[T5 = ({0, +1, x, y, -1}; +, -; 0, +1, -1)] - A New Kind of Science: The NKS ForumA New Kind of Science: The NKS Forum
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T5 = ({0, +1, x, y, -1}; +, -; 0, +1, -1)
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Posted by: Lawrence J. Thaden
Figure 1 lists a set of 27 three color rules that correspond to algebraic expressions for a ternary logic in two variables:
T5 = ({0, +1, x, y, -1}; +, -; 0, +1, -1)
where the operations, + and -, are modulo 3 sum and difference and
where the three identities are,
0: additive identity
+1: positive multiplicative identity and
-1: negative multiplicative identity.
T5 is a closed set. It is an example of a nondegenerate, complemented, distributive lattice.
These algebraic expressions can be entered in Mathematica and will perform T5 ternary logic operations provided they are specified along with a set of rules to ensure absorptive and modulo properties are preserved.
For example, if the list of algebraic expressions are assigned to the symbol t5, a list of the results for modulo 3 plus: x[] + y[[i]] for all i in t5 can be generated with rules:
{2 -> -1, -2 -> 1, 2 x -> -x, 2 y -> -y}.
Moreover, the results correspond to those obtained by use of the map function described in these posts.
Of course, the map function results in rule numbers, not algebraic expressions.
However, there is a correspondence. It is found between the mapped rule number result and the position it holds in the list of algebraic expressions.
So, if the set of rule numbers are assigned to the symbol t5Subset, executing the following code in [i]Mathematica gives {True}.
Union[(Table[(t5[[i]] + t5[[#]]&/@Range[Length[t5]]/.{2->-1, -2->1, 2 x->-x, 2 y->-y}) == (t5[[#]]&/@(Flatten[Position[t5Subset, #]&/@(map[t5Subset[[10]], t5Subset[[i]], t5Subset[[#]], dontcare]&/@Range[Length[t5Subset]])])), {i, Length[t5]}])]
where t5Subset[[10]] identifies the rule corresponding to the algebraic expression x + y.
Similar tests can be constructed for correspondence between three valued rule numbers and other simple algebraic expressions that are modulo 3 sums and differences, such as (-x + y) and (-x - y).
But what about compound algebraic expressions like (1 + x + y)?
How does one algebraically simplify something like (-x ~ (1 + x + y) ~ y), where the infix operator, (1 + x + y), is not simply modulo 3 + or -?
Personally, I do not know if there is a way to do it.
But with the mapping function we do have a means of finding all t5[[i]] ~ t5[[i]] ~ t5[[i]], where i ranges over the 27 algebraic expressions.
Figure 3 presents so called cross product projections, compilations of all possible combinations of these logic expressions.
There are 27 tables each with 27 rows and 27 columns. Each table is labeled with one of the algebraic expressions. That table contains all (t5 x t5) -> t5 when the operator is as labeled.
To simplify presentation, each algebraic expression has been assigned a color. The legend for the colors is in figure 2.
Notice that the tables for operators that have both x and y in their algebraic expression have a diagonal appearance, while those with only one of the variables has a horizontal or vertical appearance.
And these tables are related to each other through various rotations about the diagonal.
For example, (-x - y + 1) is equivalent to (x + y -1) when you rotate either one 180 degrees about the upper right to lower left diagonal.
In contrast to this, the tables with single variable expressions appear to be related by reversing the order of columns and rows.
For example, (1 - y) is equivalent to y when the order of the columns of either is reversed.
The complete list of equivalences by rotations about axes is given in the table at the bottom of this post.
Single variable algebraic expressions (horizontal and vertical appearance) are said to propagate over the columns or down the rows.
The three identities have homogeneous colors. As operators they propagate both over columns and down rows.
Note that, because these are tables, the format is such that the x axis is vertical ascending from the top to bottom of each figure. And the y axis is horizontal ascending from left to right.
Following is a table showing how the cross product projections relate to one another.
The numbers in parentheses identify the order in the list of T5 members. The first algebraic expression identifies an entire table and is the label at the top of one of the plot entries in Figure 3. “Rotation Axes” names the action to be performed on the first algebraic expression in order to convert it to the second algebraic expression. When “Horz + Vert” is specified both rotations must occur in sequence. For “Horz + Vert” the order of rotations does not matter. Nor does it matter if you start with the second algebraic expression and perform the rotations on it. You will end up with the first algebraic expression. That is, the commutative property applies.
Table of Equivalences
...Algebraic.......Rotation..........Algebraic
...Expression.......Axes.............Expression
( 2).........x - 1......Horizontal.................-x...(26)
( 3).........1 - x......Horizontal..................x...(15)
( 4).........y - 1......Vertical.....................-y...(24)
( 5)...x + y +1......Horz + Vert.........-x - y...(10)
( 6).........y - x......Horz + Vert...........x - y...( 8)
( 7).........1 - y......Vertical.......................y..(17)
( 9)...-x - y -1......Horz + Vert...........x + y..(19)
(11)......-y - 1......Vertical..................y + 1..(21)
(12)..x - y + 1.....Horz + Vert....-x + y + 1..(16)
(13).......-x - 1.....Horizontal..............x + 1..(25)
(18)..x + y - 1.....Horz + Vert.....-x - y + 1..(23)
(20)..-x + y -1.....Horz + Vert......x - y - 1...(22)
Edit:
I made a mistake in saying:
"Moreover, the results correspond to those obtained by use of the map function described in these posts. "
The correspondence is only one to one and onto if the method using symbols has (x+y) as operation and the method using the map function has (-x -y) as operation.
Posted by: Lawrence J. Thaden
One might ask: How do you algebraically express what happens when an entry in the left hand side of the table turns into the corresponding entry on the right hand side of the table?
Or in terms of the mapping function: What operation acts on the left hand side as the first operand and what is the second operand in order to map to the right hand side?
It turns out that both the first and second operands are the same, and there are three possible operations for each pair of algebraic expressions. These operations have the commutative property.
But these three possible operations are grouped into three types and each type is associated with exactly four of the pairs of algebraic expressions.
Let us call the types: alpha, beta, and gamma; and the operations alpha-1, alpha-2, etc.
Then rotations or flips are accomplished by letting one of the three operations from a given column of the following table act on the left hand side algebraic expression taken with itself.
Alpha flips:.........Beta flips:.........Gamma flips:
(2) -1 + x.............(21) 1 + y..........(10) -x - y
(4) -1 + y.............(23) 1 - x - y......(15) x
(9) -1 - x - y.........(25) 1 + x..........(17) y
Any one of the operations within the type will do.
So, for example:
(-x + y) ~(Gamma-3)~ (-x + y) -> (x - y)
says (-x + y) flips to (x - y) when ~(y)~ operates on (-x + y) taken with itself.
Here is a list of the algebraic expressions, the type of rotation and flip, and the expression after the flip.
