py_20220524

1 - maximum selfmate 4# circé

Yzarn, X.
1969

white Rc2c5 Ka5 Sf2 Ph3e4 Be6 black Pc6 Bc8 Ka3 Rg2

s#4 Maximum Circe(7+4)

Black is required to play the longest legal moves geometrically
White plays and forces Black to checkmate him
A captured unit is reborn on its home square if it is free

Lieutenant-colonel Antoine de la Vergne de Tressan (1910-1985), known as Xavier Yzarn
Solution

{

} 1.Sf2-h1 ! zugzwang. 1...Rg2-g8 2.Rc5-g5 {(ici,} 2...Rg8*g5[+wRa1] + ? {would be illegal)
} 2...Rg8-d8 3.Rc2*c6[+bPc7] 3...Rd8-d1 4.Rc6-d6 {
Justification of the key :} 4...Rd1*h1[+wSb1] {is of the same length as 4...Rd5 but is illegal because the Knight is reborn on b1.
And of course} 4...Rd1*d6[+wRa1] {is illegal as already seen.
} 4...Rd1-d5 # {: none of the 4 white units can capture this Rook which would be reborn on a8.

"A spark of originality and spirit" (H. Kluver, which awards the first prize in a French magazine that used to be of high quality).}

2 - 2# circe

Petkov, P.
1971

white Sh2b3 Bg3c4 Rf3e3 Pe2d4f4 Qb1 Kd5 black Qf2 Ke1 Pg2h3b4e4f5 Bd2d1 Sg1

#2 Circe(11+10)

Solution

{

Essais :
} 1.Bg3*f2[+bQd8] ? {is a self-check
} 1.Kd5-c5 ? Qf2*g3[+wBc1] ! {(} 2.Bc1*d2[+bBf8] ? {self-check)
} 1.Kd5-c6 ? e4*f3[+wRh1] ! {(} 2.Rh1*g1[+bSb8] ? {self-check)
} 1.Kd5-e6 ? Bd2*e3[+wRa1] ! {(} 2.Qb1*d1[+bBc8] ? {self-check)

} 1.Kd5-e5 ! {
} 1...Bd2-c1 2.Bg3*f2[+bQd8] # {
} 1...Qf2*g3[+wBc1] 2.Bc1*d2[+bBf8] # {
} 1...e4*f3[+wRh1] 2.Rh1*g1[+bSb8] # {
} 1...Bd2*e3[+wRa1] 2.Qb1*d1[+bBc8] #

3 - 2# masand

Gockel, H.
2018

white Kd8 Qf8 Bb6 Pa4d6f5 Sf1 Rc3 black Sh8g4 Rg7 Pf7b5e5 Kd5 Qe4

#2 Masand(8+8)

A checking piece causes all the pieces it "observes" to change colour

Solution

{

Essais :
} 1.Kd8-d7 ? {(threatens} 2.Qf8-a8[a4=b][h8=w] # {)
} 1...Qe4*f5[f1=b][g4=w][e5=w][f7=w] + 2.Sg4-e3[f1=w][f5=w] # 1...Qe4-c4 2.Sf1-e3[c4=w][g4=w][f5=b] # 1...f7-f6 + 2.Qf8-g8[g7=w][h8=w] # {but} 1...b5*a4 ! {is the correct defence : } 2.Qf8-a8[a4=w] + ? Qe4*a4[a8=b] + ! {

} 1.Kd8-c7 ! {(threatens} 2.Qf8-a8[a4=b][h8=w] # {)
} 1...Qe4-c4[f1=b][c3=b][a4=b][g4=w][b5=w] + 2.Sg4-e3[f1=w][c4=w][f5=b] # {
} 1...Qe4*f5 2.Sf1-e3[g4=w][f5=w] # {
} 1...f7-f6 + 2.Qf8-g8[g7=w][h8=w] # {

Mats exchanged between trial and real play.}

4 - 2# Bulgarian grasshoppers

Loerinc, J.
2021

white Ga7h7f2 Qf5 Kg2 Pe2d2c2 Ra2 black Kd6

#2 (9+1)
🨟a7, h7, f2: bulgarian grasshoppers

Bulgarian grasshoppers Bulgarian pieces (invented -- of course -- by the super-genius P. A. Petkov)
are pieces that need a sautoir to express themselves, with a condition somewhat reminiscent of the "take & make".
When they jump, the said sautoir must, as a tribute, take a step (without capturing) in accordance with the walk of the jumper.
If this is not possible, the move is illegal. When more than one move is possible, the choice is up to the jumper's side.

