py_20180417

1 - hs#2 with Nightrider, Leo, pao and Vao

Vysotska, J.
2012

white VAd8 Sg8b3 Ra8c1 Kh8 LEg2 black Pb5c7d6g7f4 Kb4 PAh1 Nh3h4

hs#2(7+9)
2 solutions
🨻h1: Pao
🨨h3, h4: Nightrider
🨶d8: Vao
🨴g2: Leo

: move as a Queen, Rook and Bishop respectively, but need a sautoir to capture
: extended knight
: white and black collaborate until the last moment where white forces black to checkmate him in 1 move

Solution

{

} 1.Sb3-d4 {Future sautoir.} Nh4-f5 + 2.VAd8-h4 f4-f3 3.LEg2-g4 + Nh3-f4 # {

} 1.Sb3-c5 Nh3-d1 + {(check by Pao h1)} 2.LEg2-h3 d6-d5 3.VAd8-e7 + Nh4-d6 # {

Double check by Pao and Nightrider. Nice harmony between the two solutions :
the NO who keeps g7 in one solution mate in the other and vice versa.
And each time, the mating NO occupies the square vacated by a black pawn.
Finally, the Leo who attacks the bK in one solution serves as a sautoir for the black Pao in the other, and similarly for the Vao.}

2 - 3# with Eagles & Paos

Rehm, HP.
2017

white Sb8f8 Pf4 Kb2 Re5c3 Bd3 black Pb6 Kd4 PAc5d6 BEf3 REd8a5

#3(7+7)
🨑d8, a5: Rook-Eagle
🨻c5, d6: Pao
🨒f3: Bishop-Eagle

: The Eagle makes a 90 degree angle to the sautoir,
so the 2 Rook-Eagles control c6: ...REa5-c5-c6 and REd8-d6-c6.
And if there is a unit on d5, the Bishop-Eagle will control c4 & e6.
Pao: see problem 1 for the definition.

Solution

{

Essais :
} 1.Bd3-e2 ? BEf3-d3 ! {
} 1.Sb8-c6 + ? REa5*c6 ! {
} 1.Sf8-e6 + ? REd8*e6 ! {
} 1.Bd3-b5 ? PAc5-d5 ! {

} 1.Bd3-f1 ! {
} 1...BEf3-h1 2.Sb8-c6 + REd8*c6 3.Sf8-e6 # {
} 1...PAc5-d5 2.Sf8-e6 + {
} 2...BEf3*e6 3.Rc3-c4 # {
} 2...REd8*e6 3.Sb8-c6 # {
} 1...PAd6-d5 2.Rc3-c4 + {
} 2...BEf3*c4 3.Sf8-e6 # {
} 2...REa5*c4 3.Sb8-c6 # {

AB-BC-CA cycle.
White occupies on the 2nd move a square on which Black has just reinforced his guard ("Keller paradox").}

3 - r#2 take & make

Laue, H.
2008

white Pg4 Sd4 Bc5h3 Ke3 Ra1 black Qg2 Pc6d6g3g5 Sb8e1 Kf1

r#2 T&M(6+8)

: After a capture (the "take"), you have to make a move without capturing (the "make") with the march of the captured unit
. (The author is precisely the inventor of the "take & make")

Solution

{

} 1.Bc5*d6-d5 ! {threatens} 2.Sd4-b5 2...c6*b5-d4 # {
} 1...c6-c5 2.Bd5-c6 2...Sb8*c6-d5 # {
} 1...c6*d5-b3 2.Sd4-c2 2...b3*c2-d4 # {
} 1...c6*d5-c4 2.Sd4-b3 2...c4*b3-d4 # {
} 1...c6*d5-f3 2.Sd4-e2 2...f3*e2-d4 # {
} 1...c6*d5-e4 2.Sd4-f3 2...e4*f3-d4 # {
} 1...c6*d5-f7 2.Sd4-e6 2...f7*e6-d4 # {
} 1...c6*d5-e6 2.Sd4-f5 2...e6*f5-d4 # {
} 1...c6*d5-b7 2.Sd4-c6 2...b7*c6-d4 # {
} 1...c6*d5-a2 2.Ra1-b1 2...a2*b1-d1=S # {(et pas} 2...a2*b1-c1=Q + 3.Bh3*g2-d2 {)

Knight's complete rose window}

4 - 2# anti-patrol (Lortap)

Kerhuel, M.
2008

white Bg8 Ke4 Qb8 Pd7e6e5 Se8d5 Rg5c4 black Ph7b5c7 Sg6 Rf8a6 Kd6

#2 Lortap(10+7)

