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Hexagram Chart 6x6?

peter

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I believe many of you know this chart with 36 hexagrams, which shows how 2 parts of "Yi Jing" were formed (30 in upper and 34 in lower). Once I looked on it and saw that we have really a cube 6x6x6 with yin or yang in every cell! reading from bottom to top or from top to bottom, we can receive all 64 hexagrams.

And I tried to read hexagrams by other directions: from front to rear or from right to left. Of course, I found coincidences of hexagrams or mirrored hexagrams. While by z axis (vertical) they don't coincide. And now I wonder - are there such cubes where we can read all 64 hexagrams by any of 3 axes? I mean that if we read along z (bottom-top and top-bottom), we find 64 hexagrams; if we read along x (front-rear and rear-front), we also find 64 hexagrams; finally, along y (left-right and right-left), we see all 64 hexagrams.

So does anybody know about such thing? It must be very wonderful and deep. Or how it can be proved that it is impossible? Maybe there are some ideas about how this cube must be made?

Here are some my thoughts so far:

1) It is comfortable to operate with 'invariants' - 36 invariants in addition to 64 hexagrams. Firstly, we must take two trigrams of a hexagram, but counting from the center, so trigrams will be 4-5-6 and 3-2-1 (lines of original hexagram). Then determine which trigram is greater (7 for Qian and 0 for Kun, 1 = Gen, 2 = Kan etc.) and calculate first item by this formula: T1 * (T1 + 1) / 2 . Then add here number of lesser trigram. Maximum is 35 and minimum is 0. 8 symmetrical hexagrams (1, 2, 27, 28, 29, 30, 61 & 62) have 1 invariant each, and the rest 56 hexagrams have 28 invariants (1 invariant for every mirror-"fan"-inversed pair).

2) Let's divide our original 6x6x6 cube on 8 lesser cubes - 3x3x3 each. Now we have 8 trigrams to place here, but 9 cells on each side! And an obvious equivalence: 8 lesser cubes - 8 trigrams. So in every cube one trigram must be basic. But! I played some time with these lesser cubes and can say that we must find 4 different (and ortogonal in some sense) principles of placing trigrams in lesser cubes, or their combinations in the original cube will repeat, and we shan't get all 36 invariants on every side of the cube.

3) I wrote a program (on C++) for blunt looking for such distribution. I made it work as fast as possible, but it has too many variants to scan... So my only real hope is with those lesser cubes.
 

mick

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Luis,

Thanks for scanning in the pages from ZD Sung's book. The cube on these pages looks different from Peter's idea of a 6x6x6 cube containing all 36 I Ching invariants.

It looks like the author is taking a 2x2x2 cube to represent the 8 trigrams. Page 44 places the wind trigram Sun in the far top right-hand corner of the cube. I think the author calls this corner D5, maybe this stands for division 5? Sun is the 5th trigram in sinking Yin sequence.

The author then expands the Sun trigram cube as a smaller 2x2x2 cube containing the 8 Hexagrams which have Sun as their lower trigram.

These are called D5S1 to D5S8 in sinking Yin sequence. Maybe these stand for division 5 sub-division 1 and so on?

Page 45 does the same expansion of the water trigram called K'am in the text. This is called D6 and the 8 hexagrams are called D6S1 to D6S8.

Presumably there are similar diagrams for each of the 8 trigrams and you end up with the I Ching in a 4x4x4 cube? Each 2x2x2 corner would contain all the hexagrams with the same trigram as their lower trigram. Sweet


The author represents Yang lines as A and Yin lines as B. But I can't work out why

(AA + BA) (AA + BA) (AB + BB) represents Hx 57
and
(AB + BB) (AA + BA) (AB + BA) represents Hx 29?

Any thoughts?

I like the expansion which maps lines 1, 2, 3 to lines 1, 3 & 5 of a new Hx and 4, 5 & 6 to lines 2, 4 & 6 of the new Hx. Is there a name for this kind of mapping?

Thanks again and best wishes,

Mick
 

peter

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Sorry for a pause. Tried to find some people that can help, asked them, but they didn't answer for now, I wait for their hints.

