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a four dimensional hypercube might be realized in 3D with a cube with the corners colored with four colors, maybe.
this is a guess that I had a few months ago, but I can't prove it and I'm not sure exactly how it would work.
the reason I'm working on this occasionally is because a 4D hypercube can also be realized with a 4variable Karnaugh map, which I used in college for digital simplification of boolean expressions.
I ask of you folks to help me iin anyway, or if not, any questions are also welcome as I might have more info I can provide if you are interested.
I forgot to mention that it is not just a plain old empty hypercube I am talking about I guess.
It is a hypercuble with the sixteen corners being colored with one of two colors (0 or 1, binary).
The reason for this is because then the hypercube expresses the output conditions of a
truth table with its colors, true or false. Each position on the hypercube (x,y,z,w) are the
input values of the variables in the boolean expression.
For example, binary output = (input x, input y, input z, input w).
So the coordinates of the corners of the hypercube are the sixteen input permutations.
And the binary output is the color you color that vertex or corner.
igloo myrtilles fourmis
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The rumor is that there are 402 ways to color a hypercubes corners with two colors.
Someone will say immediately, no there are 65536 ways, but I am counting combinations,
not permutations I guess. Like taking into account mirror images and rotational equivalents.
So I guess if this 402 is correct, if I remember correctly, then I could draw all the 4 color ones
on a cube and see how many I get. I started that a few months ago but I lost track of how to
keep track of them and categorise the shapes. Any ideas on how to keep track of things when
there is no apparent order or method? If I remember right I programmed a BASIC program
that came out to the 402 number and I also saw the 402 number on the internet a long time ago with
someones theory as to how the brain works. That made me laugh.
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I have started out with a depiction of the 22 shapes you
can make on a cube with just white and black. I have
used more colors just to make it easier to see the
grouping by the number of corners being colored.
Also note that the 5 corners and the 3 corners colored
are just inverses. And 6 and 2 corners colored are
simply inverses. And 7 and 1 corners colored are
just inverses. And the zero and all 8 corners colored
are simply inverses.
Here is a picture. This will get me started with the
next step I guess.
igloo myrtilles fourmis
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I was just pondering what I am doing and it seems to me, just casually that there are going to be more than
the 402 when 4 coloring on a cubes corners. I am even considering doing substitution
and mixing of the 4 colors so that
each color is not an absolute color, but a relative color with respect to the other 3 colors. But even so, I am
getting more and more positive in my little mind that the combinations of this will exceed 402 because the
binary number 00, 01, 10, and 11 that you can apply to an outer cube surrounding a smaller cube inside it,
which is the most typical drawing we give to the 4d hypercube (but there are others), let me rephrase that,
what I mean is that the outer cubes'corners will get the first binary digit of a double digit number and the
inner cubes'corners will get the ones place digit of the twodigit number. Now this twodigit number can attain
4 numbers = {0,1,2,3}. But the problem with me thinking I could reduce this to just the inner cube and
color it with 4 colors I believe is flawful at this point, but I will try to prove it, and that's good, because, disproving
that it could be equivalent is easier than proving the equivalency, obviously, especially just by examples,
which can have coincidental quantities and relationships that fool you into believing they are eqivalent.
But still (English is so combersome) I haven't said really why I think I am wrong about this hypercube to
cube relationship; essentially I think I am wrong is because the hypercube could be looked at from more
angles, from the left, the bottom, right, back, inside looking up, etc, and also in a torus, going around
counterclockwise, and in a stack, etc. So basically I've proven nothing yet, but I wanted to doubt that this
is going to work, just for the record. Talk later. john
I think the relative color idea is not right because if each color basically represents outer cube corner used
and inner same corner used, then mixing up these colors would make two corners used the same as one or
no corners being used. but to be more sure... say I had two reds and a green and a yellow on one plane.
Now if red means 11 and green means 01 and yellow means 00, I am pretty certain by mixing these up
the equivalent 2color hypercube would have different number of corners being utilized. Yeah, this relative
color thing is going to the back burner... (for now)
Last edited by John E. Franklin (20140129 08:06:29)
igloo myrtilles fourmis
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So let's see. I am trying to reduce a 4d, length [0,1] on each axis, to a 3d cube with some rules associated with it, essentially.
In a cube 3d, we have 3 axes, so that gives numbering skemes for the cubes corners from 0 to 7, by following axis order.
But what about rotation? I ask myself questions that make no sense. This is how I make progress when I am stuck. If you
ask questions that make no sense and are not fully developed questions, then you will by error and good fortune eventually
find something out you never would have thought of. (this is why I am not fond of the forum rules of posting)...
(to be continued)
Edit if necessary before we start solving the wrong problem!
That always bugged me, because isn't that the fun of math, running down the wrong route!!
Well for me it is, but I'm an explorer and I have no deadline for my project.
Last edited by John E. Franklin (20140129 08:20:01)
igloo myrtilles fourmis
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say you have a 2booleanvariable truth table expressed with a square.
Here is the square:
tt
tf
so three outputs are true and one is false.
the inputs are the x and y axes along the edge of the square.
x and y inputs vary from 0 to 1.
Now wikipedia does a pretty bad job explaining Karnaugh maps.
I also do a bad job because they are not easy to teach without
some interaction and a good lecture hall and an occasional question.
Why should I ramble about Karnaugh maps?
The reason why is because maybe the electrical engineers could
donate valuable information to the math guys and the mathematicians
could develop their boolean algebra a little more if they are lucky.
So here is another Karnaugh map in 2variables:
AB
CD
A,B,C,andD are just positions to talk about.
Note you can start at any location and build a tree.
A
/\
B C
The tree then closes up to D.
Basically in my mind, we are taking a 2d object, a
square, and making it into an object that is either 2d, a tree,
or 1d, the same tree, depending how you view it.
If you view the tree one vertex at a time and move through
the tree like a flowchart, following paths, then it might be a
1d object.
The reason I mentioned that is because I'm trying to take the
16 vertex graph called a hypercube, and create 3d or 2d trees
that I like so I can make comparisons either on paper or otherwise.
(again although this is "help me", I hope I'm not breaking rules here)
(basically I'm just thinking through my project and thought
someone might get interested and post a comment to spawn ideas
for me, thanks in advance, but I don't expect that to happen so
please just let me ramble since graph theory and boolean algebra
are important things to work on. Never mind knot theory!
Here is how a 4variable (hypercube) Karnaugh map is set up in 2d
for EE majors in the 1980s:
0 1 3 2