( 2).........-1 + x......Horizontal......Alpha flip...............-x...(26)
( 3).........1 - x........Horizontal......Beta flip...................x...(15)
( 4).........-1 + y.....Vertical...........Alpha flip...............-y...(24)
( 5)...1 + x + y......Horz + Vert.....Beta flip............-x - y...(10)
( 6).........-x + y.....Horz + Vert.....Gamma flip.........x - y...( 8)
( 7).........1 - y.......Vertical............Beta flip...................y...(17)
( 9)...-1 - x - y.......Horz + Vert....Alpha flip...........x + y...(19)
(11)......-1 - y........Vertical..........Gamma flip.........1 + y...(21)
(12)...1+ x - y.......Horz + Vert....Alpha flip......1 - x + y...(16)
(13)......-1 - x........Horizontal......Gamma flip.........1 + x...(25)
(18)...-1 + x + y...Horz + Vert....Gamma flip.....1 - x - y...(23)
(20)...-1 -x + y.....Horz + Vert....Beta flip........-1 + x - y...(22)
It should be noted that the operation is on all 729 cells in the table simultaneously.
That each cell flips is amazing, but why there are three possible operations per type is puzzling. Perhaps it is required in a wider algebraic context, one where it does not matter to this part but is crucial to the wider context.
Rotations or flips call to mind mechanisms for polarizing light.
Posted by: Lawrence J. Thaden
I found this passage from Lee Smolin’s book “The Trouble with Physics” apropos:
“One way to unify things that appear different is to show that the apparent difference is due to the difference in the perspective of the observers. A distinction that was previously considered absolute becomes relative. This kind of unification is rare and represents the highest form of scientific creativity. When it is achieved, it radically alters our view of the world.”
On page 883 of the NKS Book, Wolfram presented us an example of this kind of unification in his table of basic equivalences between elementary cellular automaton rules.
Although not following exactly in his footsteps, the equivalences of three color rules in this post are intended as another example along the lines of what Wolfram did.
How can this instance of Wolfram’s “scientific creativity” be said to “radically alter our view of the world”?
There are many details in NKS that are yet to be elucidated.
Posted by: Lawrence J. Thaden
Just as it is possible to flip the tables in the previous posts in this thread, it is possible to flip CA cells.
Three color CA cells have as content: {0, 1, 2}.
An arrangement of these digits makes up the initial conditions.
And another arrangement of them makes up each line of output as the rule works on them.
Since Wolfram lists the ECA rules together with their logic expression on page 884 of the NKS Book, why not be consistent and list the input to the rules as logic expressions?
For these three color rules the logic expressions of the input are:
0: Additive Identity
1: NOR
2: - NOR
And the rule acting on these inputs generates some arrangement of the same logic expressions. So the output from a three color CA is always made up of cells that contain either 0: Additive Identity, 1: NOR, or 2: - NOR.
But we have seen that conversion is possible for a set of cells by means of the Alpha, Beta, and Gamma Flips. So we can generate cell content that is not limited to {0, 1, 2}.
As an example, take rule 7348211845533: (x + y) together with this arrangement of 0: Additive Identity, 1: NOR, and 2: - NOR as initial conditions:
{2,2,2,2,2,2,2,2,2,0,0,0,0,0,0,0,0,0,1,1,1,1,1,1,1,1,1,2,2,2,1,1,1,0,0,0,1,1,1,0,0,0,2,2,2,0,0,0,2,2,2,1,1
,1,2,2,2,0,0,0,1,1,1,0,0,0,1,1,1,2,2,2,1,1,1,2,2,2,0,0,0,1,1,1,2,2,2,0,0,0,1,1,1,2,2,2,0,0,0,1,1,1,2,2,2,0
,0,0,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,0,0,0,2,2,2,2,2,2,1,1,1,0,0,0,0,0,0,2,2,2
,1,1,1,0,0,0,2,2,2,1,1,1,2,2,2,1,1,1,0,0,0,1,1,1,0,0,0,2,2,2,0,0,0,1,1,1,2,2,2,1,1,1,2,2,2,0,0,0,2,2,2,0,0
,0,1,1,1,0,0,0,0,0,0,0,0,0,2,2,2,2,2,2,2,2,2,1,1,1,1,1,1,1,1,1,2,2,2,2,2,2,2,2,2,0,0,0,0,0,0,0,0,0,1,1,1,1
,1,1,1,1,1,2,2,2,1,1,1,0,0,0,1,1,1,0,0,0,2,2,2,0,0,0,2,2,2,1,1,1,2,2,2,0,0,0,1,1,1,0,0,0,1,1,1,2,2,2,1,1,1
,2,2,2,0,0,0,1,1,1,2,2,2,0,0,0,1,1,1,2,2,2,0,0,0,1,1,1,2,2,2,0,0,0,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1
,1,1,1,1,1,1,1,1,1,1,0,0,0,2,2,2,2,2,2,1,1,1,0,0,0,0,0,0,2,2,2,1,1,1,0,0,0,2,2,2,1,1,1,2,2,2,1,1,1,0,0,0,1
,1,1,0,0,0,2,2,2,0,0,0,1,1,1,2,2,2,1,1,1,2,2,2,0,0,0,2,2,2,0,0,0,1,1,1,0,0,0,0,0,0,0,0,0,2,2,2,2,2,2,2,2,2
,1,1,1,1,1,1,1,1,1}
If you run the CA for 242 steps you get the first cycle of an endless cycling CA. (See Figure 1.)
Now what do we get from an Alpha-1 Flip on the contents of each cell?
Recall that an Alpha-1 Flip is a mapping of rule 193720085: (-1 + x) onto each cell taken with itself. After such a mapping the contents of the cells are one of the following rules:
7625597484986: -1
7625597484985: OR
7625597484984: 1 - OR.
Since ArrayPlot does not differentiate colors for such high closely spaced rule numbers, we can translate them to {0, 1, 2} by:
{7625597484984 -> 0, 7625597484985 -> 1, 7625597484986 -> 2}.
Then ArrayPlot can be used to make the Alpha-1 Flipped plot in Figure 2.
An inspection shows that the grey cells do not change when the cells are flipped and that the white and black cells exchange colors when the cells are flipped.
Stated in terms of the algebraic expressions, NOR flips to OR and holds the same pattern; namely, all the grey.
But 0 flips to -1 and - NOR flips to 1 - OR, while both swap patterns; namely all the white turns black and all the black turns white.
With Beta-1 Flips the black cells remain unchanged.
With Gamma-1 Flips the white cells remain unchanged.
The Beta-1 Flips generate rules:
3812798742492: -1 + OR
3812798742493: 1
3812798742494: 1 + NOR
The Gamma-1 Flips do not generate new rules. They just change where {0, 1, 2} go.
It seems that this knowledge could be leveraged in nanotechnology. The fact that these CAs are tubes with chirality suggests perhaps carbon nanotubes. And setting up patterns that can be “flipped” in different ways suggests reading and changing magnetic or electrical charge states.