Luis, thanks for the scan, I don't see purpose of these diagrams clearly, but I'm afraid that they won't help because they're about something different. I see some "standard" operations with hexagrams, and in my cube there are no such linear symmetry.

Mick, it is easy to complete this cubic chart. I even remember that I once created such cube 4x4x4 but hadn't seen something special about it. I mean nothing that I never seen before. Maybe I mistook...

My latest idea is to take a "seed" (cube 4x4x4) and "spread" it in my cube-6. There are 2 ways of this spreading: 1) split it on 8 cubes-2 and put these cubes-2 into vertices of the cube-6, and 2) split it on separate cells and place outer cells (56 cells which are not in the center) to outer cells of cube-6, so one shear will appear as following:

shear of cube-4:
4 4 4 4
4 4 4 4
4 4 4 4
4 4 4 4

this shear after placing it in cube-6:
4 0 4 4 0 4
0 0 0 0 0 0
4 0 4 4 0 4
4 0 4 4 0 4
0 0 0 0 0 0
4 0 4 4 0 4

And I made another program which works with this "spread seed 2" and seeks cubes.

I found some "false" cubes where we have unique hexagrams along z and x axes (I fill the cube along x axis, and along z and y check only), but not along y axis. (Yes, there also must be cubes with unique hexagram along x and y but the can simply be made from x-z, so I don't care about them). I thought that maybe I can make the cube with permutating of shears along y axis - look: if we have unique hexagrams along x and z, then we can cut this "false" cube into 6 shears and permutate them without losing uniqueness along x and z. Also we can mirror every separate shear and rotate it on 180. So we have 4 variants of 1 shear: original, mirrored along x, mirrored along z and rotated. I checked all permutations of shears including all modifications of shears, but - alas - found nothing. Or it can be said that I found that such permutations don't help much.

I run my latest program on 2 computers already (with 2 different seeds surely) and wait, wait...

Peter
 

mick

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Peter,

I agree that the 4x4x4 cube in Z D Sung's book doesn't really contain anything special. It's sort of Shao Yong's square re-arranged in 3-D.

I couldn't understand the notation Sung uses where
(AA + BA) (AA + BA) (AB + BB) represents Hx 57
and
(AB + BB) (AA + BA) (AB + BA) represents Hx 29.

I think that what you're doing is totally original and visionary. I think that just to have seen the possiblity of representing the 36 I Ching invariants as a 6x6x6 cube is quite special.

I understand your spread-seed idea. Can you just use brute force from this point? You know the polar distribution of 10-26 either way?

I feel confident that there's a solution.

By the way, (sorry to go on about this) but I've got this formula for the number of symmetrical patterns s in a set of n Yin/Yang lines.

s = 2 ^ [(n + mod2n)/2]

Then the number of invariants I is

I = s + (2^n - s)/2

I showed this to a mathematician at work, describing it in terms of binary digits, and he said that it seems to work but perhaps I'd have to find a proof by induction.

Surely some mathematicians must have been over this ground before us?

Thanks again and best wishes,

Mick
 

peter

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Mick,

Yes, your formulae are quite right.

I did put that check on pairs - I mean pairs (3-4) (inner pair), (2-5) (middle pair) and (1-6) (outer pair) - it lessened "false" cubes greatly and so I won some time that I've been loosing in older versions. This check is simply is: 1) combo "11" must be 10 strictly; 2) combo "00" must be 10 strictly; and 3) combos "10" and "01" must give 16 in sum.

About Sung's diagram - sorry, didn't examine it closely. I hope a little later.

What did you mean under "brute force"? I think I have no other way. Firstly I fill 16 hexagrams where we have 4 known digits, and then start to fill "by brute force" and sometimes check. I also check on pairs when I have not completed paired shears - only when I have 24 pairs; I check that combos "11" and "00" must meet no more than 10 times each, and sum of "10" and "01" - no more than 16 times. So I got 3 "false" cubes, where their "falsehood" is only that they have 2 similar invariants along one axis (y of course)... Too close to the solution, but not perfect still...

