...life can be translucent

Menu

Hexagram Chart 6x6?

Frankelmick

visitor
Joined
Jun 13, 1970
Messages
305
Reaction score
0
Peter,

Sounds good. This means that you now have an evaluation function for 4x4x4 cubes. A sort of measurement of "perfection".

My friend's nephew (let's call him S.) got in touch with me, but at the moment all he knows is that I'm working with someone else on an interesting and original problem.

I'll be talking to S. this weekend and I'll let you know how we get on.

Best wishes,

Mick
 

peter

visitor
Joined
Apr 12, 1970
Messages
168
Reaction score
0
Mick,

Yes, I have such a function, though it is not very specific - only 251 variants.

A very interesting thing I got just yesterday. I ran a program to find all different "basic cubes-4", which do not depend on orientation, i.e. I removed all other 47 transformations (rotations + reflections) from the total roster. If we have 41472 cubes and 48 transformations (including "zero transformation"), then we must have 41472 / 48 = 864 "basic cubes", isn't it? Nope! I've got 888 such cubes. It is because of some special cases: 840 cubes-4 are "usual" and have 48 transformations each, and 48 cubes-4 are "symmetrical" - they have only 24 transformations. It is because one of transformations of such a cube-4 leads to this cube again! This is why they're called "symmetrical", though it is not symmetry in common meaning. So 48 transformations are cut to 24. And these 48 "summetrical" cubes have different structure - I mean different fractal dimensions and dispersions.

I went further and found 24 "anti-symmetrical" cubes-4 also. Under "anti-symmetrical" I mean those which give themselves again after some transformation AND inversion ("0" to "1", "1" to "0"). So in total we have 1152 "symmetrical" cubes-4 (48 basic cubes * 24 transformations) and 1152 "anti-symmetrlcal" cubes-4 (24 basic cubes * 48 transformations), these multitudes don't cross, i.e. there are no cube which would be both "symm" and "anti-symm". 1152 is 1/36 from 41472.

I hope these cubes will be able to give some hints about the cube-6. I also expect them to be seeds for my seeking. I'll think also about symmetry (i.e. "symmetry") in these cubes-6...

One guy from "Magic cubes" runs his programs, but they seek cubes randomly, I don't rely much on such random things. Now he concentrates on "false" cubes-6 and tries to find some balanced ones. Who would think this task is too difficult to solve it quickly?..

With best regards,

Peter
 

Frankelmick

visitor
Joined
Jun 13, 1970
Messages
305
Reaction score
0
Peter,

I don't think that a random search will lead anywhere.

My friend's nephew came round today for a cup of tea and a chat. We went through your ideas from scratch and he's really interested in working on the 6x6x6 cube. We spent about an hour together.

Anyway, he took notes and diagrams with him and said that he'll be back in touch. I'll let you know what he comes back with.

I feel sure that he will find something.

Best wishes,

Mick
 

peter

visitor
Joined
Apr 12, 1970
Messages
168
Reaction score
0
Mick,

Are there no news from your friend's nephew? I have no good news too... The most recent idea is to use Prolog - a computer language made for artifical intellect developments. Some guy from a forum on this language made a program for cubes-4 and got many of them, I couldn't get so much with my own programs.

Also I recalled my triangular charts on Yi Jing and tried to make parallels with HSV (Hue-Saturation-Value) color model. In this model you have black in one corner of the triangle (Kun, evidentally), white - in another corner (Qian) and in the third - pure chosen color. But now I can't see the whole picture...

With best regards,

Peter
 

peter

visitor
Joined
Apr 12, 1970
Messages
168
Reaction score
0
YES!

I got it!