Posted by: Lawrence J. Thaden
In the previous post the initial conditions are redundant in the sense that the left and right half are duplicates of each other. Now we want to just present the left half.
Also, flipping CAs only changes the colors, not the configuration of local objects. Now we want to demonstrate how to change the configurations. Moreover, we will see that there is a cycle to the changes that brings us back to the original.
If the CA is run “horizontally” three times, its output is the same as when the CA is run once vertically.
How does one run the CA “horizontally”? Make the initial conditions be the rightmost column of the output from the vertically run CA.
Using the same rule as before, 7348211845533: (x + y), and the left half of the initial conditions:
{2,2,2,2,2,2,2,2,2,0,0,0,0,0,0,0,0,0,1,1,1,1,1,1,1,1,1,2,2,2,1,1,1,0,0,0,1,1,1,0,0,0,2,2,2,0,0,
0,2,2,2,1,1,1,2,2,2,0,0,0,1,1,1,0,0,0,1,1,1,2,2,2,1,1,1,2,2,2,0,0,0,1,1,1,2,2,2,0,0,0,1,1,1,2,2,
2,0,0,0,1,1,1,2,2,2,0,0,0,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,0,0,0,2,2,
2,2,2,2,1,1,1,0,0,0,0,0,0,2,2,2,1,1,1,0,0,0,2,2,2,1,1,1,2,2,2,1,1,1,0,0,0,1,1,1,0,0,0,2,2,2,0,0,
0,1,1,1,2,2,2,1,1,1,2,2,2,0,0,0,2,2,2,0,0,0,1,1,1,0,0,0,0,0,0,0,0,0,2,2,2,2,2,2,2,2,2,1,1,1,1,1,
1,1,1,1}
the CA comes out as pictured in figure 1. This is our starting point.
Now taking the rightmost column of the data from that CA run we have initial conditions for the first “horizontal” run:
{1,1,1,1,1,1,1,1,1,0,0,0,0,0,0,0,0,0,2,2,2,2,2,2,2,2,2,1,1,1,0,0,0,2,2,2,1,1,1,0,0,0,2,2,2,1,1,
1,0,0,0,2,2,2,2,2,2,0,0,0,1,1,1,2,2,2,0,0,0,1,1,1,2,2,2,0,0,0,1,1,1,1,1,1,2,2,2,0,0,0,1,1,1,2,2,
2,0,0,0,1,1,1,2,2,2,0,0,0,1,1,1,1,1,1,1,1,1,2,2,2,2,2,2,2,2,2,0,0,0,0,0,0,0,0,0,1,1,1,0,0,0,2,2,
2,0,0,0,2,2,2,1,1,1,2,2,2,1,1,1,0,0,0,0,0,0,2,2,2,1,1,1,2,2,2,1,1,1,0,0,0,1,1,1,0,0,0,2,2,2,2,2,
2,0,0,0,1,1,1,0,0,0,1,1,1,2,2,2,1,1,1,2,2,2,0,0,0,2,2,2,2,2,2,2,2,2,1,1,1,1,1,1,1,1,1,0,0,0,0,0,
0,0,0,0}
Run with the same rule this generates the CA as pictured in figure 2.
Again, taking the last column of data from the figure 2 CA, initial conditions for a second “horizontal” run are:
{0,0,0,0,0,0,0,0,0,2,2,2,2,2,2,2,2,2,1,1,1,1,1,1,1,1,1,0,0,0,2,2,2,1,1,1,0,0,0,2,2,2,1,1,1,0,0,
0,2,2,2,1,1,1,2,2,2,0,0,0,1,1,1,2,2,2,0,0,0,1,1,1,2,2,2,0,0,0,1,1,1,0,0,0,1,1,1,2,2,2,2,2,2,0,0,
0,1,1,1,1,1,1,2,2,2,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,0,0,0,2,2,
2,0,0,0,2,2,2,1,1,1,2,2,2,1,1,1,0,0,0,1,1,1,0,0,0,2,2,2,1,1,1,0,0,0,2,2,2,1,1,1,0,0,0,2,2,2,0,0,
0,1,1,1,2,2,2,0,0,0,1,1,1,2,2,2,0,0,0,1,1,1,2,2,2,0,0,0,0,0,0,0,0,0,1,1,1,1,1,1,1,1,1,2,2,2,2,2,
2,2,2,2}
These generate the CA in figure 3.
Finally, the last column of data from figure 3 is:
{2,2,2,2,2,2,2,2,2,0,0,0,0,0,0,0,0,0,1,1,1,1,1,1,1,1,1,2,2,2,1,1,1,0,0,0,1,1,1,0,0,0,2,2,2,0,0,
0,2,2,2,1,1,1,2,2,2,0,0,0,1,1,1,0,0,0,1,1,1,2,2,2,1,1,1,2,2,2,0,0,0,1,1,1,2,2,2,0,0,0,1,1,1,2,2,
2,0,0,0,1,1,1,2,2,2,0,0,0,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,0,0,0,2,2,
2,2,2,2,1,1,1,0,0,0,0,0,0,2,2,2,1,1,1,0,0,0,2,2,2,1,1,1,2,2,2,1,1,1,0,0,0,1,1,1,0,0,0,2,2,2,0,0,
0,1,1,1,2,2,2,1,1,1,2,2,2,0,0,0,2,2,2,0,0,0,1,1,1,0,0,0,0,0,0,0,0,0,2,2,2,2,2,2,2,2,2,1,1,1,1,1,
1,1,1,1}
which is the same as the original initial conditions used to run the starting point CA.
So a three step cycle of “horizontal” runs results in the vertical run.
In other words, an iterated use of the latest time step axis as the space axis results in an identical CA.
How is that for unifying space and time at discrete levels!
That brings me back to nanotechnology. We know that graphene sheets roll up into nanotubes and that a nanotube has hemispherical caps on each end.
But has anyone got a nanotube to join ends to make a torus?
Wikipedia reports graphene can make a wide variety of shapes such as conics, saddles, and fullerenes. But no mention is specifically made about tori.
If graphene sheets can’t form tori, but do form fullerenes, is it possible to map these CAs onto a sphere?
With such a mapping the discrete configuration changes on the surface of the sphere might be studied in connection with quantum behavior of the fullerene.
Edited: Nov. 1, 2007
The answer to the question: But has anyone got a nanotube to join ends to make a torus?
is in Journal of Applied Physics -- 15 March 2007. The article entitled “Orbiting atoms and C60 fullerenes inside carbon nanotori” is by Tamsyn A. Hilder and James M. Hill at the University of Wollongong in Australia.
Posted by: Lawrence J. Thaden
For rule 7348211845533 (x + y) there are just two intermediate phases that the CAs go through before returning to the original.
Here is another example using a different rule, 3681670999617: (-x - y), in which there are seven intermediate phases as shown in the attached figure.
