No.603 (PH)

Peter Harris
(South Africa)


Original Problems, Julia’s Fairies – 2014 (III): September – December

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


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
South Africa

original – 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 Cheylan

11 Responses to No.603 (PH)

  1. Kenneth Solja says:

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

  2. Kjell Widlert says:

    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 says:

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

  3. Seetharamanseetharaman says:

    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 !

  4. S.K.BalasubramanianS. K. Balasubramanian says:

    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!!

  5. Paul Rãican says:

    This interesting problem allows an extension:
    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
    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 says:

      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)

  6. Peter Harris says:

    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

    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 says:

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

  7. Nikola Predrag says:

    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!

  8. Peter Harris says:

    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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