A Survey of Common Running Times Part II

• Quadratic time
  • $O(n^2)$
  • When you need to compare every item to every other item.
    • And the comparison is constant.
  • Think sorting. (insertion, selection, bubble ...)
  • They pose a problem
    • Assume you have n points in a plane $P = \{p_1, p_2, ... p_n\} = \{(x_1, y_1), (x_2, y_2), .... (x_n, y_n)\}$
    • Find $p_i, p_j \in P $ such that the distance is minimal.
    • What is a brute force algorithm to do this?
    • What is the performance of this algorithm?
    • This can actually be done faster than $O(n^2)$
• Cubic time
  • Given $S = \{S_1, S_2, ... S_n\} \ni S_i \subset \{1, 2, ... , n\} ~ \forall ~ i$
  • Find a pair $S_i, S_j \in S \ni S_i \cap S_j = \{\}$

  • 
     For each pair of sets Si, Sj
         Determine if the intersection is empty.
     EndFor
     

  • $O(n)$?
    • What is line 1?
      • It is really nested for loop, so $O(n^2)$
    • What is line 2?

      • 
         For every element in Si
            Check to see if it is an element in set Sj
         EndFor
         

      • Assuming we can do line 2 in constant time this is O(n)
      • Because each set might have n/2 elements.
      • So line 1 is $O(n/2) \in O(n)$
  • So this is probably $O(n^3)$, but may be $O(n^4)$
  • Interestingly they state that there are algorithms that perform better than $O(n^3)$ but

    ... but they are quite complicated. Furthermore, it is not clear whether the improved algorithms for this problem are practical on inputs of reasonable size.

  • What does that last sentence mean?
• $O(n^k)$, where k is a constant.
  • Consider the independent subset selection problem.
  • Does an independent subset of size k exist in a graph G?
    • $G=\{v,e\}, |v| = n, n > k$

  • 
     For each subset of size k
         Check to see if S  is an independent set
     EndFor 
     

  • What is line 1 here?
    • There are ${n \choose k}$ combinations possible
    • ${n \choose k} = \frac{n!}{k! (n-k)!} $
    • $ \frac{n!}{k! (n-k)!} = \frac{n(n-1)(n-2)...(n-k+1)}{k!} $
    • $ n(n-1)(n-2)...(n-k+1) \lt n\times n\times n ... \times n $
    • The second side of this equation has k terms.
    • $ n(n-1)(n-2)...(n-k+1) \lt n^k $
    • ${n \choose k} \lt \frac{n^k}{k!} $
    • But $k!$ is a constant.
    • So line 1 is $O(n^k)$
  • Line 2 is at most $O(k^2)$, as there are k elements in each sub graph.
    • But $k^2$ is constant.
  • So the entire algorithm is $O(n^k)$ for some constant k.
  • They state that we really don't know a better algorithm.
• Hard problems
  • Not thought to be solvable in polynomial time.
  • Finding an independent subgraph of maximal size.
    • We did this for size k above.

    • 
       largest = {}
       For each subset S of nodes
          If S is independent and |S| > |largest|
             largest = S 
          EndIf
       EndFor
       

    • Uber brute force.
    • The total number of subsets for a set of n elements is $2^n$
    • So line one is $O(2^n)$
  • They point out that n! > $2^n$