0  Z A C B

1  D E G F

3  L M O N

2  H I K J
Note that E is at (1,1) input location. Output locations A, D, G, and M are
the four locations where the input coordinates vary by only one variable.
The reason I believe Maurice 1953 Karnaugh did this is because when
you combine terms or locations with a loop drawn around them, the
boolean expression naturally get smaller. For example, let's label two
boolean variables x and y and make them go horizontally 0 1 3 2 with y
as the 1's place and x as the 2's place. Now make z and w go downward
with 0 1 3 2 too(, so 00 01 11 10) for z and w making those numbers.
Now the column CGOK is when x and y make 3 and x=1 and y=1 or their
input values in the corresponding boolean truth table.
The row LMON is when z and w are both true or 1.
Location E is at (1,1), rightdown or 0101/xyzw so x=0 y=1 z=0 w=1
So for GO, xy both are true (1) and on the vertical z/w are 1/3.
zwzw
0111
So for GO the variable z is both 0 and 1, so just drop it out of the equation.
And GO has w=1 in both places so use w. So equation for GO is xyw.
A boolean truth table has an output value and likewise, the location GO
could be filled with two 1's or two 0's if you circle it together.
If two 1's, then output value is 1=xyw or OutputValue(x,y,z,w)=xyw.
The output value will be one when those three xyw variables multiply to 1,
they are all 1. They are actually AND'd together, but multiplying does the
same thing.
A little more explanation here is needed ofcourse, actually a lot, but I'm
just trying to help so you can piece it together from various sources if
you are trying to learn it.
Below I am trying to show that z is true for half of karnaugh map grid
at LMON and HIKJ see parenthesis to the right and binary inputs z,w.
Note that w is true for the middle two horizontal rows. From this info,
wz=LMON. w OR z = DEFG LMON HIKJ.
0 1 3 2
 z,w
0  Z A C B 0 0

