Week 3 notes

The primary goal of this course is to become familiar with some of the basic mathematical ideas used in programming.

Reading:

        Levin, 1.1, 1.2

Counting: Levin 1.1, on page 57

• Additive principle
• Multiplicative principle
• Set-theoretic approach
• Pies on page 64

Counting subsets of {1, 2, ..., N}, Levin 1.2 on page 70

• Counting all subsets (including ∅)
• Counting all nonempty subsets
• Counting all subsets of size 1
• Counting all subsets of size 3; = # of subsets of size 2
• Binomial coefficients
• Bit Strings, B(n,k) = # of bit strings of length n with k 1's (weight k)
• Recurrence relation: B(5,3) = B(4,3) + B(4,2)
• Lattice Paths

Wednesday

Probably the best way to understand counting all k-element subsets of {1,2,...,n} is first to count the k-element sequences (with no repeats),
using the multiplication principle, and then divide by k! because one set of size k corresponds to k! sequences.

Levin does this in Section 1.3.

Recurrence relationships:

    factorial: fact(1) = 1, fact(n+1) = (n+1)*fact(n)

    fib(0) = fib(1) = 1, fib(n+1) = fib(n) + fib(n-1)

Now for C(n,k), or (n choose k). Let us define it to be n! / (n-k)!k!, or n*(n-1)*...*(n-k+1) / k!.

Claim: C(n,0) = C(n,n) = 1, C(n+1,k) = C(n,k) + C(n,k-1)

Claim: number of k-weight bitstrings of length n is equal to the number of k-sized subsets of {1,...,n}.

Proof: Natural one-to-one correspondence of bitstring s to {i<=n | s(i) = 1}, that is,  <0,1,1,0,0,0,1,0> matches to {2,3,7}
(wait, are bitstrings indexed starting at 0 or starting at 1?)

Recurrence relation for bitstrings: B(n,0) = B(n,n) = 1, B(n+1,k) = B(n,k) + B(n,k-1)

Therefore B(n,k) = C(n,k)

This is an alternative proof!

Pascal's triangle, using the recurrence rule

Binomial theorem: expansion of (x+y)ⁿ  

Counting lattice paths: same as bitstrings, where 1 = horizontal and 0 = vertical

Pascal's Triangle

                         1
                       1   1
                     1   2   1
                   1   3   3   1
                 1   4   6   4   1
               1   5  10  10   5   1
             1   6  15  20  15   6   1
           1   7  21  35  35  21   7   1
         1   8  28  56  70  56  28   8   1

Pascal's triangle modulo 2:

See also here.

Counting the single value at the top as row 0 (so the row (1 3 3 1) is row 3), we can prove by induction that all entries in row 2ⁿ of Pascal's Triangle are odd, and therefore in the mod-2 version are all 1. From this we can then prove the repeating pattern evident above.