Hexagram Chart 6x6?

[Frankelmick]

Arien/Kevin,

Chapter Two

Peter has devised a formula that numbers the invariants from 0 to 35 and converts a hexagram into its invariant number.

The formula splits the hexagram into two trigrams, the upper trigram (lines 4,5,6) and the mirror image of the lower trigram (lines 3,2,1).

The trigrams are numbered in sinking Yang sequence

Earth = 0
Mountain = 1
Water = 2
Wind = 3
Thunder = 4
Fire = 5
Lake = 6
Heaven = 7

The first step in Peter?s formula is to compare the upper trigram and the mirror-lower trigram and see which is the higher number. Let?s call the higher numbered one Max. Let?s call the smaller number Min.

Then use this formula

Invariant = [(Max x (Max + 1))/2] + Min

Using this formula, any hexagram is converted into an invariant numbered from 0 to 35.

For example, to convert Hx 3 100010 into its invariant number the Upper Trigram is Water = 2. The Mirror-Lower is Mountain = 1. So Max = 2, Min =1.

The Hx 3 Invaraint = [(2 x (2+1))/2] + 1
= [(2 x 3)/2] + 1 = 3 + 1 = 4

Notice that when you convert Hx 4 010001 using the same formula, you get the same invariant number. This time Upper = Mountain = 1, Mirror-Lower = Water = 2. So Max = 2, Min =1 again.

The symmetrical Hexagrams are mirror-doubled Hexagrams. For example, Hx 27 has Mountain as the Upper Trigram and Mountain as the Mirror-Lower trigram. In this case, Max = 1 and Min = 1. The Invariant is [(1 x 2)/2 ] + 1 = 2.

I thoroughly recommend that you have a go at putting together a spreadsheet that converts Hexagrams into Invariants.

Set up an 8 by 8 square of Trigrams and you?ll find that the invariants form a beautiful pattern.

Best wishes,

Mick

 

[cheiron]

Thanks Mick

Lovely and clear - wonderful.

With Homework too...

(Grin)

I will do that on a spread sheet... what type of value should each cell show? (binomial number? Hx. Number?)

This might be a silly question: Why only work with the invariants?

Also - Will this 6X6X6 cube demonstrate something, will it represent something? Sorry to ask that, I know that maths looks at patterns and not always at meanings.

Thanks

--Kevin

 

[Frankelmick]

Kevin,

Thanks very much

I put the 8 trigrams numbered 0 to 7 across the top and down the side of an 8 x 8 grid. Then I applied Peter's formula to each of the 64 cells in the 8 x 8 grid and I ended up with a number between 0 and 35 in each cell.

It gave me a nice way-in to Peter's ideas.

You said, "This might be a silly question: Why only work with the invariants?"

I dunno. This thread is about the 36 invariants. It seems to be enough for the moment

Then you ask, "Will this 6X6X6 cube demonstrate something, will it represent something? Sorry to ask that, I know that maths looks at patterns and not always at meanings."

No need to apologise. I think I agree with Peter when he says, "I don't play with meanings, so while I'm on firm mathematical ground, I feel good."

For the moment, this is really a Pure Maths problem.

Hope this helps. I'm working on Chapter Three.

Best wishes,

Mick

 

[cheiron]

Mick

That's great... Looking forward to chapter three.

Thanks for your effort and patience on this.

--Kevin

 

[peter]

Mick,

Thank you for interpreting my ideas! I have little time for it, while I'm deep in them and sometimes can't see what is wrong with my explanations.

I have no clear answer on the question "what is this cube for?" For me it is just interesting to find one (at least one, but from it we automatically can create others by rotating and reflecting) and examine it. Maybe I'll find some things that in their turn will be more "mathematical" than "intuitional" or what else.

About cubes 4x4x4: I found that there are exactly 41472 such "perfect" cubes. I'll specify amount of variations (rotating and reflecting) and tell how many different configurations we have among them.

With best regards,

Peter

 

[Frankelmick]

Peter,

Over 41,000 perfect 4x4x4 cubes! Surely there must be a perfect 6x6x6 cube.

I enjoy trying to explain your ideas, also it helps to clarify them in my mind. I just talked about your ideas to a friend of mine who's a Maths teacher, he knows someone who's a Maths graduate from Cambridge University currently doing research in London. I'll be very interested in what he has to say.

Kevin,

Here's Chapter Three.

