1. Analysis of Algorithms CS 477/677 Lecture 8 Instructor: Monica Nicolescu

2. CS 477/677 - Lecture 82 Quick Announcement Hand imaging collection (hand shape) Reza Amayeh: amayeh@cse.unr.edu Lab address: LME building room 314 Time: –Mon, Wed, Fri: 10:00 am until 5:00pm –Thu, Thr: 3:00 pm until 5:00pm

3. CS 477/677 - Lecture 83 How Fast Can We Sort? Insertion sort, Bubble Sort, Selection Sort Merge sort Quicksort What is common to all these algorithms? –These algorithms sort by making comparisons between the input elements To sort n elements, comparison sorts must make  (nlgn) comparisons in the worst case  (n 2 )  (nlgn)

4. CS 477/677 - Lecture 84 Decision Tree Model Represents the comparisons made by a sorting algorithm on an input of a given size: models all possible execution traces Control, data movement, other operations are ignored Count only the comparisons Decision tree for insertion sort on three elements: node leaf: one execution trace

5. CS 477/677 - Lecture 85 Counting Sort Assumption: –The elements to be sorted are integers in the range 0 to k Idea: –Determine for each input element x, the number of elements smaller than x –Place element x into its correct position in the output array 30320352 12345678 A 77422 12345 C 8 0 53332200 12345678 B

6. CS 477/677 - Lecture 86 Analysis of Counting Sort Alg.: COUNTING-SORT(A, B, n, k) 1.for i ← 0 to k 2. do C[ i ] ← 0 3.for j ← 1 to n 4. do C[A[ j ]] ← C[A[ j ]] + 1 5. C[i] contains the number of elements equal to i 6.for i ← 1 to k 7. do C[ i ] ← C[ i ] + C[i -1] 8. C[i] contains the number of elements ≤ i 9.for j ← n downto 1 10. do B[C[A[ j ]]] ← A[ j ] 11. C[A[ j ]] ← C[A[ j ]] - 1  (k)  (n)  (k)  (n) Overall time:  (n + k)

7. CS 477/677 - Lecture 87 Analysis of Counting Sort Overall time:  (n + k) In practice we use COUNTING sort when k = O(n)  running time is  (n) Counting sort is stable –Numbers with the same value appear in the same order in the output array –Important when satellite data is carried around with the sorted keys

8. CS 477/677 - Lecture 88 Radix Sort Considers keys as numbers in a base-R number –A d -digit number will occupy a field of d columns Sorting looks at one column at a time –For a d digit number, sort the least significant digit first –Continue sorting on the next least significant digit, until all digits have been sorted –Requires only d passes through the list

9. CS 477/677 - Lecture 89 RADIX-SORT Alg.: RADIX-SORT (A, d) for i ← 1 to d do use a stable sort to sort array A on digit i 1 is the lowest order digit, d is the highest-order digit

10. CS 477/677 - Lecture 810 Analysis of Radix Sort Given n numbers of d digits each, where each digit may take up to k possible values, RADIX- SORT correctly sorts the numbers in  (d(n+k)) –One pass of sorting per digit takes  (n+k) assuming that we use counting sort –There are d passes (for each digit)

11. CS 477/677 - Lecture 811 Correctness of Radix sort We use induction on number of passes through each digit Basis: If d = 1, there’s only one digit, trivial Inductive step: assume digits 1, 2,..., d-1 are sorted –Now sort on the d -th digit –If a d < b d, sort will put a before b : correct a < b regardless of the low-order digits –If a d > b d, sort will put a after b : correct a > b regardless of the low-order digits –If a d = b d, sort will leave a and b in the same order and a and b are already sorted on the low-order d-1 digits