I asked some specialists on puzzles in hope that they would interest in this truly puzzling problem.

Peter
 

mick

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Peter,

You are getting so close.

Thanks for checking my formulae.

But I know that my mathematician colleague will say, "But how can you prove that this formula is true for patterns of 10 lines or 10,000,000 lines?" Any thoughts on how to respond politely?


By "brute force" I meant going through every possible combination. As you say, you have no other way.

I just thought this might now be a bit less time-consuming since you already have 64 of the 216 cells filled by the exploded 4x4x4 seed cube.

Here's my understanding of the steps involved in finding a solution by brute force.

1) Distribute the 36 invariants

2) Convert them into a stream of 216 bits

3) Check the 36 Hexagrams formed by these groups of bits (1,7,13,19,25,31); (2,8,14,20,26,32)up to (186,192,198,204,210,216) and if these contain all 36 invariants...

4) Check the 36 Hexagrams formed by (1,37,73,109,145,181); (2,38,74,110,146,182) up to (36,72,108,144,180,216) and if they contain all 36 invariants - open the Champagne


I make it that there are about 3 x 10^41 possible distributions of the 36 invariants. Is that right?

By starting with the 4x4x4 seed cube and checking the combos after 24 pairs, you are way ahead of this crude strategy.

Best wishes,

Mick
 

peter

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Mick,

I made more simple algorithm, so to say. I appplied a very interesting thing that is used in programming - a container with key value. So when I run cycles on filling, I put one hexagram, then take this list of hexagrams and delete hexagrams with identical invariant, so they won't appear in further filling. I make this filling along x axis, and then make interim checks along y and z axes. It works not too slow, but not very quick either - not so quick as I want, of course.

Explanation of your formulae is very easy. When you calculate symmetrical numbers, you see, that there are equal amounts of them for 3 and 4, 5 and 6, 7 and 8 etc. It is their feature. And when you calculate the total amount of invariants, you simply subtract symmetrical ones from amount of numbers, and then the remaining you divide by 2, and add symmetrical to it. It can be easily read from your formulae, so I don't understand your fears, honestly. Every good matematician will read them without comments.

Peter
 

mick

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Peter,

Thanks for that. Your container with key value sounds like an excellent short-cut.

I thought of another short-cut. Is this included in your algorithm?

Let's take a simplified 2x2x2 cube.

For example, when you fill with

00 11
01 10

I think that you are also checking the "back-to-front" version. I think your short-cut takes care of this.

01 10
11 00

But you are also checking the Yin/Yang opposite version

11 00
10 01

Some fills will be the same back-to-front as Yin/Yang opposite.

Does this idea help at all?

Thanks again and best wishes,

Mick
 

mick

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Peter,

I also thought that when you check for the invariants numbered 0 to 35, you could maybe first check that the total is 630?

I have access to a lot of processing power on my IBM mid-range computer at work. I might have a go at a very crude, brute force algorithm that I can leave to run overnight.

But first I'll check with my manager to estimate how many variations I'll be able to hammer through in 12 hours


Cheers,

Mick
 

peter

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Mick,

I didn't catch your idea with 2x2x2 cubes... I don't check them while they are too many. As I mentioned, I check symmetrical pairs: for full set it is (10; 10; 16), and for partial - (< 11; < 11; < 17).

Most checks are partial, so I cannot apply total sum there. With my filling along x axis I ensure that they'll be unique along it, and checks are made for y and z axes. After some 7 milliard of checks (they differ in "levels", as I called it - simply a number of current cycle where this check failed, so when it fails on 17th level, it rejects more dead-ends that when it fails on 23rd level) I got about 5 "false" cubes for 3 seeds. These "false" cubes, as I said, have 108 "1" and 108 "0", and all previous checks to the very level of 35 were passed, and along z axis there are unique hexagrams, but not along y... So I assume that these checks are severe enough, while in my previous version I got many thousands "false" cubes, not so strict, of course.

