Complexity Present sorting methods. Binary search. Other measures. Homework: [presentations]. Postings.

Big Oh mechanics Big Oh defines an upper bound. You get to pick a coefficient c (multiplier) and you get to pick an n beyond which the bound holds. Example: f(n) = n +1000 is bounded by n and also by n2. For O(n), try c = 1001 and n0 = 1. (n+1000) < 1001* (n) For O(n2), try c=1001 and n0 = 1 (n+1000) < 1001*(n2)

Big Oh But can we find a c and a n0 to make n2 be gounded by n? Answer: no. Because no matter what the coefficient c, let n be bigger than it, and n2 > n.

Another example n2+n+1 is O(n2) set c =4, n0 = 1, then (n2+n+1) < 4 * (n2) or set c = 3, n0 = 2, then (n2+n+1) < 3 * (n2) starting with n = 2 NOTE: don't worry about the c and the n0 being the best (tightest) bounds.

Bounds Distinction sometimes made between: mean bound average bound achieved bound There also is omega notation. posting opportunity

Refresher on log Definition: logbN is defined to be the value e such that be is equal to N. When using log for Big Oh, it doesn't matter what the base is because logs to different bases differ by a coefficient. log210 * log10N is equal to log2N. Proof: raise 2 to each Right side: 2 log210 * log10N is (2 log210 ) log10N is 10 log10N is N is the same as left side.

Compare logN and N log N is bound by N, that is, always less than N. Order O(c) constant less than O(logN) less than O(N) etc.

Sorts Sorts that are O(n2) when n is the size of the data grow much faster than sorts that are O(N * logN ) Demonstrate?

O(N) = linear algorithm Data set size 10 goes to size 100 An algorithm that is O(N) would go from time proportional to 10 to time proportional to 100, a factor of 10 increase.

O(N * log N) Data set size 10 goes to size 100 An algorithm that is O(N * log N) would go from time proportional to 10 * log 10 to 100 * log 100. Choose base 10, go from 10 * 1 to 100* 2, 10 to 200 is increase factor of 20.

O(N2) Data set size 10 goes to size 100 An algorithm that is O(N2) would go from time proportional to 100 to 100*100, increase by a factor of 100.

Compare Data set size 10 goes to size 100 An algorithm that is O(N) would go from time proportional to 10 to time proportional to 100, a factor of 10 increase. An algorithm that is O(N * log N) would go from time proportional to 10 * log 10 to 100 * log 100. Choose base 10, go from 10 * 1 to 100* 2, 10 to 200 is increase factor of 20. An algorithm that is O(N2) would go from time proportional to 100 to 100*100, increase by a factor of 100.

Sorting presentations heap sort quick sort merge sort

Heapsort http://www.ee.ryerson.ca/~courses/coe428/sorting/heapsort.html Method: represent a tree structure using a linear area. Repeated step: make the tree a heap, where heap means that the parent node is bigger than either of the child nodes The root (top) node will be the maximum, so swap it to the last position and then repeat for all up to but not including that last position. That swap probably means that it is no longer a heap, so heapify again and keep going.

Quicksort http://www.sorting-algorithms.com/quick-sort http://en.wikipedia.org/wiki/Quicksort Method: choose a value, called the pivot and then compare and divide the whole set into less than pivot and greater than pivot. Repeat for each part.

Merge sort http://en.wikipedia.org/wiki/Merge_sort http://www.sorting-algorithms.com/merge-sort Method: divide into sub lists (typically 2, but if distributed computing is available, would do more). Sort each list, possibly using merge sort. Combine by advancing (my word) the one that is the least.

Sorting Stability: does sort preserve original order for elements that are in order? This clearly is required for sorting on multiple fields. Is method easily (productively) parallelizable? Space: can it be done in-place?

Binary search Have a sorted set (key + data) with M entries. Have new data and want to find it in the set and replace OR insert in correct position. How to do it? compare key to item in the middle. If a match, okay. If not, if new key is less, repeat for the upper half, else repeat for the lower half. How many compares?

Answer Say M = 2N, then answer is N. Put another way, answer is ceiling(log2(M)) So binary search is O(log2(M)) This is less than linear… If size of set goes up factor of 4, then time for search goes up by factor of 2.

Other measures Space requirements Note: some sorts are done 'in place' Focus on specific operations, such as file reads or writes ?

Homework Extra credit: present another (very different) sort or search or … Postings Presentations