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Computer Science 101. Fast Algorithms. What Is Really Fast?. n O(log 2 n) O(n) O(n 2 ) O(2 n ) 2 1 2 4 4 4 2 4 16 64 8 3 8 64 256 16 4 16 256 65536
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Computer Science 101 Fast Algorithms
What Is Really Fast? n O(log2n) O(n) O(n2) O(2n) 2 1 2 4 4 4 2 4 16 64 8 3 8 64 256 16 4 16 256 65536 32 5 32 1024 4294967296 64 6 64 4096 19 digits 128 7 128 16384 whoa! 256 8 256 65536 512 9 512 262144 1024 10 1024 1048576
Logarithms • Definition: The logarithm to base 2 of n is a number k such that 2k=n, so log2(n) = k • When we say log(n), we mean log2(n) • Example: 25 = 32 so log(32) = 5 • Another way to think of this is that log(n) is the number of times we must divide n by 2 until we get 1 • Note: log(n) is usually a fraction, but the closest integer will work for us
Improving Efficiency • Tweaking does not improve the big-O, so the change is not noticeable when N becomes very large • If we can go from O(n) to O(logn), we have really accomplished something
Example: Sequential Search set Current to 1 Set Found to false while Current <= N and not Found do if A(Current) = Target then set Found to true else increment Current if Found then output Current else output 0 If the data items are in random order, then each one must be examined in the worst case Complexity of sequential search = O(n)
Searching a Sorted List • When we search a phone book, we don’t begin with the first name and look at each successor • We skip over large numbers of names until we find the target or give up
Binary Search • Strategy • Have pointers marking left and right ends of the list still to be processed • Compute the position of the midpoint between the two pointers • If the target equals the value at midpoint, quit with the position found • Otherwise, if the target is less than that value, search just the positions to the left of midpoint • Otherwise, search the just the positions to the right of midpoint • Give up when the pointers cross 14 16 22 32 34 66 80 90 66 target left mid right
Binary Search • Strategy • Have pointers marking left and right ends of the list still to be processed • Compute the position of the midpoint between the two pointers • If the target equals the value at midpoint, quit with the position found • Otherwise, if the target is less than that value, search just the positions to the left of midpoint • Otherwise, search the just the positions to the right of midpoint • Give up when the pointers cross 14 16 22 32 34 66 80 90 66 target left mid right
The Binary Search Algorithm set Left to 1 Set Right to N Set Found to false while Left <= Right and not Found do compute the midpoint if Target = A(Mid) then set Found to true else if Target < A(Mid) then search to the left of the midpoint else search to the right of the midpoint if Found then output Mid else output 0
The Binary Search Algorithm set Left to 1 Set Right to N Set Found to false while Left <= Right and not Found do set Mid to (Left + Right) / 2 if Target = A(Mid) then set Found to true else if Target < A(Mid) then search to the left of the midpoint else search to the right of the midpoint if Found then output Mid else output 0
The Binary Search Algorithm set Left to 1 Set Right to N Set Found to false while Left <= Right and not Found do set Mid to (Left + Right) / 2 if Target = A(Mid) then set Found to true else if Target < A(Mid) then set Right to Mid – 1 else search to the right of the midpoint if Found then output Mid else output 0
The Binary Search Algorithm set Left to 1 Set Right to N Set Found to false while Left <= Right and not Found do set Mid to (Left + Right) / 2 if Target = A(Mid) then set Found to true else if Target < A(Mid) then set Right to Mid – 1 else set Left to Mid + 1 if Found then output Mid else output 0
Analysis of Binary Search • On each pass through the loop, ½ of the positions in the list are discarded • In the worst case, the number of comparisons equals the number of times the size of the list can be divided by 2 • O(logn) algorithm! However, must have a sorted list
Improving on N2 Sorting • Several algorithms have been developed to break the N2 barrier for sorting • Most of them use a divide-and-conquer strategy • Break the list into smaller pieces and apply another algorithm to them
Quicksort • Strategy - Divide and Conquer: • Partition list into two parts, with small elements in the first part and large elements in the second part • Sort the first part • Sort the second part • Question - How do we sort the sections?Answer - Apply Quicksort to them • Recursive algorithm - one which makes use of itself to solve smaller problems of the same type
Quicksort • Question - Will this recursive process ever stop? • Answer - Yes, when the problem is small enough, we no longer use recursion. Such cases are called base cases
Partitioning a List • To partition a list, we choose a pivot element • The elements that are less than or equal to the pivot go into the first section • The elements larger than the pivot go into the second section
Partitioning a List 19 8 15 5 30 20 10 1 28 25 12 Pivot is the element at the midpoint Partition 10 1 19 8 15 5 12 20 30 28 25 Sublist to sort Sublist to sort Data are where they should be relative to the pivot
The Quicksort Algorithm if the list to sort has more than 1 element then if the list has exactly two elements then if the elements are out of order then exchange them else perform the Partition Algorithm on the list apply QuickSort to the first section apply QuickSort to the second section
Partitioning: Choosing the Pivot • Ideal would be to choose the median element as the pivot, but this would take too long • Some versions just choose the first element • Our choice - the median of the first three elements
Partitioning a List 19 8 15 5 30 20 10 1 28 25 12 Pivot is median of first three items Partition 10 8 12 5 1 15 20 30 28 25 19 The median of the first three items is a better approximation to the actual median than the item at the midpoint and results in more even splits
The Partition Algorithm exchange the median of the first 3 items with the first set P to first position of list set L to second position of list set U to last position of list while L <= U while A(L) A(P) do set L to L + 1 while A(U) > A(P) do set U to U - 1 if L < U then exchange A(L) and A(U) exchange A(P) and A(U) A The list P The position of the pivot element L Probes for items > pivot U Probes for items <= pivot
Quicksort: Rough Analysis • For simplification, assume that we always get even splits when we partition • When we partition the entire list, each element is compared with the pivot - approximately n comparisons • Each of the halves is partitioned, each taking about n/2 comparisons, thus about n more comparisons • Each of the fourths is partitioned,each taking about n/4 comparisons - n more
Quicksort: Rough Analysis • How many levels of about n comparisons do we get? • Roughly, we keep splitting until the pieces are about size 1 • How many times must we divide n by 2 before we get 1? • log(n) times, of course • Thus comparisons n Log(n) • Quicksort is O(n Log(n)) – in the ideal or best case
Quicksort: Best Case/Worst Case • List is already sorted and we choose the pivot from the midpoint – leads to even splits all the way down, log(n) total splits • List is already sorted and we choose the pivot from the first position – leads to the worst splits at each level, n total splits • O(nlogn) in best case, O(n2) in worst case