Notice that some of the phases in the figure look the same. However, only the original and horizontal 8 are identical.
Initial conditions for the original are two concatenated copies of digit expansions of T5 rules. Initial conditions for all the other phases are the cells from the last column of the previous CA.
There are 1458 time steps for each CA, the number required to complete the cycle for the original.
Since there are only 729 digits in the concatenated digit expansions for T5 rules, it is necessary to replicate them for the original CA’s initial conditions.
This is required to assure square shaped CAs from which the last column is selected as input to each next phase.
Posted by: Lawrence J. Thaden
Attached is a graphic of the CA phases rule 7479339588409: (1 - x - y) cycles through before repeating.
Again, although some of the phases look alike, only horizontal 8 is identical to the original.
Structurally, these are the same as for 3681670999617: (-x - y). However, the colors are different.
Posted by: Lawrence J. Thaden
Whereas rules and expressions:
3681670999617: (-x - y)
7348211845533: (x + y)
7479339588409: (1 - x - y)
have CA phases with (3 x 3), (2 x 2), and (3 x 3) matrices,
the next three rules and expressions:
146257896577: (1 + x + y)
277385639453: (-1 - x - y)
3943926485369: (-1 + x + y)
have CA phases with (2 x 2), (3 x 3), and (2 x 2) matrices.
It appears that the matrix size is (2 x 2) whenever the expression contains (x + y) and (3 x 3) whenever it contains (-x - y).
Although the original figure in the phase matrix for rule 277385639453: (-1 - x - y) does not appear to resemble the originals for the other two rules, a likeness is seen when the original figure for rule 277385639453 is enlarged by, say, 300 percent. Clusters in the original figure for rule 277385639453 then appear similar to the original figures for rules 146257896577 and 3943926485369.
Attached are CA phase matrices for rules 146257896577, 277385639453, and 3943926485369.
Posted by: Lawrence J. Thaden
I would like to quote from Wolfram’s keynote address given at the NKS 2007 Conference in Burlington, Vermont this last summer:
“It's interesting how the tradeoffs for creating things change when one is mining the computational universe rather than doing traditional engineering.
Traditional engineering puts a great premium on copying what one's done before. On reusing the same ideas of wheels and levers and half-adders and whatever.
But in mining the computational universe, creativity in coming up with new mechanisms is in a sense free.
And what tends to make the most sense is not to start from the devices one's already built, but instead to take the pure raw material available in some particular system, then to search the computational universe for ways to use it.
Like take nanotechnology. Where the traditional approach is to take mechanisms we know work in our traditional engineering, and to try to shrink them down to molecular scales.
Well, it's often rather difficult to make things like wheels out of molecules. But what the NKS approach says it that that's not what one needs to do.
Instead, one can take whatever is easy to create with raw molecules, then just search the abstract computational universe for how to use it to achieve some particular functional purpose.
Coming up with ... engineering structures that look nothing like our traditional engineering structures.”
As an example of the traditional approach consider IBM’s announcement in 2004 that they had succeeded in creating a carbon nanotube transistor.
They laid a bundle of nanotubes across a substrate of silicon and draped gold leads over it. Thus, to use Wolfram’s words, they took “mechanisms we know work in our traditional engineering” and shrank “them down to molecular scales”.
A problem with this approach is that the leads and the carbon nanotubes are disproportionate in size. With 50,000 nanotubes fitting within a human hair, how thin must the leads be so as not to have the nano device looking like a CERN detector?
As early as 1999 David Tomanek and his collaborators showed that leads are not required to communicate with nanotubes:
“For carbon nanotubes... third-order nonlinearity provides a method for current injection without contacts. It has proven experimentally difficult to fabricate low-resistance electrical contacts with carbon nanotubes by conventional submicron lithographic methods.” Coherent control of photocurrents in graphene and carbon nanotubes, E. J. Mele, Petr Kral, and David Tomanek, Phys. Rev. B, 61, 11 (15 Mar 2000)
So let’s continue to “search the abstract computational universe for how to use [current injection without contacts] ... to achieve some particular functional purpose.” Namely, a carbon nanotube transistor.
With that lengthy introduction, I want to continue this thread by introducing some findings on alternating phase CAs, where one phase appears to be on the outside of the CA tube and the other on the inside.
So here goes.
Previous posts featured CAs involving rules in terms of both x and y.
This post treats each separately.
There are two sets of three expressions for each variable taken separately.
For x the first set has:
(15) 3812992433055: (x)
(25) 7625210074339: (1 + x)
(26) 7625403764901: (-x)
and the second set for x has:
(2) 193720085: (-1 + x)
(3) 387410647: (1 - x)
(13) 3812605051931: (-1 - x)
For y the first set has:
(17) 3943753520967: (y)
(21) 7348577761291: (1 + y)
(24) 7479532539765: (-y)
and the second set for y has:
(4) 146064945221: (-1 + y)
(7) 277019723695: (1 - y)
(11) 3681843964019: (-1 - y)
In this post we restrict the discussion to the six expressions for variable x, all of which manifest a chiral behavior. We will continue investigation later on about the six expressions for variable y. They manifest a vertical behavior which on closer examination has zig-zag properties.
Now to consider each of these six sets for x.
First {x, 1 + x, and -x}.
CAs for these are in Figures 1 - 3 of the attached graphic.
Both vertical and horizontal CAs for these three rules are chiral along the primary diagonal. The cell counts for each CA has the same number of 0s, 1s, and 2s. Cycle lengths for x and 1 + x are both 1458 time steps, while that for -x is 729 time steps. All three have one horizontal CA which, when repeated, generates the original vertical CA. That is, all three alternate between two phases. In each case the horizontal CA is related to the original vertical CA. Here is the relation for each of the three sets:
(x): Rotate original vertical CA 180 degrees about the primary diagonal and then replace all cells in the primary diagonal, which are 0s, with alternating 2s and 1s. The result equates to the horizontal CA.
(1 + x): Rotate original vertical CA 180 degrees about the primary diagonal and then replace all cells in the primary diagonal, which are alternating 0s and 1s, with 2s. The result equates to the horizontal CA.
(-x): Rotate original vertical CA 180 degrees about the primary diagonal and then replace all cells in the primary diagonal, which are all 0s, with 2s. The result equates to the horizontal CA.
Figures 4 - 6 depict enlargements of the upper left corner for each of the three rules before and after rotation.
Notice that the barred effect in figures 4 and 5 is “lost” when the CAs are shrunk down to the size displayed in figures 1 and 2. Actually, there are also other “hidden effects” in the full sized graphics for figures 1 and 2 that are “lost”, such as staggered bar.
Now consider the second set for x: {(-1 + x), (1 - x), (-1 - x)}.
CAs for these are in Figures 7 - 9 of the attached graphic.
Due to areas of relatively even distribution of cell values details in gray scale images are quite indistinguishable. Therefore, these figures were generated with ColorRules -> {0 -> Red, 1 -> Green, 2 -> Blue}. Then they were converted in Adobe Photoshop to grays.