The cube is here:
1 1 1 1 1 0
1 1 1 1 0 1
1 1 0 1 1 1
1 0 0 1 0 1
1 0 0 0 0 1
0 0 0 0 0 0

1 1 1 0 0 0
1 1 0 0 0 1
1 0 0 1 1 0
1 0 1 0 1 1
0 0 0 1 0 0
1 1 0 0 1 1

1 1 0 0 0 0
1 0 0 1 0 0
0 0 1 1 1 0
1 1 1 0 0 1
1 1 1 1 1 1
1 0 0 0 1 0

1 1 1 1 0 0
0 0 1 0 1 0
0 1 1 0 1 0
1 0 0 0 0 0
0 1 0 0 1 1
1 1 1 0 1 0

1 0 1 0 0 0
0 0 1 1 0 1
0 0 1 0 1 1
0 1 1 0 1 1
0 0 1 1 0 0
1 0 1 1 0 1

0 1 1 1 1 0
0 1 0 1 0 1
0 0 0 1 1 0
0 0 0 0 1 0
0 1 1 1 0 1
0 1 0 0 1 0

And there must be more of them (I mean other configurations, that cannot be got with shears permutating). The last idea was cool enough: to seek through cubes with symmetry along a triagonal ("3D diagonal", so to say) - triple symmetry, of course. The funnest thing is that the man who gave me this piece of advice is a Prolog programmer.

You can congradulate me.

With best regards,

Peter
 

Frankelmick

visitor
Joined
Jun 13, 1970
Messages
305
Reaction score
0
Peter Peter Peter,

Congratulations!!!!!

I'm really really pleased that your perseverance paid off.

Fantastic
happy.gif


Mick
 

Frankelmick

visitor
Joined
Jun 13, 1970
Messages
305
Reaction score
0
Peter,

My mathematician friend sends his congratulations and he's very interested in your solution.

I'll send him a print-out of the cube.

He's also going to write out a technical proof of my formula for the number of invariants in the set of all n-digit binary numbers.

I've really enjoyed this thread and I'm delighted that you found a solution.

Thanks and best wishes,

Mick
 

peter

visitor
Joined
Apr 12, 1970
Messages
168
Reaction score
0
Mick,

I've got already... mmm... about 30 these cubes, I think I'll have about 1000 of them finally - but I have to wait for some time. I'm very excited, but I see now that cubes I'm receiving are not balanced very well. They're not bad, no at all, but... magine: if we have too much Yang in one corner, they repulse (as similar charges, like two electrons or two protons). Now it is interesting to me to find a most balanced cube, where these "tensions" will be minimal. I'll use multifractal analysis for them too.

And what proof do you need? As I've said, it is evident (for me, at least), I don't see mistakes in it. For what do you need this formula?

With best regards,

Peter
 

Frankelmick

visitor
Joined
Jun 13, 1970
Messages
305
Reaction score
0
Peter,

Yes. The formula is evident to me as well. I don't need proof as proof. My mathematician friend just offered to write out an explanation in mathematical language that's all.

I don't need the formula for anything.

I think that it will be fascinating to see the most balanced cubes.

Best wishes,

Mick
 

peter

visitor
Joined
Apr 12, 1970
Messages
168
Reaction score
0
Kevin, thanks!

Now I think about such multitude of crystals already got (and more of them that will be got)... Even when I'll find most "homogeneous" cube (that with minimal "tensions"), others will be as perfect as that, but less homogeneous only. What is the meaning of such multitude? Yes, it is a philosophical question, I know, maybe it is more hard to answer it than to find a crystal...

BTW, the name of "crystal" to some people seems not very exact. While there are no evident "crystallinity" in this configuration. Maybe there are some suggestions what this figure can be? A "maze"? (Remember Zelazny's "Maze" - or "Labyrinth"? - in his "Amber" cycle. There was also a version of the Labyrinth in Oberon's Ruby.)