Square numbers are numbers that can be calculated by multiplying a number by itself. For example, 4 is a square number because it?s 2 times 2.

Triangular numbers are numbers that are the sum of a sequence of numbers starting from 1.

Snooker players know that 15 is a triangular number because there are 15 reds that you set up in a triangular shape at the start of a frame.
15 = 1 + 2 + 3 + 4 + 5

36 is a square number because 6 x 6 = 36 and 36 is also a triangular number because
1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 = 36

This means that the 36 invariants can be arranged in a triangle which Peter has done in his post on Monday, January 12, 2004 - 08:22 pm

Peter also extended this to a triangular arrangement of the I Ching in his post on Saturday, January 17, 2004 - 03:18 pm

The triangular arrangement of the invariants gave us some clues about trying to create an algorithm to find the 6x6x6 cube.

Peter noticed something about the pairs of lines 3 & 4 in the triangular arrangement. There are 10 invariants where lines 3 & 4 are both Yang. There are 10 invariants where lines 3 & 4 are both Yin. And there are 16 invariants where lines 3 & 4 have 1 Yin line and 1 Yang line.

I noticed that this property was true for line pairs 2 & 5 and 1 & 6 as well. 10 invariants where the pair has 2 Yang lines, 10 where the pair has 2 Yin lines and 16 where the pair has 1 Yang line and 1 Yin line.

This means that on any axis of the 6x6x6 cube, the cells that represent a line must contain at least 10 Yang cells and at least 10 Yin cells. The most uneven distributions possible are 10 Yin and 26 Yang or 10 Yang and 26 Yin lines. Peter calls these the polar distributions.

So as you try to build a 6x6x6 cube, as soon as you find 27 Yin cells or 27 Yang cells on one line of any axis, you know that this version of the cube won?t contain all 36 invariants along this axis. It can?t be a perfect cube.

This avoids a lot of dead ends but Peter has still not been able to find a perfect 6x6x6 cube.

There are millions and millions of possible 6x6x6 cubes so to try to find a short-cut, Peter worked on finding a perfect 4x4x4 cube and then seeing if the 4x4x4 cube might somehow form the basis of a 6x6x6 cube.

That's pretty much where we are now in this thread. Hope this makes sense.

Best wishes,

Mick

 

[peter]

Mick,

It is very interesting to me too - to know what that mathematician will say! It's a pity that I don't know anybody among my friends who'd know a professional mathematician...

Yes, this number of 41472 can seem to be large, but all variants of cubes 4x4x4 are about 20 trillions (millions of millions), so these 40,000 are a mere 2e-7% (i.e. 2 * 10^-7 %). Interesting result also, that you can find all of them having simply one perfect cube 4x4x4 - by permutating shears and reflecting. One guy from "Magic cubes" already got all of them (because I'd sent him my first 1080 cubes, and he derived all the rest).

About my program with a seed: well, it tends to be very long to seek... I estimated that it must seek 2 years of continuous work.

I want also to remind about "quadruplets" - four digits that are made of 2 symmetrical pairs, say (3&4) and (2&5). There must be exactly 3 of each symmetrical quadruplet and not more than 4 of invariants of all the rest 12 non-symmetrical quadruplets.

With best regards,

Peter

 

[Frankelmick]

Peter,

I suppose 41,472 isn't really that big a number then. But 2 years does sound like a long time.

Let's hope the mathematicians come up with something.

In the mean time, I thought I'd try to test out the computer power at work by writing a simple program to loop through all the possible permutations of 36 numbers. Then I'll run the program overnight to see how far it gets.

At least that might give me a benchmark for how long it might take using "brute force".

Thanks for the reminder about the "quadruplets". I'll try to write up an explanation with examples.

By the way, after writing up my notes on the polar distribution, I worked out that the lower number, 10, is the number of invariants <FONT FACE="times new roman">(I)</FONT> at the 4-level

I think that for the invariants with n lines, the lower number in the polar distribution is <FONT FACE="times new roman">I(n-2)</FONT>.

Best wishes,

Mick

 

[Frankelmick]

Kevin,

Here's Chapter Four. This ones quite tricky to explain. Let me know how you get on.

Let?s take a step back to 4-line invariant patterns. There are 16 possible 4-line patterns, 4 of which are symmetrical.

0000, 0110, 1001 and 1111.