Both vertical and horizontal CAs for these three rules are chiral along the primary diagonal. The cell counts for each CA has the same number of 0s, 1s, and 2s. Cycle length for (-1 + x) is 1458 time steps, while that for (1 - x) and (-1 - x) is 729 time steps. Only the first has a single horizontal CA which, when repeated, generates the original vertical CA. That is, (-1 + x) alternates between two phases. The horizontal CA is related to the original vertical CA as follows:
(-1 + x): Rotate each row of the original vertical CA to the right one cell while replacing all 0 cells with 2s. The result equates to the horizontal CA. (See figure 7 and enlargement figure 10.)
So far, all of the behaviors discussed appear to represent events happening on either surface of the CA tubes. So if we posit that the original is the outside surface of the CA, then the surface on the inside is the corresponding horizontal CA. This conclusion is based upon the reasoning that a rotation of 180 degrees about the primary diagonal puts you on the backside of the CA or when rolled up on the inside of the tube.
Now we step out of line to see a novel behavior with more than just two alternating phases.
The case for the other two expressions, (1 - x) and (-1 - x) is different from that for (-1 + x). In fact it came as a surprise, for there are five horizontal phases before repeating the original vertical one.
As for (-1 + x), so also here, ColorRules are employed to bring out the details. However, because of the extremely even distribution of the cell values, even with ColorRules -> {0 -> Red, 1 -> Green, 2 -> Blue} only gray scale is seen.
To see how evenly distributed the cell values are examine figure 11. It shows the original vertical CA for (1 - x) along with three masked images of the same CA. In the first image only 0s appear and 1s and 2s are masked out. In the next image only 1s appear and the 0s and 2s are masked out. In the last image only the 2s appear and the 0s and 1s are masked out. Cell value distribution is so similar that the reds, greens and blues merge into shades of gray.
The complete set of phases for (1 - x) is shown in figure 8.
It is difficult to see how the original for (1 - x) relates to horizontal 1. To begin with, rows of the original appear to have long runs of same valued cells whereas horizontal 1 has short runs. It is as though the original is more x-like and horizontal 1 is more y-like.
This is also manifest in the plots. There are equivalent local objects in the original that that have a horizontal alignment, while in horizontal 1 the alignment is vertical.
I do not know of an operation or combination of operations that equates the two.
However, the original, horizontal 2, and horizontal 4 bear a resemblance as do horizontal 1, horizontal 3, and horizontal 5.
For (1 - x) the original with rule changes: {0 -> 2, 1 -> 0, 2 -> 1} is equivalent to horizontal 2.
And horizontal 2 with the same rule changes: {0 -> 2, 1 -> 0, 2 -> 1} is equivalent to horizontal 4.
When these three are animated, there appears to be a flow of equivalent local objects along the direction of the diagonal starting from the bottom right and proceeding to the top left.
Likewise, horizontal 1 with the same rule changes: {0 -> 2, 1 -> 0, 2 -> 1} is equivalent to horizontal 3.
Finally, horizontal 3 with the same rule changes: {0 -> 2, 1 -> 0, 2 -> 1} is equivalent to horizontal 5.
And when these three are animated, there also appears to be a flow of equivalent local objects along the direction of the diagonal starting from the bottom right and proceeding to the top left.
Now let’s consider the last form for x.
It is (13) 3812605051931 (-1 - x).
The complete set of phases for (-1 - x) is shown in figure 9.
Things here are comparable to what occurs for (1 - x), only the rule changes are: {0 -> 1, 1 -> 2, 2 -> 0} instead of {0 -> 2, 1 -> 0, 2 -> 1}. That is, the rule statements reverse directions: {0 -> 1} instead of {1 -> 0}, {1 -> 2} instead of {2 -> 1}, and {2 -> 0} instead of {0 -> 2}.
Because of the difficulty I have in attaching graphics of different size, please bear with me for only attaching figures 1 through 7 and 10 to this post. The next post has figures 8 and 9. The post after that has figure 11. And a final post has a notebook of pictures which you can animate to see the flows for (1 - x) and (-1 - x).
Posted by: Lawrence J. Thaden
Attached are Figures 8 and 9.
Posted by: Lawrence J. Thaden
Attached is Figure 11.
Posted by: Lawrence J. Thaden
Attached is the notebook.
Posted by: Lawrence J. Thaden
In the last post the notebook contains an error. It is not in the code of the notebook, but in the last comment where I said the channel either side of the primary diagonal wraps twice. It does not wrap twice. It is parallel and approximately opposite on the cross section of the tube from the corresponding position on the primary diagonal. If the tube were transparent the channels would appear to cross each other like the bars on the letter x.
You may want to replace the entire comment with:
Now select cells for Original, Horizontal 2, and Horizontal 4. Then animate them and observe the interference pattern flowing from lower right to upper left along and parallel to the primary diagonal. Because the CA forms a tube, there are two channels in which the flow is observed. The channel along the primary diagonal wraps the tube once. The channel on either side of the primary diagonal wraps the tube once starting 27 cells short of halfway around the tube and parallel to the primary diagonal.
Now select cells for Original, Horizontal 2, and Horizontal 4. Then animate them and observe the interference pattern flowing from lower right to upper left along and parallel to the primary diagonal. Because the CA forms a tube, there are two channels in which the flow is observed. The channel along the primary diagonal wraps the tube once. The channel on either side of the primary diagonal wraps the tube once starting 27 cells short of halfway around the tube and parallel to the primary diagonal.
This difference of 27 cells suggests that the cross section of the tube may be an ellipse.
Notice that the equivalent local objects that flow for Original, Horizontal 2, and Horizontal 4 are biased toward a horizontal orientation while the equivalent local objects that flow for Horizontal 1, Horizontal 3, and Horizontal 5 are biased toward a vertical orientation. Together they form a perpendicular arrangement.
Compared with the local objects for (1 - x), the Original for (-1 - x) appears to be concave and wider on the right, while the Original for (1 - x) appears to be convex and wider on the left.
It might be that Original, Horizontal 2, and Horizontal 4 describe what is happening on the outside of the CA tube while Horizontal 1, Horizontal 3, and Horizontal 5 account for what is happening on the inside surface of the CA tube.
----------
Regarding the relation between the original and horizontal 1 for (1 - x) I stated:
“I do not know of an operation or combination of operations that equates the two.”
Well, they are related. Perhaps the most efficient way to state the relation is in Mathematica code.
Let the values of the cells for (3) Rule 387410647: (1 - x) Original be assigned to the symbol t5r3.
And let the of the cells for (3) Rule 387410647: (1 - x) Horizontal 1 be assigned to the symbol t5r3p2.
Then the following will be True.
