Download Other Approaches to 102 Linear algebra, Groups and polynomials

Other Approaches
to 102
Linear algebra, Groups and
polynomials
What We’ve Done So Far

Discovered some cool properties of Rule 102.


Used straight-forward brute-force proofs.
But we can take advantage of techniques from
other branches of mathematics.




Linear Algebra
Group theory.
Polynomials.
Other... (you discover it!)
Optional Sections

The sections on linear algebra and group theory
may be skipped.




Depends on time constraints.
Depends on student background and interest.
Neither section is critical to later developments.
The section on polynomials and derivatives is
important to later developments.
Linear Algebra


Basic idea is to use matrix multiplication to get
the derivative.
Consider
1

0
0
D

0

1

1 0 0  0 0

1 1 0  0 0
0 1 1
0 0




0 0 0
1 1
0 0 0  0 1 
First Derivative Matrix

What happens if multiply D by x?

Get first derivative!
Try it!
These give the sum of the
second two elements.
These give the sum of the
first two elements.
D
1

0
0
Dx  
0
0

1

1
1
0
0
0
0
x
0
1
1
0
0
0
0
0
1
1
0
0
0
0
0
1
1
0
0  0 
 
d(x)
0  1 
0  0 
   1 1 1 0 1 0
0  1 



1  1 
1  0 
Kth Derivative Matrix

So call D the first derivative
matrix.

What if I want the kth
derivative?




Apply first derivative k times.
D(Dx) = 2nd derivative.
D(D(Dx) = 3rd derivative.
That’s just matrix
multiplication. The kth
derivative is just
1

0
0
k
D 

0

1

1 0 0  0 0

1 1 0  0 0
0 1 1
0 0



0 0 0
1 1 
0 0 0  0 1 
k
2nd Derivative Matrix

What’s the second derivative matrix look like?
1

0
0
2
D 

0

1

1 0 0  0 0 2  1 0


1 1 0  0 0
0 1
0 0
0 1 1
0 0
 



1 0
0 0 0
1 1



0 1
0 0 0  0 1

1 0  0
0 1  0
1 0
0

0 0
0
0 0  0
Can you guess the pattern?
0
0
0

1
0
0

0
0


0

1 
Pascal Triangle Gives kth
Derivative

The rows of the kth derivative matrix are just the kth
derivative of Pascal’s triangle.


Example, 4th row of Pascal is





Modified for finite length.
So
4th
0001
0011
0101
1111
1

0
0
4
D 

0

1

derivative matrix is
1 0 0  0 0 4  1 1


1 1 0  0 0
0 1
0 0
0 1 1
0 0
 




1 1
0 0 0
1 1


1 1
0 0 0  0 1 

1 1
1 1
1 1
0
1
1

0 0 0
1 0 0
 0 0

 0 0
 0 0



 1 1

 0 1 
Eigenvector Solutions

So now can use all the power of linear
algebra.

Example: Recall eigenvectors?

Eigenvectors are vectors x such that Dx = l x for
some scalar l and l is the eigenvalue).
Eigenvectors Give Cycle Lengths

If we want the cycle length Just want the
derivative that gives back x. i.e., dkx = x.

Theorem (eigenvector cycle lengths):
The eigenvectors x associated with
eigenvalue 1 of the kth derivative matrix
are those vectors with cycle length k (i.e.,
vectors such that dkx = x).
Eigenvector Implication

Implication: If we want to find the vectors
that have cycle length k, we just have to
find the eigenvectors of Dk.

So use linear algebra to find those
eigenvectors!

I leave that to you.
The Identity Matrix


If Dkx = x, then that means Dk is some kind of
identity matrix.
Is it
1
0
0

0
0 0
1 0
0 1
0
0
0

0 0 0
0
0
0
1
?
Not the Real Identity

No! The real identity matrix would imply
that any and every vector cycles.

For example, consider 000001.
Does not cycle.
 Why not? The “odd order corollary” says its cycle
length is zero.
 But if Dk is the standard identity then that implies
00001 will multiply times Dk to give itself. i.e., has
cycle length k. Oops!

Even Identity

Is there a matrix that lets 000011 cycle but not
000001?
0

1
I  1


1


Yes! Let

That’s our identity!
1 1  1

0 1  1
1 0
1

 
1 1  0 
Group Theory

And now with an appropriate identity we can
take a small diversion.

Some of you may not have taken group theory.

Don’t panic!
I will explain what you need.
We will barely use it, and I will not test you on it.
Just want you to consider something delicious and cool.

Sit back and enjoy the ride.

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
What’s a Group?

Any collection of objects G with an
operation “.” that satisfies the following
properties.
1.
2.
3.
4.
Closure: if a, b are in G, then a.b is also in G.
Associative: a.(b.c) = (a.b).c.
Identity: There exists and identity e such that a.e
= e.a = a for all a in G.
Inverse: For every a in G there exists a-1 such
that a-1.a = e.
Example of a Group

The set of all integers ..., -2, -1, 0, 1, 2, ...
with operation “+”.
i.e., a.b = a+b.
 Closure obvious.
 Associative obvious.
 0 is the identity.
 -a is the inverse of a.


This is an infinite sized group.
Example of Finite Sized Group

The numbers 1 and -1 with the operation
multiplication.
Identity is 1
 Inverse of 1 is 1. Inverse of -1 is -1.


Has size or order of 2.
Example of Geometric Group

Rotations.

Consider a triangle.



Can rotate 120 degrees three times before back at beginning
position.
 0th rotation is the identity e. (No rotation.)
 1st rotation is a.
 2nd rotation is a2
 3rd rotation is a3 = e
So we have a group {e, a, a2}.
 What’s the inverse of each element?
Consider a pentagon.

Rotations create group {e, a, a2 , a3 , a4}.
Rotations and Reflections Group

Consider triangle with rotation “a” AND reflection
“b” about vertical axis.



Have e, a, a2 for rotations.
And e, b for reflections (two reflections give back e).
Also have ab (rotation followed by reflection).
 And ba, a2b, ba2.
 Note ab does not equal ba! But ba = a2b and ab = ba2.

Try it!
 Draw a triangle.
 Label vertices 1, 2, and 3.
 Now try combinations of rotations and reflections.

Whole group is {e, a, a2, b, ab, ba}.
 Other combinations give one of the elements listed above.
Cyclic Groups

Consider an integer n. Construct a group of
order n by considering any symbol “a” and the
symbols a0, a1, a2, a3, ...an-1 where we insist that
a0 = an = e, ai . aj = ai+j mod n.

Called cyclic groups. “a” is the generator.

Example: triangle rotation group!

Example: integers modulo n with addition.

Consider modulo 5. Have 0, 1, 2, 3, 4.
 Generator is 1.
 Identity is 0.
 Inverse is the number plus 5 (ai+5 . aj = 0).
Rule 102 Derivative Matrix Group

Now coolness! The kth derivative matrices form a cyclic
group.

Theorem (derivative matrix group): Suppose Dkx = x.
Then D1, D2, ...Dk are a cyclic group of order k.

Proof:



Identity is I = Dk (the “even” identity matrix).
Inverse of Di is Di+(k-1).
Associative and closure are obvious.
Who Cares if it is a Group?

Can now use group theory to prove results
about rule 102!

Example:

Subgroups are subsets of a group that also form their own
group.
 0, 1, 2, 3 with addition mod 4. Has subgroup 0, 2.

Subgroups always have a size that divides the size of their
parent group.
 Whoa, is that cool or what!
Cycle Length Divisors

So consider subgroups of the “derivative matrix
group”.

Their size will divide k (the cycle length).
Can prove from this that any vector x will have a cycle length
that divides the cycle length of 000...011.

That’s the divisor theorem again!

But can now see how to expand the divisor theorem to find
out exactly what vectors will be in each cycle.

An Aside on Other CA and Groups

I find group theory to be beautiful. If you do too,
you may want to explore the following.

Many CA correspond to a group!

How? Treat each state of the CA as a group
element.

E.g., 0 and 1


But call them a and b.
Or 0, 1, and 2

But call them a, b, and c.
Group Z2

Now consider rules for their combination.

E.g., for rule 102




aa = a
ab = b
ba = b
bb = a
(0+0 = 0)
(0+1 = 1)
(1+0 = 1)
(1+1 = 0)

Note the identity, associativity, inverse, closure.

This group is called Z2.

The integers 0, 1, ..., n with sum mod n form the “cyclic
groups” Zn.
Evolution of Group Elements

Now consider the time evolution as a CA.





a
a
aa
b
ab
aaab
aaababba
a
ba
abba
...
...
...
...
The patterns that we see in the CA (e.g., fractal)
reflect underlying structures in the groups!
CA with “Reflection”

So now treat a and b as rotation and
reflection of a triangle!

Then have other states as well (besides a and b).
 Remember {e, a, a2, b, ab , ba}?
 So have 6 states.
 Each one of these states should get a different color.

Now can look at the CA picture.
a
a
aa = a2
b
ab
ba
aaab = b
abba = a2
aaababba = ab
a
...
...
...
...
Explore CA With Groups

Fascinated?




Explore what CA correspond to what groups.
For a given CA, explore what group elements form different
visual patterns.
Find subgroups, cyclic subgroups, normal subgroups, cosets,
etc. in the visual patterns.
Group theory is deep and powerful. There are
many other possibilities for exploration.

Perhaps good idea for class final project...
One More Approach: Polynomials

Just briefly, another avenue is to treat each
vector as a polynomial.

Consider a = (a0, a1, a2, a3, ..., an). Write this
as a polynomial
i n
i
i
i 0
p( x)   a x
Derivative as Polynomial
Multiplication

The derivative is found by multiplying by
1
x 1

Why?
Multiplying a polynomial by xj shifts each coefficient j places to
the right.
 Multiplying a polynomial by x-j shifts each coefficient j places to
the left.


So x-1+1 adds the (i+1)st polynomial term to the ith term.
The term to the right has
been shifted to the left.
Polynomial Example
p( x)  1011  1x 0  0 x1  1x 2  1x 3
d ( x)  (1x 0  0 x1  1x 2  1x 3 )( x 1  1)
1
 (1x  0 x  1x  1x ) 
0
1
2
(1x 0  0 x1  1x 2  1x 3 )
1
 1x  1x  1x  0 x  1x
0
1
2
3
Term due to x-1.
Term due to 1.
But wait! Now, my polynomial is
too long. See next slide.
But basic idea is to wrap around
that term and add to the end.
Boundary Conditions

How do you get wrap around?

Take the polynomial modulo (xn+1).

In other words, the remainder after dividing by
xn+1.

Yeah, slight pain in the rear.
Advantage of Polynomial
Approach?

Easily generalized.

Suppose we are using rule 90.


That’s the same as adding the left and right neighbors mod
2.
Try it!





Try writing out rule 90.
000 goes to what?
001 maps to what?
Etc.
Now the derivative is found by multiplying by
x 1  0  x
Other Rules

Can look at any other additive rules by
changing the derivative polynomial.

For example, what rule number has a
derivative that is found by multiplying by
1 x
1
Reference

The polynomial approach was used by
Stephen Wolfram.

See his 1984 paper, Algebraic properties of
cellular automata, in Communications in
Mathematical Physics, vol. 93, pp. 219-258.

Good stuff.
Try Exploring!

I leave it to you to explore with
polynomials.
And groups
 And linear algebra
 And ...!
