Fairy chess composition

# No.603 (PH)

 No.603 Peter Harris(South Africa) Original Problems, Julia’s Fairies – 2014 (III): September – December →Previous ; →Next ; →List 2014(III) Please send your original fairy problems to: julia@juliasfairies.com

No.603 by Peter Harris – A dynamics of Isardam and Anti-Circe effects! (JV)

Definitions:

Isardam: Any move, including capture of the King, is Isardam illegal if a Madrasi-type paralysis would result from it.

Madrasi: Units, other than Kings, are paralysed when they attack each other. Paralysed units cannot move, capture or give check, their only power being that of causing paralysis.

Anti-Circe Cheylan: When a piece captures (including King), it must come back to its rebirth square. If this square is occupied, the capture is forbidden. A Pawn capturing on its promotion rank promotes before it is reborn. The captures on the rebirth square are forbidden.

 No.603 Peter HarrisSouth Africaoriginal – 21.09.2014 Solutions: (click to show/hide) white kh8 qd5 rd1 bc6d6 pf6 black kh1 qa1 rd8 bc4 hs#2,5         b) Bc6→f3           (6+4)Anti-Circe CheylanIsardam a) 1...Rd8-d7 2.Kh8-g8 Qa1-a6 3.Qd5-d3 + Qa6*c6[bQc6->d8] # b) wBc6-->f3 1...Qa1-c1 2.Bd6-g3 Qc1-h6 3.Qd5-d3 + Rd8*d3[bRd3->a8] # { (C+ by Popeye 4.69)} Why Kh8-g8,Qa1-a6(-a4?) and Bd6-g3(-e5?)? Qd5-d3+ unifies the solutions. In one mate the checking piece is captured, in the other a blocked line is re-opened and a square vacated. (Author)

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Kenneth Solja
September 21, 2014 17:44

I must say that I don’t understand this problem at all.

Kjell Widlert
September 21, 2014 18:39

To spare you some hard thinking, I can give you my own answers to the composer’s three questions:
* Why Kh8-g8? To turn Qxc6 into a double check from bQ and bB; a single check is not enough as both wBd6 and wQ can defend a single check from bQd8.
* Why Qa6(-a4?)? To avoid Qd5-d3 being a self-check from bBc4.
* Why Bg3(-e5?)? To stop the defence 3. – Qh3!, which would make Bf3xKh1?? illegal.

Kjell Widlert
September 21, 2014 18:40

“you” in my post does not refer specifically to Kenneth (I hadn’t even seen his post), but to all of you!

seetharaman
September 21, 2014 19:49

This combination of these two fairy conditions makes one think hard to understand what is happening. While Individually Anticirce and Isardam are difficult to comprehend, their combination is a heady mix !

S. K. Balasubramanian
September 26, 2014 20:13

No doubt it is difficult to understand, even the diagram position where the Rooks on d1 & d8 do not check their opponent kings and the wQd5 does not check Kh1. But what an imagination! Hats off to Haris!!

Paul Rãican
September 28, 2014 23:12

This interesting problem allows an extension:
Pieces
White Kh8 Pa7 Bc6 Bd6 Pg6 Qd5 Rd1
Black Rd8 Bc4 Qa2 Kh1
Stipulation Hs#2.5
Condition AntiCirce cheylan isardam
b) pa7->b5 c) Bc6->f3 d)=c) & a7->d7
Solutions:
a) 1…Rd8-d7 2.Kh8-g8 Qa2-a6 3.Qd5-d3 + Qa6*c6[bQc6->d8] #
b) 1…Qa2-a7 2.Bd6-h2 Qa7-h7 3.Rd1-e1 + Rd8*d5[bRd5->a8] #
c) 1…Qa2-h2 2.Bd6-g3 Qh2-h6 3.Qd5-d3 + Rd8*d3[bRd3->a8] #
d) 1…Rd8-e8 2.Bf3-g2 Re8-e7 3.Qd5-d3 + Qa2*g2[bQg2->d8] #
What do you say, Peter?

Nikola Predrag
September 29, 2014 01:24

Very interesting but d) is the same as a), only poorer due to the dropped pin of Bc4 by 2…Qa6!

b) has a very nice 2.Bh2! while 3.Re1+ prevents the capturing moves and rebirths of both Kings.
This is a kind of compensation for dropping the excellent real/virtual Isardam pins of bBc4/wBf3 in a)/c)

Peter Harris
September 29, 2014 10:40

Good man Paul.

I remember of old how often you improved the problems I sent you for Quartz.

In the present instance I found MANY solutions – but sought to get two which I thought went best together. Of which intention Nicola would approve.

In particular: I wanted one solution which captures the checking piece and one which blocked a line – with one capture by the Q and one by the R.

And the pair you see in the problem was the best I could do.

Looking at your 4 solutions they are not such a pretty sight.

But an extraordinary thing is that with your bQa2 and not my a1 then:

if the wPa7 is simply absent 2 solutions appear!

beg pie
whi kh8 qd5 rd1 bc6d6 pg6
bla kh1 qa2 rd8 bc4
stip hs#3 opt whi
cond isardam anticirce cheylan
end

1…Qa2-a7 2.Bd6-h2 Qa7-h7 3.Rd1-e1 + Rd8*d5[bRd5->a8] #
1…Rd8-d7 2.Kh8-g8 Qa2-a6 3.Qd5-d3 + Qa6*c6[bQc6->d8] #

[In the bQa7 solution the inability of the bQh7 to observe d1 is a nice touch.]

So well done Paul.

If you wish Julia may show this as a version with PH & PR as joint authors.

Paul Rãican
September 29, 2014 22:22

After Nikola’s notifications, my intention is actually to remove the d phase (and to maintain the a/b/c phases). Do you agree?

Nikola Predrag
September 29, 2014 15:11

I am not not so much interested in the mere play of solution(s).
The logic behind the play creates the depth of a problem.
The real and virtual play make only the superficial appearance of underlying deeper logic.

The relation between wQ&bQ is perfectly clear and essentially affects the play in both twins of No.603.
During the solutions, bQa6-bBc4-wQd3 becomes visible in a) and wQd3-wBf3-bQh3 remains invisible in b).

But I claim that both effects do exist and govern the complete solution of the problem as a whole!

Actually, the solution 1…Qa7 is hardly acceptable, bBc4 and wBc6 are superfluous. bBc4 only legalizes the relative positions of bQa2/wQd5!

Peter Harris
September 29, 2014 22:42

No Paul I do not agree.

I will agree only to what I show in my last post.

Otherwise my problem is to stand as is.

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