This means that the 12 non-symmetrical patterns can be grouped in pairs which are the upside-down versions of each other.

This makes 10 4-line invariants in all; 4 symmetrical and 6 non-symmetrical.

I thought about this and came up with this formula which seems to work nicely.

For the set of all patterns of n lines, the number of symmetrical patterns (s) is

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

The ^ symbol means raised to the power of. The mod2n term will be zero if n is an even number and 1 if n is an odd number. For example, for the trigrams where n=3,
s = 2 ^ [(3+1)/2] = 2 ^ 2 = 4

The 4 symmetrical trigrams are Earth, Heaven, Fire and Water.

From this formula, the number of invariants <FONT FACE="times new roman">(I)</FONT> in the set of all patterns of n lines is

<FONT FACE="times new roman"> I = s + (2^n ? s)/2</FONT>

So, let?s take a symmetrical 4-line invariant. For example 1111.

The core of a hexagram is lines 2,3,4 & 5. So the four lines 1111 form the core of 3 6-line invariants. 2 symmetrical invariants Hx 1 and Hx 28 and 1 non-symmetrical invariant Hx 43 which is the same pattern as Hx 44. (From now on I?ll only name the first one in King Wen sequence.)

But, a non-symmetrical 4-line invariant forms the core of 4 6-line invariants, all of which are non-symmetrical.

For example the four lines 0111 form the core of these 4 non-symmetrical 6-line invariants.

Hx 13 101111, Hx 31 001110, Hx 33 001111 and Hx 49 101110.

This gives Peter another check to avoid going down too many dead-ends.

A quadruplet represents 2 pairs of lines such as 3 & 4, 2 & 5. As he builds the 6x6x6 cube, Peter checks that there?s exactly 3 of each symmetrical quadruplet and not more than 4 of the other non-symmetrical invariants.

I hope this makes sense.

Best wishes,

Mick

 

[cheiron]

Thanks Mick

Need to catch this one tomorrow..

Cheers

--Kevin

 

[cheiron]

Hi Mick

Chapter 3 ? yes lovely.

Chapter 4 ? Hmm? I can see what you mean about checks and the problem Peter is trying to shortcut? Good enough for me.

It is a very interesting problem? But I think it does lead somewhere too.

If you can tolerate some non math thought here? and I do not mean to distract you guys from your task.

It seems to me that the Yijing is a lot more than the linear representation which a book might suggest. If it represents a reality then it must have a shape. Otherwise life would be random? maybe it is? Maybe that is what you guy?s are working out.

When you get through with this piece of work could one of you, or both, comment on the idea that the Yijing is perhaps representative of or might be modelled by a wave.

I sometimes imagine it as a wave with interference patterns and harmonics with the particular Hexagrams representing 64 points along it? sort of static snapshots of the continuum.

But as I say? this is an aside to your current task.

I wish you both well with this and thanks for you patience.

Looking forward to hearing your results.

Thanks

--Kevin

 

[citizenkoine]

I'll go out on a limb, and say that I have only scanned this thread, but more doggedly than its predecessor. There are many beautiful ideas farting off in the wake of this thing.

Rather than solid cubes, could you approach this as a set of three shells?: a 2x2x2cel cube, a 4x4x4cel shell, and a 6x6x6cel shell?

Another consideration, could be that your cube, whatever you end up with, will just be a frozen instance within a continuous matrix; i.e, a 2x2x2cel-metacel occupying the 'top/right/front' corner of your cube could also be occupying the 'bottom/left/back' corner, or the 'center/top/front' vertex-position of another/other 3x3x3metacel-cube-recombinations, which surround each cel/cel-combination in three dimensions. Long sentence. I am dubious that the Yi-space can be 'frozen' into a single, 'magic', crystalline cube. Show me my folly (besides having not enough patience to read algorithms).

It does occur to me though, from a visual apparatus, to 'check' your cube from one corner to its opposite, diagonally, across the center of the cube, or 'orthogonally'. x8(corner-directions).

In the spirit of a hot-poker-to-the-scapula,
-koine

 

[citizenkoine]

meant edge-, not vertex-.

-koine

 

[peter]

Kevin,

When speaking about "Yi Jing wave", maybe you mean that "yang"s are crests and "yin"s - troughs? We can create a pattern in different ways. I've ssen somewhere similar idea, but it was not specific too.