This problem was awarded in the competition organised by PAP in memory of his mother Danka Petkova (1920-1993)

Solution

{

} 1.Gh7-e4[-f5][+wQf1] {(blocus, leaving the Black King with a complete rosace.
Note that the bGf2 must control d2, which it does very well because, on a possible} 1...Kd6*d2 {, she would play} 2.Gf2*d2[-e2][+wPh2] {(Pe2 on h2).

} 1...Kd6-c7 2.Ga7-d7[-c7][+bKc1] # {
} 1...Kd6-d7 2.Ga7-e7[-d7][+bKd1] # {
} 1...Kd6-e7 2.Ge4-e8[-e7][+bKe1] # {
} 1...Kd6-c6 2.Ge4-b7[-c6][+bKc1] # {
} 1...Kd6-e6 2.Ge4-e7[-e6][+bKe1] # {
} 1...Kd6-c5 2.Ga7-d4[-c5][+bKc1] # {(et non} 2.Gf2-b6[-c5][+bKc1] Kc1*d2! {)
} 1...Kd6-d5 2.Ge4-c6[-d5][+bKd1] # {
} 1...Kd6-e5 2.Ge4-e6[-e5][+bKe1] # {

Comment from G. Foster : "The key seems exceptionally generous, giving 5 escapes and exiling the Queen on f1.
But on the other hand, mates require that this Queen be on the first line, so that the key is, in the end, quite strong".}

5 - h#2 2 solutions with bulgarian lions

Gadjanski, B.
2021

white LIa8 Kc6 Sd5 black Kg8 Ra4

h#2 2sol(3+2)
🨴a8: bulgarian Lion

bulgarians lions as above, but instead of Grasshopper, it's a Lion

Solution

{

} 1.Ra4-a3 ! LIa8-h8[-g8][+bKa2] 2.Ka2-a1 Sd5-c3 # ! {
It's a real checkmate, because after } 3...LIh8*a1 {, Knight can play on c7 or c8.

} 1.Kg8-h8 ! LIa8-a1[-a4][+bRe8] 2.Re8-f8 Sd5-f6 # ! {
It's checkmate, because after} 3...LIa1*h8 {, Knight can play on b6 or a6}

6 - 2# breton

Loustau, JM.
2018

white Bc8h2 Qf7 Se6d2 Rc5f1 Pa4d4h3 Ka2 black Qa7 Bf6 Kf5 Rf4 Pe5 Sf2g3

#2 breton(11+7)

breton after each capture, a unit of the same nature must disappear (if possible) from the capturing side.
If there is more than one, the choice is up to the capturing side.
Thus, the move ...♛xa4+ is illegal, as it would have to remove ♟e5, leading to a self check.

Solution

{

Essais :
} 1.Bc8-d7 ? {(threatens} 2.Qf7-h7 # {)} 1...Rf4-h4 ! {(} 1...Rf4-g4 ? 2.Rc5*e5[-h3] # {using g4 self-blocking))
} 1.Rc5*e5[-h3] ? Kf5-g4 ! {or} 1.Rc5*e5[-d4] ? Kf5*e5 ! {or} 1.Rc5*e5[-a4] {illegal
} 1.d4*e5[-e5] ? Qa7*c5[-f4] ! {