1  D E G F \ 0 1
 w
3  L M O N \ / 1 1
 z
2  H I K J / 1 0
w XOR z = DEFG OR HIKJ.
If you don't know this stuff you kind should piece it together from many
angles since few people have time to post a comprehensive explanation,
but I'll do piecemeal at times to aid a little.
Last edited by John E. Franklin (20140129 11:27:53)
igloo myrtilles fourmis
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0 1 3 2
 z,w
0  Z A C B 0 0

1  D E G F \ 0 1
 w
3  L M O N \ / 1 1
 z
2  H I K J / 1 0
LO, EI, MN, MG, OJ, KN, DH all have equations that are identical looking except the variables are switched, but the form is the same.
Note LO and MN are disjoint on a row but OJ and KN are diagonal. This is the only flaw I know of with the 2d representatiion over
using a 4d hypercube. But some comprimise was needed to make it a 2d array.
Another neat thing I learned is that if you rotate the four quadrants of the 4d Kmap, it is still equivalent, but the variables change.
So
0 1 3 2
 z,w
0  Z A C B 0 0 rotate stuff D Z B F
 E A C G
1  D E G F \ 0 1
 w M I K O
3  L M O N \ / 1 1 L H J N this rotation creates a similar equation with different variables.
 z it is like a "gear"rotation among the four quadrants and helps to realize
2  H I K J / 1 0 shapes that are not easily realized in this 2d format.
Notice the gear stuff on the right, I made that up to compensate for the 4d in 2d losses.
Last edited by John E. Franklin (20140129 11:47:46)
igloo myrtilles fourmis
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In graph theory,
the vertices are
drawn as dots, but
I have expanded this
for my drawing of a
cube in 2D.
I am allowing the
zero ring to be equivalent
to a dot to allow the image
to be symmetrical in 2D.
igloo myrtilles fourmis
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Perhaps this 3d
drawing will
assist me in my
blurry project
about boolean
algebra, karnaugh maps
and cubes and hypercubes
and maybe logic circuit
minimization.
igloo myrtilles fourmis
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The Karnuagh map intended originally to
produce SOP and POS boolean equations
is shown below using graph theory to
emphasize how the adjacent boxes in
the karnaugh map differ by one binary
digit (chiffre en francais). Notice also
that the Karnaugh map "wraps"around
like the old Atari Astroids game.
It cannot be overemphasized that this
Karnaugh map graph is exactly a
hypercube graph. Most hypercube
drawings don't look this flat but it
still wraps around vertically and horizontally.
Last edited by John E. Franklin (20140222 12:36:20)
igloo myrtilles fourmis
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Now a hypercube of any
dimension can be engraved on
a number line with rainbow
curved lines to interconnect the
vertices. Here is a rough
sketch to give the impression
of what is happening.
As each dimensiion is added,
we clone the below and put
it above and make the many
interconnections in a simple
way.
igloo myrtilles fourmis
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Can the faces of the hypercube be represented as a planar graph?
'And fun? If maths is fun, then getting a tooth extraction is fun. A viral infection is fun. Rabies shots are fun.'
'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'
'Humanity is still kept intact. It remains within.' Alokananda
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The following is just info on the cuff with little proven:
I am persuaded to believe there are 24 faces and 8 adjacent faces to each face. This I see with the tesseract hypercube drawing and also with the karnaugh map 4x4 array drawn on a torus or doughnut shape. The torus one is harder to see, but you get 16 small squares followed by 4 big rings and 4 waistrings for the 4 edges of the faces. (of course I may not be right on this because they are just two examples in 3d) To make a face from the binary numbers, just choose 2 of the digits and do KT, TK, KT, TK, where T=toggle and K=keep same. As for drawing this new graph, I think it is more complicated than the hypercube at first glance because each vertex has eight exiting edges instead of four.
igloo myrtilles fourmis
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Following from post#3, I have changed the order of the 22 shapes
so that many letters of the alphabet remind me of the shapes, for
example the Q, the R, the M, the N, the I, the braille letter F and braille C.
And it's nice that V and A are upsidedown of each other with a not sign in the A.
So here is the next drawing I will build upon later, some in my head.
igloo myrtilles fourmis
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I spent a couple days drawing this out on a pad of paper and then I entered it into the computer.
It is relativecolors, 1 to 4 colors on the corners of a simple cube or icecube shape.
The use of this is unknown.
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