Union[Table[(RotateLeft[Flatten[Partition[Transpose[t5r3p2][], 27]]] == t5r3[[i]]), {i, Length[t5r3p2]}]]
Edited 11/21/2007 8:53 PM EST
This same line of code relates each of the phases to the next phase. Just keep changing the last phase as input to the Transpose function and pointing to the next phase after the equals sign.
Edited 11/22/2007 1:00 PM EST
A similar line of code relates each of the phases with the next phase for (13) Rule 3812605051931 (-1 - x).
Assign the cells for the next phase to the symbol nextPhase and the cells for the present phase to the symbol presentPhase. Then the following line of code returns {True}.
Union[Table[(RotateLeft[Flatten[Partition[Transpose[nextPhase][[i]], 27]]] == presentPhase[i]]), {i, Length[nextPhase]}]]
Transpose on an entire CA of 27 rows and 27 columns in effect rotates the CA 180 degrees about the primary diagonal.
But here we are using transpose on each row of the CA starting from the first row and rippling down to the last. This suggests that the effect being modeled is wave-like.
I do not understand exactly how or why Transpose allows a flat list as input in this case.
I expected [i]Mathematica to return:
Transpose::nmtx : The first two levels of the one-dimensional list {2,1,0,2,1,0,2,1,0,2, <<719>>} cannot be transposed.
Perhaps someone at Wolfram Research can explain what is happening.
Edited 11/24/2007 5:35 AM EST
My bad! Transpose was applied to the entire phase and then the rows had RotateLeft applied to them. It still ripples.
----------
Another interesting relation is that between (3) Rule 387410647: (1 - x) Original and
(3) Rule 387410647: (1 - x) Horizontal 1 rotated 180 degrees about the primary axis.
When the two ArrayPlots for these are animated, the image is the same except for a slight horizontal vibration.
To produce the two images in Mathematica code use:
ArrayPlot[t5r3] and ArrayPlot[Transpose[t5r3p2]]
Posted by: Lawrence J. Thaden
Now let’s consider the Sierpinski triangle rule numbers in T5.
To start off, I have attached Carbon Nanotechnologies, Inc.’s photo of a 100 element single walled carbon nanotube rope. The triangular lattice packing visible in the cross section calls to mind the Sierpinski triangle. (See Figure 1.)
While this thread makes no claim that these T5 CAs are exact computations corresponding to single walled carbon nanotube ropes, it does introduce a powerful tool for exploring computational behavior that may inspire insights into the physical properties useful in nanotechnology research. That tool is the simple process of repeatedly taking the last column of CA output and using it as the initial conditions for a subsequent CA.
And the CAs do not necessarily have strictly cyclic properties. Indeed those featured in this post break the pattern seen in the previous posts of this thread. They have startups and all eventually either “ground” or “short” out in rows of homogeneous cell values.
When I saw the Sierpinski triangle like lattice packing in the photo of the carbon nanotube ropes (Figure 1), it made me think of the mollusc photos on page 423 in the NKS book. It is as though chemistry were pointing the “mouse” at itself and saying, “come and click here.”
So I did just that. And what turned up? Among other things, pages of information on Erik Winfree’s work on self assembly of DNA tilings.
Using a four-armed double crossover molecule called DX and Wang tile theory, he developed an algorithm that mimics the operation of a Turing Machine.
But what is relevant here is found on page 66 of his doctoral thesis, where with the same Wang tile theory he illustrates the growth of the Sierpinski Triangle.
So I asked: Is there something similar to Wang tile rules causing the Sierpinski triangular lattice packing in the Carbon Nanotechnologies’ photo?
This motivated me to look further into the T5 CA rules that generate variations of the Sierpinski triangle.
Here is what I found.
There are two sets of three T5 algebraic expressions corresponding to three color rules that generate variations of Sierpinski triangle CA output.
The first set is:
(6) 146430861993 (-x + y)
(16) 3943560596989 (1 - x + y)
(20) 7348404768497 (-1 - x + y)
Each CA in this set has four phases, the original and three horizontal phases, before reproducing the original phase. (See Figure 2. It is attached to this post.)
The second set is:
(8) 277192716489 (x - y)
(12) 3682036887997 (1 + x - y)
(22) 7479166622993 (-1 + x - y)
Each CA in this set has only three phases, the original and two horizontal phases, before reproducing the original phase. (See Figure 3. It is attached separately to the next post.)
Unlike the rules considered thus far, the originals for each of these six CAs generate a startup of just over 250 time steps before going into a cycle length of 1. I selected a time step limit of 728 so that (1) the images would be comparable with those from previous rules, and (2) to accommodate some of the horizontal CAs in this set that have startups of 729 time steps.
Now let’s consider each one.
(6) 146430861993 (-x + y) has four phases, the original vertical phase and three horizontal phases, before reproducing the original. Of these only horizontal 3 is not some variation of the Sierpinski triangle. Also horizontal 3 has a startup of 729 steps. That is, the whole CA is a startup. If we continue to run it beyond 729 time steps, it has a cycle length of 1 and all the cells are 0s. It “grounds” out.
(16) 3943560596989 (1 - x + y) has four phases, the original vertical phase and three horizontal phases, before reproducing the original. Of these only horizontal 3 is not some variation of the Sierpinski triangle. Also horizontal 3 has a startup of 728 steps before it begins a cycle of 1 time step, and this time step has all cell values of 1. It positively “shorts” out.
(20) 7348404768497 (-1 - x + y) has four phases, the original vertical phase and three horizontal phases, before reproducing the original. Of these only horizontal 3 is not some variation of the Sierpinski triangle. Also horizontal 3 has a startup of 702 steps before it begins cycles of 1 time step, and these are all cell values of 2. It also “shorts” out, but negatively.
(8) 277192716489 (x - y) has three phases, the original vertical phase and two horizontal phases, before reproducing the original. All three phases are variations of the Sierpinski triangle. The startups are as follows: original: 253 steps; horizontal 1: 693 steps; and horizontal 2: 729 steps. After that, each “grounds” out with rows of all zeros. However, note that the original has all 2s for time step 253 and horizontal 2 has all 2s for time step 729.
(12) 3682036887997 (1 + x - y) has three phases, the original vertical phase and two horizontal phases, before reproducing the original. All three phases are variations of the Sierpinski triangle. But horizontal 2 has slight imperfections compared to an ideal. This was also the case with Winfree’s Kinetic Assembly Model of the Sierpinski triangle sited previously. However, Winfree’s defects were limited to the baseline, while these defects are also in the alignments of triangular structures above the baseline triangles. The startups are as follows: original: 253 steps; horizontal 1: 729 steps; and horizontal 2: 729 steps. After that, each “shorts” out positively with rows of all 1s. However, note that the original has all 0s for time step 253, horizontal 1 has all 0s for time step 729, and horizontal 2 has all 2s for time step 729.
