Notes.txt

Slide 1: Hi, I am going to talk about the patterns formed by the pascal's triangle
Slide 2: 
Slide 3:
Slide 4: It is a triangular array of binomial coefficients 
Slide 5: It observed by coloring odd cells black and even cells white,
Slide 6:
Slide 7: We can look at the binomial expansion of 1+x, we know each coefficient is given by the formula --
Slide 8: But this is a inefficient was computing whether the coefficient is divisible by 2 or not since
Slide 9: The idea of coloring cells, depending on the states of it's neighbors is the basic idea of cellular automata, which I am going to talk about in more detail
Slide 10: // read the slides nothing much here
Slide 11: As you can see, each new generation of cells is only depended on the previous generation. This is the basic idea of Cellular automata.
Slide 12: Let's dig a bit more into 1D CA. The simplest kind is Elementary CA. Next generation of cells depend only cell itself, and it's left and right neighbors. If you order all the rules, you can number each rule. You need 8 rules to completely define the CA, so the total number of rules are 256. 
Slide 16: Let's play a Game, The Game of life! Each cell has 8 neighbors. The state of the cell is completely defined by the number of alive neighbors. This is an important note, the relative position of the alive and dead cells don't matter for this specific rule. It's also known as a totalistic CA 
Slide 18: Show one out of eight rule 
Slide 19: Remember, Game of life is just one set of rules. "Show majority rule"
Slide 20: Let's now look at a different set of neighbors, instead of 8, we will only look at 4 neighbors

Slide 23: Let us explore the connection between CA and polynomials a bit further. Let's look at a simple example. So technically a(n) gives the nth row in CA. Looking at the divisibility property of a(n) with 2 gives 2 possibilities. Adding them according to recursive formula in mod 2 arithmetic gives -, but this is exactly same as our previous look-up table
Slide 24: So this figure indirectly shows the binomial coefficient modulo 2, isn't that a lot of information in a small picture. I find it amazing that you can find if a number is divisible by an integer even without knowing the number, you can also get the remainder. 

Slide 25: So how can we generalize this ?
Slide 29: Since the new generation is only depended on it's "neighbors", we can make a look-up table like before and run a cellular automata. 

Slide 31: We looked how divisibility of a cell or number is depended on it's neighbors in the pascal triangle. We will look at different approach. It turns out the location of the cell itself can say if it is divisible by p or not 

Slide 34: Before we move on, let us formally define our problem in terms of sets. 

Slide 38: Remember (n,k) is the coordinate of (n+k)C(k). 

Slide 39: So Kummer's result says that the number of prime factors in (n+k)C(k) is equal to number of carries in the p-adic addition of n and k. In other words, the prime facrotization of (n+k)C(k) contains exactly cp factors of p. Looking at the example, Kummer's result implies the following result 

Slide 40: So for a binomial coordinate to have no factors of p, it must have no carries in the p-adic addition of it's coordinates. ...... It is the same mod-p condition that we used to define our figures


Slide 42: To have a global view of the patterns generated by Pascal triangle, we fall upon to Iterated function system. 

Slide 21: Till now we looked at the remainders of binomial coefficients with an integer p. We will again shift back to to the problem of coloring pascal triangle with just 2 colours. 


I need to think about transistions