Some Additional Notes

• Just remember the notation
  • $f(n) = \sqrt(n) \in O(n)$ therefore $f(n)$ has polynomial running time
• Logs
  • $\log_b n = x $ means that $b^x = n$
  • We frequently deal with $\log_2 n$ which is sometimes written $\lg n$
  • $\log_e n$ is frequently written $\ln n$
  • The relationship between logs in different bases
    • $$\begin{eqnarray} \log_a n = x \\ a^x =n \\ \log_ba^x = \log_bn \\ x \log_b a = \log_b n \\ \log_a n \log_b a = \log_b n \\ \log_a n = \frac{\log_b n}{\log_b a} \\ \end{eqnarray}$$
    • So the logs in different bases are just a constant $\frac{1}{\log_b a}$ of another.
    • Because of this we will generally work in base 2.
    • Base 2 is natural for computing
    • So $\log_a n \in \Theta(\log_b n)$
    • And we generally write that something is $O\log n$, even if we worked in another base.
  • They have a discussion of implementing GS Stable Matching.
    • you should read this.
    • Note the input will be $O(n^2)$ because there are $n$ men, each with a preference list containing $n$ items.
    • As we discussed before, you need to be careful about implementation or you might be $O(n^3)$ or higher.
    • In my implementation, i used a index list.
      • I used a 0 based ranking system (so array indexes worked)
      • If woman w ranks five men 3 1 0 2 4

      • Man ID	Rank$_0$
        3	0
        1	1
        0	2
        2	3
        4	4

      • So I buit an index array (IA)

        Index	0	1	2	3	4	Man I
        value	2	1	3	0	4	Has rank

      • This way I can determine the preference of man n in constant time.
      • Does she prefer man 2 or man 3?
        • Man 2's preference IA[2] = 3
        • Man 3's preference IA[3] = 0
        • IA[3] < IA[2]
        • She prefers man 3.
      • I am trading space (store a second array) for time.