free squares for bK} 1...Qg2*g3 {(to parry} 2.Se6-f4 + ? {by} Sf2*g4 ! {)} 2.Se6-d4 # {because after} 2...Sf2*g4 {, the black King would have the square h5.
} 1...Sf2*g4 {(to parry} 2.Se6-f4 + ? {by} Qg2*g3 ! {)} 2.Se6-c7 # {because after} 2...Qg2*g3 {, the black King would have the square h5.}

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