Original Problems (pages 101-110)

Original Problems (page 101)

Original fairy problems published during 2012 will participate in the informal tourney JF-2012

The site is mostly about fairies, but h# and s# are also welcomed for publication! Please send your problems to my e-mail: julia@juliasfairies.com

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No.166 – Construct an “Illegal cluster”! by Seetharaman Kalyan 

(08.11.2012) The author offers you a problem to solve. Book prize is offered. Solving time – 1 month from publication date, till 10-Dec-2012. I won’t show the solutions for now, but will show it by author’s request or after 1 month. Please send your solutions to seetharamankalyan@gmail.com. I wish luck to all solvers!  (JV)

(13.12.2012) Solving time is over. So, author writes: “My problem No.166 in JuliasFairies.com and the construction Task challenge received correct solutions and responses from six talented composers.”  See the results in  Response-to-Illegal-Cluster-Problem ! (JV)


Definition (by author):

“Illegal Cluster”:  An illegal position which becomes legal on removing any of the pieces (or pawns).   This is an old Fairy condition made popular by T.R.Dawson and others, in the 1920s.


No.166 Seetharaman Kalyan
India
original-08.11.2012
 
 
Construct an “Illegal Cluster” by adding the following:
 
  • (a) WS, BR & BS
  • (b) WB, BR & BS
  • (c) WP, BR & BS
  • (d) WQ, WS & BS
  • (e) WR, WS, BS & BB.

I offer a book prize to anyone who sends me the best set of solutions and also constructs an “Illegal cluster” by adding FIVE or more pieces to the above position, with an unique solution (that is, the position of none of the added pieces can be changed). The term for solving is 1 month from publication day.   (Author)

 

The diagrams are made on WinChloe and its Echecs font is used for Logo design

12 Responses to Original Problems (pages 101-110)

  1. SeetharamanSeetharaman says:

    An example of “Illegal cluster” can be easily constructed in the aboove position by “adding Three WPs & one bB’. One position is to have WPs at a2,b2,c2 & bBb1. Removing any unit will make this position legal. As you can see this stipulation has no unique solution. The three pawns can be anywhere on the second rank with bishop below the middlw pawn. Good luck !

  2. SeetharamanSeetharaman says:

    I have received a welcome set of solutions from Mr.Vlaicu Crisan, the talented composer from Romania. I clarify the book prize goes in case of a tie to the most complete set of solutions (pointing out more than one solution, if any, to the twins) AND to the most complex Illegal Cluster possible ie.containing the MOST NUMBER OF PIECES!!

  3. SeetharamanSeetharaman says:

    Two other similar problems quoted here.
    http://matplus.net/pub/start.php?px=1352658432&app=forum&act=posts&fid=tt&tid=1149&pid=9112#n9112
    Both very good. Have fun solving !

  4. SeetharamanSeetharaman says:

    Alex Levit has accepted the challenge to produce a complex Illegal Cluster! (as only he can !). This is getting very interesting. Keep trying !

  5. SeetharamanSeetharaman says:

    I have received some nice efforts from Alex Levit, Vlaicu Crisan, Thomas Brand and Mario Richter !

    I think it will be impossible to check the alternate settings! So I am no longer insisting on Unique setting. Any sound setting with maximum number of pieces will be welcome ! Please try your hand !

    • Paul Raican says:

      Pietro’s solution could be improved, I guess. Add wS, 5 wPs and 13 black units for an IC. Solution (don’t read it if you want to resolve the problem) : wSf6 wPb6 d5 e4 g4 h5 bPa7 b7 c7 d7 e7 f7 g7 h6 bSc8 bRd8 bRh8 bQe8 bBf8.

      • Paul Raican says:

        Without wSf6, the position is still illegal (the presence of bKg8 cannot be explained). Must try someone else ….

        • Paul Raican says:

          Something else is:
          Add 18 units: 5 wPs, 1wS, 2bQ, 2bB, 1bR, 7bPs. Intention: wPc7 d5 e4 g4 h5 wSf6 bQd8 bQe8 bBc8 bBf8 bRh8 pPa7 b7 d7 e7 f7 g7 h6. (-1)Sh7xSf6 retro-stalemate, (-1)Sh7xRf6? illegal because the second bR was captured on a8 or b8. But this position remains illegal without bPd7: double retro-check!

          • Paul Raican says:

            Finally I hope that the following position is a rally Illegal cluster.
            Add 19 units: White S B 6P Black 2Q 2B R 6P
            Intention: wSf6 wBe8 wPc7 d5 e4 e6 g4 h5 bQd8 bQh8 bBc8 bBf8 bRg6 bPa7 b7 d7 e7 g7 h6.
            Retro: 1.Sh7xSf6? retro-stalemate. An uncapture of P, Bishop, Rook or Queen on f6 involves too many black promotions (Rg6 and one Q are promoted). Don’t works a switch of places between Pa7-Pc7 or Pb7-Pc7, because Rg6 could be the original Ra8 (and then an uncapture of R(Q,P) on f6 becomes legal). Similarly, a switch of places between Ph5-Ph6 don’t works, because Rg6 could be the original Rh8!

            • Themis Argirakopoulos says:

              Please, have a look at the following setting :
              White Se8 Bf7 Pg6 Kf5 Pa4 Pa3 Pf3 Pa2 Pf2
              Black Qb8 Kg8 Pc7 Pd7 Pg7 Pb6 Bd5 Ph4 Bg3 Pe2 Qa1 Rg1

              If white was able to play Be6xf7+, position is legal. But, white need four captures to get his pawns at a3,a4 and f3…
              So, removing wSe8 or wPg6 we have a legal position, as wB can be placed at f7, using e8-h5 diagonal.
              Removing any other white pawn, wB can capture lets say a bS at f7. Same by removing any black unit, then only 11 black men are on board and white can execute five captures, four for the pawns and one more on f7.
              And of course, if I am right, we can easily have millions of similar positions.

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