# Original Problems (58)

## Original Problems (page 58)

Original fairy problems published during 2012 will participate in the informal tourney JF-2012

The site is mostly about fairies, but h# and s# are also welcomed for publication! Please send your problems to my e-mail: julia@juliasfairies.com

### Go to →List of Problems ; →Page 57 ;  →Page 59

The first time Romanian author on the site! Warm welcome to Paul Răican!

No.114 – hs#5,5* – Paul Răican  –  Just four pieces but rich play with 4 solutions in both phases! (JV)

Definitions:

Circe Parrain – After a capture, the captured piece is reborn only after another piece of its own side has moved. The line between capturing square and rebirth square is parallel with and of same direction and length as the move of this other piece. Pawns can be reborn on 1st and 8th rank. From their own base rank, they may move one-step; if reborn on the promotion rank, the Pawn at once promotes, the promotion piece being determined by the Pawn side.

Einstein chess: all units (Ks excluded) change their type when they move, according to a precise pattern. For non-capture moves: Q>R, R>B, B>S, S>P, P remains P. For capture-moves: R>Q, B>R, S>B, P>S, Qremains Q.

Republican Chess Type 2 : There are no Kings : if the side which has played can put the opposite King on a square where it would be legally mate, then the opposite King is put on such a square. The opposite side can then put itself the other King on a square where it is mated.

You can сlick on “Solutions” to show or hide the solutions!

 No.114 Paul Răican Romania original – 23.08.2012   hs#5,5**       2 solutions             (2+2) Circe Parrain Einstein Chess Republican Chess Type 2     Solutions: (click to show/hide)   Setplays: I. 1… … 2.f3-f4 Sd4-b5=P 3.f4-f5 Be4*f5=R 4.Rd5*b5=Q [+wPd5] Rf5-f8=B [+bPb8] 5.d5-d6 Bf8*d6=R 6.Qb5-a5=R [+wPc6][+wbKa8] + Rd6*c6=Q [+wKa6] # II. 1… … 2.f3*e4=S Sd4-c2=P [+bBd2] 3.Se4*d2=B c2-c1 [+bBd1] 4.Bd2*c1=R Bd1-e2=S [+bPd2] 5.Rd5*d2=Q Se2*c1=B [+bPb1] 6.Qd2-h6=R [+wRg5][+wbKh3] + Bc1-f4=S [+wKh5]  # Real solutions: I. 1…Sd4-c6=P 2.f3*e4=S c6-c5 [+bBe3] 3.Se4*c5=B Be3*c5=R [+bPa7] 4.Rd5*c5=Q [+wBb5] a7-a6 [+bRc4] 5.Bb5*a6=R Rc4-d4=B [+bPb6] 6.Qc5-b4=R [+bKa2] + Bd4-c3=S [+wKa4] # II. 1…Be4*f3=R 2.Rd5*d4=Q [+wPf2] Rf3-g3=B [+bSe4] 3.Qd4-d6=R Bg3*d6=R 4.f2-f3 [+wRd7] Rd6*d7=Q 5.f3*e4=S [+wRc8] Qd7-f7=R [+bSg4] 6.Se4-f6=P [+bKf8] + Sg4*f6=B [+wKd8] #

The diagrams are made on WinChloe and its Echecs font is used for Logo design

### 2 Responses to Original Problems (58)

1. seetharaman says:

With more than one Fairy condition, I find it difficult to compose. I admire the capability of this and other authors who conceive ideas in their mind with so many conditions operating !! I think it is a special skill like juggling requiring lot of practice !

• paul says:

I must admit that this work is mainly a cooperation between man and computer. The problem is C+ Popeye.