Presentation is loading. Please wait.

Presentation is loading. Please wait.

Recursion CSC 172 SPRING 2002 LECTURE 8 Visual Proof 123.n0 1 2 3 n.

Published by Modified over 9 years ago

Similar presentations

Presentation on theme: "Recursion CSC 172 SPRING 2002 LECTURE 8 Visual Proof 123.n0 1 2 3 n."— Presentation transcript:

1

2 Recursion CSC 172 SPRING 2002 LECTURE 8

3 Visual Proof 123.n0 1 2 3 n

4 1..(n+1)/2..n0 1 2 3 n

5 Visual Proof 1..(n+1)/2..n0 1 2 3 n

6 Visual Proof 1..(n+1)/2..n0 1 2 3 n

7 Visual Proof 1..(n+1)/2..n0 1 2 3 n

8 Visual Proof 1..(n+1)/2..n0 1 2 3 n+1 n

9 Visual Proof 1..(n+1)/2..n0 1 2 3 n+1 n n/2

10 Visual Proof 1..(n+1)/2..n0 1 2 3 n+1 n

11 Visual Proof 1..(n+1)/2..n0 1 2 3 n+1 n

12 Visual Proof 1..(n+1)/2..n0 1 2 3 n+1 n

13 Recursion: Basis & Induction Usually thought of as a programming technique Can also serve as a method of definition

14 Recursive Definition of Expressions Example: expressions with binary operators Basis: An operand (variable or constant) is an expression - x1, y, 3, 57, 21.3

15 Recursive Definition of Expressions Induction: 1. If E is an expression, then (E) is an expression 2. If E 1 and E 2 are expressions, and ° is a binary operator (e.g. +, *), then E 1 ° E 2 is an expression Thus we can build, x, y, 3, x+y, (x+y), (x+y)*3

16 An Interesting Proof S(n): An expression E with binary operators of length n has one more operand than operators. Proof is by complete induction on the length (number of operators, operands, and parentheses) of the expression

17 S(n): An expression E with binary operators of length n has one more operand than operators. Basis: n=1 E must be a single operand Zero operators

18 S(n): An expression E with binary operators of length n has one more operand than operators. Induction (complete): Assume S(1), S(2),... S(n). Let E have length n+1 > 1 How was E constructed? Rule 1, or Rule 2?  If by rule 1, then E = (E 1 )  E1 has length n-1  BTIH: S(n-1) has one more operand than operators  E and E1 have the same number of operands & operators  So, S(n+1) will hold under construction by rule 1

19 S(n): An expression E with binary operators of length n has one more operand than operators. If by rule 2, then E = E 1 ° E 2 Both E 1 and E 2 have length <= n, because ° is one symbol length(E 1 ) + length(E 2 ) = n Let E 1 and E 2 have a and b operators BTIH E 1 and E 2 have a+1 and b+1 operands Thus, E has (a+1)+(b+1) = a+b+2 operands E has a+b+1 operators (the “+1” is for the “°”)

20 S(n): An expression E with binary operators of length n has one more operand than operators. E has (a+1)+(b+1) = a+b+2 operands E has a+b+1 operators Note: we used all of S(1),S(2),...,S(n) in the inductive step The fact that “expression” was defined recursively let us break expressions apart. We don’t know how it broke up, but we have all the cases covered.

21 Recursion: an organization of process A style of programming and problem-solving where we express a solution in terms of smaller instances of itself. Uses basis/induction just like inductive proofs Basis: needs no smaller instances Induction: solution in terms of smaller instances

22 Sorting Lists with MergeSort Note: a list of length 1 is sorted If two lists are sorted, how can you combine them into one sorted list? How can you divide one list into two?

23 Recursive Mergesort Basis: A list of length 1 is sorted Induction: for lists of >= 1 element 1. Split the list into two equal (as possible) parts 2. Recursively sort each part 3. Merge the result of the two recursive sorts One at a time, select the smaller element from the two fronts of each list Physical demo

24 Simplified MergeSort A linked list is a list of nodes with Node getNext(), setNext() methods. and getData(), setData methods.

25 Node split public static Node split(Node head){ Node secondNode; if (head == null) return null; else if (head.getNext() == null) return null; else { secondNode = head.getNext(); head.setNext(secondNode.getNext()); secondNode.setNext(split(secondNode.getNext()); return secondNode; }

26 A linked List 6 1 3 2 5 9 

27 A linked List in Split 6 1 3 2 5 9  Split head

28 A linked List in Split 6 1 3 2 5 9  Split 0 head secondNode secondNode = head.getNext();

29 Split 0 head secondNode A linked List in Split 6 1 3 2 5 9  head.setNext(secondNode.getNext());

30 Split 0 head secondNode First recursive call in Split 6 1 3 2 5 9  secondNode.setNext(split(secondNode.getNext());

31 Split 1 head secondNode New Split 6 1 3 2 5 9  split(secondNode.getNext());

32 Split 1 head secondNode New Split 6 1 3 2 5 9  head.setNext(secondNode.getNext());

33 Split 1 head secondNode Second recursive Split 6 1 3 2 5 9  secondNode.setNext(split(secondNode.getNext());

34 Split 2 head secondNode Second recursive Split 6 1 3 2 5 9  secondNode.setNext(split(secondNode.getNext());

35 Split 2 head secondNode Second recursive Split 6 1 3 2 5  9  head.setNext(secondNode.getNext()); secondNode.setNext(split(secondNode.getNext()); ? What gets returned? return secondNode;

