Drexel dragonThe Math ForumDonate to the Math Forum

Ask Dr. Math - Questions and Answers from our Archives
_____________________________________________
Associated Topics || Dr. Math Home || Search Dr. Math
_____________________________________________

Rainbow Logic


Date: 13 Jan 1995 12:15:47 -0500
From: Cathy Brady
Subject: combinations problem

In the Family Math book, there is an activity called Rainbow Logic.  It's 
like battleship in that one player sets up a matrix and the other 
player(s) have to match their matrices by asking questions.  The matrix, 
if it's 3x3, has three pieces of each color which can be put on the board 
in any position, just so long as each square shares a border with a 
square with the same color.  The players ask "What colors are in column 
one? (or row three, or column three...)" and the person who is it tells 
them the color(s) but not in any order.  The game can also be played with 
a 2x2 grid (with 2 colors) or a 4x4 grid (with 4 colors).

By drawing the grids, I found that the number of possible layouts for 2x2 
is 4, and the number of possible for 3x3 is 72, and for 4x4 - well, I got 
tired of drawing, but it's over 250.  So my question is: Is there a 
formula for the number of layouts for an nxn grid with n colors where 
each color is connected by at least one border?

Cathy Brady


Date: 17 Jan 1995 00:42:38 GMT
From: Dr. Math
Subject: Re: combinations problem

Hello there!

I've got to ask you a question about this before I dive in: when you say
each color is connected by at least one border, what exactly does that
mean?  Does that mean that each region of the same color must be
connected?

We await your reply.


Date: 16 Jan 1995 19:48:53 -0500
From: Cathy Brady
Subject: Re: combinations problem

YES


Date: 17 Jan 1995 02:34:53 GMT
From: Dr. Math
Subject: Re: combinations problem

Okay, we've been doing some thinking.  We think we've tackled the 4x4
question, and we've come up with a little on the general formula.  It
looks like finding a closed form general formula might get kind of hairy. 
Also, we've come up with something different than you got for the 3x3
case.

Anyway, the first thing we (fellow Math Doctor Ethan Magness and I) did
was to eliminate all the different rotations and color changes, and call
them all essentially the same pattern.  Once you figure out what the
"fundamental" patterns are, you can rotate them four different ways, and
then choose the colors two different ways.

In general, with an nxn array, you'll always figure out how many different
fundamental layouts there will be, and then you'll rotate it four ways and
choose the colors n! (which means n(n-1)(n-2)...(2)(1)) ways.  So if there
are f different fundamental arrays, there are 4fn! total different layouts
for your final answer.

But there's a complication, because sometimes you'll end up duplicating a
pattern by rotating it.  For instance, in the 2x2 case, there's only one
fundamental array you can make, one that has one color on the top row and
the other color on the bottom row.  When you rotate this array though,
you'll see that rotating it 180 degrees will produce the same array as not
rotating it at all.  So we've got to get rid of half of these, so we say
that there are .5 different fundamental layouts, and then we multiply by
4n!, i.e. 8, and we end up with your answer, 4.

In the 3x3 case, we can have three different fundamental layouts. 
However, if you look at the one that has three straight pieces arranged in
columns, you'll notice that again, rotating it 180 degrees has the same
effect as not rotating it at all.  So really, we'll say we have 2.5
different fundamental arrays, and that the total is 2.5 x 4 x 6 = 60
different layouts.

We did this for the 4x4 case too, and we found that there were 28.25
different fundamental arrays you could have (we simplified the process a
little by noticing that some of the arrangements were just reflections of
each other, and by first looking at arrangements that contained a long
four-in-a-row piece across the top).  So we found that you would have
28.25 x 4 x 24 = 2712 different total arrangements.

Anyway, this is as far as we've gotten, and getting much farther would
probably require a lot more insight than we've gotten so far.  But it's a
pretty cool question, and I encourage you to keep thinking about it!  If
you find anything else interesting about it, let us know!

-Ken "Dr." Math
    
Associated Topics:
College Modern Algebra

Search the Dr. Math Library:


Find items containing (put spaces between keywords):
 
Click only once for faster results:

[ Choose "whole words" when searching for a word like age.]

all keywords, in any order at least one, that exact phrase
parts of words whole words

Submit your own question to Dr. Math

[Privacy Policy] [Terms of Use]

_____________________________________
Math Forum Home || Math Library || Quick Reference || Math Forum Search
_____________________________________

Ask Dr. MathTM
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/