From: umatf071@unibi.hrz.uni-bielefeld.de (sio)
Subject: Re: Game of pentominos
Date: Fri, 18 Dec 92 11:33:18 GMT
Organization: Universitaet Bielefeld

The 4th Pentacube Contest 92/93

  Build the figure "volcano" with all 29 pentacubes.

    1 1 1 1 1 1 1 1 1        Each number says, how
    1 2 2 2 2 2 2 2 1        many cubes are stacked.
    1 2 3 3 3 3 3 2 1
    1 2 3 2 2 2 3 2 1
    1 2 3 2 1 2 3 2 1
    1 2 3 2 2 2 3 2 1
    1 2 3 3 3 3 3 2 1
    1 2 2 2 2 2 2 2 1
    1 1 1 1 1 1 1 1 1

  There is a published solution already. If you send it, you will get another
  problem, which is more difficult. You will see, the above problem is easy
  enough to be solved by yourself (or your computer). -> (*)

Send your solution to:
  Ekkehard Kuenzell,
  PENTAKUBENSPIELE, Schoenrathstr. 79, D-5100 Aachen, Germany
  Tel.: +49-241-58881

Deadline: >>>>> 30.04.1993 <<<<<

Prizes:
  1. prize: 200 DM  (ca. 140 $)
  2. prize: 100 DM  (ca.  70 $)
  3. prize: 100 DM  (ca.  70 $)

All participants will have equal probability winning a prize, independent of
the number of solutions (>=1 of cause) s/he has sent.

The Outcome of the contest will be sent to all participants till 31.05.93.

Notation for the Pentacubes:
 The notation follows a mnemonic system and is easily recognized by the
 solution of the following pentacube problem. The system is explaned in
 the book "Pentakubenspiele", which is avalable by the author E. Kuenzell.

  2 1 2 1 2 1 2 1 2 1 2 1 2 1 2    a pentacube problem
  2 1 2 1 2 1 2 1 2 1 2 1 2 1 2
  2 1 2 1 2 1 2 1 2 1 2 1 2 1 2
  2 1 2 1 2 1 2 1 2 1 2 1 2 1 2
  2 1 2 1 2 1 2 1 2 1 2 1 2 1 2
  2 2 2 2 2 2 2 2 2 2 2 2 2 2 2

 11 12 22 22 22 72 72 72 90 51 51 51 71 71 71  one solution
 10 12 42 42 22 34 34 72 20 20 51 82 82 71 70  (bottom layer)
 10 12 12 33 41 41 34 34 90 20 35 35 70 70 70
 10 12 60 33 33 36 36 30 30 20 20 35 50 70 40
 10 37 60 81 33 36 32 32 30 30 31 50 50 50 40
 10 37 60 81 81 81 21 32 32 30 31 31 50 61 61

 11    22    41    72    90    51    82    71  (top layer)
 11    42    41    34    90    35    82    40
 11    42    41    36    90    35    82    40
 11    42    33    36    80    31    13    40
 37    60    21    32    80    31    13    61
 37 37 60 81 21 21 21 80 80 80 13 13 13 61 61

The list of figures of the last three contests:

 1) number of participants: 10

       2 2 2   2 2 2       b = 11
       2 4 4 4"4 4 2       n"= the two bottom cubes are missing of
       2 4 6 6"6 4 2           the pile of height n.
         4"6"b"6"4"
       2 4 6 6"6 4 2
       2 4 4 4"4 4 2
       2 2 2   2 2 2

 2) number of participants: 5

             2
           2 2 2
         2 2 2 2 2
       2 2 2 1 2 2 2
     2 2 2 1 1 1 2 2 2
   2 2 2 1 1 1 1 1 2 2 2
 2 2 2 1 1 1 1 1 1 1 2 2 2
   2 2 2 1 1 1 1 1 2 2 2
     2 2 2 1 1 1 2 2 2
       2 2 2 1 2 2 2
         2 2 2 2 2
           2 2 2
             2

 3) number of participants: 6

       5 5 5 5 5 5 5
       5 1 1 1 1 1 5
       5 1 1 1 1 1 5
       5 1 1 1 1 1 5
       5 1 1 1 1 1 5
       5 1 1 1 1 1 5
       5 5 5 5 5 5 5

Good luck, Torsten Sillke.
I've sent my solution already.

(*)  This is the more difficult problem

   1 2 3 4 5 6 5 4 3 2 1          n'= the bottom cube is missing of
   1 2 3'4'5'5'5'4'3'2 1              the pile of height n.
   1 2 3'4 4 4 4 4 3'2 1
   1 2 3'3'3'3'3'3'3'2 1          the figure must be stable.
   1 2 2 2 2 2 2 2 2 2 1
   1 1 1 1 1 1 1 1 1 1 1