|
No.603 |
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 Harris |
Solutions: (click to show/hide) |
|
white kh8 qd5 rd1 bc6d6 pf6
black kh1 qa1 rd8 bc4
hs#2,5 b) Bc6→f3 (6+4) |
|
I must say that I don’t understand this problem at all.
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.
“you” in my post does not refer specifically to Kenneth (I hadn’t even seen his post), but to all of you!
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 !
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!!
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?
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)
Good man Paul.
I remember of old how often you improved the problems I sent you for Quartz.
I become much immersed in my problems – trying to get the best out of them.
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.
After Nikola’s notifications, my intention is actually to remove the d phase (and to maintain the a/b/c phases). Do you agree?
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!
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.