No.983,984 (EB)

 No.983, 984  Erich Bartel (Germany) Original Problems, Julia’s Fairies – 2015 (II): July – December    →Previous ; →Next ; →List 2015(II) Please send your original fairy problems to: julia@juliasfairies.com

No.983, 984 by Erich Bartel – A long march of the white King with echo stalemates in No.983 and 4 corners cycle in No.984. (JV)

Definitions:

Circe Equipollents: After a piece is captured, it is immediately replaced on the square which is the same distance and direction from the square of its capture, as was that square from the square upon which its captor commenced its move. (If Qf7 captures a Pawn on e7, it is reborn on d7, because d7 is a same distance and direction from e7 as e7 is from f7. Similarly, if Qg7 captures a piece on ‘e7’ its rebirth square is ‘c7’). If the rebirth square is occupied the captured piece disappears. Castling with replaced Rook is permitted. Pawns may be reborn on the 1st and 8th ranks. Pawns reborn on the 8th rank are promoted as part of rebirth and the promotion is chosen by the player who makes the capture, i.e., if White captures a black Pawn, and the black Pawn is reborn on the 8th rank, White (not Black) decides what Black’s Pawn will be promoted to. Pawns reborn on the 1st or 8th rank can make only one-square move (for example black Pc8 can play only on c7 or it can capture an enemy units on ‘b7’ or ‘d7’). During the en passant capture, the Pawn is replaced on the rank opposite of the captor. For example, if Black moves c7-c5, White Pawn on ‘b5’ captures en passant, moves to ‘c6’, while the black Pawn is reborn on ‘d6’.

Lion(LI): Moves along Queen lines over another unit of either color to any square beyond that unit. A capture may be made on arrival, but the hurdle is not affected.

Grasshopper(G): Moves along Q-lines over another unit of either color to the square immediately beyond that unit. A capture may be made on arrival, but the hurdle is not affected.

 No.983 Erich Bartel Germany original – 28.12.2015 Solution: (click to show/hide) white Ka2 black Pc5 LIg7 Ser-s=19      b) Pc5→g2           (1+2) Circe Equipollents Lion g7 (no bK) a) 1.Ka2-b3 2.Kb3-c4 3.Kc4-d5 4.Kd5-e6 5.Ke6-f7 6.Kf7-g8 7.Kg8*g7[+bLIg6] 8.Kg7-h7 { } 9.Kh7*g6[+bLIf5] 10.Kg6*f5[+bLIe4] 11.Kf5*e4[+bLId3] 12.Ke4-d5 13.Kd5-c6 { } 14.Kc6*c5[+bPc4] 15.Kc5*c4[+bPc3] 16.Kc4*d3[+bLIe2] 17.Kd3-c2 18.Kc2-b1 19.Kb1-a1 c3-c2 {=} b) bPc5-->g2 1.Ka2-b3 2.Kb3-c4 3.Kc4-d5 4.Kd5-e6 5.Ke6-f7 6.Kf7-g8 7.Kg8*g7[+bLIg6] 8.Kg7*g6[+bLIg5] { } 9.Kg6-h6 10.Kh6*g5[+bLIf4] 11.Kg5*f4[+bLIe3] 12.Kf4*e3[+bLId2] 13.Ke3-f2 { } 14.Kf2-g1 15.Kg1*g2[+bPg3] 16.Kg2-h3 17.Kh3*g3[+bPf3] 18.Kg3-h2 19.Kh2-h1 f3-f2 {= (C+ by Popeye 4.73)} No.984 Erich Bartel Germany original – 28.12.2015 Solution: (click to show/hide) white Kh1 black Ba1 Pg6h6 Ga7 Sh8 Ser-s#36                                   (1+5) Circe Equipollents Grasshopper a7 (no bK) 1.Kh1-g2 2.Kg2-f3 3.Kf3-e4 4.Ke4-d5 5.Kd5-c6 6.Kc6-b7 7.Kb7-a8 8.Ka8*a7[+bGa6] { } 9.Ka7*a6[+bGa5] 10.Ka6*a5[+bGa4] 11.Ka5*a4[+bGa3] 12.Ka4*a3[+bGa2] { } 13.Ka3-b3 14.Kb3-c2 15.Kc2-b1 16.Kb1*a1 17.Ka1-b2 18.Kb2-c3 19.Kc3-d4 20.Kd4-e5 21.Ke5-f6 { } 22.Kf6-g7 23.Kg7*h8 24.Kh8-g7 25.Kg7*g6[+bPg5] 26.Kg6-h7 27.Kh7*h6[+bPh5] { } 28.Kh6*h5[+bPh4] 29.Kh5-g6 30.Kg6*g5[+bPg4] 31.Kg5*g4[+bPg3] 32.Kg4-h5 { } 33.Kh5*h4[+bPh3] 34.Kh4*h3[+bPh2] 35.Kh3-g2 36.Kg2-h1 g3-g2 # { (C+ by Popeye 4.73)} 4 corners cycle by white king. (Author)

3 Responses to No.983,984 (EB)

1. Bartel Erich says:

pdb.dieschwalbe.de/search.jsp?expression=PROBID=’P1314511′

pdb.dieschwalbe.de/search.jsp?expression=PROBID=’P1314512′

2. Luce Sebastien says:

Nice n°984 !
It seems the first four royal corners monosolution in Circe equipollent.

In Winchloe I found only the following :

White : Moose d5
Black : Kd4 Double Grasshopper ç7

Harald GRUBERT
ChessProblems.ca 2014
sh=9 (1+2) C+
b) ç7>ç1
c) d5>d3
d) d5>é3
Circé équipollent
nul moves prohibited
QG=Double-Grasshopper
EL=Moose

a) 1.K×d5(ELd6) 2.QGç5 4.K×d6(ELç7) 5.QGb7 6.K×ç7(ELb8) 9.Ka8 ELç6=
b) 1.Kç4 2.QGé5 3.K×d5(ELé6) 4.K×é6(ELf7) 5.QGg7 6.K×f7(ELg8) 9.Kh8 ELf6=
c) 1.Ké5 2.QGd6 4.K×d3(ELç2) 5.QGb2 6.K×ç2(ELb1) 9.Ka1 ELç3=
d) 1.K×é3(ELf2) 2.Kf4 3.QGé5 4.Ké3 5.QGg2 6.K×f2(ELg1) 9.Kh1 ELf3=

SL

3. Bartel Erich says: