Chapter 1 Introduction Definition of Algorithm An algorithm is

Chapter 1 Introduction Definition of Algorithm An algorithm is a finite sequence of precise instructions for performing a computation or for solving a problem. Example 1 Describe an algorithm for finding the maximum (largest) value in a finite sequence of integers. Solution 1. Set the temporary maximum equal to the first integer in the sequence. 2. Compare the next integer in the sequence to the temporary maximum, and set the larger one to be temporary maximum. 3. Repeat the previous step if there are more integers in the sequence. 4. Stop when there are no integers left in the sequence. The temporary maximum at this point is the maximum in the sequence. Algorithm 1. Finding the Maximum Element in a Finite Sequence //Input : n integers a1,a2 ,.,an // Output : max (the maximum of a1,a2 ,.,an ) max : a1 ; for i: 2 to n if max ai then max : ai return max Instead of using a particular computer language, we use a form of pseudocode.

Algorithmic Problem Solving Understand the problem Design an algorithm and proper data structures Analyze the algorithm Code the algorithm Ascertaining the Capabilities of a Computational Device Choosing between Exact and Approximate Problem Solving Deciding on Appropriate Data Structures Algorithm Design Algorithm Analysis Coding

Important Problem Types Sorting Searching String processing (e.g. string matching) Graph problems (e.g. graph coloring problem) Combinatorial problems (e.g. maximizes a cost) Geometric problems (e.g. convex hull problem) Numerical problems (e.g. solving equations )

Fundamental Data Structures 1. Array a[1] a[2] a[3] 2. Link List Head of list a1 a[n] a3 a2 an 10 3. Binary Tree 6 4 1 9 7 17 13 4 9 20

Fundamental Abstract Data Structures 1. Stack an Operations a stack supports: last-in-first-out (pop, push) Implementation: using an array or a list (need to check underflows and overflows) a2 a1 2. Queue Operations a queue supports: first-in-first-out (enqueue, dequeue) Implementation: using an array or a list (need to check underflows and overflows) am am 1 am 2 front an rear 3. Priority Queue (each elements of a priority queue has a key) Operations a priority queue supports: inserting an element, returning an element with maximum/minimum key. Implementation: heap

Assignment (1) Read the sections of the text book about array, linked list, stack, queue, heap and priority queue. (2) (i) Implement stack S and queue Q using both of array and linked list. (ii) Write a main function to do the following operations for S: pop, push 10 times, pop, pop, repeat push and pop 7 times. Do the following operations for Q: dequeue, enqueue 10 times, dequeue, depueue, repeat enqueue and dequeue 7 times. When you use array, declare the size of the array to be 10. Printout all the elements in S and in Q.

Some Advanced Data Structure 1. Graph Operations a graph support: finding neighbors Implementation: using list or matrix G (V , E ) V (a, b, c, d , e, f ) E {(a, c ), (b, c ), (d , a), (c, e), (e, c), (b, f ), (d , e), (e, f )} a c b d e f c a c f b e c a e d f c e f Adjacent lists a b c d e f a 0 0 1 0 0 0 b0 0 1 0 0 1 c0 0 0 0 1 0 d1 0 0 0 1 0 e0 0 1 0 0 1 f 0 0 0 0 0 0 Adjacent matrix

2. Binary Search Tree Definition 1 Binary Search Tree is a Binary Tree satisfying the following condition: 15 (1) Each vertex contains an item called as key which belongs to a total ordering set and two links to its left child and right child, respectively. (2) In each node, its key is larger than the keys of all vertices in its left subtree and smaller than the keys of all the vertices in its right subtree. 6 3 1 18 7 17 13 4 9 Operations a binary search tree support: search, insert, and delete an element with a given key. How to select a data structure for a given problem? 20

Example: Select a data structure for supporting Dynamic Dictionary Definition 2 A Dynamic Dictionary is a data structure of item with keys that support the following basic operations: (1) Insert a new item (2) Remove an item with a given key (3) Search an item with a given key What data structure is the best?

Chapter 2 Fundamentals of the Analysis of Algorithm Implementation and Empirical Analysis Challenge in empirical analysis: Develop a correct and complete implementation. Determine the nature of the input data and other factors influencing on the experiment. Typically, There are three choices: actual data, random data, or perverse data. Compare implementations independent to programmers, machines, compilers, or other related systems. Consider performance characteristics of algorithms, especially for those whose running time is big. Can we analyze algorithms that haven’t run yet?

Example 1 Describe an algorithm for finding an element x in a list of distinct elements a1 , a2 ,., an . Algothm 1 The linear Search Algorithm. Procedure linear search ( x : integer, a1,a2 ,., an : distinct integers) i: 1; while (i n and x ai ) i: i 1; if i n then location: i else locatiton: 0; {location is the subscript of term that equals x, or is 0 if x is not found}

Algorithm 2 The Binary Search Algorithm. Procedure binary search ( x : integer, a1,a2 ,., an : increasing integers) i: 1; j: n; while (i j ) begin m: (i j)/ 2 ; if x am then i: m 1 else j: m; end; if x ai then location: i else location: 0; {location is the subscript of term that equals x, or is 0 if x is not found}

Linear Search and Binary Search written in C 1. 2. Linear search int lsearch(int a[], int v, int l, int r) { for (int i l; i n; i ) if (v a[i]) return i; return -1; } Binary search int bsearch(int a[], int v, int l, int n) { while (r l) {int m (l r)/2; if (v a[m]) return m; if (v a[m]) r m-1; else l m 1; } return -1; }

Complexity of Algorithms Assume that both algorithms A and B solve the problem P. Which one is better? Time complexity: the time required to solve a problem of a specified size. Space complexity: the computer memory required to solve a problem of a specified size. The time complexity is expressed in terms of the number of operations used by the algorithm. Worst case analysis: the largest number of operations needed to solve the given problem using this algorithm. Average case analysis: the average number of operations used to solve the problem over all inputs.