: a piece can only capture if it is not patrolled (observed) by a piece on its side
. Thus, the only escape for the bK is e7; d7 would be another escape if there were no pawn on that square

Solution

{

Try :
} 1.Qb8-b7 ? b5*c4 !{gives the escape square c5

} 1.Qb8-a7 ! {threatens} 2.d7-d8=Q # {
} 1...b5*c4 2.Qa7-c5 # {
} 1...c7-c5 2.d7-d8=S # {
} 1...c7-c6 2.d7-d8=R # {et pas} 2.d7-d8=Q + Kd6*e5 ! {Rg5 being now "patrolled" by Qd8
} 1...Rf8-f7 2.d7-d8=B #{

AUW.}

5 - h==4,5 mars circe

In tribute to a great composer who has just passed away.

Smits, G.
2014

white Pe7f7h5g3c4 Kg5 Bh7 black Qe1 Pg4f2e2d2c5 Ke3

==4.5 mars Circe(7+7)

: to capture, a piece virtually passes through its native square

Solution

{

the h6 square being controlled by g4 via g7, the white king cannot go there immediately
} 1.f7-f8=B Qe1-f1 2.Bf8-g7 e2-e1=R 3.Kg5-h6 Ke3-e2 4.e7-e8=Q + {check via d1} d2-d1=S + {check via g8} 5.Qe8-g8! {

double stalemate with 6 pinned units Qg8 is pinned by Nd1 which is itself nailed by Qg8
Bh7 is pinned by Re1 which is itself nailed by Kh6
Qf1 is pinned by Bh7 and of course Bg7 which is pinned by the pg4
AUW.
Solution without capture.}

6 - s#5 with camel and Leo

Petkov, P.
2003

white LEc1 CAd3 Qe4 Pg3h2d7c6 Sc8 Ba7 Rh7e8 Kg1 black Qb1 Pa2b2c2c3e3h3g5 Kc7 Ba1

s#5(12+10)
🨴c1: Leo
🨷d3: Camel

: bouncer (1,3) (for information, Knight is a bouncer (1,2)

Solution

{

} 1.CAd3-e6 ! {
} 1...e3-e2 {ou g4} 2.Qe4-e5 + 2...Kc7*c6 3.Qe5*c3 + 3...Kc6-d5 4.Qc3-c5 + 4...Kd5-e4 5.LEc1-e1 + {(d3 & f3 are monitored by CAe6)} 5...Qb1*e1 # {

} 1...b2*c1=BL 2.d7-d8=B + 2...Kc7*c8 3.Bd8-e7 + 3...Kc8-c7 4.Be7*g5 + 4...Kc7-d6 5.Bg5-f4 + 5...BLc1*f4 # {
} 1...b2*c1=B 2.d7-d8=CA + 2...Kc7*c8 3.CAd8-g7 + 3...Kc8-c7 4.CAg7-h4 + 4...Kc7-d6 5.CAh4*e3 + 5...Bc1*e3 # {
} 1...b2*c1=BC 2.d7-d8=S + 2...Kc7*c8 3.Sd8-f7 + 3...Kc8-c7 4.Sf7-e5 + 4...Kc7-d6 5.Qe4-d4 + 5...BCc1*d4 # {
} 1...b2*c1=S 2.d7-d8=LE + 2...Kc7*c8 3.LEd8-d7 + 3...Kc8-c7 4.LEd7-d3 + 4...Kc7-d6 5.Ba7-d4 + 5...Sc1*d3 # {

cyclic Babson LB-BCh-ChN-NL.}

7 - h#3 annan chess

Argirakopoulos, T.
2013

white Pd6 Kc6 Sc8 black Pd3g2h4 Kh7 Bg8 Sc5

h#3 Annan Chess(3+6)
b) ♝g8-->e7

: is also called "southern chess", probably because a piece adopts the march of the one located in the south...
of its own.
Note that South is not at the same place for White and Black !

Solution

{

} a) {
} 1.g2-g1=B d6-d7 2.Bg1-d4 Sc8-d6 3.Bd4-h8 d7-f8=S # {
(the Kh7 having the march of the Bh8 can unfortunately for him only go on the squares g6, f5 or e4)

} b) bBg8-->e7 {Fg8-->e7
} 1.g2-g1=R d6*e7 2.Rg1-g8 e7-e8=Q 3.Rg8-h8 Qe8-g6 # {
(As Kh7 has the Rh8 march, he has no more escape square)}

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