CSCS-200 Data Structure and Algorithms. Lecture

Transcription

1 CSCS-200 Data Structure and Algorithms Lecture

2 Recursion

3 What is recursion? Sometimes, the best way to solve a problem is by solving a smaller version of the exact same problem first Recursion is a technique that solves a problem by solving a smaller problem of the same type

4 When you turn this into a program, you end up with functions that call themselves (recursive functions) public int f(int x) { int y; } if(x==0) else { } return 1; y = 2 * f(x-1); return y+1;

5 Problems defined recursively There are many problems whose solution can be defined recursively Example: n factorial n!= 1 if n = 0 (n-1)!*n if n > 0 (recursive solution) n!= 1 if n = 0 1*2*3* *(n-1)*n if n > 0 (closed form solution)

6 Coding the factorial function Recursive implementation public int Factorial(int n) { if (n==0) // base case return 1; else return n * Factorial(n-1); }

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8 Coding the factorial function (cont.) Iterative implementation public int Factorial(int n) { int fact = 1; for(int count = 2; count <= n; count++) fact = fact * count; } return fact;

9 Recursion vs. iteration Iteration can be used in place of recursion An iterative algorithm uses a looping construct A recursive algorithm uses a branching structure Recursive solutions are often less efficient, in terms of both time and space, than iterative solutions Recursion can simplify the solution of a problem, often resulting in shorter, more easily understood source code

10 How do I write a recursive function? Determine the size factor Determine the base case(s) (the one for which you know the answer) Determine the general case(s) (the one where the problem is expressed as a smaller version of itself) Verify the algorithm (use the "Three-Question-Method")

11 Three-Question Verification Method 1. The Base-Case Question: Is there a non-recursive way out of the function, and does the routine work correctly for this "base" case? 2. The Smaller-Caller Question: Does each recursive call to the function involve a smaller case of the original problem, leading inescapably to the base case? 3. The General-Case Question: Assuming that the recursive call(s) work correctly, does the whole function work correctly?

12 Binary Search public int ispresent(int *arr, int val, int N) { int low = 0; int high = N - 1; int mid; while ( low <= high ){ mid = ( low + high )/2; if (arr[mid]== val) return 1; // found! else if (arr[mid] < val) low = mid + 1; else high = mid - 1; } } return 0; // not found

13 Recursive Binary Search What is the size factor? The number of elements in (arr[first]... arr[last]) What is the base case(s)? (1) If low > high, return false (2) If val==arr[mid], return true What is the general case? if val < arr[mid] search the first half if val > arr[mid], search the second half

14 Recursive Binary Search public int BinarySearch(int *arr, int val, int low, int high) { int mid; if(low > high) // base case 1 return false; else { mid = (low + high)/2; if(val < arr[mid]) return BinarySearch(arr, val, low, mid-1); else if (val == arr[mid]) { // base case 2 val = arr[mid]; return true; } else return BinarySearch(arr, val, mid+1, high); } }

15 How is recursion implemented? What happens when a function gets called? public int a(int w) { return w+w; } int b(int x) { int z,y; // other statements z = a(x) + y; } return z;

16 What happens when a function is called? (cont.) An activation record is stored into a stack (runtime stack) 1) The computer has to stop executing function b and starts executing function a 2) Since it needs to come back to function b later, it needs to store everything about function b that is going to need (x, y, z, and the place to start executing upon return) 3) Then, x from a is bounded to w from b 4) Control is transferred to function a

17 What happens when a function is called? (cont.) After function a is executed, the activation record is popped out of the run-time stack All the old values of the parameters and variables in function b are restored and the return value of function a replaces a(x) in the assignment statement

18 What happens when a recursive function is called? Except the fact that the calling and called functions have the same name, there is really no difference between recursive and nonrecursive calls int f(int x) { int y; } if(x==0) return 1; else { y = 2 * f(x-1); return y+1; }

19 2*f(2) 2*f(1) 2*f(1) =f(0) =f(1) =f(2) =f(3)

20 Tree Data Structure

21 Tree Data Structures There are a number of applications where linear data structures are not appropriate. Consider a genealogy tree of a family. Mohammad Aslam Khan Sohail Aslam Javed Aslam Yasmeen Aslam Haaris Saad Qasim Asim Fahd Ahmad Sara Omer

22 Tree Data Structure A linear linked list will not be able to capture the treelike relationship with ease. Shortly, we will see that for applications that require searching, linear data structures are not suitable. We will focus our attention on binary trees.

