# Original Problems (25)

## Original Problems (page 25)

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 24 ;  →Page 26

I believe that Peter Harris has became one of the most active authors at the site!! And as always presents some new condition to learn!  Thank you, Peter!

No.65 – hs#2.5 by Peter Harris – An exotic opus of Mr.Harris! Pay attention to the fact that in the initial setting moves Kc5xd4(+w.Pc5) and Kxc5(+bl.Pd4) are impossible because of the illegal Isardam-paralysis. Therefore, the position of Kc5 and Kd4 is legal here! (JV)

No.66 – h#4.5 by Peter Harris А nice problem in typical Peter’s style! The initial setting is curious – the both Kings are not under check because of the Anti- Circe rules. (JV)

Definitions:

Sentinels Pion advers:   When a piece (not a Pawn) moves, a Pawn of the colour of the opposite side appears on the vacated square if it is not on the first or the last rank, and if there are less than 8 Pawns of that colour on the board.

Isardam: The moves causing a Madrasi – like paralysis are illegal. This holds right up to the capture of the mated King. This is standard form of Isardam.

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.

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

 – No.65 Peter Harris South Africa original-04.08.2012 – hs#2.5         2 solutions         (2+2+1) Sentinels Pion Advers Isardam   – Solutions: (click to show/hide) I.1…nSa2-c3[+wPa2] 2.e3-e4 + nSc3-d5[+wPc3] 3.e4xd5 b6-b5 # II.1…nSa2-b4[+wPa2] 2.nSb4-d5[+bPb4] nSd5-c7[+wPd5] 3.nSc7-a8[+bPc7] b6-b5 # – The problem serves as an introduction to Isardam – that solvers may enjoy. (Author) – No.66 Peter Harris South Africa original-05.08.2012 – h#4.5           b) LOa6->a5           (3+5) White Maximummer Anti-Circe Locust a6   – Solutions: (click to show/hide) a) 1…Kd3-c2 2.h2-h1=B (!) Bc8-b7 3.Bh1-d5 Bb7-a8 4.Kd7-c7 Ba8-c6 5.Bf1-d3 + Kc2xd1[wKd1->e1] # b) 1…Kd3-e4 2.h2-h1=L+ (!) Ke4-f5 3.Kd7xc8[bKc8->e8] Qd8xd1[wQd1->d1] 4.Bf1-h3 + Kf5-g6 5.Ke8-f8 Qd1-d8 # – The whitemaxi condition plays an interesting role. (Author)

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

### 3 Responses to Original Problems (25)

1. seetharaman says:

No.66.. Very complex problem again from Mr.Harris. Even after deep thought (!!) I could not understand why black bishop should play to h3 (and not to any other square) in the second solution. So I give up 🙂 The need for Locust promotion is too deep to grasp immediately !!

• Dmitri Turevski says:

This is how i got it:

2.h1=L+ (!) (2.h1=Q/B? do not check the white king – white would be then forced to play unwanted longest move 2…Qh4)

4.Bh3+ (this is a doublecheck as wB no longer occupies c8 so 4. B~? would be simple check by locust a5 and white would be forced to play 4…Qd5 – longest way to parry the check)

• seetharaman says:

You are right Mr.Dmitri! I overlooked that in the second solution after 1…K.e4 the white queen is no longer pinned !