12. CS 477/677 - Lecture 812 Bucket Sort Assumption: –the input is generated by a random process that distributes elements uniformly over [0, 1) Idea: –Divide [0, 1) into n equal-sized buckets –Distribute the n input values into the buckets –Sort each bucket –Go through the buckets in order, listing elements in each one Input: A[1.. n], where 0 ≤ A[i] < 1 for all i Output: elements a i sorted Auxiliary array: B[0.. n - 1] of linked lists, each list initially empty

13. CS 477/677 - Lecture 813 BUCKET-SORT Alg.: BUCKET-SORT(A, n) for i ← 1 to n do insert A[i] into list B[  nA[i]  ] for i ← 0 to n - 1 do sort list B[i] with insertion sort concatenate lists B[0], B[1],..., B[n -1] together in order return the concatenated lists

14. CS 477/677 - Lecture 814 Example - Bucket Sort.78.17.39.26.72.94.21.12.23.68 0 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9 10.21.12 /.72 /.23 /.78.94 /.68 /.39 /.26.17 / / / /

15. CS 477/677 - Lecture 815 Example - Bucket Sort 0 1 2 3 4 5 6 7 8 9.23.17 /.78 /.26 /.72.94 /.68 /.39 /.21.12 / / / /.17.12.23.26.21.39.68.78.72.94 / Concatenate the lists from 0 to n – 1 together, in order

16. CS 477/677 - Lecture 816 Correctness of Bucket Sort Consider two elements A[i], A[ j] Assume without loss of generality that A[i] ≤ A[j] Then  nA[i]  ≤  nA[j]  –A[i] belongs to the same group as A[j] or to a group with a lower index than that of A[j] If A[i], A[j] belong to the same bucket: –insertion sort puts them in the proper order If A[i], A[j] are put in different buckets: –concatenation of the lists puts them in the proper order

17. CS 477/677 - Lecture 817 Analysis of Bucket Sort Alg.: BUCKET-SORT(A, n) for i ← 1 to n do insert A[i] into list B[  nA[i]  ] for i ← 0 to n - 1 do sort list B[i] with insertion sort concatenate lists B[0], B[1],..., B[n -1] together in order return the concatenated lists O(n)  (n) O(n)  (n)

18. CS 477/677 - Lecture 818 Conclusion Any comparison sort will take at least nlgn to sort an array of n numbers We can achieve a better running time for sorting if we can make certain assumptions on the input data: –Counting sort: each of the n input elements is an integer in the range 0 to k –Radix sort: the elements in the input are integers represented with d digits –Bucket sort: the numbers in the input are uniformly distributed over the interval [0, 1)

19. CS 477/677 - Lecture 819 A Job Scheduling Application Job scheduling –The key is the priority of the jobs in the queue –The job with the highest priority needs to be executed next Operations –Insert, remove maximum Data structures –Priority queues –Ordered array/list, unordered array/list

20. CS 477/677 - Lecture 820 Example

21. CS 477/677 - Lecture 821 PQ Implementations & Cost Worst-case asymptotic costs for a PQ with N items InsertRemove max ordered array ordered list unordered array unordered list N N N N 1 1 1 1 Can we implement both operations efficiently?

22. CS 477/677 - Lecture 822 Background on Trees Def: Binary tree = structure composed of a finite set of nodes that either: –Contains no nodes, or –Is composed of three disjoint sets of nodes: a root node, a left subtree and a right subtree 2 148 1 16 4 3 910 root Right subtree Left subtree

23. CS 477/677 - Lecture 823 Special Types of Trees Def: Full binary tree = a binary tree in which each node is either a leaf or has degree exactly 2. Def: Complete binary tree = a binary tree in which all leaves have the same depth and all internal nodes have degree 2. Full binary tree 2 148 1 16 7 4 3 910 12 Complete binary tree 2 1 16 4 3 910

24. CS 477/677 - Lecture 824 The Heap Data Structure Def: A heap is a nearly complete binary tree with the following two properties: –Structural property: all levels are full, except possibly the last one, which is filled from left to right –Order (heap) property: for any node x Parent(x) ≥ x Heap 5 7 8 4 It doesn’t matter that 4 in level 1 is smaller than 5 in level 2 2

