No.578 (LP)

LadislavPackaNo.578, 578.1 
Ladislav Packa


Original Problems, Julia’s Fairies – 2014 (II): May – August

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No.578 by Ladislav Packa – A nice play by magic squares! (JV)

No.578.1 by Ladislav Packa – AUW with a use of 5 magic squares! A new version of No.578 created during discussion to No.578 with Dominique Forlot. (JV)


Magic Square: If a piece (excluding Kings) moves on a Magic Square – it changes its color.

No.578 Ladislav Packa

original – 25.07.2014

Solutions: (click to show/hide)

White Ra7 Pd3 Ke2 Pf4g4g5 Black Pf2 Kg6

h#2         3 solutions        (6+2)
Magic Squares e3 f3 g2 f1

No.578.1 Ladislav Packa

version of No.578 – 28.07.2014
Dedicated to Dominique Forlot

Solutions: (click to show/hide)

White Kc2 Rg7 Pd4e3e5g4 Black Ke6 Pd2d5

h#2         4 solutions        (6+3)
Magic Squares c3 d3 f3 e2 d1

14 Responses to No.578 (LP)

  1. Dominique ForlotDominique Forlot says:

    thank Ladislav for this interesting problem with an unusual condition.
    with the same matrix and 3 magics squares more we can have an AUW
    the difficulty to obtain the queen’s promotion with a variant different from tour’s solution and bishop’s solution. A pawn move saves us from this dilemma
    White : Kc2 Rg7 Pe5c4d4g4
    Black : Ke6 Pd2
    h‡2 4 solutions (6+2)
    7 Magic Squares a1,b1,d1,e2,c3,d3,b6

    1.d1=Q[c1=w] Qa1[a1=b] 2.Qc3[c3=w] d5‡
    1.d1=R[c1=w] Rb1[b1=b] 2.Ra1[a1=w] Ra6‡
    1.d1=B[c1=w] Be2[e2=b] 2.Bd3[d3=w] Bf5‡
    1.d1=S[c1=w] Sc3[c3=b] 2.Se2[e2=w] Sf4‡

    Can be that it makes too many magic cases and the accommodate of c6 (as forbidden compartment) is it inelegant? Too many Magic Squares? Surely.

    Friendly … a try, only for the fun :o)

    • Laco Packa says:

      This would be the way to expand the content:
      Kc2, Rg7, Pd4, e3, e5, g4 – Ke6, Pd2, d5
      MagicSquares c3d3e2f3d1, 4 solutions
      C+ by Popeye v.4.67

      1.d2-d1 = S = w Sd1-c3 = b 2.Sc3-e2 = w Se2-f4 #
      1.d2-d1 = R w = Rd1-d3 = b 2.Rd3-c3 = c3-c6 w #
      1.d2-d1 = B = w Bd1-2.Be2 e2 = b = w-d3-Bd3 f5 #
      1.d2-d1 = Q = w QD1-f3 = b 2.Qf3-e2 = w QE2-a6 #
      Cyclic content is maintained in the first three solutions, the fourth solution completes AUW. What do you think?

      • Laco Packa says:

        Copying is also sometimes magical. Corrected solution:

        1.d2-d1=Q=w Qd1-f3=b 2.Qf3-e2=w Qe2-a6 #
        1.d2-d1=S=w Sd1-c3=b 2.Sc3-e2=w Se2-f4 #
        1.d2-d1=R=w Rd1-d3=b 2.Rd3-c3=w Rc3-c6 #
        1.d2-d1=B=w Bd1-e2=b 2.Be2-d3=w Bd3-f5 #

      • Dominique ForlotDominique Forlot says:

        It is, by far, the best presentation, economic and thematic, beautiful work!

        • Laco Packa says:

          We should publish such joint work. Do you agree?
          There is still a possibility to save wPe3 using twins:
          a) MagicSquares d1c3d3 (solution with Rook)
          b) MS d1c3e2 (Knight)
          c) MS d1d3e2 (Bishop)
          d) MS d1e2f1 (Queen)

          • Dominique ForlotDominique Forlot says:

            i think it’s your composition .
            My role here was only that of “mouche du coche”…

            I prefer your presentation from : July 26, 2014 at 13:24 more economic.

            I think each twin would be used only one magicSquare move, no more. Even with a pawn of more the precedent adorned me more attractive. It is subjective of course!

            but, your idea of twin show some strange arrangements of magicSquares all resumed in this net :

            / \
            d1 c3
            \ /

            I do not know that to think of part that of it it adorned me singular.

            • Dominique ForlotDominique Forlot says:

              Oops! the parser eats all my space caracters.
              peraps by this way:

              • Laco Packa says:

                There is always a change of only one magic square in continuous twins.
                d1 is a magic square always, I will not mention it more.
                a) MS c3 d3 (Rook)
                b) d3-> e2 (Knight)
                c) b) + c3-> d3 (Bishop)
                d) c) + d3-> f1 (Queen)

                A pity that you do not add to the co-authorship. I think for joint work is not the most important deal of work, but also the inspiration for its improvement. And without your ideas fourth solution would not be created.
                In any case, I hope that you will accept at least the dedication.

  2. Laco Packa says:

    Thank you, Dominique, for an interesting analysis. You are absolutely right, you have shown the way which can be achieved AUW. I have to say (and I had mention it into comment to problem, mea culpa) that I wanted to do cycle squares that enter pieces during solutions. Transforming on Queen is in this case rather disturbing element.
    I think your composition is completely original, despite the use of the mechanism of my problem.

  3. Dominique ForlotDominique Forlot says:

    Thank you so much, Ladislav for the dedication. It was a real pleasure to exchange our ideas here.

  4. seetharaman says:

    Three minor promotions with cyclic pattern of magic squares seemed compact and complete. But anyway adding the Queen phase is also interesting. Nice idea Mr.Dominique !

  5. Sebastien Luce says:

    Dear Ladislav,
    your AUW with magic squares is really beautiful !
    I saw also in Strategems 68 oct 2014 another very pretty composition from you :
    White : Kf6 Black : Ke8 Ra2 Bc5 Sd1 magic square f2
    sh=3 Anticirce
    zeroposition & a) f6->g7 b)c) e8->h5/f4
    I tried to realize something of the same level with this kind of square but it is very difficult !! If you want to exchange ideas with me, here is my mail:

    • Laco Packa says:

      Thank you for the compliment, such words are read very well 🙂 With magic squares, I did not deal very systematically, but I still have some ideas in the archive. I’ll send you mail.

  6. Kenneth Solja says:

    This is lovely problem, but I was thinking almost immeadiatedly that is this possible to do with Anti-Andernach -condition.
    You see, in Anti-Andernach fairy condition a piece changes it colour after every non-capturing move.
    I think it could be possible, but the problem won’t be so economical like Ladislav’s problem here.
    Really enjoyable problem!

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