COMS 482: unofficial class blog

• 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:

1. Give an algorithm
2. Give a proof of correctness
3. 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 r^th interval is at least as good as the r^th interval in the optimal set”

A contains no overlapping intervals. Let {i₁, … , i_k} be the set of segments added to A, in order, and let {j₁, … , j_k} 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(i_r-1) < = f(j_r-1). We know that f(j_r-1) < = s(j_r) because all the segments in S are compatible. With our inductive assumption, we can write f(i_r-1) < = s(j_r), thus j_r has not been already selected or eliminated when we choose i_r. Since the greedy algorithm chooses the interval with the smallest finish time, we pick one equal or smaller to j_r: i_r < = j_r.

“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+1^th request after both i_k and j_k end. This means that after deleting all requests that are incompatible with {i₁, … , i_k} the set of possible requests R still contains j_k+1. But the greedy algorithm stops on request i_k, and it only stops when R is empty–a contradiction!

This entry was posted on Friday, January 28th, 2005 at 11:29 am and is tagged with greedy algorithms, greedy algorithm, stable marriage, time interval, course packets, optimal solution, finish time, proof of correctness, contradiction, segments, running time, intervals, assumption, lt, conflicts, segment, bookstore, nbsp, marriage. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback.

One Response to 'Lecture 3: Greedy Algorithms'

1. 
   The COMS 482 Blog » Blog Archive » Final Review: Chapters 1 to 4 said:

   on May 14th, 2005 at 9:42 pm

   […] ts until done. Proof proceeds by “stays ahead” argument. See lecture 3 for a proof of optimality. Interval P […]

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