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Conference rusure::math

Title:Mathematics at DEC
Moderator:RUSURE::EDP
Created:Mon Feb 03 1986
Last Modified:Fri Jun 06 1997
Last Successful Update:Fri Jun 06 1997
Number of topics:2083
Total number of notes:14613

260.0. "More Problems from Omni" by BEING::POSTPISCHIL () Wed Apr 17 1985 12:45

Now that I've sent in my answers to the Omni test, I can discuss them.  Is
anyone else entering?  Does anyone else think they have more than half the
problems correct?

I'm going to enter some of the questions I could only guess at or am unsure of.
I would like to see other people's answers to these.  If anybody is interested
in my answers to other problems, write a note with the problem and I will
reply, if I have an answer.  If someone doesn't have the magazine and wants to
see some questions, send me mail and I may type in more. 


Problem 34.

Suppose that each side of a cube is painted a single uniform color -- red,
blue, or yellow -- such that any two sides painted red are chromatically
indistinguishable, as are any two painted blue or any two painted yellow.
When all six sides are visualized simultaneously, they constitute a color
pattern for the cube.  Two color patterns are mutually indistinguishable
whenever one can be made to coincide with the other by suitable rigid
rotations.  For example, there is only one distinguishable color pattern
consisting of one blue side and five yellow sides.  How many distinguishable
color patterns can a cube have, counting all six sides in each pattern and
assuming that each side must be painted red, blue, or yellow?


My answer to this is in the first reply.  Additional problems are in following
notes.

				-- edp
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260.1BEING::POSTPISCHILFri Apr 26 1985 14:5065
My solution to Omni problem 34 is shown in the table below.  First, the
possible colorings are broken into groups by partitioning the number of sides
to be colored (six).  Thus, all six sides can be colored one color, or five
sides can be one color and one another, et cetera.  Each partitioning is then
divided into patterns that can be made with the partitioning.  For the first
two partitions, only one pattern is possible, but others have multiple
patterns.  For example, if two sides are to be colored the same color, they
can be adjacent sides or opposite sides.  Since this cannot be changed by
rotation, they are different patterns.

The third column in the table shows the number of colorings possible for the
selected pattern.  This is obtained by multiplying by three if one color is
used, then two when another color is used (because only two choices are left),
and by one when the third color is used.  Then, if two colors of the pattern
would be indistinguishable because they occur the same number of times and they
could be exchanged by rotation, the number of colorings is divided by two.  In
the cases where all three colors occur equally and are indistinguishable
because of rotation, the number of colorings is divided by three and two.  The
product of all of this is shown in the fourth column.  A running sum is shown
in the fifth column.

Some of the terms in the Pattern column need explaining.  "Minors adjacent"
means the two colors that only occur once are on adjacent sides.  "Minors
opposite" means they occur on opposite sides.  "Lines" and "corners" refer to
the placement of three sides of identical colors.  Three sides can either wrap
around the cube to form a U or line shape ("lines"), or they can have a vertex
in common ("corners").  A number in the Pattern column, such as 3 or 2, refers
to the 3 or 2 sides all to be given the same color, except the terms "3 pairs"
or "1 pair" which are simply referring to the numbers of pairs.  The last two
entries refer to "pattern A" and "pattern B".  In these patterns, there are
two sides of each color, and same-colored sides are adjacent.  Since all colors
appear, imagine the two red sides in this position:  One is on the front of the
cube, and one is on the bottom.  Now, whatever color is on the back of the cube
must also appear on the left or right side (if it were on top, the left and
right sides would be the last color, forming a "1 pair opposite" pattern, but
this is supposed to be a "0 pair opposite" pattern).  If it is on the left
side, this is pattern A; if it is on the right, it is pattern B.  This will not
change even if the two ride sides are reversed, so that the other non-red color
is now on the back of the cube.

I hope this is not too confusing; send mail or enter replies if you have any
questions or comments which could help (or corrections).

Partition	Pattern			Colorings	Product	Running Sum
---------	-------			---------	------- -----------
      6		unique			    3		   3	     3
    1 5		unique			   3*2		   6	     9
    2 4         2 adjacent		   3*2		   6	    15
		2 opposite		   3*2		   6	    21
  1 1 4		minors adjacent		 3*2*1/2	   3	    24
		minors opposite		 3*2*1/2	   3	    27
    3 3		lines			  3*2/2		   3	    30
		corners			  3*2/2		   3	    33
  1 2 3		3 in a line, 2 opposite	  3*2*1		   6	    39
		3 in a line, 2 adjacent	  3*2*1		   6	    45
		3 in a corner		  3*2*1		   6	    51
  2 2 2		3 pairs of opposites	3*2*1/3/2	   1	    52
		1 pair of opposites	 3*2*1/2	   3	    55
		0 opposites, pattern A	3*2*1/3/2	   1	    56
		0 opposites, pattern B  3*2*1/3/2	   1	    57

Thus, the grand total is 57.


				-- edp (WHOAREYOU note 329)
260.2GOLLY::GILBERTSat Apr 27 1985 05:252
Note that if Polya's counting (coloring) theorem is applied, the problem
should be trivial.  FWIW, your result looks correct.