(22) 7479166622993 (-1 + x - y) has three phases, the original vertical phase and two horizontal phases, before reproducing the original. All three phases are variations of the Sierpinski triangle. The startups are as follows: original: 253 steps; horizontal 1: 729 steps; and horizontal 2: 702 steps. After that, each “shorts” out negatively with rows of all 2s. However, note that the original has all 1s for time step 253, horizontal 1 has all 0s for time step 729 and horizontal 2 has all 0s for time step 702.
The fact that there is no persistent cycle in these behaviors indicates that if they do have a “real world” analog, it will be fleeting and thus hard to detect, unless they, as it were, leave “tracks” like those triangular lattice packing clues visible in the Carbon Nanotechnologies’ photo.
I find encouragement in this regard from what Wolfram sees in the chemistry of the mollusc shell. Those events are also fleeting and momentary. (See page 424-425 of the NKS Book.)
Posted by: Lawrence J. Thaden
Figure 3 is attached.
Posted by: Lawrence J. Thaden
We now turn attention to six T5 expressions for the variable y. As with variable x, these are grouped in two sets of three expressions.
The first set is:
(17) 3943753520967: (y)
(21) 7348577761291: (1 + y)
(24) 7479532539765: (-y)
and the second set for y is:
(4) 146064945221: (-1 + y)
(7) 277019723695: (1 - y)
(11) 3681843964019: (-1 - y)
Whereas expressions for the variable x generate CAs that have the chiral property, expressions for the variable y generate CAs that have the appearance of being vertical.
However, on closer inspection, their properties are more complicated than they first appear.
(17) 3943753520967: (y)
Let’s begin with the expression for rule 3943753520967: (y).
The first phase, the original CA, has a two step cycle. However, at first glance it appears to be a series of parallel vertical bars. (See figure 1a.)
Actually it is more complex. All odd rows are the same and all even rows are the same. But odd rows differ from even rows. Wherever there is a cell with a 1 in the odd row there is a corresponding cell with a 2 in the even row. And conversely, wherever there is a cell with a 2 in the odd row there is a cell with a 1 in the even row. But 0s remain the same for both odd and even rows. (See figure 1b for a close up view of this zigzag pattern. It is from the upper left hand corner of the CA output.)
The second phase, the horizontal CA, also has a cycle with just two time steps. In repeated fashion the time steps appear as a checkerboard pattern of cells with 2s and 1s. (See figure 2 for a close up view of the checker board pattern. It is from the upper left hand corner of the CA output.)
If you attempt to run a third phase, it just replicates the horizontal phase.
(21) 7348577761291: (1 + y)
The original for (1 + y) is similar to the rule for (y): all odd lines the same and all even lines the same. Only in this case it is 0s and 1s that get switched while 2s are the same for both. But the horizontal is not a checker board as it is for (y). Rather, it is a CA with every cell value a 2.
Obviously, attempting to run a third phase only results in this same CA with every cell value a 2.
(24) 7479532539765: (-y)
The original for (-y) has the same pattern on every row. So its CA output is truly vertical bars. (See figure 3.) The horizontal CA is the same as for (1 + y); namely, all cells values are 2.
Again, there is no third phase. It just results in another CA with all cell values a 2.
Now for the second set.
(4) 146064945221: (-1 + y)
This is similar to (y) and (1 + y), only here for the original the 0s and 2s get switched on odd and even rows while the 1s are the same for odd and even rows. (See figure 4.) The horizontal is a checkerboard of 2s and 0s.
Again, that is as far as it goes. There are just these two phases: the original and the checker board horizontal. For when the last column from the checkerboard CA is used as initial conditions with (-1 + y) as rule it produces itself. (See figure 5 for a close up view of the checker board pattern. It is from the upper left hand corner of the CA output.)
(7) 277019723695: (1 - y)
For this rule there is an original and six horizontals. The six horizontals constitute a never ending cycle of phases which, when animated, appear as a flow.
Let’s start with the original. Unlike the other y variable CAs this one has a complex vertical pattern. The others had what appeared to be vertical bars. This one looks more like vertical interference patterns interspersed with diagonal bars. (See figure 6a.)
But where the other originals for y variables have an easily explained pattern switching between odd and even rows, this one has a pattern that repeats every three rows. The pattern expressed in terms of Mathematica rules is:
Row 1 /. {0 -> 1, 1 -> 2, 2 ->0} == row 2 and row 2 /. {0 -> 1, 1 -> 2, 2 ->0} == row 3.
Then row 4 is the same as row 1, etc. (See figure 6b for a close up view of the three step cycle. It is from the upper left hand corner of the CA output.)
Now for the horizontals. The first horizontal has cycles of three rows, just as the original has. Their pattern expressed in terms of Mathematica rules is the same as for the original CA.
Moreover, there is a pattern within each row. It repeats every three columns. For row 1 it is: {2, 0, 1}. For row 2 it is: {0, 1, 2}. And for row 3 it is: {1, 2, 0}. In other words it shifts left.
Finally, the corresponding rows and columns are the same. So it is symmetric through transposition. In Mathematica code: Transpose[horizontal 1] == horizontal 1. That is, it is invariant through rotations of 180 degrees about the primary diagonal.
The second horizontal is related to the first as expressed in this Mathematica code:
Union[Table[(RotateLeft[Flatten[Partition[Transpose[second horizontal][], 27]]] == first horizontal[[i]]), {i, Length[second horizontal]}]]
The third horizontal is related to the second in the same way. And the relation passes down through to the sixth horizontal. After the sixth horizontal it cycles back through to the first horizontal.
It is interesting to note that this same [i]Mathematica code expresses the relation between the horizontals for (3) 387410647: (1 - x).
Figure 7 shows all six horizontals.
When the three even horizontals are animated, there appears to be a flow of rainbow-like objects with a chiral direction perpendicular to that observed for the x variable. The flow is along the primary diagonal of the table. And it begins at the upper left corner and proceeds to the lower right corner. When the three odd numbered horizontals are animated, the same behavior is observed. This direction is the opposite of what it is for (1 - x) and (-1 - x).
However, if all six horizontals for (1 - y) are selected and animated, the flow is observed to start from the lower right corner and proceed to the upper left corner. (I don’t know if this change in the direction of flow is an optical illusion, a quirk of my graphics card, or what.)
Based on the hypothesis that the odd numbered horizontal phases are on the outside surface of the tube and the even are on the inside of the tube, and keeping in mind that each horizontal phase is invariant with respect to rotations of 180 degrees about the primary diagonal, each phase appears the same whether inside or outside the tube. The only way to tell behavior on the outside from behavior on the inside is by the displacement indicated in the Mathematica command RotateLeft.
(11) 3681843964019: (-1 - y)
As with the previous rule: (7) 277019723695: (1 - y), for this rule there is an original and six horizontals. The six horizontals constitute a never ending cycle of phases which, when animated, appear as a flow.
Figures 8a and 8b present graphs for the original. As with rule: (7) 277019723695: (1 - y), both a masked set of images and a close up of the upper left corner are presented. Clearly, this rule also has a three step cycle in the original phase even though its overall appearance is a complex vertical interference pattern.
