Category Archives: Asides

Mathematical Aside: Golden Mean Shift and Pascal’s Triangle (Part 3)

As was derived in Part 2, |c_{m,n}|= {n-(m-1)}\choose{m} where c_{m,n} denotes the set of binary strings of length n with exactly m non-adjacent ones.

Today we will analyze c_{n} the set of binary strings of length n with any number of non-adjacent ones. We will need some additional notation.

  • c_{n,0} denote the the subset of c_{n} ending with 0
  • c_{n,1} denote the the subset of c_{n} ending with 1

After some thought, we realize that c_{n,0} =c_{n-1} and c_{n-2}. We can then derive the recurrence |c_{n}| = |c_{n,0}|+|c_{n,1}|=|c_{n-1}|+|c_{n-2}|. After verifying the base case, we can conclude |c_{n}| = F_{n+2}, the Fibonacci Numbers!

Another way to compute |c_{n}| is |c_{n}| = \sum_{m=0}^{\infty} {n-(m-1)}\choose{m}.

Therefore F_{n+2} = \sum_{m=0}^{\infty} {n-(m-1)}\choose{m}.

With some shifting, we can solve for the n^{th} Fibonnaci number :

F_{n} = \sum_{m=0}^{\infty} {n-m-1}\choose{m}

While this formula can be quite easily verified inductively, its nice to construct it explicitly.

 

Mathematical Aside: Golden Mean Shift and Pascal’s Triangle (Part 2)

As we discussed in Part 1, we noticed a common pattern in the the m^{th} element in the n^{th} diagonal, d_{m,n},  of Pascal’s Triangle and the function c_{m,n}, denoting the set of binary strings of length n with exactly m non-adjacent ones. Today we derive an explicit formula for c_{m,n} in two ways.

Firstly, some more notation. The element in r^{th} row and k^{th} column in Pascal’s Triangle is denoted by {r}\choose{k}. Then we can see d_{m,n} = {n+m-1,m}. Unfortunately, c_{m,n} is at a slight shift. To see this, we think about how long the string needs to be just so can fit in m ones. The answer is 2(m-1). While the d_{m,n} is non-zero straight away, it takes c_{m,n} 2(m-1) increments to the length before it is non-zero. Therefore, |c_{m,n}|=d_{m,n-2(m-1)}= {n-(m-1)}\choose{m}.

Another way to see this requires knowledge of combinations.  If we consider the unconstrained system, the clearly we have u_{m,n} {n}\choose{m} options. We can construct a bijection from u_{m,n-(m-1)} to c_{m,n} as follows:

  • For every element in u_{m,n-(m-1)}, we can add append a 0 to all but the rightmost 1 and construct a unique element in c_{m,n}
  • For every element in c_{m,n}, we can add delete a 0 from the right side of all but the rightmost 1 and construct a unique element in u_{m,n-(m-1)}

Thus |c_{m,n}| = |u_{m,n-(m-1)}|= {n-(m-1)}\choose{m} as required.

We will use this to find an intriguing connection to the Fibonacci numbers in Part 3!

Mathematical Aside: Golden Mean Shift and Pascal’s Triangle

I noticed something curious when I was working on my summer project last year. Basically, you consider the set c_n of binary strings (strings of length n where each symbol is either a 1 or a 0 and 1‘s are non-adjacent).  So we construct a table like

 

Length 1 2 3 4 5 6
Strings 0;1 00;01;10 000;001;010;100;101 0000;0001;0010;0100;0101;1000;1001;1010

And then count the number of elements of length n with m ones.

Number of ones [m] (below) Number of strings (below)
Length of string [n] (right) 1 2 3 4 5 6
0 1 1 1 1 1 1
1 1 2 3 4 5 6
2 0 0 1 3 6 10
3 0 0 1 4
4 0 0

 

Do you see Pascal’s Triangle hiding inside? Starting from the top left and diagonally down to the left. First one is 1,2,1, then is 1,3,3,1 etcThe explanation is simple enough. We are using the fact that the subset of  c_n composed of strings with m ones (lets call this c_{n,m}) can be made by taking an element in c_{n-1,m} and appending a zero to it or an element in c_{n-2,m-1} and appending a zero-one to it. We end up with the recurrence c_{n,m}= c_{n-1,m}+ c_{n-2,m-1}. If we consider d_{n,m} as the n^{th} element in the m^{th} diagonal, then we realize that d_{n.m} follows the same recurrence. This reccurence is well documented in the literature for calculating |c_n|, here we are just using this idea for subsets of c_n with exactly m ones to connect it with Pascal’s Triangle.

The adventure continues in Part 2!