23 Binary Tree A binary tree is a finite set of elements that is either empty or is partitioned into three disjoint subsets. The first subset contains a single element called the root of the tree. The other two subsets are themselves binary trees called the left and right subtrees. Each element of a binary tree is called a node of the tree.

24 Binary Tree Binary tree with 9 nodes. A B C D E F G H I

25 Binary Tree root A B C D E F G H I Left subtree Right subtree

26 Binary Tree Recursive definition root A B C D E F Left subtree G H I Right subtree

27 Binary Tree Recursive definition A B root C D E F G H I Left subtree

28 Binary Tree Recursive definition A B C D E F G root H I

29 Binary Tree Recursive definition A root B C D E F G H I Right subtree

30 Binary Tree Recursive definition A B C root D E F G H I Left subtree Right subtree

31 Not a Tree Structures that are not trees. A B C D E F G H I

32 Not a Tree Structures that are not trees. A B C D E F G H I

33 Not a Tree Structures that are not trees. A B C D E F G H I

34 Binary Tree: Terminology parent A Left descendant B C Right descendant D E F G H I Leaf nodes Leaf nodes

35 Binary Tree If every non-leaf node in a binary tree has non-empty left and right subtrees, the tree is termed a strictly binary tree. A B C D E J F G K H I

36 Level of a Binary Tree Node The level of a node in a binary tree is defined as follows: Root has level 0, Level of any other node is one more than the level its parent (father). The depth of a binary tree is the maximum level of any leaf in the tree.

37 Level of a Binary Tree Node A 0 Level 0 B 1 C 1 Level 1 D E F Level 2 G H I Level 3

38 Complete Binary Tree A complete binary tree of depth d is the strictly binary all of whose leaves are at level d. A 0 B 1 C 1 D 2 E 2 F 2 G 2 H 3 I J 3 K L M 3 3 N O 3 3

39 Complete Binary Tree A Level 0: 2 0 nodes B C Level 1: 2 1 nodes D E F G Level 2: 2 2 nodes H I J K L M N O Level 3: 2 3 nodes

40 Complete Binary Tree At level k, there are 2 k nodes. Total number of nodes in the tree of depth d: d = 2 j = 2 d+1 1 d In a complete binary tree, there are 2 d leaf nodes and (2 d - 1) non-leaf (inner) nodes. j=0

41 Complete Binary Tree If the tree is built out of n nodes then n = 2 d+1 1 or log 2 (n+1) = d+1 or d = log 2 (n+1) 1 I.e., the depth of the complete binary tree built using n nodes will be log 2 (n+1) 1. For example, for n=100,000, log 2 (100001) is less than 20; the tree would be 20 levels deep. The significance of this shallowness will become evident later.

42 Operations on Binary Tree There are a number of operations that can be defined for a binary tree. If p is pointing to a node in an existing tree then left(p) returns pointer to the left subtree right(p) returns pointer to right subtree parent(p) returns the father of p brother(p) returns brother of p. info(p) returns content of the node.

43 Operations on Binary Tree In order to construct a binary tree, the following can be useful: setleft(p,x) creates the left child node of p. The child node contains the info x. setright(p,x) creates the right child node of p. The child node contains the info x.

44 Applications of Binary Trees A binary tree is a useful data structure when two-way decisions must be made at each point in a process. For example, suppose we wanted to find all duplicates in a list of numbers: 14, 15, 4, 9, 7, 18, 3, 5, 16, 4, 20, 17, 9, 14, 5

45 Applications of Binary Trees One way of finding duplicates is to compare each number with all those that precede it. 14, 15, 4, 9, 7, 18, 3, 5, 16, 4, 20, 17, 9, 14, 5 14, 15, 4, 9, 7, 18, 3, 5, 16, 4, 20, 17, 9, 14, 5

46 Searching for Duplicates If the list of numbers is large and is growing, this procedure involves a large number of comparisons. A linked list could handle the growth but the comparisons would still be large. The number of comparisons can be drastically reduced by using a binary tree. The tree grows dynamically like the linked list.

47 Searching for Duplicates The binary tree is built in a special way. The first number in the list is placed in a node that is designated as the root of a binary tree. Initially, both left and right subtrees of the root are empty. We take the next number and compare it with the number placed in the root. If it is the same then we have a duplicate.

48 Searching for Duplicates Otherwise, we create a new tree node and put the new number in it. The new node is made the left child of the root node if the second number is less than the one in the root. The new node is made the right child if the number is greater than the one in the root.