----------------

koine,

these sets of shells are under my sight now. No specific 2x2x2 inner cube can be made - they seem to be all equal, I believe it simply must be balanced (4 "1" & 4 "0") or near-balanced (3 or 5 "1" & 5 or 3 "0"). About 4x4x4 cube - I try now to receive some results with some perfect cubes-4 as seeds, as you can see in this thread, but in vain so far. I think I must firstly check these cubes-4 somehow, maybe with fractal analysis (does anybody here have some experience in such analysis?)... And then run seeking with some particular seeds, not intuitively picked.

Why do you think that this crystal is "frozen"? Personally I believe that if this crystal will be made (or "when be made"?), we'll be able to say that the life and existence are highly organized things. Doesn't it nears us to God? (At least those of us who believe they're not with God.) As you may know, "under" human life ancient alchemists also spoke about "animal life", "plant life" and "crystal life" (or "mineral life"). Nothing prevents crystals to be "living", though it is not a form of life that is "near" to ours.

BTW, just now I found an interesting link about crystals - diamonds especially: http://cfa-www.harvard.edu/press/pr0407.html (a whole star of diamond about 4000 km in diameter).

With best regards,

Peter

 

[citizenkoine]

Peter,
-Processing.
Will rejoin at the next opportunity.
thanx.
-koine

 

[cheiron]

Thanks Mick - Perhaps - Line positions need accountingo too - I will think on it - Thanks for taking the trouble to explain.

Now is the time for me to listen.

Wishing you the best

--Kevin

 

[cheiron]

Oh, Sorry - That was from Peter.

Van Den Berghe makes a sort of a stab at it.

http://www.ping.be/icrea/explan.html

All the best

--Kevin

 

[yellowblue]

Brad has a wonderful way with teaching...and with sites and ambiance...

You have always been a great source of inspiration for me Brad. Most times, and especially in the beginning off the beaten path... but that is the catalyst for learning and becoming...

Thanks for all of it,

Deb

 

[peter]

Kevin,

I've seen this stuff. Interesting, yes, but not thorough. Gives some food for thoughts nevertheless.

We can transform Yi Jing sequence into wave form by different ways. Then examine it using Fourier analysis and usual mathematical apparatus for such cases...

Later today I'll post one of "false" cubes - which have all hexagrams along 2 of 3 axes.

Peter

 

[peter]

So, as I promised, I post here this picture.

Top side is the upper square 6x6 (black colour) - lines were gathered into hexagrams from bottom to top. Front side is the lower square (blue colour) - lines here were gathered into hexagrams from front to rear. Along horisontal axis there are repeats - just look at the black diagram, and you see that top lines in the first hexagram row form hex #17 (or #18 if you read from left to right), and top lines n the second row give us the same hexagram. But repeats here are not often - at least, not so often as it were in versions of my program with less severe checks. So, it is just a particle example of what I seek.

Peter

 

[peter]

Oh, I mistook a bit - that diamond star is only 1500 km in diameter. Big nevertheless.

Mick, did you receive any response after that talking with your friend - a math teacher?

With best regards,

Peter

 

[Frankelmick]

Peter,

Thanks for posting the near-miss cube. I see where Hx 18 is repeated.

I'll talk to my friend this weekend.

I've been trying to follow the discussions on the Magiccubes Yahoo group. It's hard for me to understand but are some of them saying that they don't think that a solution is possible?

Best wishes,

Mick

 

[peter]

Mick,

I asked one guy from Novosibirsk - he studies mathematics good, and in Novosibirsk there is a powerful scientific school. I hope he'll be able to find some hints.

Really, there are only 2 active members (except me) in "Magic cubes" group, as you could see. They simply don't see why we should prefer any variant - whether it exists or not. I see that there are only people who use combinatorical ways, just like me. I really like those variants of perfect cubes-4 and 48 transformation matrices, but... They don't near me significantly. Gave some useful instruments though.

My latest thing found: I applied that idea about rotating and permutating shears along already perfect axis, and got an evident provement of idea of symmetrical shears - only those cubes kept uniqueness along this axis intact, which kept symmetry of originally symmetrical shears. In original cube-6 we have symmetrical pairs (0-5), (1-4) and (2-3), and they go as 0-1-2-3-4-5; Then another 7 cubes with shear 0 in the first place will preserve uniqueness along this axis:
0-1-3-2-4-5
0-4-2-3-1-5
0-4-3-2-1-5
0-2-1-4-3-5
0-2-4-1-3-5
0-3-1-4-2-5
0-3-4-1-2-5
You see? Pairs don't part, they only change positions. So if we now multiply this result on 6 (amount of shears), we'll get 48 - total amount of permuted, but not less unique cubes.

