Lecture 3: Greedy Algorithms
- The schedule is now online.
- HW1 is due on Friday
- If the bookstore is out of course packets, put in a request for one so that they’ll print more. Otherwise you might not get one.
(4.1 - 4.6) Greedy Algorithms:
- Always choose a local optimum
- Stable Marriage is an example of a greedy algorithm
Solving an algorithms problem:
- Give an algorithm
- Give a proof of correctness
- Analyze running time
Interval Scheduling:
Some suggestions for choosing the interval segments. Choose the:
- First to finish
- x Shortest one
- x First left endpoint
- x One with fewest conflicts
S = {segments}
A = {}
while S != {} do
choose s in S with minimum right endpoint
A = A U {s}
delete all s’ in S that overlap with s
end
“The rth interval is at least as good as the rth interval in the optimal set”
A contains no overlapping intervals. Let {i1, … , ik} be the set of segments added to A, in order, and let {j1, … , jk} be the set of segments in O, the optimal solution. Since by our rule f(i1) < = f(j1), we see that we "stay ahead" of the optimal solution at each step by choosing the "shortest" possible segment. This allows us to assume that f(ir-1) < = f(jr-1). We know that f(jr-1) < = s(jr) because all the segments in S are compatible. With our inductive assumption, we can write f(ir-1) < = s(jr), thus jr has not been already selected or eliminated when we choose ir. Since the greedy algorithm chooses the interval with the smallest finish time, we pick one equal or smaller to jr: ir < = jr.
“A is an optimal solution”
By contradiction, if A isn’t optimal, then the optimal set O will have more requests. This would mean that (a) there must be m > k requests, and (b) O makes a k+1th request after both ik and jk end. This means that after deleting all requests that are incompatible with {i1, … , ik} the set of possible requests R still contains jk+1. But the greedy algorithm stops on request ik, and it only stops when R is empty–a contradiction!
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