Held: Wednesday, 2 November 2005
Summary:
Today we conclude our introduction to algorithm analysis.
Related Pages:
Notes:
- Some concern has been raised about the daily assignments. We will discuss those concerns.
- Expect to receive a reading for Friday's class.
- Exam 2 will be distributed on Friday.
Overview:
- Analyzing search methods.
- Recurrence relations, revisited.
- Your homework for today was to finish
the lab on algorithm analysis.
- How did you test sequential search? (Or, how might we test sequential search?)
- How did you build a sorted vector for testing binary search?
- How did you test binary search? (Or, how might we test sequential search?)
- What did you find?
- Recurrence relations provide a method of analyzing recursively defined
methods.
- We begin by defining a function, t(n), that gives the
running time of sum for a list of length n.
- We usually have two cases for t(n).
- A base case that corresponds to the worst of the base cases for the method.
- A recursive case that corresponds to the worst of the recursive cases for the method.
- Our goal is to remove the recursion from the definition.
- The strategy:
- Repeatedly expand a recursive call in the recursive definition.
- Look for a pattern.
- Express that pattern in terms of a variable, k. (k represents how
many times we've expanded).
- (For more advanced students: Prove the correctness of the pattern.)
- Determine for which k we can use the base definition of t.
- Substitute into the pattern.
- For example, let's consider the recursive sum procedure
- t(0) = c, for some constant c, because we just have do do the test and return 0.
- t(n) = d + t(n-1), because we need to do a test, an addition, a car, and a cdr plus a recursive call with a list of size n-1.
- Let's repeatedly expand
- t(n) = d + t(n-1)
- t(n) = d + d + t(n-2) = 2d + t(n-2)
- t(n) = 2d + d + t(n-3) = 3d + t(n-3)
- t(n) = 3d + d + t(n-4) = 3d + t(n-4)
- I see the pattern t(n) = k*d + t(n-k)
- Determine when we can eliminate the t(n-k). When n-k = 0, t(n-k) = t(0) = c.
- Substituting k for n, t(n) = n*d + c
- Therefore, t(n) is in O(n).
- Here are some other recurrences to study
- t(n) = d + n + t(n-1)
- t(n) = d + t(n/2) // Binary search
- t(n) = d + 2*t(n/2) // Generic divide-and-conquer
- t(n) = 2*t(n/2)
- t(n) = d + n + t(n/2) // ???
- t(n) = d + 2*t(n-1) // Upper bound on Fibonacci
- t(n) = d + 2*t(n-2) // Lower bound on Fibonacci
;
Check with Bobby