Julia's Fairies

Original Problems (19)

Original Problems (page 19)

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 >>> Page 18 ;  >>> Page 20

3 problems by Peter Harris! Mr.Harris has suggested to publish problems without solutions first – thinking of solvers. Here I’ve added an option to hide (show) the solution. 1 my own problem, Anti-Andernach this time.

No.47 – hs#6.5 by Peter Harris – this problem has an Orphan piece which is author’s favorite piece!

No.48 – hs#2.5 by Peter Harris – Author: I do not know whether it would be easy or difficult for a solver!

No.49 – h#2 by Julia Vysotska – Anti-Andernach – my first try with this condition! Double interchange of functions.

No.50 – h#2 by Peter Harris – Author: Locusts are strange pieces and stranger still with Transmuting Kings!

I’ll put the Definitions right here for now (later will be added to the Terms page):

Orphan: An orphan can move only when observed by an enemy piece; when so observed it can move like the piece(s).

Maximummer – Black must play the geometrically longst move or may choose from among longest moves of equal length, distances being measured from the center of each square. Diagonal and oblique distances are measured from the orthogonal coordinates by using Pythagora’s theorem (take the square root of the sum of the squares of the orthogonal distances). All other orthodox chess rules apply.

White Maximummer – Only White must play the geometrically longst moves, Black plays orthodoxal.

Patrol Chess: A piece can capture or give check only if it is observed by a piece of its own side.

Anti-Andernach – A piece (excluding King) changes its color after any non-capturing move. After capture, the piece retains its color. Rooks on a1, h1, a8 and h8 can be used for castling, provided the usual other rules for that move are satisfied. After castling, Rooks do not change color, If White makes a non-capturing move with neutral or halfneutral piece, that piece becomes black and vice versa.

Transmuted Kings: When in check a King moves like the checking piece(s).

Lortap: A piece cannot capture or check if it is observed by a piece of its own side.

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

No.47 Peter Harris
(South Africa)
hs#6.5                                           (2+3)
Orphan h2
White Maximummer
b) + ChameleonChess
Solutions: (click to show/hide)

No.48 Peter Harris
(South Africa)
hs#2.5           2 solutions            (3+4)
Patrol Chess
Solutions: (click to show/hide)

No.49 Julia Vysotska
h#2                2 solutions            (3+6)
Nightriders: f6,h4
Solutions: (click to show/hide)
Сreation of reciprocal batteries N/B and B/N; double interchange of function between: Nf6/Bf4 and Nh4/Bf1; thematic sacrifices on g5/d5; model mates. (Author)

No.50 Peter Harris
(South Africa)
h#2            b) Sb4->d8              (4+2)
Transmuted Kings
Solutions: (click to show/hide)
You may find part (b) difficult to understand. Locusts are strange pieces and stranger still with Transmuting Kings. (Author)


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

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July 21, 2012 21:37

It is a worthwhile effect for those who would like to solve the problems. In 47 and 48, the solutions are visible by default. I think a better way is to hide the solutions and show it to those who click “Solution”.

July 22, 2012 10:49
Reply to  Julia

Works nicely. I too think that most do not try solve. But with this option, one or two might try !

July 24, 2012 09:27

Indeed N.50 is very nice and difficult to understand. In (b) I assumed that the first move 1.Kg5 is a check as the Locust is no longer supported, but realised after sometime that due to Transmuting King condition, the black king is still supporting the Locust (having bishop power) !
It is even more interesting why white cant mate by 2…Bc3 (after 1…Bd2+). I leave it to the other readers to guess !!

Nikola Predrag
Nikola Predrag
July 24, 2012 15:31

No.50. A solver which for the first time encounters these combined conditions may spent some time to solve a) – it is nice but not very hard, after all. For solving of b) the combined conditions must be fully comprehended. Therefore the solving for me actually started when I had finally surrended and looked at the solution of b). And to ‘solve’ only why the given solution indeed works, was quite enough for me, quite a magical mystery tour:)

Wel, after even more thinking, I tend to believe that the author’s solution of b) (as well as the Popeye’s) is ILLEGAL!

The LEGAL solution should start with 1.Kg5+!!!
Or perhaps I must continue to ‘solve’ the given solution:(

Bartel Erich
Bartel Erich
July 31, 2015 19:03
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