If you want, I can mail you my program, so you'll have to run it with a seed and go home. Then in the morning you'll come, stop them, and in the evening start them again from the very place they stopped (I foresaw that they'd work for long, so I provided a manual stop with saving of current session). I made different folders for every seed, and now I have some 4-5 sessions on my comp, and many those cube-4 seeds.

This is strange that I'm getting too close with my "false" cubes. I can say that it frightens me, while I don't see why "perfect" cubes-4 wouldn't serve as seeds for "perfect" cubes-6.

And I also want to find how many variations of the same cube-4 I can get. I already told about them, here are they once more:
1) rotation - gives us 24 cases, i.e. the original cube and 23 variants; I calculated it with the following thought: take a cube, make every side as front and rotate it so this side remains in the front; 6 sides * 4 rotations gives 24 cases;
2) reflection - select a side on the cube and reflect the cube on (with? by? at?) it; reflections on opposite sides will be identical, so we have 4 cases here (original and 3 its reflections);
3) inversion - change all "1" to "0" and vice versa, 2 cases in total.

So in total we have 24 * 4 * 2 = 192 cases? it seems so... And here is the question: which cube-4 we should take to get this "perfect" cube-6 as quickly as it is possible???

Peter
 

mick

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Peter,

My idea with the 2x2x2 cubes was to try to demonstrate checking for inversion.

But you already do that anyway. You've covered rotation and reflection (using?) as well. I think I'm getting desperate to try to help you find possible short-cuts.

It is strange that you're getting false cubes. It feels like there should be a solution.

Thanks very much for offering to send me your program. You mentioned C++ earlier. Is it a C++ program? I'll find out if I can run C++ programs.

Best wishes,

Mick
 

peter

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Mick,

Don't you think I feel desperate seeing that I cannot find a solution for more than a month?

My program is a simple console application - it's a-la DOS program, but it runs under Windows only. If you have Windows (98, 2000, XP), you can run my program easily, so I can send you an ".exe" file. And instructions to it, of course. It loads CPU for 100%, so sometimes the system may hesitate to response.

How many computers do you have? I mean - how many seeds I must prepare for you?

Peter
 

peter

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And about total sum of 630 - it is not sufficient. Look: if we miss say pair (1; 34), but have instead of it (12; 23), it gives us the same sum 35 and 630 in total also, but we don't meet all variants here. So we must check also sum of squares of these numbers. For the set of 36 unique invariants this sum will be 14910. So we must check both numbers for (630; 14910). It is only full check, but also not a small one, so it can be used for post-factum checks, but not for interim ones.

Peter
 

mick

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Peter,

Yes I understand about the 630 check. I meant it as a negative test. If the sum is not 630 then it's not worth doing the full check. That's all.

I have access to one very large Mid-range IBM computer. Not PCs.

If the program was in C++ then we could run it.

What a shame! I feel your desperation and I share it.

Maybe I'll have to try to write a crude brute-force program to just hammer through the combinations giving myself re-start points like you do. It might take months.

Mick
 

peter

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Mick,

My program is in C++, maybe you should do slight changes to make it work under your system. But I'm not satisfied with my present algorithm still. Now I correspond with some mathematicians, hope they help me to create a good algorithm for seeking. What are characteristics of your IBM computer? I mean CPU frequency, RAM size etc.

Peter
 

mick

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Peter,

Please let us know if the mathematicians are able to help you to create the seeking algorithm.

The IBM computer has 60 Gb of memory and 4 processors.

Best wishes,

Mick
 

peter

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Mick,

Are you sure? 60 Gb of RAM memory???

But it is of not much interest. The main thing is the CPU frequency, or with what equivalent frequency these processors will work? 3 GHz? 8 GHz?

Peter
 

peter

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Mick,

Well, I subscribed on yahooGroup "Magic cubes" at here: http://groups.yahoo.com/group/magiccubes . I didn't mention Yi Jing and presented it as a pure mathematical problem. We'll examine for some time results that I already have - cubes-4 and "false" cubes-6. And I think this evening I'll send you my program in one C++ file (matr366.cpp) with some instructions.