Example 2 Analyze the time complexities of linear search algorithm and binary search algorithm Linear Search Algorithm. Procedure linear search ( x : integer, a1,a2 ,., an : distinct integers) i: 1; while (i n and x ai ) i: i 1; 1 2(n 1) n if i n then location: i else locatiton: 0; {location is the subscript of term that equals x, or is 0 if x is not found} 2 3n 5

Binary Search Algorithm. Procedure binary search ( x : integer, a1,a2 ,., an : increasing integers) i: 1; j: n; 1 1 while (i j ) 1 begin m: (i j)/ 2 ; 1 if x am then i: m 1 else j: m; 2 Let n 2 k. It repeats at most k (k log 2 n) times. end; if x ai then location: i else location: 0; 2 {location is the subscript of term that equals x, or is 0 if x is not found} Number of operations 4 log n 4 2

Orders of Growth 6 Running time for a problem with size n 10 Running Time Operation Per second necessary lg n operations n n 2 2n 6 instant 1 second 11.5 days Never end 259350 days 12 instant Instant 1 second Never end 259340 days 10 10 Using silicon computer, no matter how fast CPU will be you can never solve the problem whose running time is exponential !!!

Asymptotic Notations: O-notation Definition 2.1 A function t(n) is said to be O(g(n)) if there c0 0 and n0 0 such that t (n) c 0 g (n) for all n n0 . exist some constant c0 g ( n ) t (n) n0 If lim n n t ( n) c (c 0 is a constant ) , then t(n) O(g(n)). g ( n)

Example 3 Prove 2n 1 O(n) Example 4 Prove 10n 2 12n 5 O(n 2 ) Example 5 List the following function in O-notation in increasing order: lg n, n, n 2 , n lg n, n 3 , n!,2 n. Example 6 What is the big - oh of the following functions? 5n 5 100n 2 1000, n lg n 2 100n lg n 1000n, 0.0001 2 n n

Asymptotic Analysis of algorithms (using O-notation) Example 7 Analyze the time complexities of linear search algorithm and binary search algorithm asymptotically. Linear Search Algorithm. Procedure linear search ( x : integer, a1,a2 ,., an : distinct integers) i: 1; while (i n and x ai ) i: i 1; It repeats n times. if i n then location: i else locatiton: 0; {location is the subscript of term that equals x, or is 0 if x is not found} Totally(addition): O(n)

Binary Search Algorithm. Procedure binary search ( x : integer, a1,a2 ,., an : increasing integers) i: 1; j: n; while (i j ) begin m: (i j)/ 2 ; if x am then i: m 1 else j: m; Let n 2 k. It repeats at most k (k log 2 n) times. end; if x ai then location: i else location: 0; {location is the subscript of term that equals x, or is 0 if x is not found} Totally(comparison): O(log n)

Example 8 Analyze the time complexities of following algorithm asymptotically. Matrix addition algorithm Procedure MatricAddition(A[0.n-1,0.n-1],B[0.n-1,0.n-1]) for i 0 to n-1 do for j 0 to n-1 do C[i,j] A[i,j] B[i,j]; return C; Repeat n times Repeat n times 2 Totally(addition): O (n )

Recursive Algorithms Example 9 Computing the factorial function F(n) n!. F(n) can be defined recursively as follows: F (0) 1 F (n) F (n 1) n Factorial Algorithm Procedure factorial(n) Algorithm factorial calls itself in its body! if n 0 return 1 else return factorial(n-1) * n; Time complexity(multiplication): T(0) 0 T(n) T(n-1) 1 when n 0 recurrence

Basic Recurrences Example 10 Solving the following recurrence T(0) 1 T(n) T(n-1) 1 T(n) T(n-1) 1 n 0 Example 11 Solve the following recurrence Tn 2Tn / 2 n n 1 T1 c' T (n) 2T (n / 2) cn 2(2T (n / 2 2 ) c(n / 2)) cn 2 2 T (n / 2 2 ) cn cn 2 3 2 3 3 T(n-2) 1 1 T(n-2) 2 2 (2T (n / 2 ) c(n / 2 )) cn cn 2 T (n / 2 ) 3cn . T(n-3) 1 2 T(n-3) 3 2i T (n / 2i ) icn . T(n-i) i 2 k T (n / 2 k ) kcn (k log n) nT1 cn log n c' cn log n O(n log n) T(n-n) n n

Example 12 Solve the recurrence Example13 Solve the recurrence T (n) T (n 1) n n 1 T (n) T (n / 2) 1 n 1 T1 1 T (1) 1 (Assume that n is a power of 2. T (n) T (n 1) n T (n 2) (n 1) n That is, n 2 k , where k log n). T (n 3) (n 2) (n 1) n . T (n) T (n / 2) 1 T 1 2 3 . (n 2) (n 1) n T (n / 23 ) 1 1 1 . 1 2 3 . n n(n 1) / 2 T (n / 2 2 ) 1 1 T (n / 2i ) i . T (n / 2 k ) k (k log n) 1 log n