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Some additional material to ponder…. The additional slides with example were found at: www.cs.utoronto.ca/~tabrown/csc263/week1.ppt The introduction to Big O, Omega, and Theta was found at: https://www.youtube.com/watch?v=6Ol2JbwoJp0 Insertion sort discussion (tie in with the lab this week):
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Some additional material to ponder… The additional slides with example were found at: www.cs.utoronto.ca/~tabrown/csc263/week1.ppt The introduction to Big O, Omega, and Theta was found at: https://www.youtube.com/watch?v=6Ol2JbwoJp0 Insertion sort discussion (tie in with the lab this week): https://www.youtube.com/watch?v=c4BRHC7kTaQ http://bigocheatsheet.com/
Analyzing algorithms using big-O, omega, theta • First, we analyze some easy algorithms. • Most of them have the same running time for all inputs of length n.
Example 1 char A[100] for i = 1..100 print A[i] • Running time: 100*c = Theta(1) (why?) • O(1), Omega(1) => Theta(1)
Example 2 char A[n] for i = 1..n print A[i] • Running time: n*c • O(n), Omega(n) => Theta(n)
Example 3 int A[n] for i = 1..n binarySearch(A, i) • Remember: binarySearch is O(lg n). • O(n lg n)
Example 4 char A[n] for i = 1..n for j = 1..n print A[i] • O(n^2), Omega(n^2) => Theta(n^2)
Example 5 char A[n] char A[n] for i = 1..n >= for i = n/2..n for j = 1..i for j = 1..n/2 print A[j] print A[j] • Big-O: i is always <= n, so j always iterates up to at most n, so this is less work than n*n, which is O(n^2). • Big-Omega: • Useful trick: consider the iterations of the first loop for which i >= n/2. • In these iterations, the second loop iterates from j=1 to at least j=n/2. • Therefore, the time to perform just these iterations is at least n/2*n/2 = (1/4) n^2, which is Omega(n^2). • The time to perform ALL iterations is even more than this so, of course, the whole algorithm must take Omega(n^2) time.
Example 6 char A[n] char A[n] for i = 1..n >= for i = n/2..n for j = 1..i for j = n/4..n/2 for k = 1..j for k = 1..n/4 for l = 1..7 for l = 1..7 print A[k] print A[k] • Big-O: i <= n, and j <= n, so running time < n*n*n*7 = c*n^3 = O(n^3). • Big-Omega: • Consider only the iterations of the first loop from n/2 up to n. • For these values of i, the second loop always iterates up to at least n/2! • Consider only the iterations of the second loop from n/4 up to n/2. • For these values of j, the third loop always iterates up to at least n/4! • Consider only the iterations of the third loop up to n/4. • The fourth loop always iterates up to 7. • Total iterations: n/2 * n/4 * n/4 * 7 = (7/32) n^3 = Omega(n^3)
Analyzing algorithms using big-O, omega, theta • Now, we are going to perform some slightly different analysis. • The next two algorithms’ running times vary widely for different inputs of length n. • For some length n inputs, they terminate very quickly. • For other inputs n inputs, they can be slow.
Example 1 LinearSearch(A[1..n], key) for i = 1..n if A[i] = key then return true return false • Might take 1 iteration… might take n... • Big-O: at most n iterations, constant work per iteration, so O(n) • Big-Omega: can you find a bad input that makes the algorithm take Omega(n) time?
Example 2 • Your boss tells you that your group needs to develop a sorting algorithm for the company's web server that runs in O(n lg n) time. • If you can't prove that it can sort any input of length n in O(n lg n) time, it's no good to him. He doesn't want to lose orders because of a slow server. • Your colleague tells you that he's been working on this piece of code, which he believes should fit the bill. He says he thinks it can sort any input of length n in O(n lg n) time, but he doesn't know how to prove this.
Example 2 continued WeirdSort(A[1..n]) last := n sorted := false while not sorted do sorted := true for j := 1 to last-1 do if A[j] > A[j+1] then swap A[j] and A[j+1] sorted := false end if end for last := last-1 end while [Discuss… result is:] O(n^2), Omega(n^2) => Theta(n^2)!