The steps of the cycle are related as:
Row 1 /. {0 -> 2, 2 -> 1, 1 ->0} == row 2 and row 2 /. {0 -> 2, 2 -> 1, 1 ->0} == row 3.
Then row 4 equals row 1, etc.
Now for the horizontals. The first horizontal has cycles of three rows, just as the original has. Their pattern expressed in terms of Mathematica rules is the same as for the original CA.
Moreover, there is a pattern within each row. It repeats every three columns. For row 1 it is: {2, 1, 0}. For row 2 it is: {1, 0, 2}. And for row 3 it is: {0, 2, 1}. In other words it shifts left.
Finally, the corresponding rows and columns are the same. So it is symmetric through transposition. In Mathematica code: Transpose[horizontal 1] == horizontal 1. That is, it is invariant through rotations of 180 degrees about the primary diagonal.
The second horizontal is related to the first as expressed in this Mathematica code:
Union[Table[(RotateLeft[Flatten[Partition[Transpose[second horizontal][], 27]]] == first horizontal[[i]]), {i, Length[second horizontal]}]]
The pattern within each row picks up where it left off in the last row of the first horizontal. Again it repeats every three columns. For row 1 it is the same as for the last row in the first horizontal: {0, 2, 1}. For row 2 it is: {2, 1, 0}. And for row 3 it is: {1, 0, 2}. In other words it shifts left.
The third horizontal has a similar set of relations within rows and columns and it is also related to the second as the second is to the first. And these relations pass down through to the sixth horizontal. After the sixth horizontal it cycles back through to the first horizontal.
Figure 9 shows a close up of the upper left corners for all of the horizontals. The pattern is a series of parallel diagonals running perpendicular to the primary diagonal. They cycle through blue green and red, representing cell values 2, 1, and 0 respectively. Notice that each next horizontal graph continues the same pattern left off at the lower right corner of the previous horizontal.
Figure 10 shows all six horizontals.
When the three even horizontals are animated, there appears to be a flow of rainbow-like objects with a chiral direction perpendicular to that observed for the x variable. The flow is along the primary diagonal of the table. And it begins at the upper left corner and proceeds to the lower right corner. When the three odd numbered horizontals are animated, the same behavior is observed.
However, if all six horizontals are selected and animated, the flow is observed to start from the lower right corner and proceed to the upper left corner.
Moreover, the rainbow-like objects are mirror images of those observed for (7) 277019723695: (1 - y).
This completes a basic coverage of the expressions for T5. Every rule except the three rules for the identities has been considered.
Here is a summary of findings for the y variable expressions.
(17) 3943753520967: (y)
2 phases
1st has alternating rows
2nd is checkerboard and this phase cycles
(21) 7348577761291: (1 + y)
2 phases
1st has alternating rows
2nd shorts out with -1
(24) 7479532539765: (-y)
2 phases
1st is vertical bars
2nd shorts out with -1
(4) 146064945221: (-1 + y)
2 phases
1st has alternating rows
2nd is checkerboard and this phase cycles
(7) 277019723695: (1 - y)
7 phases
1st has rows cycling every third time step
2nd through 7th has rows cycling every third time step
and the set of phases form a cycle
rows and corresponding columns same in each phase
rainbow-like objects perpendicular to primary diagonal
animated flow up primary diagonal
(11) 3681843964019: (-1 - y)
Same as for (7) 277019723695: (1 - y) except rainbow-like objects are mirror images
Although I am still getting familiar with these findings, I would say the most astonishing one is that the rows and corresponding columns of horizontals for (1 - y) and (-1 - y) are the same, so that if you rotate any one of the six horizontals for either of these expressions 180 degrees about the primary diagonal you get the same thing.
The second most astonishing thing is how animations of all of the horizontals for (1 - y) and (-1 - y) proceed in the opposite direction as the animations of just odd or even numbered horizontals.
Finally, the fact that so many of the CAs have [i]Mathematica expressions relating them to other CAs. And many times the same rules are applicable to the original and an entire set of horizontals.
Figure 9 is attached to the following post. Figures 7 and 10 are attached to the one after that. All the rest of the figures are attached to this post.
Posted by: Lawrence J. Thaden
Figure 9 attached.
Posted by: Lawrence J. Thaden
Figures 7 and 10 attached.
Posted by: Lawrence J. Thaden
I am working now on T5 = ({0, +1, x, z, -1}; +, -; 0, +1, -1).
Recall that 27 three color rule numbers are one to one and onto with the algebraic expressions in T5. The significance here is that the y in {0, +1, x, y, -1} is not an arbitrary symbol and so it differs from z in ({0, +1, x, z, -1}. Or simply put: the rule number for y is not the same as the rule number for z.
So there will also be a T5 = ({0, +1, y, z, -1}; +, -; 0, +1, -1).
When I finish working on these, I will post a summary of findings together with notebooks for each.
However, I cannot wait to share a discovery about the graphs for phases of the expression for rule 1466461054805: (-1 + z). They do not look very interesting from the perspective of Wolfram’s four-fold classification. However, from the standpoint of supersymmetry in physics, they are very curious.
Each phase has 1458 steps and geometrically forms the surface of a torus. That is, it cycles.
Moreover, the phases themselves cycle after twice as many steps; namely, 2916.
But what is curious is that each phase differs from the previous phase in that it has rows and columns switched and then rows shifted left one cell. So somehow the space and time axes are indistinguishable.
Because this interchange of rows and columns amounts to a rotation of 180 degrees about the primary diagonal, it certainly appears that the phases are alternating between the inside and outside of the torus.
But here is the connection to supersymmetry in physics:
"Van Nieuwenhuizen explained that supersymmetry can be seen as a deepening of the symmetries of space. This is because of a profound and beautiful property: If you change all the fermions into bosons and then change them back again, you get the same world you had before but with everything moved a little bit in space." (The Trouble with Physics, Lee Smolin, Houghton Mifflin Company, 2006, page 93.
I am not advocating supersymmetry. But I am suggesting there may be a “logical” explanation for some of its conclusions.
If I were a physicist, I would try to relate all of this to Gerard ‘t Hooft’s holographic principle that states everything happens on surfaces surrounding space rather than in space.
Posted by: Lawrence J. Thaden
Space-time is mirror imaged whenever these CA phases are rotated 180 degrees about the primary axis. But in physics mirror images of space time define matter and antimatter. An antimatter particle goes backward in time compared to its mirror imaged particle traveling forward in time.
Thus the surface of one phase describes matter while the surface of the subsequent phase describes antimatter.
But these surfaces are the inside and outside of space time.
So, you don’t need supersymmetry. Matter and antimatter are sufficient.
However, this does not explain the shift left in going from one phase to another.
Moreover, many a physicist and cosmologist would shoot me for saying such a thing, since it defies the time honored notion that there is much more matter in the universe than antimatter.
I had better retreat to logic where I belong.
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