36 Split 1 head secondNode Backing up 6 1 3 2 5  9  secondNode.setNext(split(secondNode.getNext());

37 Split 1 head secondNode Backing up 6 1 3 2 5  9  secondNode.setNext(split(secondNode.getNext()); What gets returned? return secondNode;

38 Split 1 head secondNode Backing up 6 1 3 2 5  9  secondNode.setNext(split(secondNode.getNext()); What gets returned? return secondNode;

39 Split 1 head secondNode Backing up 6 1 3 2 5  9  secondNode.setNext(split(secondNode.getNext()); What gets returned? return secondNode;

40 Node Merge public static Node merge(Node list1, Node list2){ if (list1 == null) return list2; else if (list2 == null) return list1; else if (list1.getData.compareTo(list2.getData()) < 1) { list1.setNext(merge(list1.getNext(),list2); return list1; } else { list2.setNext(merge(list1,list2.getNext()); return list2; }

41 Two linked Lists 1 3 7 2 8 9  

42 Two linked Lists in merge 1 3 7 2 8 9   list1 merge 0 list2

43 Two linked Lists in merge 1 3 7 2 8 9   list1 merge 1 list2 0 th

44 Two linked Lists in merge 1 3 7 2 8 9   list1 merge 2 list2 0 th 1 st

45 Two linked Lists in merge 1 3 7 2 8 9   list1 merge 3 list2 0 th 1 st 2 nd

46 Two linked Lists in merge 1 3 7 2 8 9   list1 merge 4 list2 0 th 1 st 2 nd  3 rd

47 Two linked Lists in merge 1 3 7 2 8 9  list1 merge 3 list2 0 th 1 st 2 nd

48 Two linked Lists in merge 1 3 7 2 8 9  list1 merge 2 list2 0 th 1 st 2 nd

49 Two linked Lists in merge 1 3 7 2 8 9  list1 merge 2 list2 0 th 1 st 2 nd

50 Two linked Lists in merge 1 3 7 2 8 9  list1 merge 0 list2 0 th 1 st 2 nd

51 Node MergeSort public static Node mergeSort(Node list){ Node secondList; if (list == null) return null; else if (list.getNext() == null) return list; else { secondList = split(list); return merge(mergeSort(list),mergeSort(secondList)); }

52 Mergesort 31415926

53 Mergesort: Split 3141592 6

54 314159 2 6

55 31415 9 2 6

56 3141 5 9 2 6

57 314 1 5 9 2 6

58 31 4 1 5 9 2 6

59 3 1 4 1 5 9 2 6

60 3 1 4 1 5 9 2 6

61 3 1 4 1 5 9 2 6

62 3 1 4 1 5 9 2 6

63 3 1 4 1 5 9 2 6

64 3 1 4 1 5 9 2 6

65 3 1 4 1 5 9 2 6

66 3 1 4 1 5 9 2 6

67 Mergesort: Merge 3 1 4 1 5 9 2 6

68 3 1 4 1 5 9 2 6

69 3 1 4 1 5 9 2 6

70 Mergesort: Split 3 1 4 1 5 9 2 6

71 3 1 4 1 5 9 2 6

72 3 1 4 1 5 9 2 6

73 Mergesort: Merge 3 1 4 1 5 9 2 6

74 3 1 4 1 5 9 2 6

75 3 1 4 1 5 9 2 6

76 3 1 4 1 5 9 2 6

77 3 1 4 1 5 9 2 6

78 3 1 4 1 5 9 2 6

79 3 1 4 1 5 9 2 6

80 Mergesort: Split 3 1 4 1 5 9 2 6

81 3 1 4 1 5 9 2 6

82 3 1 4 1 5 9 2 6

83 3 1 4 1 5 9 2 6

84 3 1 4 1 5 9 2 6

85 3 1 4 1 5 9 2 6

86 3 1 4 1 5 9 2 6

87 Mergesort: Merge 3 1 4 1 5 9 2 6

88 3 1 4 1 5 9 2 6

89 3 1 4 1 5 9 2 6

90 Mergesort: Split 3 1 4 1 5 9 2 6

91 3 1 4 1 5 9 2 6

92 3 1 4 1 5 9 2 6

93 Mergesort: Merge 3 1 4 1 5 9 2 6

94 3 1 4 1 5 9 2 6

95 3 1 4 1 5 9 2 6

96 3 1 4 1 5 9 2 6

97 3 1 4 1 5 9 2 6

98 3 1 4 1 5 9 2 6

99 3 1 4 1 5 9 2 6

100 3 1 4 1 5 9 2 6

101 3 1 4 1 5 9 2 6

102 3 1 4 1 5 9 2 6

103 31 4 1 5 9 2 6

104 3141 5 9 2 6

105 31415 9 2 6

106 31415926

Download ppt "Recursion CSC 172 SPRING 2002 LECTURE 8 Visual Proof 123.n0 1 2 3 n."