Peter
 

peter

visitor
Joined
Apr 12, 1970
Messages
168
Reaction score
0
I hopw you can imagine the following thing: I already got 200,000 such cubes! Yes, 200 thousand, even a little more. So now I try to sort them and find out the most interesting configurations. Here are my criteria:
1) balance between two halves:
1.a) each half has exactly 54 "yang" and "yin", so the whole cube is very "even", so to say;
1.b) maybe there are some ratio between "yin" and "yang" in each half (in another one it will be inverse)? I didn't examine these things closely yet;
2) balance between layers - the most interesting here I think is the equal amount of "yin" and "yang" in each layer; of course, cubes with all balanced layers will be automatically balanced by halves;
3) trigrams in halves:
3.a) while we have 9 of each trigram (remember, we read trigrams from the center) in a cube, and in every half there must be at least one of each trigram, the requisition of balance is that we must have 4 or 5 of each trigram in one half;
3.b) it is also rather interesting to find cubes where we have the "straight": I mean that 8 trigrams in one half meet in unique amount, EG 1 trigram "Kun", 2 trigrams "Li", 3 trigrams "Dui", ..., 8 trigrams "Kan"; in this case we can take hexagrams of this cube and make a "pyramid" with them (see my pyramid in an earlier post), I plan to run such checks tomorrow;
4) while these cubes are symmetrical by one of triagonals (cubic diagonal in 3 dimensions) - I call it "the main triagonal", it is interesting to find some cubes where we have "pipes": pure hexagrams ("Qian" and "Kun") on this main triagonal, so these hexagrams meet in one cell on the triagonal; there are even cubes with both pure hexagrams on the man triagonal, and in some cubes - in corresponding layers (they are 1-6, 2-5 and 3-4);
5) we can analyze also octants of these cubes (sub-cubes of the order of 3) on their "yin" or "yang" nature - there will be no "neutral" cubes while we have the odd amount of cells in them, cubes-3 with 13 "yang" or less are "Yang" (due to the Chinese principle of "the lesser rules the greater"), and with 14 "yang" and more (13 "yin" and less, respectively) are "Yin"; now I don't see a rule by which we can assign a trigram to each octant.

I got 24 cubes where we have the full balance - exactly 54 "yang" (and "yin") in each half, and amounts of trigrams in each half are not less than 4 and not greater than 5, and layers are also totally balanced. Yes, these cubes are only 24 for now (from those 200 thousand).

I tried to find a cube which consists of hexagrams of King Wen's chart 6x6 (which inspired me), but no - it seems that there are no such cube, I checked those cubes and found, that at least 3 hexagrams do not enter this multitude (when other 33 enter).

Also, I checked invariants (as I calculated them) whether they make a magic square on a side of a cube, and also got nothing - there are no cube with such feature (and all those cubes have all 6 sides with invariants identical).

Oh, and the last feature, that really derives from the symmetry: we can have only one of 22 hexagrams on the main triagonal. It is because we have the triple symmetry - amount of "1" in the cells "around" the main triagonal (not belonging to it) must be divisible by 3 without a remainder (remainder = 0), and then amount of "1" on the main triagonal also must obey the same rule. So we can have only 0, 3 or 6 "1" (yang) on the main triagonal. "Even" hexagrams (3 yang - 3 yin) are 20, and also add here "Qian" and "Kun" - 22 exactly. (Maybe somebody can see here some links with Tarot and Qabbalah.)

Do you have ideas about what can be interesting in these cubes?

With best regards,

Peter
 

Frankelmick

visitor
Joined
Jun 13, 1970
Messages
305
Reaction score
0
Peter,

Fantastic. Plenty of food for thought.

Does Triagonal mean the long diagonal? For example from the far north-west corner to the near south-east corner?

Thanks very much,

Mick
 

peter

visitor
Joined
Apr 12, 1970
Messages
168
Reaction score
0
Mick,

Yes, it is just the "long diagonal" - from one cube corner (vertex) to the opposite.

Can you imagine some cube "in correspondence with I Ching" another than balanced? Some that reflects "Tai Chi" symbol?

With best regards,

Peter
 

Frankelmick

visitor
Joined
Jun 13, 1970
Messages
305
Reaction score
0
Peter,

Thanks for that.

I was thinking of trying to analyse the cube using your invariant function. Maybe there's a cube with the numbers 0 to 35 in some kind of sequence? Perhaps there might even be some relationship to the I Ching invariant number sequence?

Just a thought.

Best wishes,

Mick
 

peter

visitor
Joined
Apr 12, 1970
Messages
168
Reaction score
0
Hello Mick,

Well, I just checked all my cubes on your idea: invariants following those in King Wen sequence. I found no such cubes, but I checked "directly", not by halves (as we can see on that chart 6x6), in 4 directions - straight by x, straight by y, reversed by x and by y. I'll check also cubes using that diagram later.

