Boxes of Arrows

John E. Franklin · 2010-08-12 06:49:51

I was trying to come
up with something
new to work on, logical,
but not numerical. So
I started making these
drawings and switching
the arrows different
ways. (See pic below, click on)

Can someone give me
ideas as to what I might do
next with this and what
branch of mathematics
do you think this might be
close to.

Also imagine a cube with
8 boxes in 3-D and having
arrows go in six coordinate
directions.

Thanks for helping me
from boredom to the pursuit
of awesome things...

John E. Franklin · 2010-08-13 03:42:24

If the arrows are allowed to go East, West, South and North.
Then there are 4^4 or 256 permutations.
Then if you account for quarter rotations 4 times around,
then there are 70 shapes according to my BASIC program
I wrote this morning. Still I have to draw them out from
the program results to check for bugs.
Interesting that 70 is greater than 256 / 4.
I suspected that from other similar projects, such as
shapes in boolean algebra cubes of 3-D. (3 variables)

mathsyperson · 2010-08-13 03:53:49

There are more than 64 combinations because rotating doesn't always get you 4 different things.

For example, Up Right Left Down wouldn't be affected by rotation at all.

Now I'm curious as to how many distinct combinations there would be if you counted two things as the same when they're a reflection of each other.

soroban · 2010-08-14 03:56:32

. .

. . You go ahead . . . I'll wait in the car.
.

John E. Franklin · 2010-08-14 11:17:18

To mathsy,
Your question about the reflection reduction is
a good one. That is the same as looking through from the
backside of the piece of paper of the drawings.
My groupings on a large pad of paper show there are 45
of these arrangements. And 45 > (70 / 2).

(Hi soroban, I might tackle your questions next week.... Sweet ideas!)

Avon · 2010-08-15 11:11:58

I would usually use Burnside's lemma to count such things.

In the first example we are letting the rotation group of the square act on the drawings. This group has four elements: the identity, the quarter-turn clockwise, the quarter-turn anticlockwise, and the half-turn.

The identity fixes all of the 256 drawings.
The quarter-turns each fix 4 drawings.
The half-turn fixes 16 drawings.

Hence the number of different drawings if two are considered the same if one is a rotation of the other is
(256+4+4+16)/4 = 70.

If we also want to allow reflections then there are four more elements in our group: the two reflections in the diagonals of the square, the reflection in the vertical line through the centre of the square, and the reflection in the horizontal line through the centre of the square.

The two diagonal reflections don't fix any of the drawings.
The reflections in the vertical and horizontal lines each fix 16 drawings.

Hence the number of different drawings if reflections and rotations are considered the same is
(256+4+4+16+0+0+16+16)/8 = 39.
I notice that this is not the same as what John counted.

The 2x2x2 cube could be handled similarly, but I don't really have the time to do it right now.

John E. Franklin · 2010-08-16 09:26:28

Fascinating Avon!! This is amazing
there are ways to calculate this count
from the 1800's. I don't get it yet
though, but thanks!!

John E. Franklin · 2010-08-17 05:09:03

Avon, I found
an easy non-mathematician's
article on burnside's lemma.
http://baxterweb.com/puzzles/burnside5.pdf

John E. Franklin · 2010-08-17 08:55:37

My 45 count was incorrect, done too hastily, and I missed 6 pairs.
Avon is correct about the 39 arrangements, which includes
reducing by rotations and reflexions (reflections).

hero_hont · 2010-08-19 02:46:26

John E. Franklin wrote:

My 45 count was incorrect, done too hastily, and I missed 6 pairs.
Avon is correct about the 39 arrangements, which includes
reducing by rotations and reflexions (reflections).

hih, i count 44.. but i don't know how do this exercises do solution?
have you reply total this exeercises solution...i can reference it?
thanks......

John E. Franklin · 2010-08-21 07:29:46

Maybe if I get the time, I will
draw up or use LaTeX to
draw up the arrangements.
If I do that, I will try
to put pairs that are
reflections beside each
other, so you will see how
the 70 pair up to 39.

hero_hont · 2010-08-21 19:08:33

hih.............:P
I will have started making these
drawings and switching
the arrows different
ways.
hih, try work............

John E. Franklin · 2010-08-24 05:39:54

Yes, I hope you
draw them out and
get 39 and 70, because I
am busy with reading
"The Trachtenberg
Speed System Of Basic Mathematics"
book right now...

John E. Franklin · 2010-08-26 06:49:39

I'll try a little LaTeX, hope I can get the 2 by 2 array up...
Okay, I got the LaTeX to work.
The eight boxes of arrows below are
the only eight ones from the 70 that
cannot be paired up with a mirror
image one. So there are 31 pairs (62),
plus 8 non-pairs. 31 + 8 = 39 arrangements.
I have not drawn out the 31 pairs yet.
Hope I don't have to...
(Keep in mind we are doing quarter rotations 3 times, and
consider those all the same with the 70 count)

Last edited by John E. Franklin (2010-09-02 01:13:35)

John E. Franklin · 2010-08-26 08:19:39

So it appears the numbers
70 and 39 are a trend in
this thread. Does anyone
know something in
chemistry that has these
two numbers and look
similar?

Math Is Fun Forum

#1 2010-08-12 06:49:51

Boxes of Arrows

#2 2010-08-13 03:42:24

Re: Boxes of Arrows

#3 2010-08-13 03:53:49

Re: Boxes of Arrows

#4 2010-08-14 03:56:32

Re: Boxes of Arrows

#5 2010-08-14 11:17:18

Re: Boxes of Arrows

#6 2010-08-15 11:11:58

Re: Boxes of Arrows

#7 2010-08-16 09:26:28

Re: Boxes of Arrows

#8 2010-08-17 05:09:03

Re: Boxes of Arrows

#9 2010-08-17 08:55:37

Re: Boxes of Arrows

#10 2010-08-19 02:46:26

Re: Boxes of Arrows

#11 2010-08-21 07:29:46

Re: Boxes of Arrows

#12 2010-08-21 19:08:33

Re: Boxes of Arrows

#13 2010-08-24 05:39:54

Re: Boxes of Arrows

#14 2010-08-26 06:49:39

Re: Boxes of Arrows

#15 2010-08-26 08:19:39

Re: Boxes of Arrows

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