With best regards,

Peter
 

mick

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Peter,

I checked the yahoo link but I don't understand a word that they're saying. You're way out of my league now.

I work for a major Investment Bank. They have one of the largest IBM mid-range installations in Europe. Their test machine has 60 gigabytes of memory.

IBM mid-range have their own standard measurement of CPU power called CPW. This is a commercial workload figure. (There is a CPU clock-speed but it's not used to measure mid-range Processing Power).

The test computer has over 650 CPW.

Thank you very much for offering to send me your C++ program. I'll let you know how I get on.

Best wishes,

Mick
 

peter

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Mick,

Don't worry, there are some specialized mathematical terms, but I already get used to them. I'll continue this thread though, so you (and silent readers too) won't be in a "deaf zone".

Now, while computers bluntly check and refuse lots of dead-end variants, I plan to examine and classify all cubes-4. Firstly, I have to pick "original" cubes-4 (and get rid of their rotations and reflections - 95 in total), and then examine them closely. I hope to find the most "perfect", so to say, seed for the cube-6.

With best regards,

Peter
 

arien

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hu.. justa quick question: what on earth are you guys talking about?

you seem to be having fun, though, grant you that
 

peter

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Arien, did you read the archive? We had a large conversation here. Do you know "hexagram chart 6x6" made on the base of King Wen hexagram sequence, or have I to put here this diagram?

Peter
 

mick

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Arien,

I was hoping that someone would ask this. Thank you


Here's an introduction to Peter?s ideas. Please let me know if you have any questions.

- - -

The 64 I Ching Hexagrams are patterns of Yin and Yang lines. They can be written as binary numbers so that, for example, Hx 1 could be written 111111, Hx 2 000000, Hx 3 100010, Hx 4 010001 and so on.

Eight of the Hexagrams are symmetrical. These are Hx 1 111111, Hx 2 000000, Hx 27 100001, Hx 28 011110, Hx 29 010010, Hx 30 101101, Hx 61 110011 and Hx 62 001100

Notice that each of the symmetrical Hexagrams is created by having the same trigram mirrored. So, for example, Hx 27 has the Mountain trigram on top and the mirror-image of the Mountain trigram below.

There are 56 non-symmetrical Hexagrams which can be arranged in pairs. For example, Hx 3 100010 and Hx 4 010001 are really the same pattern upside-down.

This means that the 56 non-symmetrical Hexagrams are really 28 basic patterns. These basic patterns are what Peter Manyakhin calls Invariants.

There are 36 invariants in the I Ching. 8 symmetrical patterns and 28 non-symmetrical patterns.

- - -

Since each invariant is a pattern of six lines, Peter has visualised a 6x6x6 cube which contains the 36 invariants.

Imagine that the cube is made up of transparent cells and each cell contains either Yin or Yang (0 or 1).

Peter is trying to find a cube that contains all 36 invariants whichever way you look at it.

Imagine a 2x2x2 cube. There are four 2-line patterns or Bigrams 00,01,10 and 11

So there are 3 invariants 00, 10 and 11. (The cube contains four groups of two so at least one invariant must be repeated in this cube.) We want to create a cube where all three invariants can be read whichever way you read the cube.

The cube can be read in 3 different ways. Each way corresponds to a different axis.

- - -

Let?s fill the front of the 2x2x2 cube from top to bottom like this

00
01

And the back of the cube from top to bottom like this

10
11

So looking at it and reading from left to right, the cube contains all 3 invariants

But you can also read the cube from top to bottom

Working from left to right, it now reads 00, 01, 11, 01 when you read from top to bottom.

So it contains all 3 invariants along this axis as well.

Finally, you can read the cube from front-to-back.

Again, working front-to-back from left to right it reads 01, 00, 01, 11 when you read from top to bottom

So this cube contains all 3 invariants whichever way you read it.

Peter is trying to find a 6x6x6 cube that contains all 36 invariants whichever way you read it.

- - -

I hope this gives you a way in to understanding Peter's ideas.

Let me know if you'd like me carry on with "Chapter 2".