As I said, I also checked rotations of shears, but they led nowhere - all cubes with unique hexagrams permuted had shears rotated in the same way. So I got 192 "intact" cubes, and the total amount of checks was 2949120 (it is 6! * 4^6), percent is horribly tiny: 0,00065%.

Well, if I try now to make some permuted cubes and then try to gain a "perfect" cube with them?..

With best regards,

Peter

P.S. When I wrote a letter to that mathematician, I found that my previous formulations was not good... No wonder that it could bot be understood by many people...

 

[peter]

Well, my attempt ended with nothing, because I simply checked symmetrical shears along "bad" axis - along which we don't have unique hexagrams. Of course, there weren't necessary pairs: 10 of "00" and 10 of "11" (the rest 16 will occur automatically).

Peter

 

[Frankelmick]

Peter,

Let's hope the mathematicians come back with some ideas. Novosibirsk sounds like the perfect place to find a solution to the 6x6x6 cube

By the way, please feel free to make use of my explanations to Kevin if that would help to introduce your ideas to other people.

Best wishes,

Mick

 

[peter]

Mick,

How is your friend? I understand that he had little to say, so you didn't post his answer here. That mathematician from Novosibirsk is silent too.

If we'll find at least 1 cube, we'll automatically get about 5 millions of them using various transformations (it is 48^4 - 48 for all reflections/rotations and 48 for shears' mutating along each axis independently).

I didn't sort those cubes-4 yet, plan to do it on the next week, but I have an idea how to analyze them, so I'll simply program it soon. And give results of it - which cube is the most homohenous and which is the least.

With best regards,

Peter

 

[Frankelmick]

Peter,

I saw my friend last weekend but the Mathematician (his nephew) is away at a conference at the moment.

I should see my friend this weekend and I'll let you know what happens.

Novosibirsk must be pretty cold at this time of year. Maybe you'll hear something in the Spring

In the meantime, I thought I'd write a benchmark program to test my computer's processing power.

The program just does lots of nested loops like this

Do 36 times
Do 35 times
Do 34 times...
Do 2
what's the time?
Enddo
Endoo etc...

It's been running for a day and a half and so far it's used 131,355,898 CPU milliseconds

Looking forward to seeing the Cube-4 results.

Best wishes,

Mick

 

[Frankelmick]

Peter,

I didn't get a chance to talk to my Maths teacher friend this weekend.

It sounds like one of the people on the MagicCubes list thinks that there is no solution to the 6x6x6 cube. That would be a shame.

If that's true, I suppose the mathematicians will want proof that there's no solution.

What's your impression? Do you think that a solution is still possible?

Best wishes,

Mick

 

[peter]

Mick,

Well, I didn't think that they'd be as eager in finding the solution as me. I already gave those figures: 1e50 cubes totally, maybe 1e30 cubes with 2 axes "perfect", but nobody knows how many there must be totally "perfect" cubes. But the percent will be very-very small.

Of course, I'd like to see a strict proof that there are no solution of this problem. Without it I'll continue seeking the solution. I simply believe that the solution exists. Yes, it is hard to find, but I'm not disappointed yet.

With best regards,

Peter

 

[peter]

Mick,

Well, I made some calculations for these cubes-4 and found 251 variants. They don't depend on orientation of these cubes (but somewhy amounts of cubes of each variant are multiples to 24, not 48 as I expected - 48 is amount of all transformations including rotations and reflections). Firstly, for all these cubes there are only 5 variants of dispersion from average amount of "1" in cubes-2 (I took 8 cubes-2 - in every vertex), which is "4" for all. Dispersion is "(mean - s1)^2 + (mean - s2)^2 + ... + (mean - s8)^2". Then I examined 8 cubes-3 - in every vertex too, and all three criterions: dispersion of cubes-2, mean of cubes-3 and dispersion of cubes-3 - gave me those 251 variants. And now I want to find those variants which will give me my "perfect" cube-6 as quickly as possible, and their oprientations too.

Well, if you need more details on these statistics, we can correspond privately (if nobody else needs this mathematics).

With best regards,

Peter