No.750 (JV)

Julia Vysotska


Original Problems, Julia’s Fairies – 2015 (I): January – June

   →Previous ; →Next ; →List 2015(I)

Please send your original fairy problems to:

No.750 by Julia Vysotska –  A combination of Back-To-Back and Take & Make. (JV)


Back-To-Back: When pieces of opposite colors stand back-to-back with each other on the same file (white piece is on the top of black!), they exchange their roles. A pawn on the first rank cannot move. Any piece can make an en passant capture when it has got a role of Pawn by Back-To-Back.

Take&Make: Having captured, a unit must immediately, as part of its move, play a non-capturing move in imitation of the captured unit from the capture-square. If no such move is available, the capture is illegal. Promotion by capture occurs only when a pawn arrives on the promotion rank as the result of a take&make move. Checks are as in normal chess: after the notional capture of the checked K, the checking unit does not move away from the King’s square.

Lion(Li): Moves along Queen lines over another unit of either color to any square beyond that unit. A capture may be made on arrival, but the hurdle is not affected.

Rabbit(RT): Moves along Q-lines over 2 hurdles of either color (which may or may not stand on adjacent squares) to any square beyond the 2nd unit. A capture may be made on arrival, but the hurdles are not affected. (same as Kangaroo-Lion(KL))

No.750 Julia Vysotska

original – 13.03.2015

Solutions: (click to show/hide)

White Ke8 LIH5 Pe4 Black RTH4 Ka3 Be6 Bh2 Sc3 Pb3 Pd4 Pf3 Ph3 Pa2 Pe2 Pf6

h#2              b) Bh2→d1          (3+12)
Take & Make
Lion h5
Rabbit h4

12 Responses to No.750 (JV)

  1. Luce Sebastien says:

    Dear Julia,
    I am happy you found time to enjoy us with some jumps of Rabbit & Lion !
    Very original !

  2. Luce Sebastien says:

    …but one bad news.
    some undesired solutions on Winchloe :
    1.Sd5 LI×h2(LIé5) 2.KALa4 é×d5(é7)‡
    1.Sd5 é×d5(f6) 2.Bf5 f×h4(a4)‡
    1.Sd5 é×d5(é7)+ 2.Bd6 é×h4(a4)‡
    1.Bd5 LI×d1(LIç2) 2.KALb4 é×d5(ç4)‡
    1.Sd5 é×d5(f6) 2.Bf5 f×h4(a4)‡

    Julia you have to buy Winchloe ! Much more easy to use
    and much more fairy pieces & conditions !
    You can also translate automatically the solutions and so on…

    • JuliaJulia says:

      Thanks, Sebastien! I’ve checked the “undesired” solutions – WinChloe is right here! +bPf6 makes a problem C+ by WinChloe. Seems like Popeye “thinks” that on h4 a pawn captures a usual Rabbit, but in reality the Rabbit has Lion’s power at that moment (by BTB). I’ll correct published problem, so number of pieces will be 3+12, C+ by the both solving programs – Popeye & WinChloe.

      Sebastien, but of course, I have WinChloe! I’ve bought it in a few months after had started to compose, in 2011. I just find Popeye faster and more convenient in many cases. I also use Popeye’s notation/solution for animation on JF. But I test with WinChloe those problems for publication which have no solutions or cooks in Popeye, and I use it for my own composing in a case if Popeye doesn’t have the fairy element I need. Echecs database is also a great thing! (I should mention, that this year Sebastien adds all problems from JF to WinChloe’s Echecs database. Thank you, Sebastien, for doing this work for us!!)

      • JuliaJulia says:

        One more comment about Popeye’s interpretation of combination BTB+TM: it looks like in all cases in a case of TM capturing of the one piece in BTB pair, the capturing piece doesn’t take a role of another piece of that pair. I believe, it’s either a bug or incorrect interpretation.
        The simple examples:
        1) Capturing a black piece in BTB pair.

        Popeye Windows-64Bit v4.69 (5740 MB)
        beg sti #1
        pie whi qf1 bh4
        bla ka8 pa7 sh3
        con backtoback take&make end
        |                                   |
        8  -K   .   .   .   .   .   .   .   8
        |                                   |
        7  -P   .   .   .   .   .   .   .   7
        |                                   |
        6   .   .   .   .   .   .   .   .   6
        |                                   |
        5   .   .   .   .   .   .   .   .   5
        |                                   |
        4   .   .   .   .   .   .   .   B   4
        |                                   |
        3   .   .   .   .   .   .   .  -S   3
        |                                   |
        2   .   .   .   .   .   .   .   .   2
        |                                   |
        1   .   .   .   .   .   Q   .   .   1
        |                                   |
          #1                          2 + 3