Best wishes,

Mick
 

chrislofting

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You need to zoom down to the level of the dodecagrams (4096). When these are compressed into hexagrams we get 64 pure hexagrams (and so each reflecting a single dodecagram) and the remainder of the dodecagrams are represented by changing line patterns in each hexagram.

Apply the same rules and the 64 hexagrams compress into 8 pure trigrams, with changing line patterns in trigrams representing all of the other hexagrams.

Thus there is a line of continuity, of 'purity' for eight dodecagrams where they serve as both descriptors of a particular but also serve as summarising all of the rest in general - they are like hubs or 'strange attractors'.

When we follow the threads we end up with eight dodecagrams that compress into eight hexagrams that compress into eight trigrams (that can be compressed into yin and yang!) - and so a continuity in meaning at the particular besides the general.

Thus the QUALITATIVE roots of the trigrams are not in only the 'doubled trigram' hexagrams but are in fact:

111 - 111111 - hex 01
000 - 000000 - hex 02
110 - 111100 - hex 34
001 - 000011 - hex 20
101 - 110011 - hex 61
010 - 001100 - hex 62
100 - 110000 - hex 19
011 - 001111 - hex 33

If you focus at the hexagram only level, mapping the hexagrams down to trigrams will give 27 categories, 8 of which are the above and 19 are 'superpositions', made-up of hexagrams that share the same compression space. Draw this pattern out and you get - a wave interference pattern:

http://pages.prodigy.net/lofting/wave.jpg

this is also created in our experiments in quantum mechanics - simply put, use recursion and dichotomisation and indeterminacy and out pops these patterns - nothing to do with scale etc - the methods create the patterns.

Note that in recursion so each row maps to a geometric form in that each row contains qualities that serve as sources of meaning. Geometrically, to ensure the meaning, so the qualities are distanced from each other as best as possible and out of that growth dynamics comes tetrahedrons to dodecahedrons ;-) - distorted a little in that they are derived from the dynamics and as such are not 'ideal' forms but they are still 'archetypal'.

Chris.
 

peter

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Chris,

Are you sure you post it in the right thread? What here is about hexagram chart 6x6? I mean no offense, just why did you write it?

With best regards,

Peter
 

arien

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Mick, thanks for the lengthy explanation, I now understand what it is all about (without having to read the whole archive, as Peter sugests, sorry for being so lazy
)

I see what you term invariants to be the same 36 options that form the basis of the division of the hexagrams (1-30, 31-64), as Steve Marshall's refered article explained some time ago

So, Peter, is your program tring all possibilities in order to find one solution?
It sounds interesting, and who knows, sometimes great discoveries are made this way, by just asking "what if..."

I too understand that mathematical curiosity, and am prone to this sort of stuff, and although Im also of the opinion that the Yi is NOT mathematical, nor should be approached mathematically in its conventional use, this is still an interesting challenge, and if it wasnt there wouldnt be so many studiers doing all sorts of diagrams from the beginning of the times.

One of the interesting possibilities, is to find out that the Yi's structure is mathematically harmonious, even if that wasnt envisioned nor intended by its creators, who surely werent even aware of binary thinking. It still passes the test, somehow, like all masterpieces

Anyway, thanks once more for putting up with me, and please let us know of anything interesting you happen to find
 
C

cheiron

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Hi Mick

That was wonderfully clear even to a numbers dummy like myself.

Any chance of Chapter Two?

Thanks

--Kevin
 

mick

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Arien/Kevin,

Thanks very much. Chapter two is on its way soon.

Yes the invariants are the same 36 options that divide the Upper and Lower Canon as in Steve Marshall's article.

Peter's program doesn't hammer through all the possibilities. Peter is trying to find an algorithm that will search for a perfect cube without going down too many dead-ends.

I'll try to write some more tomorrow.

