py_20160607

h#2 with Moose and Eagle

Klaus WENDA
2015

white Pe7f7 Kc2 black Pc6c4e5f5 Kd4 mc5 EAg4

h#2(3+7)
b) ♙c4
🨐c5: Moose
🨺g4: Eagle

: a kind of Grasshopper that deflects 45° after the sautoir
. : a kind of Grasshopper that deflects 90° after the sautoir

Solution

{
Beware, it is a moose on c5 and an eagle on g4. The moose, playing as a grasshopper along the path of a Queen, makes a 45 degree angle from the sautoir, relative to the finishing square of a grasshopper. For example the c5 moose can go to f6 or f4 via the e5 sautoir (f5-e5-f6 = 45 degrees), to d3 or e4 via the d4 sautoir (the latter move like a knight, but it would go to e2 or f3 if the bK were on e3, as f2-e3-e2 = 45 degrees). The eagle is at 90 degrees, so the g4 eagle can go to d3 or d5 via the d4 sautoir, or to e4 or g6 via the f5 sautoir.
} a) {
} 1.e5-e4 {
Try 1...EAd5? 2 Kd2 Me4 3 e8=M+ EAe6!.
} f7-f8=WE 2.EAg4-e3 {
EAgle pass through e4.
} e7-e8=EA # {
by double check: the eagle e8 via e4 (and it controls d5 via c6), the eagle f8 via c5 (and it controls e5 via f5). Without double check it would not be mate, the EAf8 check being parried by any move of Mc5, the EAe8 check being parried by EAe3-e5, via d4.

} b) +wPc4 {wPc4
} 1.EAg4-e4 {
Eagle arrives via f5
} f7-f8=WM 2.f5-f4 e7-e8=WM # {
but this time the checkmate is simple, from Me8 (via e5), which cannot be intercepted; Mf8 defends the wPc4 (via c5) which controls d5 ; this same Mf8 controls e3 (via f4) ; this mate is not a model one because Me8 controls c5}

h#2 with Moose and Eagle

Nimrod GUTTMAN
2015

white Pe7 Kf7 Mh1 EAh7 black Pf6f5e5e3g3h3 Kf4 Ba8h8

h#2(4+9)
b) ♟g3-->g2
🨴h7: Eagle
🨊h1:: Moose

Solution

{

} a) {

} 1.Ba8-c6 Mh1-b6 { via c6} 2.Bc6-e4 e7-e8=M # ! { via e5. EAh7 controls g4 via f5, the EAb6 controls g5 via f6 & f3 via e3.
Model Mat

} b) bPg3-->g2 {

} 1.Ba8-f3 {
would be impossible with the bP at g3, because of the check of the moose h1 via f3
} EAh7*f6 + {
via f7, the eagle checks via e5 ;
} 2.e5-e4 { (sautoir moves)} e7-e8=EA # {! mate via e4; e5 and g5 are controlled by the eagle f6, via f5; g3 & g4 are controlled by the moose h1, via g2 & h3 respectively.
Model mat}

h#4 Köko

Anatoly STYOPOCHKIN
2014

white Kc6 Re2 Be1 black Pf2 Kh5

#2 Köko(3+2)
b)♗e1-->d7
c)♗e1-->g2
d)♚h5-->g7

: at the end of a move, a piece must be in contact with another one

Solution

{

} a) {

} 1.f2-f1=B Re2-g2 2.Bf1-e2 Rg2-g5 3.Kh5-h6 Be1-d2 4.Be2-h5 Rg5-g4 # {

} b) wBe1-->d7 {Be1-->d7

} 1.f2-f1=R Kc6-d6 2.Rf1-f3 Re2-e6 3.Kh5-g4 Kd6-d5 4.Kg4-f5 Re6-e4 # {

} c) wBe1-->g2 {Be1-->g2

} 1.f2-f1=S Re2-e1 2.Sf1-g3 Re1-h1 3.Kh5-g4 Bg2-h3 + 4.Kg4-h4 Bh3-d7 # {

} d) bKh5-->g7 {Kh5-->g7

} 1.f2-f1=Q Be1-d2 2.Qf1-f7 Re2-e7 3.Kg7-f6 Kc6-d6 4.Qf7-g6 Kd6-e5 # {

Weak discriminants, but 3 out of 4 mats batteries, accompanying the AUW. A Seneca landing in our time would have removed the solution with Queen promotion, getting: 3 batteries, 3 minor promos, good discriminants.}

h#2 Exchange Circe (PWC)

Mario PARRINELLO
2015

white Pf3e3b5 Sb3 Bc8 Ka8 black Bd4 Kd5 Qb6 Pd6g7 Sf4 Re6g8

h#2 Exchange Circe (PWC)(6+8)
b) ♗c8=♘

(PWC) : A captured piece is reborn at the start of the captured piece's move

Solution

{

} a) {

} 1.Re6-e4 e3*d4[+bBe3] 2.Sf4-e6 Bc8*e6[+bSc8] # {

} b) +wSc8 {

} 1.Qb6-c7 e3*f4[+bSe3] 2.Bd4-b6 Sc8*b6[+bBc8] #

h=5 Circe

Unto HEINONEN
2015

white Bg7 Kd7 Qh4 Pf4e6 Sd5 Rh5 black Bf2 Kf3 Qd4 Pg5e7 Se2 Rf6

h=5 Circe(7+7)

: A captured piece is reborn on its native square if free

Solution

{

} 1.Bf2*h4 [+wQd1] Qd1-f1 + 2.Qd4-f2 Sd5*f6 [+bRh8] {
} 3.Rh8*h5 [+wRh1] Bg7-h6 4.Se2-g3 Pf4*g5 [+bPg7] {
} 5.Pg7*h6 [+wBc1] Rh1-h3 {

Clerk's note: The problem would have been perfect if Nf6 had been "en prise". By adding a pbe6 and a pne7, the job is done !
The fact that Nf6 may be captured adds a finesse: ...exf6 (wNg1) illegal.
}

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