        While Popeye gives no solutions here, WinChloe shows 1.Qxh3->c8#

        2) Capturing a white piece in BTB pair.

        beg sti h#1
        pie whi se1f5
        bla ka1  pa2b2 sc1 rh5 bf4
        con backtoback take&make end
        |                                   |
        8   .   .   .   .   .   .   .   .   8
        |                                   |
        7   .   .   .   .   .   .   .   .   7
        |                                   |
        6   .   .   .   .   .   .   .   .   6
        |                                   |
        5   .   .   .   .   .   S   .  -R   5
        |                                   |
        4   .   .   .   .   .  -B   .   .   4
        |                                   |
        3   .   .   .   .   .   .   .   .   3
        |                                   |
        2  -P  -P   .   .   .   .   .   .   2
        |                                   |
        1  -K   .  -S   .   S   .   .   .   1
        |                                   |
          h#1                         2 + 6

        No solutions by Popeye, and the solution by WinChloe: 1.Rxf5->b1 Sc2 #

        • Kjell Widlert says:

          I believe this is not exactly a bug, but rather a question of interpretation. But having said that, I must say that WinChloe’s interpretation is clearly the most natural one: if BTB gives a piece a different power of movement, then a piece capturing that piece should perform its T&M “make” move according to that power of movement – not according to the “normal” power of movement, which has been temporarily suspended by BTB.

          • S.K.BalasubramanianS K Balasubramanian says:

            Dear Widlert,

            I fully agree with you that if BTB gives a piece a different power of movement for a piece, then if that piece is captured, it should perform its T&M “make” move according to that power of movement which it has gained due to BTB and not according to its “normal” power. I have already done some research on this combination and also have composed a problem with this combination which is yet to be published. It is not only that Popeye interprets differently for BTB+Take& Make but also for BTB+Madrasi combination. I would like to give the following examples where POPEYE either interprets differently or behaves erratically for these combinations.

            Example 1:
            stip h#1
            wh ka7 rg1 rg2
            bl kh8 rh6
            Popeye gives the following results:
            1.Rh6-a6 Rg1-h1 # 1.Rh6-a6 Rg2-g8 # 1.Rh6-a6 Rg2-h2 #
            But when Ra6 moves under wKa7, then why is wK not in check?

            Example 2:
            stip h#0.5
            wh kf8 rh2 ba8
            bl kh8 rf2
            Popeye give the following result:
            1…Ba8-f3 #
            Here why bRf2 behaving like a bB cannot defend by Ra2-h4!

            Example 3:
            stip h#0.5
            wh kf8 rg6 bb8
            bl kh8 pf5 pg5
            It should give the solution as:

            But Popeye does not give any solution.
            Probably the bPg5 after capturing wRg6 moves like a rook. But it should move like a pawn as the wRg6 behaves like a pawn and not like a rook due to BTB effect.


  3. seetharaman says:

    1….Lixh2-e5.? Can this be right? I thought Lions cannot jump beyond the second hurdle….

    • JuliaJulia says:

      Lions jump over just one hurdle, of course, but here by BTB Lion has a power of Rabbit (LIh5 & RTh4 – BTB pair), so it jumps over two hurdles. Same, a pawn mates on the last move using a power (role) of B/S.

  4. Mike Neumeier says:

    Solved it. Very nice twin. I, too, agree with transference of BTB powers. I am sure a two-solution setting could be found, keeping the B/N sacrifices to wP, keeping Lion play (changed guard of flight square) , keeping Rabbit play (changed self-block), yet extend to H#3 whereby key moves are by a bQ to the two sacrificial squares (removing dark-square bB). The added white move could be, possibly, a different wK move (involving that piece more prominently) or wP move. Definitely keep the BTB plays! Already delightful twinning, so these are ideas for possible further development of your matrix, only. Interesting blend are these conditions, with many strategies out there.

  5. Nicolas Dupontdupont says:

    Yes, I also agree that the “make” move should follow the moving possibilities of the captured piece at the time it is captured.

    Interesting too is when the “take” move is creating a BTB situation, such as in the scheme wPa2 bSb2 bBb3. After the take axBb3, the make should be Bishop-like according to T&M, but Knight-like according to BTB.

    Both Popeye and WinChloe are adopting here the Bishop-like move, which seems to be the best option – after axBb3 the T&M move is not over, so that wPb3 is not really BTB with bSb2, but only BTB “en passant”!

    Btw, dear Julia, it seems that a plural should be used at the end of the Rabbit definition – “the hurdles are not affected”.

    • JuliaJulia says:

      Nicolas, I agree with you about the interpretation that in the middle of T&M a capturing piece can’t create a BTB pair! (the definition of Rabbit is corrected, thanks!)

Leave a Reply

Your email address will not be published. Required fields are marked *

You can add images to your comment by clicking here.