Best wishes,

Mick
 

chrislofting

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Peter,

you wrote:
>
> Chris,
>
> Are you sure you post it in the right thread? What here is
> about hexagram chart 6x6? I mean no offense, just why did
> you write it?
>

LOL! ;-) Perhaps you need to focus a bit more.... consider:

(1) What I sent, when compared to your initial conditions, showed TWO different perspectives giving the same initial 56/8 pattern in the context of 'pure' forms (8) vs mixed (56). (my post reflects the core trigram meaning generated from natural recursion of yin/yang. From there comes variations on the theme - as in your symmetry focus... see more below)

(2) ANY focus on pairs will reflect the use of recursion to derive those pairs. Your focus on symmetry in the image of the elements of a pair to allow you to compress information (two is mapped into one) shows that focus.

(3) In the I Ching, the binary sequence comes in pairs (using traditional numbers moving from 111111 to 000000 : 1,43 14,34 9,5 26,11 .... 20,8 23,2) as does the traditional sequence (1,2 3,4 5,6 .... 63,64). The manner of derivation is recursion applied to a particular dichotomy from a *semantic* viewpoint - thus the binary stems from qualities in pure yang vs pure yin, qualities *represented* by the hexagrams 01/02. The traditional stems from the qualities derived from recursing the qualities expressed in 01/64 (and so 01 is to 02 as 01 is to 64 and visa versa) - you can derive thousands of these sequences and in all will be the 56/8 pattern where for each sequence 8 will reflect 'pure' expressions, 56 will reflect mixed. As such, all of these sequences reflect 'variations on the theme' of the binary sequence - 111111....000000 that is itself a compression of the binary sequence at the 2^12 level (dodecagrams) : 111111111111 .... 000000000000.

(3a) In passing we note that each level of recursion comes with a geometric representation based on the number of qualities expressed at that level. Thus level 2 has four qualities (digrams/bigrams) that map to a tetrahedron where each quality, to assert its uniqueness, maximises the distance to the other qualities but cannot exceed the whole that encapsulates those qualities - thus four qualities will form a 'meaning space' that, observed from the outside takes on the form of a tetrahedron (pyramid) where this form maximises 'meaning' (the meaning space is 'within' the bounds of the walls) but without breaking the set of all relationships (although in nature we can distort - thus in taste the tetrahedron formed from sweet, sour, bitter, salt has 'distortions' on some of the surfaces - these forms derive from 'mindless' growth dynamics but the forms have then been idealised by us!)

(4) In the context of the I Ching, starting at the bottom, 4096 dodecagrams, as we can shift perspectives from 2^12 (where, BTW we have 4032/64) to 2^6 (hexagrams) so we can shift perspectives from working with 56/8 to move up a level in the recursion from 2^6 to 2^5. Thus we move to relationships that would appear to compress into 28/4 (not 28/6). Each SYMBOL you use reflects a quality and as such is NOT the quality, it is a metaphor - thus be wary about playing with bit patterns (0s, 1s) as if they are the thing, they are not. If you do you will get a skewed perspective of the IC.

(5) The focus on encoding qualities and using analogies etc means that, for example, in the traditional sequence the PAIR of 1,2 folds into a relational space at the 2^5 level that QUALITATIVELY reflects them sharing the same space (as compared to being in opposite spaces in the binary sequence). Ignoring this, and taking the IMAGE of 111111 vs 000000 as not being related as compared to the rotation of IMAGE of 3/4 etc distorts the qualitative focus that is the core focus in the IC. (if you want to do the reduction based on image rotation concept (and so 3,4 compress into one space) then you have to extend the 64 to 72 and work from there - THAT gives you the compression to 36 etc (and your dynamic in exclusion etc has led to that level regardless ;-)) but THAT is a distortion of the IC - you are trying to impose a triadic perspective on a diadic expression and so introduce distortion. Is there value there? ummm.. sort of ...