49 Searching for Duplicates 14 14, 15, 4, 9, 7, 18, 3, 5, 16, 4, 20, 17, 9, 14, 5

50 Searching for Duplicates , 4, 9, 7, 18, 3, 5, 16, 4, 20, 17, 9, 14, 5

51 Searching for Duplicates , 4, 9, 7, 18, 3, 5, 16, 4, 20, 17, 9, 14, 5

52 Searching for Duplicates , 9, 7, 18, 3, 5, 16, 4, 20, 17, 9, 14, 5

53 Searching for Duplicates , 9, 7, 18, 3, 5, 16, 4, 20, 17, 9, 14, 5

54 Searching for Duplicates , 7, 18, 3, 5, 16, 4, 20, 17, 9, 14, 5

55 Searching for Duplicates , 7, 18, 3, 5, 16, 4, 20, 17, 9, 14, 5

56 Searching for Duplicates , 18, 3, 5, 16, 4, 20, 17, 9, 14, 5

57 Searching for Duplicates , 18, 3, 5, 16, 4, 20, 17, 9, 14, 5

58 Searching for Duplicates , 3, 5, 16, 4, 20, 17, 9, 14, 5

59 Searching for Duplicates , 3, 5, 16, 4, 20, 17, 9, 14, 5

60 Searching for Duplicates , 5, 16, 4, 20, 17, 9, 14, 5

61 Searching for Duplicates , 5, 16, 4, 20, 17, 9, 14, 5

62 Searching for Duplicates , 16, 4, 20, 17, 9, 14, 5

63 Searching for Duplicates , 16, 4, 20, 17, 9, 14, 5

64 Searching for Duplicates , 4, 20, 17, 9, 14, 5

65 Searching for Duplicates , 4, 20, 17, 9, 14, 5

66 Searching for Duplicates , 20, 17, 9, 14, 5

67 Searching for Duplicates , 17, 9, 14, 5

68 Searching for Duplicates , 17, 9, 14, 5

69 Searching for Duplicates , 9, 14, 5

70 Searching for Duplicates , 9, 14, 5

71 Searching for Duplicates , 14, 5

72 Implementation public class TreeNode { public: // constructors TreeNode() { this.object = NULL; this.left = this.right = NULL; }; TreeNode( int object ) { }; this.object = object; this.left = this.right = NULL;

73 Implementation public int getinfo() { return this.object; }; void setinfo(int object) { this.object = object; }; public TreeNode getleft() { return left; }; void setleft(treenode left) { this.left = left; };

74 Implementation public TreeNode getright() { return right; }; void setright(treenode right) { this.right = right; }; public int isleaf( ) { if( this.left == NULL && this.right == NULL ) return 1; return 0; };

75 Implementation private: int object; TreeNode left; TreeNode right; }; // end class TreeNode

76 Implementation import "TreeNode.java" public static void main(string args []) { } int x[] = { 14, 15, 4, 9, 7, 18, 3, 5, 16,4, 20, 17, 9, 14,5, -1}; TreeNode root = new TreeNode(); root.setinfo( x[0] ); for(int i=1; x[i] > 0; i++ ) { } insert(root, x[i] );

77 Implementation void insert(treenode root, int info) { TreeNode node = new TreeNode(info); TreeNode p, q; p = q = root; while( info!= (p.getinfo()) q!= NULL ) { p = q; if( info < (p.getinfo()) ) q = p.getleft(); else q = p.getright(); }

78 Implementation if( info == (p.getinfo()) ){ print "attempt to insert duplicate: ; } else if( info < (p.getinfo()) ) p.setleft( node ); else p.setright( node ); } // end of insert

79 Trace of insert 17 p q , 9, 14, 5

80 Trace of insert 17 p 14 4 q , 9, 14, 5

81 Trace of insert p q , 9, 14, 5

82 Trace of insert p q , 9, 14, 5

83 Trace of insert p q , 9, 14, 5

84 Trace of insert p 18 7 q , 9, 14, 5

85 Trace of insert p q , 9, 14, 5

86 Trace of insert p q 17, 9, 14, 5

87 Trace of insert p node 17 17, 9, 14, 5 p.setright( node );

88 Cost of Search Given that a binary tree is level d deep. How long does it take to find out whether a number is already present? Consider the insert(17) in the example tree. Each time around the while loop, we did one comparison. After the comparison, we moved a level down.

