CS 112 Introduction to Programming Sorting of an Array Debayan Gupta Computer Science Department Yale University 308A Watson, Phone:

Sorting 2

3 Roadmap: Arrays  Motivation, declaration, initialization, access  Reference semantics: arrays as objects  Example usage of arrays m Tallying: array elements as counters m Keeping state  Manipulating arrays m Sorting an array

4 Sorting an Array  The process of arranging an array of elements into some order, say increasing order, is called sorting  Many problems require sorting m Google: display from highest ranked to lower ranked m Morse code

5 Sorting in CS  Sorting is a classical topic in algorithm design

6 Sorting an Array  How do we sort an array of numbers? int[] numbers = { 3, 9, 6, 1, 2 };

Many Sorting Algorithms  Insertion sort  Selection sort  Bubble sort  Merge sort  Quick sort  … 7

8 Insertion Sort  Basic idea: divide and conquer (reduction) m reduce sorting n numbers to sort the first n-1 numbers insert the n-th number to the sorted first n-1

9 insertPos = insertPos =3 insertPos =2 insertPos =1

10 Insertion PseudoCode // assume 0 to n – 1 already sorted // now insert numbers[n] // insertPos = n; // repeat (number at insertPos-1 > to_be_inserted) { // shift larger values to the right // numbers[insertPos] <- numbers[insertPos-1]; // insertPos--; // numbers[insertPos] <- to_be_inserted;

11 // insertPos = n; // repeat (number at insertPos-1 > to_be_inserted) { // shift larger values to the right // numbers[insertPos] <- numbers[insertPos-1]; // insertPos--; // numbers[insertPos] <- to_be_inserted; 0123

12 Refinement: Insertion PseudoCode // assume 0 to n – 1 already sorted // now insert numbers[n] // insertPos = n; // repeat (insertPos > 0 && number at insertPos-1 > to_be_inserted) { // shift larger values to the right // numbers[insertPos] <- numbers[insertPos-1]; // insertPos--; // numbers[insertPos] <- to_be_inserted;

13 Insertion Sort Implementation public static void sort (int[] numbers) { for (int n = 1; n < numbers.length; index++) { int key = numbers[n]; int insertPos = n; // invariant: the elements from 0 to index -1 // are already sorted. Insert the element at // index to this sorted sublist while (insertPos > 0 && numbers[insertPos-1] > key) { // shift larger values to the right numbers[insertPos] = numbers[insertPos-1]; insertPos--; } numbers[insertPos] = key; } // end of for } // end of sort

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Analysis of Insertion Sort  What is algorithm complexity in the worst case? 20 int[] numbers = { 3, 9, 6, 1, 2 };

21 Sorting Arrays: Bubble Sort  Scan the array multiple times m during each scan, if elements at i and i+1 are out of order, we swap them  This sorting approach is called bubble sort m  Remaining question: when do we stop (the termination condition)?

Sorting: Bubble Sort 22 public static void sort (int[] numbers) { boolean outOfOrder = false; do { outOfOrder = false; // one scan for (int i = 0; i < numbers.length-1; i++) { if (numbers[i] > numbers[i+1]) { // out of order // swap int x = numbers[i]; numbers[i] = numbers[i+1]; numbers[i+1] = x; outOfOrder = true; } // end of if } // end of for } while (outOfOrder); } // end of sort

23 Selection Sort  For the i-th iteration, we select the i-th smallest element and put it in its final place in the sort list

24 Selection Sort  The approach of Selection Sort: m select one value and put it in its final place in the sort list m repeat for all other values  In more detail: m find the smallest value in the list m switch it with the value in the first position m find the next smallest value in the list m switch it with the value in the second position m repeat until all values are placed

25 Selection Sort  An example: original: smallest is 1: smallest is 2: smallest is 3: smallest is 6:

26 Sorting: Selection Sort public static void sort (int[] numbers) { int min, temp; for (int i = 0; i < numbers.length-1; i++) { // identify the i-th smallest element min = i; for (int scan = i+1; scan < numbers.length; scan++) if (numbers[scan] < numbers[min]) min = scan; // swap the i-th smallest element with that at i temp = numbers[min]; numbers[min] = numbers[i]; numbers[i] = temp; } // end of for } // end of sort

Analysis of Selection Sort  What is algorithm complexity in the worst case? 27 int[] numbers = { 3, 9, 6, 1, 2 };

Roadmap  Both insertion and selection have complexity of O(N 2 )  Q: What is the best that one can do and can we achieve it? 28

Sorting: Merge Sort  Split list into two parts  Sort them separately  Combine the two sorted lists (Merge!)  Divide and Conquer! 29

Sorting 30 public void sort(int[] values) { numbers = values; // numbers has been previously declared mergesort(0, number - 1); } private void mergesort(int low, int high) { // check if low is smaller then high, if not then the array is sorted if (low < high) { // Get the index of the element which is in the middle int middle = low + (high - low) / 2; mergesort(low, middle); // Sort the left side of the array mergesort(middle + 1, high); // Sort the right side of the array merge(low, middle, high); // Combine them both }

Merging 31 private void merge(int low, int middle, int high) { // Copy both parts into the helper array for (int i = low; i <= high; i++) { helper[i] = numbers[i]; } int i = low, j = middle + 1, k = low; // Copy the smallest values from either side while (i <= middle && j <= high) { if (helper[i] <= helper[j]) { numbers[k] = helper[i]; i++; } else { numbers[k] = helper[j]; j++; } k++; } // Copy the rest of the left side of the array into the target array while (i <= middle) { numbers[k] = helper[i]; k++; i++; }

Merging

Sorting: Quick Sort  Select a random element  Compare it to every other element in your list to find out its rank or position  You have now split the list into two smaller lists (if a > x and x > b, then we know that a > b – we don’t need to compare!) 33

Quicksort 34 private void quicksort(int low, int high) { int i = low, j = high; // Get the pivot element from the middle of the list int pivot = numbers[low + (high-low)/2]; // Divide into two lists while (i <= j) { // If the current value from the left list is smaller then the pivot // element then get the next element from the left list while (numbers[i] < pivot) { i++; } // If the current value from the right list is larger then the pivot // element then get the next element from the right list while (numbers[j] > pivot) { j--; }

Quicksort.. Contd. 35 // If we have found a values in the left list which is larger then // the pivot element and if we have found a value in the right list // which is smaller then the pivot element then we exchange the // values. // As we are done we can increase i and j if (i <= j) { exchange(i, j); i++; j--; } // Recursion if (low < j) quicksort(low, j); if (i < high) quicksort(i, high); } private void exchange(int i, int j) { int temp = numbers[i]; numbers[i] = numbers[j]; numbers[j] = temp; }

What’s the best we can do?  N 2 ?  N log N  Why? 36

N log N  N! possible outcomes  If we compare two numbers, there are only 2 possible combinations that we can get  So, if we have x steps, then we can produce a total of 2 x combinations  To get 2 x > N!, we need x > N log N 37

Questions? 38