(6) Each LEVEL of meaning, each level of recursion, comes with a geometric representation based initially on maximising distances of one quality from all others. This is 'archetypal' expression, competitive (reflects leaves on plants forming into fibonacci patterns due to the geometry of trying to maximise exposure to sunlight etc In the IC the same pattern exists but the spirals are of squares! - all due to the rigid yin XOR yang focus - IOW the dynamics of fibonacci are reflected in the IC with this focus on maximising 'meaning' and overall reflects a bias to general recursive processes in any development)

(7) There is ONE focus that introduces distortion but also seems to retains 'meaning' and that is in qualities that form into pairs, and so elements are cooperative (we move from light vs dark to male + female), where the geometric distance between these qualities is reduced to emphasise the paired-ness but not too much as to de-emphasise the differences in expression. This gets into fractal dimensions in that the focus is on the space inbetween the levels of recursion. Thus at level 5 we have 32 qualities. At level 6 we have 64. Pairedness, represented geometrically, means a focus on a geometric representation in the 'middle' of 32/64 or more so showing the 64 but distorted into 32 obvious pairs. (that said, if you split the 32/64 we have 48 - and so we enter the realm of threes - IOW any focus on distortion will move into the realm of triadic as compared to diadic interpretations as a matter of method alone.

(7a) The problem here is that the 48 position reflects not pairs but -- 3/2 relationships. This is close to what we want geometrically but we have to retain the 64 and yet catch their dynamics to maximise distance just in time to represent their semantic paired-ness geometrically - 3/2 does not reflect the 'cut' between the elements of the pairs that ensures their unique expressions as well as retain their pairedness and so we have to reflect this in fractal forms - it is like trying to differentiate the cells from meitosis but retain their qualitative link as they being one cell overall but seen as if two and yet, taking one of those I can regenerate the whole.

(8) Recursion ensures that the traditional sequence appears to reflect the qualities encoded into the 1/64 dichotomy such that, when we focus on the PURE forms of yang(1)/yin(2), that 1/64 relationship is represented in the local ordering represented by 1, 2. (and this also leads to the PAIRS 1,2 and 3,4 reflecting this relationship where now (1,2) is like 1, (3,4) is like 64 etc etc etc - for the numbering, recursion etc see http://pages.prodigy.net/lofting/cracked.html)

(9) The traditional sequence overall is of analogies, of 'like' relationships, such as of 1 to 2, 3/4, 5/6 and on into 63/64. (qualities included in 1/64 are individual/social, pure/mixed and synonyms - in this sequence 1,2 are cooperative, in the binary they are interpretable as competitive, the distance between them is 62 hexagrams as compared to the trad where there is no hexagram between them).

(10) the binary sequence is also about analogies and maps the universal relationship of 1/2 where the similar, local, relationships are 1/43, 14/34 .... 20/8, 23/2 - the competition is in the subtle differences, thus 1 is differentiating, 43 is integrating and so reflects qualities of 2 but still in a general context of yang etc etc. Visually, using recursion you can 'see' the pairing even when looking at an individual hexagram. THAT needs to be represented geometrically rather than in idealised, archetypal forms, of maximising distance for all symbols.

Now, all of this in fact reflects the compression of 4096 forms into 64 such that applying geometry at the level of 64, and distorting it a touch, will distort the 'big picture'. Working with recursion means you retain the overall structure but just get more general as you move away from the base 4096. Any geometric representations must capture all of the above, you cannot wander-off into triadic interpretations and build a geometry from there, it will not be 'complete'. At the same time, sticking rigidly to dyadic interpretations, disallowing for close associations and so a degree of distortion, means you can also miss elements of the IC. ... so... my comments on needing to go to the level of dodecagrams and work from there, understanding recursion and its mapping geometrically and how to take the 'root form' and 'compress' it without distorting it to a degree where you lose perspectives - all you should lose is resolution in perspectives, not the perspectives outright.

Chris.
 

peter

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Chris,

I almost didn't doubt that you'd finish with calling to your favourite dodecagrams. But sorry, I have nothing to do with them. Triadic approach instead of diadic? Well, 2^6 already gives me this triadness. And 2^6 fits this cube.

About missing some elements: well, I don't want you to get rid of all the traditional knowledge, and also of many "new age" additions. But if youwant to find something completely new, you simply have to put this heritage aside for a while. I wand to find some "new dimension". I don't play with meanings, so while I'm on firm mathematical ground, I feel good.

Please, don't write so much text without direct connection to the topic.

With best regards,

Peter
 

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