89 Cost of Search With the binary tree in place, we can write a routine find(x) that returns true if the number x is present in the tree, false otherwise. How many comparison are needed to find out if x is present in the tree? We do one comparison at each level of the tree until either x is found or q becomes NULL.

90 Cost of Search If the binary tree is built out of n numbers, how many comparisons are needed to find out if a number x is in the tree? Recall that the depth of the complete binary tree built using n nodes will be log 2 (n+1) 1. For example, for n=100,000, log 2 (100001) is less than 20; the tree would be 20 levels deep.

91 Cost of Search If the tree is complete binary or nearly complete, searching through 100,000 numbers will require a maximum of 20 comparisons. Or in general, approximately log 2 (n). Compare this with a linked list of 100,000 numbers. The comparisons required could be a maximum of n.

92 Binary Search Tree A binary tree with the property that items in the left subtree are smaller than the root and items are larger or equal in the right subtree is called a binary search tree (BST). The tree we built for searching for duplicate numbers was a binary search tree. BST and its variations play an important role in searching algorithms.

93 Traversing a Binary Tree Suppose we have a binary tree, ordered (BST) or unordered. We want to print all the values stored in the nodes of the tree. In what order should we print them?

94 Traversing a Binary Tree Ways to print a 3 node tree: (4, 14, 15), (4,15,14) (14,4,15), (14,15,4) (15,4,14), (15,14,4)

95 Traversing a Binary Tree In case of the general binary tree: N node L left subtree right subtree R (L,N,R), (L,R,N) (N,L,R), (N,R,L) (R,L,N), (R,N,L)

96 Traversing a Binary Tree Three common ways N node L left subtree right subtree R Preorder: (N,L,R) Inorder: (L,N,R) Postorder: (L,R,N)

97 Traversing a Binary Tree public void preorder(treenode treenode) { if( treenode!= NULL ) { print(treenode.getinfo()); preorder(treenode.getleft()); preorder(treenode.getright()); } }

98 Traversing a Binary Tree void inorder(treenode treenode) { if( treenode!= NULL ) { inorder(treenode.getleft()); print(treenode.getinfo()); inorder(treenode.getright()); } }

99 Traversing a Binary Tree void postorder(treenode treenode) { if( treenode!= NULL ) { postorder(treenode.getleft()); postorder(treenode.getright()); print(treenode.getinfo()); } }

100 Traversing a Binary Tree Preorder:

101 Traversing a Binary Tree Inorder:

102 Traversing a Binary Tree Postorder:

103 Recursive Call Recall that a stack is used during function calls. The caller function places the arguments on the stack and passes control to the called function. Local variables are allocated storage on the call stack. Calling a function itself makes no difference as far as the call stack is concerned.

104 Stack Layout during a call Here is stack layout when function F calls function F (recursively): Parameters(F) Parameters(F) Parameters(F) Local variables(f) Local variables(f) Local variables(f) sp Return address(f) Parameters(F) Return address(f) Parameters(F) Local variables(f) sp Return address(f) sp Return address(f) At point of call During execution of F After call

105 Recursion: preorder preorder(14) 14..preorder(4) 4...preorder(3) 3...preorder(null)...preorder(null)...preorder(9) 9...preorder(7) 7...preorder(5) 5...preorder(null)...preorder(null)...preorder(null)...preorder(null)

106 Recursion: preorder preorder(15) 15...preorder(null)...preorder(18) 18...preorder(16) 16...preorder(null)...preorder(17) 17...preorder(null)...preorder(null)...preorder(20) 20...preorder(null)...preorder(null)

107 Recursion: inorder inorder(14)..inorder(4)...inorder(3)...inorder(null) 3...inorder(null) 4...inorder(9)...inorder(7)...inorder(5)...inorder(null) 5...inorder(null) 7...inorder(null) 9...inorder(null) 14

108 Recursion: inorder inorder(15)...inorder(null) 15...inorder(18)...inorder(16)...inorder(null) 16...inorder(17)...inorder(nul l) 17...inorder(nul l) 18...inorder(20)...inorder(null) 20...inorder(null)

109 Non Recursive Traversal We can implement non-recursive versions of the preorder, inorder and postorder traversal by using an explicit stack. The stack will be used to store the tree nodes in the appropriate order. Here, for example, is the routine for inorder traversal that uses a stack.

110 Level-order Traversal There is yet another way of traversing a binary tree that is not related to recursive traversal procedures discussed previously. In level-order traversal, we visit the nodes at each level before proceeding to the next level. At each level, we visit the nodes in a left-to-right order.

111 Level-order Traversal Level-order: