Solving Recurrence Relations

Transcription

1 Solving Recurrence Relations T(n)= 2T(n/2)+n = 2*(2T(n/4)+n/2)+n = 4*T(n/4) +2*n = 8*T(n/8) + 3*n = 2 (log n) * T(1) + (log n) * n = n * 1 + n log n O(n log n)

2 T(n) = T(n/2) + O(1) = T(n/4)+ O(1) + O(1) = T(n/8)+O(1) + O(1)+ O(1) = O(1) + (log n)*o(1) = O(log n)

3 Prime numbers Previously we checked for primality of an integer n by dividing it by all integers up to n We only need to divide by the primes up to n Use an array to remember the primes seen so far

4 Prime numbers class PrimeNumbers { // Find all primes up to n public static void main (String arg[]) { int n = 10000; // Assume n > 1 // Use Dusart s bound (just using n wastes // too much of memory for large n) int maxnumprimes = (int)((n/math.log(n))*( /math.log(n))); int primes[] = new int[maxnumprimes]; int numprimesfound = 0; int i; // continued on next slide

5 Prime numbers for (i=2; i<=n; i++) { if (CheckPrimality (i, primes, numprimesfound)) { primes[numprimesfound] = i; numprimesfound++; PrintPrimes(primes, numprimesfound); // end main // continued on next slide

6 Prime numbers public static boolean CheckPrimality (int n, int primes[], int count) { int i; for (i=0; (i<count) && (primes[i] <= Math.sqrt(n)); i++) { if ((n%primes[i])==0) { return false; return true; // continued on next slide

7 Prime numbers public static void PrintPrimes (int primes[], int count) { int i, j=0; for (i=0; i<count; i++) { System.out.print (primes[i] + ); j++; if (j==10) { // print 10 primes per line System.out.print ( \n ); j=0; System.out.print ( \n ); // end class

8 Multi-dimensional arrays Allows you to work with matrices int array[][] = new int[20][30] Int [][]array= new int[20][30] First dimension is number of rows Second dimension is number of columns Can have more than two dimensions Allocated row-pointer in memory A row can be stored anywhere in memory Elements in a row are stored contiguously Can visualize as array of arrays 8

9 Matrix multiplication class MatrixMultiply { public static void main (String arg[]) { int m = 20, n = 30, p = 40; double A[][] = new double[m][n]; double B[][] = new double[n][p]; double C[][]; InitializeArray (A, m, n); InitializeArray (B, n, p); C = Multiply (A, B, m, n, p); // continued in next slide 9

10 Matrix multiplication public static void InitializeArray (double x[][], int m, int n) { int i, j; for (i=0;i<m;i++) { for (j=0;j<n;j++) { x[i][j] = i+j; 10

11 Matrix multiplication public static double[][] Multiply (double x[][], double y[][], int p, int q, int r) { double z[][] = new double[p][r]; int i, j, k; for (i=0;i<p;i++) { for (j=0;j<r;j++) { z[i][j] = 0; for (k=0;k<q;k++) { z[i][j] += (x[i][k]*y[k][j]); return z; 11 // end class

12 Initializing a multi-dimensional array int[][]array={{1,2,3,{4,5,6,{7,8,9{10,11,12; int[][]array=new int[4][3]; array[0][0]=1; array[0][1]=2; array[0][1]=3; array[1][0]=4; array[1][1]=5; array[1][2]=6; array[2][0]=7; array[2][1]=8; array[2][2]=9; array[3][0]=10; array[3][1]=11; array[3][2]=12;

13 Obtaining lengths of two dimensional arrays int [][] x; x.length will give the length of array x (i.e., number of rows). Each of these rows itself is an array and the length of individual array can be obtained using x[i].length. Remember that unlike matrix all these arrays need not have equal lengths.

14 Allocating a triangular matrix May save memory for symmetric matrices Allocate space to store the starting addresses of m rows Then allocate the m rows themselves int m = 20; double[] triangle[] = new double[m][]; int i; for (i=0;i<m;i++) { // allocate lower triangle triangle[i] = new double[i+1]; 14

15 int[][] trianglearray={ {1,2,3,4,5, {2,3,4,5, {3,4,5, {4,5, {5, If we do not know the values but only sizes of a ragged array: int[][] raggedarray = new int[5][]; raggedarray[0]= new int[5]; raggedarray[1]= new int[4]; raggedarray[2]= new int[3]; raggedarray[3]= new int[2]; raggedarray[4]= new int[1];

16 Strings

17 Strings Can view as array of characters Implemented as a class with its own methods Simplest operation on strings is concatenating two strings String x = Good ; String y = Morning! ; String z = x + + y; String w = x + + y; // also works Possible to concatenate strings to variables of other types e.g., int, float, etc. Internally converted to string

18 Methods of string class Determining the length of a string String x = abc ; int lengthofx = x.length(); Determining the character at a position String x = abc ; char cat2 = x.charat(2); char first = x.charat(0); char last = x.charat(x.length()-1); Case conversion String x = Ashwin ; String y = x.touppercase(); String z = x.tolowercase(); // y is ASHWIN // z is ashwin

19 More on concatenation Concatenating strings is different from concatenating characters String x = a ; String y = b ; String z = x + y; char x1 = a ; char y1 = b ; char z1 = x1 + y1; // z is ab // z is not ab For concatenating two strings you can use the concat method also String z = to.concat( get ).concat( her );

20 Extracting substrings Can extract the substring starting at a position String x = together ; String y = x.substring(2); // y is gether String z = x.substring(3); // z is ether String w = x.substring(0); // same as x String u = x.substring(x.length()-1); // r String v = x.substring(x.length()); // blank Can extract substring between two positions String y = x.substring(0, 5); // y is toget String z = x.substring(5, x.length()); // her String w = x.substring(5, 6); // w is h String u = x.substring(5, 5); // u is

21 Searching in a string Finding the first occurrence of a character or a substring within a string String x = abracadabra ; int k = x.indexof( a ); // k is 0 int p = x.indexof( a, 1); // p is 3; search // begins from pos 1 int t = x.indexof( e );// t is -1 int q = x.indexof( ra ); // q is 2 int s = x.indexof( ra, 3); // s is 9 Possible to find the last occurrence also int p = x.lastindexof( r ); // p is 9

22 String Comparison Can not use == operator to find if two strings are same. Use operator equals string1.equals(string2)

23 Comparison of strings Compares strings in dictionary order (also known as lexicographical order) Returns zero if strings are equal String x = abc ; String y = abcd ; String z = ab ; String w = abd ; int p = x.compareto(y); int q = x.compareto(x); int r = x.compareto(z); int s = y.compareto(w); // p is negative // q is zero // r is positive // s is negative This comparison is case sensitive Use comparetoignorecase otherwise

24 Some more useful methods Removing leading and trailing whitespaces String x = abc ; String y = x.trim(); Test for prefix // y is abc String x = Ashwin ; boolean y = x.startswith( Ashw ); boolean z = x.startswith( win, 3); Test for suffix String x = Canada ; boolean y = x.endswith( ada ); // y is true // y is true // z is true

25 Some more useful methods Substitute all occurrences of a character with another character String x = deer ; String y = x.replace( e, o ); String z = x.replace( a, o ); Partial string match // y is door // z is deer String x = abracadabra ; boolean y = x.regionmatches(true, 2, bracket, 1, 3); // y is true The first argument should be set to true if the match is intended to be case ignorant

26 Converting string to integer class StringToInt { public static void main (String arg[]) { // Assume arg[0] is the input string int result = 0, pos; int len = arg[0].length(); char c; for (pos = 0; pos < len; pos++) { c = arg[0].charat(len-pos-1); if ((c >= 0 ) && (c <= 9 )) { result += ((c- 0 )*Math.pow(10, pos)); // continued in next slide

27 Converting string to integer else if ((c== - ) && (pos==len-1)) { result = -result; else { System.out.println( Invalid input: + arg[0]); break; // end for if (pos==len) { System.out.println( Integer value: + result);

28 Reversing a string class StringReverse { public static void main (String arg[]) { // Assume that the input is arg[0] String reversed = ; int pos; for (pos=arg[0].length()-1; pos >= 0; pos--) { reversed += arg[0].charat(pos); System.out.println ( Original: + arg[0] +, Reversed: + reversed);

29 Sorting a list of names class NaiveDictionarySort { public static void main (String arg[]) { // Assume that the names are in arg int n = arg.length; // list size String sortedlist[] = new String[n]; boolean indexarray[] = new boolean[n]; int k, j, runningindex=0, minindex; for (k=0; k<n; k++) { indexarray[k] = false; // continued on next slide

30 { Sorting a list of names while (runningindex < n) { for (k=0; k<n; k++) { if (!indexarray[k]) break; sortedlist[runningindex] = arg[k]; minindex = k; for (j=k+1; j<n; j++) { if (indexarray[j]) continue; if (sortedlist[runningindex].compareto(arg[j]) > 0) sortedlist[runningindex] = arg[j]; minindex = j; // continued on next slide

31 Sorting a list of names indexarray[minindex] = true; runningindex++; System.out.println ( Sorted list: ); for (k=0; k<n; k++) { System.out.println (sortedlist[k]);

32 String argument Strings are objects Passed by reference value But strings cannot modified Every string operation creates a new string Consider the following Java class class StringTester { public static void main (String arg[]) { String s = abcd ; ModifyString (s); System.out.println ( From main: + s); // Continued in next slide

33 String argument public static void ModifyString (String s) { s += e ; System.out.println ( From ModifyString: + s); The output is as follows From ModifyString: abcde From main: abcd

34 String argument The change is reflected only within ModifyString The concatenation operation creates a new string and stores abcde in that Assigns this new string to s Addresses of s before and after the concatenation are different Original string s remains unchanged with contents abcd