} 1.Bc8-b7 ! {(threatens} 2.Bb7-e4 # {because the capture of the Bishop, leading to the disappearance of Bf6, is illegal.)
} 1...Sg3-e4 {display-departure-file} 2.Rc5*e5[-d4] # {because} 2...Kf5*e5[-f4] {is illegal ; but not} 2.Bb7*e4[-d2] + ? Kf5*e4 {or} 2.Bb7*e4[-e6] + ? Kf5-g5 {according to the White Rider who was removed
} 1...Sf2-e4 {display-departure-file} 2.d4*e5[-e5] # {because} 2...Qa7*c5[-f4] {and} 2...Se4*c5[-f4] {are illegal
} 1...Qa7*c5[-f4] {, does not counter the threat)} 2.Bb7-e4 # {
} 1...Rf4-e4 {(does not counter the threat)} 2.Bb7*e4[-f1] # {or} 2.Bb7*e4[-c5] # {
} 1...Qa7*b7[-f6] { illegal}

7 - h#2 2 solutions mirror circe

Dragoun, M.
2003

white Kb2 Rg1 Sc6 Bg6 Pg2 black Qd1 Be8 Re6h1 Kf6

h#2 mirror Circe(5+5)
2 solutions

a unit is reborn as if it had changed colour: a wQ in d8, a bQ in d1.

Solution

{

} 1.Rh1-h3 g2*h3[+bRh1] 2.Rh1*h3[+wPh7] h7-h8=Q # {(et} 3.Rh3*h8[+wQd8] {is impossible)

} 1.Qd1-f3 g2*f3[+bQd1] 2.Qd1*f3[+wPf7] f7*e8=S[+bBf1] # {(et} 3.Re6*e8[+wSg8] {is impossible)

Round trip in each solution.}}

8 - h#2 circé assassin avec moa rider

Uitenbroek, H.
2018

white oad5 Sa5 Kh1 black Pc6b5b3b2c2d2 Kc8

h#2 Circe assassin(3+7)
🨷d5: Moa rider
b) ♚c8-->b8

a captured piece returns to its home square, even if that square is occupied, in which case the occupant disappears
moa extended in successive moa movements
As a reminder, the Moa is a creeping knight performing first a bishop step and then a rook step

Solution

{

} a) {
} 1.c2-c1=S Sa5*b3[+bPb7] 2.b2-b1=Q OAd5-e7 # {
It is well checkmate: the bK cannot go to b8} 3.Kc8-b8 {because of} Sb3*c1[+bSb8] {, neither on d8} 3.Kc8-d8 {because of} OAe7*b1[+bQd8] {
, nor on c7} 3.Kc8-c7 {because of} OAe7*c6[+bPc7] {, nor on d7} 3.Kc8-d7 {because of} Sb3*d2[+bPd7] {

} b) bKc8-->b8 {Kc8-->b8} 1.d2-d1=B OAd5-c3 2.b2-b1=R Sa5*c6[+bPc7] # {
bK cannot go to a8} 3.Kb8-a8 {because of} OAc3*b1[+bRa8] {forcing the revival on a8 which pulverizes the Black King,
on c8} 3.Kb8-c8 {because of} OAc3*d1[+bBc8] {nor on b7} 3.Kb8-b7 {because of} OAc3*b5[+bPb7] {
Noter that} 1...c6-c5 ? {does not work fine :} 2.Sa5-c6 + Kb8-a8 {and} 3.OAd5-c7 {is illegal, which explains the choice of this creature rather than a simple nightrider}

9 - h#2 set play sentinelles

Tura, W.
2018

white Sc5 Bb8h5 Kc4 black Qb6 Ke7 Sf8

h#2 Sentinelles(5+5)
Jeu Apparaent

each playing figure leaves a pawn on its starting square (except 1st or 8th row),
under the condition that the number of pawns of the same side does not exceed 8

Solution

{

J.A.
} 1...Bb8-e5 2.Qb6-d8[+bPb6] Be5-f6[+wPe5] # {

Essai :
} 1.Qb6-d8[+bPb6] ? Bb8-e5 {(no waiting move because ...b5 checks)} 2...Be5-f6[+wPe5] # ! {
} 1.Qb6-f6[+bPb6] ? Bb8-c7 2.Qf6-d6[+bPf6] Bc7-d8[+wPc7] 3.Qd6*d8[+bPd6] {
} 1.Qb6-a5[+bPb6] ? {so no Qd8

} 1.Qb6-d6[+bPb6] Bb8-c7 2.Qd6-f6[+bPd6] Bc7-d8[+wPc7] #

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