I found 120 cubes which can form a pyramid from their hexagrams, but sequences of trigrams in them are strange. (And there are no King Wen sequence of trigrams there, and Fu Xi too.)

So my thoughts of approaching to the traditional sequence are still open...

With best regards,

Peter

Update: I checked all cubes on that chart invariants - nothing. Can't predict whether there will be some cubes with such order...
 

Frankelmick

visitor
Joined
Jun 13, 1970
Messages
305
Reaction score
0
Peter,

Thanks for checking!

I'll let you know if I can think of another possible approach.

Best wishes,

Mick
 

peter

visitor
Joined
Apr 12, 1970
Messages
168
Reaction score
0
Here is one "crystal" that can be remade to a pyramid:

2061.jpg


Peter
 

peter

visitor
Joined
Apr 12, 1970
Messages
168
Reaction score
0
Well, I saw that another pyramid can be made using 36 numbers, see below:

2096.jpg


I can't imagine how 36 hexagrams must be distributed here though. What can be these 6 groups by 6 hexagrams? We can get 6 groups theoretically by uniting a) Sun and Dui trigrams and b) Zhen and Gen trigrams (Qian, Kun, Li and Kan are symmetrical and each refers to its own group). But I haven't found a cube where these groups were presented - I mean, 6 hexagrams with Qian upper trigram, 6 - with Kun, 6 - with Kan and 6 with Li, and other 2 groups were made of hexagrams with Dui or Sun as the upper trigram (6 in total), and with Zhen or Gen as the upper trigram. I'll check their palaces and generations also, but later... Does anybody have an idea?

Peter
 

Frankelmick

visitor
Joined
Jun 13, 1970
Messages
305
Reaction score
0
Peter,

My mathematician friend got back in touch.

He says that he's also found a solution to the cube problem.

Would you be interested in seeing what he's found?

Best wishes,

Mick
 

peter

visitor
Joined
Apr 12, 1970
Messages
168
Reaction score
0
Hello, Mick!

What solution did he find? Yes, it is still interesting to me. Has he found at least one cube with his solution?

Though I think now about meaning of these cubes. And using too.
 

Frankelmick

visitor
Joined
Jun 13, 1970
Messages
305
Reaction score
0
Peter,

Sorry I misunderstood. It turns out that he just had this formula for the number of invariants in the set of numbers with n binary digits:

2<SUP>(n-1)</SUP> + 2<SUP>ROUNDDOWN((n-1)/2)</SUP>

How are you using the 6x6x6 cubes?

Best wishes,

Mick
 

peter

visitor
Joined
Apr 12, 1970
Messages
168
Reaction score
0
Well, it is not very interesting, brcause I know number of invariants for n=6, and other cases are not interesting to me at all.

I don't know how to use them - sometimes I try to contemplate on one chosen cube, but, pitifully, I seldom choose time to do it.
 

Frankelmick

visitor
Joined
Jun 13, 1970
Messages
305
Reaction score
0
Peter,

My friend and I talked about the idea of trying to build a 3-D model of the 6x6x6 cube.

Did you ever do this?

Best wishes,

Mick
 

peter

visitor
Joined
Apr 12, 1970
Messages
168
Reaction score
0
Mick,

Firstly I thought about creating a cube like balls connected by sticks - something like models of molecules. But I didn't like it. The last good idea was to make 36 pillars made of 6 parallelepipeds each. So it can be made with stones - say a square 6x6 meters and each pillar 5-6 meters in height.

Peter
 

Frankelmick

visitor
Joined
Jun 13, 1970
Messages
305
Reaction score
0
Peter,

Sounds like a piece of Modern Art.

You should apply for a grant
happy.gif


Best wishes,

Mick
 

peter

visitor
Joined
Apr 12, 1970
Messages
168
Reaction score
0
Grant? Sorry, I'm weak in such things - do you really think that I can gain a grant? Or what? How?
 

Clarity,
Office 17622,
PO Box 6945,
London.
W1A 6US
United Kingdom

Phone/ Voicemail:
+44 (0)20 3287 3053 (UK)
+1 (561) 459-4758 (US).

Top