25. CS 477/677 - Lecture 825 Definitions Height of a node = the number of edges on a longest simple path from the node down to a leaf Depth of a node = the length of a path from the root to the node Height of tree = height of root node =  lgn , for a heap of n elements 2 148 1 16 4 3 910 Height of root = 3 Height of (2)= 1 Depth of (10)= 2

26. CS 477/677 - Lecture 826 Array Representation of Heaps A heap can be stored as an array A. –Root of tree is A[1] –Left child of A[i] = A[2i] –Right child of A[i] = A[2i + 1] –Parent of A[i] = A[  i/2  ] –Heapsize[A] ≤ length[A] The elements in the subarray A[(  n/2  +1).. n] are leaves The root is the maximum element of the heap A heap is a binary tree that is filled in order

27. CS 477/677 - Lecture 827 Heap Types Max-heaps (largest element at root), have the max-heap property: –for all nodes i, excluding the root: A[PARENT(i)] ≥ A[i] Min-heaps (smallest element at root), have the min-heap property: –for all nodes i, excluding the root: A[PARENT(i)] ≤ A[i]

28. CS 477/677 - Lecture 828 Operations on Heaps Maintain the max-heap property –MAX-HEAPIFY Create a max-heap from an unordered array –BUILD-MAX-HEAP Sort an array in place –HEAPSORT Priority queue operations

29. CS 477/677 - Lecture 829 Operations on Priority Queues Max-priority queues support the following operations: –INSERT (S, x) : inserts element x into set S –EXTRACT-MAX (S) : removes and returns element of S with largest key –MAXIMUM (S) : returns element of S with largest key –INCREASE-KEY (S, x, k) : increases value of element x ’s key to k (Assume k ≥ x ’s current key value)

30. CS 477/677 - Lecture 830 Maintaining the Heap Property Suppose a node is smaller than a child –Left and Right subtrees of i are max-heaps Invariant: –the heap condition is violated only at that node To eliminate the violation: –Exchange with larger child –Move down the tree –Continue until node is not smaller than children

31. CS 477/677 - Lecture 831 Maintaining the Heap Property Assumptions: –Left and Right subtrees of i are max-heaps –A[i] may be smaller than its children Alg: MAX-HEAPIFY( A, i, n ) 1.l ← LEFT( i ) 2.r ← RIGHT( i ) 3.if l ≤ n and A[l] > A[i] 4. then largest ← l 5. else largest ← i 6.if r ≤ n and A[r] > A[largest] 7. then largest ← r 8.if largest  i 9. then exchange A[i] ↔ A[largest] 10. MAX-HEAPIFY( A, largest, n )

32. CS 477/677 - Lecture 832 Example MAX-HEAPIFY(A, 2, 10) A[2] violates the heap property A[2]  A[4] A[4] violates the heap property A[4]  A[9] Heap property restored

33. CS 477/677 - Lecture 833 MAX-HEAPIFY Running Time Intuitively: –A heap is an almost complete binary tree  must process O(lgn) levels, with constant work at each level Running time of MAX-HEAPIFY is O(lgn) Can be written in terms of the height of the heap, as being O(h) –Since the height of the heap is  lgn 

34. CS 477/677 - Lecture 834 Building a Heap Alg: BUILD-MAX-HEAP (A) 1.n = length[A] 2. for i ←  n/2  downto 1 3. do MAX-HEAPIFY (A, i, n) Convert an array A[1 … n] into a max-heap ( n = length[A] ) The elements in the subarray A[(  n/2  +1).. n] are leaves Apply MAX-HEAPIFY on elements between 1 and  n/2  2 148 1 16 7 4 3 910 1 23 4567 89 4132169101487 A:

35. CS 477/677 - Lecture 835 Readings Chapter 8, 6