Math 298 -- Readings in Combinatorics -- Summer 2006


Course Overview

This summer reading course in combinatorics will cover basic topics in combinatorics, using the books: We will be roughly following the outline:
  1. Counting
    • Counting Handshakes and other simple things
    • Subsets and strings
    • Binomial Coefficients and Pascal's triangle
    • Fibonacci Numbers and Generating Functions
  2. Graphs and Trees
    • Matchings
    • Colorings
    • Spanning Trees and the Travelling Salesman problem (i.e. finding the shortest path)
  3. Geometry
    • Regular Solids
    • Projetive geometry and Error Correcting Codes

Outline of Meetings

  1. Friday 5/19/06 4:30-5:30pm (my office)
    • Overview of Combinatorics (counting handshakes, Pascal's triangle, graphs, colorings, travelling salesman)
    • Discuss Books, Syllabus, and decide on a final exam for evaluation.
    • Reading for next time: Ch5 of Tucker, Ch1 and 3 of Lovasz et. al.

  2. Monday 5/22/06 2-3:50pm (Vesic library)
    • Ask what they learned.
    • # of handshakes is a binomial coefficient, which counts (unordered) subsets of a set.
    • Justify/Prove the formula (n, k) = n! / ((n-k)! k!) for the binomial coefficients in general.
    • Discuss the fastest way of computing it, and why we write it this way.
    • Ask if they have any nice properties -- They said (n, k) = (n, n-k)
    • Can you prove it? They gave two proofs (by formula, or by a bijective correspondence)!
    • How would you get your students to discover this for themselves? They said compute (7, k) for all k.
    • Can easily see it from Pascal's triangle! =)
    • What other properties do the bimonial coefficients have? (i.e. identites can you see?)
    • They conjectured:
      • (n, k) = (n-1, k-1) + (n-1, k)
      • (n, k) = (n, n-k)
      • (n, 1) = (n, n) = n; (n, 0) = (n, n) = 1
      • (n, 0) + (n, 1) + ... + (n, n) = 2n
      • If p is prime, then p divides (p,k) for all k except k = 0 and k = p.
    • For next time: Be able to prove these identities (sometimes in 2 ways), and do all of the examples in the readings from last time. Also, try to look at problems (especially counting and binomial coefficient identities) to see how well you understand what is going on.

  3. Thursday 5/25/06 10:30am - 12:30pm (Vesic library)
    • Asked how the reading went, and then starte to try some problems from Chapter 5 of Tucker (2nd ed.)
      • p169, 1a: 4 * 3 = 12
      • p169, 1b: Two ways to do it, depnding on how we interpret the "Choices" in the question:
        8*7 = 56 (if the choices of the same letter are indistinguishable by their position)
        5*10 + 3 * 9 = 77 (otherwise, where the first term counts letters appearing once
      • XXXXXXXXXXXXXXXXXXXXXXX
      • XXXXXXXXXXXXXXXXXXXXXXX
    • XXXXXXXXXXXXXXXXXXXXXXXXX
    • XXXXXXXXXXXXXXXXXXXXXXXXX

  4. Monday 5/29/06 2:00--4:00pm (Vesic library)
    • They wanted to talk about
    • XXXXXXXXXXXXXXXXXXXXXXXXX
    • XXXXXXXXXXXXXXXXXXXXXXXX

  5. Monday 6/1/06 10:40--11:25am (Vesic library)
    • XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX
    • XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX

  6. Wednesday 6/7/06 10:30--12:15am (Vesic library)
    • Defined the Fibonacci numbers in terms of the 2-step recursion and F0 = F1 = 1.
    • Discussed how to write this as a generating function
      (G(x) = 1 + x + x2 + 3x3 + 5x4 + 8x5 + 13x6 + ...)
    • Described two ways of expressing the recursion relation
      (as Fn = Fn-1 + Fn-2 or as Fn - Fn-1 - Fn-2 = 0)
    • Found 2 ways of writing a generating function H(x) whose n-th coefficient is Fn - Fn-1 - Fn-2 :
      • H(x) = 1 by extending the Fibonacci numbers so F-1 = F-2 = ... = 0 and computing all coefficients of xn.
      • H(x) = (1-x-x2) G(x) by trying to express the recursion in terms of power series multiplication.
    • Using the formulas for H(x) above, we find that G(x) = 1/(1-x-x2).
    • Discussed that in general we can use this idea to write a generating function for any fixed lendth linear recurrence relation, and this generating function is a rational function!
    • Asked them to find the generating function of the "super-Fibonacci" numbers, defined by:
      • F0 = F1 = F2 = 1
      • Fn = Fn-1 + Fn-2 + Fn-3
    • After some discussion they computed it as G(x) = (1-x2)/(1-x-x2-x3). Yea! =)
    • Described the partial fraction/geometric series method for finding and explicit formula for the n-th Fibonacci number, and then we computed the general formula.
    • Mentioned that while this is not a great way to compute Fn, this formula tells us interesting things like:
      • Fn is very close to ((1+√5)/2)n+1 as n → ∞
      • Fn/Fn-1 is very close to (1+√5)/2 as n → ∞
    • For next time: Read Chapters 6.4 (exp generating functions) and 7 (recurrunce relations), and I will assign review problems tomorrow.

  7. Monday 6/12/06 10:30--XXXXam (Vesic library)
    • XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX
    • XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX