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by: Taylor Kahl

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Week 9 Notes CSC 2310

Taylor Kahl
GSU
GPA 4.21

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Notes on Chapter 13, Searching and Sorting
COURSE
Princliples of Computer Programming
PROF.
Kebina Manandhar
TYPE
Class Notes
PAGES
5
WORDS
KARMA
25 ?

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This 5 page Class Notes was uploaded by Taylor Kahl on Saturday March 12, 2016. The Class Notes belongs to CSC 2310 at Georgia State University taught by Kebina Manandhar in Winter 2016. Since its upload, it has received 12 views. For similar materials see Princliples of Computer Programming in ComputerScienence at Georgia State University.

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Date Created: 03/12/16
Chapter 13: Searching and Sorting  Sequential search – examines every value in an array of list from start to finish until it locates the target element o returns the index of the target element o returns –1 if element not found  Binary search – locates target element by eliminating half of the array or list at a time o have to import java.util.* or java.util.Arrays to use methods from the Array class o can only use on sorted lists o compares the target element to the middle element of the list  if target > middle, eliminate 1 half, search 2 ndhalf  if target < middle, eliminate 2 ndhalf, search 1 half o repeats the process until it returns the index of the target element o if the element is not found, returns –(insertionPoint + 1)  insertionPoint is the index where the target element would appear if it were in the sorted array o Example: int[a] = [1, 4, 17, 18, 20]. Target element = 6. 6 is not in the array, but if it were, it would be between 4 and 17, at index 2 binarySearch will return –(2 + 1) = – 3  Binary search is usually more efficient than a sequential search  Useful methods from the Arrays class: binarySearch(array, value) searches entire array binarySerach(array, minIndex, searches a range of indexes in the maxIndex, value) array, from minIndex(inclusive) to maxIndex(exclusive) copyOf(array, length) returns a copy of an array with a new length equals(array1, array2) returns true if arrays contain the same elements in the same order fill(array, value) sets every element in the array to given value sort(array) sorts elements toString(array) “[10, 30, 2]” Remember that these are static methods, so the syntax is: Arrays.methodName(parameters)  Binary search code: public static int binarySearch(int[] a, int target) { int min = 0; int max = a.length – 1; while (min <= max) { int mid = (min + max) / 2; if (a[mid] < target) { //too small min = mid + 1; } else if (a[mid] > target) { //too big max = mid – 1; } else { //a[mid] = target return mid; } } return – (min + 1) //if target isn’t found } Here’s how this code would use binarySearch(a, 6); for my example int[] a = [1, 4, 17, 18, 20] min = 0 max = a. length – 1 = 5 – 1 = 4 min <= max mid = (min + max) / 2 = (0 + 4) / 2 = 2 a[2] = 17 17 > 6 max = mid – 1 = 2 – 1 = 1 min <= max, go through the loop again mid = (min + max) / 2 = (0 + 1) / 2 = 0 a[0] = 1 1 < 6 min = mid + 1 = 0 + 1 = 1 min <= max mid = (min + max) /2 = (1 + 1) / 2 = 1 a[1] = 4 4 < 6 min = mid + 1 = 1 + 1 = 2 min > max, end the loop return –(min + 1) = – 3  A recursive method for binarySearch: o if the target is not found, return -1 o this method has a helper method public static int binarySearch(int[] a, int target) { return binarySearch(a, target, 0, a.length – 1); //calls the helper method } private static int binarySearch(int[] a, int target, int min, int max) {//extra parameters min & max if (min > max) { return -1; } else { int mid = (min + max) / 2; if (a[mid] < target) { //too small return binarySearch(a, target, mid + 1, max);//change min } else if (a[mid] > target) { //too big return binarySearch(a, target, min, mid – 1); //change max } else { return mid; //a[mid] = target } } }  binarySearch can also be used on an array of Strings Strings o strings have a natural ordering  they can be arranged alphabetically  a String is “less than” another if it comes first alphabetically o binarySearch uses the compareTo method o A.compareTo(B) returns:  < 0 if A before B  > 0 if A after B  0 if A == B Runtime efficiency:  Efficiency: a measure of computing sources required to run your code o usually refers to runtime  To calculate efficiency, assume that: o Every 1 java statement has the same runtime o A method’s runtime = total # of statements in its body o A loop’s runtime, if the loop runs N times = N * # of statements in its body  Example: Statement1; Statement2; 2 for (int i = 1; i <= N, i++) { statement3; 2N statement4; } for (int i = 1; i <= N; i++) { for (int j = 1; j <= N; j++) { 2 N*N = N statement5; N } } 2 This code will take N + 2N + 2 units of time to run  Growth rate – change in runtime as N changes. Runtime is proportional to the input data size, N  If N is extremely large, use the highest order of N to determine growth rate o Ignore constants or coefficients of N – we’re only concerned with how quickly runtime grows in proportion to N. The actual value doesn’t matter 2  Example: N + 2N + 2 o N dominates the overall run time  because if N is extremely large, 2N and 3 are so small compared 2 to N that we won’t count them 2 o This algorithm runs “on the order of” N o This algorithm is “Big-Oh” of N , written as O(N )2 3 2  Another example: 0.4N + 25N + 8N + 17 o This is O(N ). N is the highest order of N. Ignore the 0.4  Complexity class – category of algorithm efficiency based on algorithm’s relationship to N o an algorithm in the lowest complexity class (with the smallest order of N) is the most efficient and will run fastest Class Big-Oh If you double N, the growth rate constant O(1) doesn’t change logarithmic O(log 2) increases slightly linear O(N) doubles log-linear O(N log 2) slightly more than 2 doubles quadratic O(N ) quadruples cubic O(N ) multiplies by 8 exponential O(2 ) multiplies drastically – AVOID! this could take years to run  What’s the complexity class of binarySearch? o for an array of size N, it eliminates ½ of the elements until 1 remains o The number of elements goes in a series: N, N/2, N/4…4, 2, 1 o From the other direction, how many times do you need to multiply by 2 to reach N?  1, 2, 4…N/4, N/2, N  xxnumber of multiplications  2 = N (after multiplying by 2 x times, arrive at N)  x = log2N  logarithmic complexity class Happy Spring Break!!!

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