Similar presentations

Chapter 14 Recursion Lecture Slides to Accompany An Introduction to Computer Science Using Java (2nd Edition) by S.N. Kamin, D. Mickunas,, E. Reingold.

Chapter 14 Recursion Lecture Slides to Accompany An Introduction to Computer Science Using Java (2nd Edition) by S.N. Kamin, D. Mickunas,, E. Reingold.
22C:19 Discrete Structures Induction and Recursion Spring 2014 Sukumar Ghosh.

22C:19 Discrete Structures Induction and Recursion Spring 2014 Sukumar Ghosh.
Linked Lists. 2 Merge Sorted Lists Write an algorithm that merges two sorted linked lists The function should return a pointer to a single combined list.

Linked Lists. 2 Merge Sorted Lists Write an algorithm that merges two sorted linked lists The function should return a pointer to a single combined list.
22C:19 Discrete Structures Induction and Recursion Fall 2014 Sukumar Ghosh.

22C:19 Discrete Structures Induction and Recursion Fall 2014 Sukumar Ghosh.
A simple example finding the maximum of a set S of n numbers.

A simple example finding the maximum of a set S of n numbers.
1 Divide & Conquer Algorithms. 2 Recursion Review A function that calls itself either directly or indirectly through another function Recursive solutions.

1 Divide & Conquer Algorithms. 2 Recursion Review A function that calls itself either directly or indirectly through another function Recursive solutions.
September 12, 20021 Algorithms and Data Structures Lecture III Simonas Šaltenis Nykredit Center for Database Research Aalborg University

September 12, Algorithms and Data Structures Lecture III Simonas Šaltenis Nykredit Center for Database Research Aalborg University
1 Divide-and-Conquer CSC401 – Analysis of Algorithms Lecture Notes 11 Divide-and-Conquer Objectives: Introduce the Divide-and-conquer paradigm Review the.

1 Divide-and-Conquer CSC401 – Analysis of Algorithms Lecture Notes 11 Divide-and-Conquer Objectives: Introduce the Divide-and-conquer paradigm Review the.
Recurrences. What is a Recurrence Relation? A system of equations giving the value of a function from numbers to numbers in terms of the value of the.

Recurrences. What is a Recurrence Relation? A system of equations giving the value of a function from numbers to numbers in terms of the value of the.
Algorithms Recurrences. Definition – a recurrence is an equation or inequality that describes a function in terms of its value on smaller inputs Example.

Algorithms Recurrences. Definition – a recurrence is an equation or inequality that describes a function in terms of its value on smaller inputs Example.
Recursion.

Recursion.
CSE115/ENGR160 Discrete Mathematics 04/03/12 Ming-Hsuan Yang UC Merced 1.

CSE115/ENGR160 Discrete Mathematics 04/03/12 Ming-Hsuan Yang UC Merced 1.
Divide-and-Conquer1 7 2  9 4  2 4 7 9 7  2  2 79  4  4 9 7  72  29  94  4.

Divide-and-Conquer1 7 2  9 4   2  2 79  4   72  29  94  4.
Divide-and-Conquer1 7 2  9 4  2 4 7 9 7  2  2 79  4  4 9 7  72  29  94  4.

Divide-and-Conquer1 7 2  9 4   2  2 79  4   72  29  94  4.
Recursive Definitions Rosen, 3.4. Recursive (or inductive) Definitions Sometimes easier to define an object in terms of itself. This process is called.

Recursive Definitions Rosen, 3.4. Recursive (or inductive) Definitions Sometimes easier to define an object in terms of itself. This process is called.
Data Structures, Spring 2004 © L. Joskowicz 1 Data Structures – LECTURE 3 Recurrence equations Formulating recurrence equations Solving recurrence equations.

Data Structures, Spring 2004 © L. Joskowicz 1 Data Structures – LECTURE 3 Recurrence equations Formulating recurrence equations Solving recurrence equations.
Data Structures, Spring 2006 © L. Joskowicz 1 Data Structures – LECTURE 3 Recurrence equations Formulating recurrence equations Solving recurrence equations.

Data Structures, Spring 2006 © L. Joskowicz 1 Data Structures – LECTURE 3 Recurrence equations Formulating recurrence equations Solving recurrence equations.
Runtime with Functions CSC 172 SPRING 2002 LECTURE 9.

Runtime with Functions CSC 172 SPRING 2002 LECTURE 9.
CSE115/ENGR160 Discrete Mathematics 03/31/11

CSE115/ENGR160 Discrete Mathematics 03/31/11
Recursion. Definitions I A recursive definition is a definition in which the thing being defined occurs as part of its own definition Example: A list.

Recursion. Definitions I A recursive definition is a definition in which the thing being defined occurs as part of its own definition Example: A list.

Similar presentations

Ads by Google