Programming in Java

http://www.csie.nctu.edu.tw/~tsaiwn/java/ Programming in Java More examples, … 蔡文能交通大學資訊工程學系 tsaiwn@csie.nctu.edu.tw

Agenda • Computing factorial • Loop method • Recursive method • Computing primes • Sieve method • Sorting an array • Selection Sort • Insertion Sort • Bubble Sort • Quick Sort • Merge Sort

Computing Factorial • The factorial of an integer is the product of that number and all of the positive integers smaller than it. • - 0! = 1 • - 1! = 1 • . . . • - 5! = 5*4*3*2*1 = 120 • - 6! = 6*5*4*3*2*1 = 720 • . . . • - 50! = • 30414093201713378043612608166064768844377641568960512000000000000

Factorial -- Iteration • public class Factorial { • public static long factorial(int n) { • long fact = 1; • for (int i =2; i <= n; i ++) //for Loop • fact *= i; //shorthand for fact=fact*i; • return fact; • } • } In separate files • public class ComputingFactorial { • public static void main(String arg[ ]) { • long a = Factorial.factorial(Integer.parseInt(arg[0])); • System.out.println("Factorial of " + arg[0] + "=" + a); • } • }

Recursion 遞迴概念 • 很久很久以前, 這裡有一座山, 山裡有一座廟, 廟裡有個老和尚和小和尚, 小和尚要老和尚說故事給他聽。老和尚就說:『很久很久以前, 這裡有一座山, 山裡友一座廟, 廟裡有個老和尚和小和尚, 小和尚要老和尚說故事給他聽。..... • 莊子與惠子游於濠梁之上。莊子曰：「鯈於出游從容，是魚之樂也。」惠子曰:『子非魚焉知魚之樂?』莊子曰:『子非我焉知我不知魚之樂?』惠子曰:『子非我焉知我不知你不知魚之樂?』莊子曰:『子非我焉知我不知你不知我知魚之樂?』惠子曰:『子非我焉知我不知你不知我不知你不知魚之樂?』莊子曰:『子非我焉知 ... !@#$%^&*? ...

Factorial – Recursion, in Java • /** • * This class shows a recursive method to compute factorials. This method • * calls itself repeatedly based on the formula: n! = n* (n-1)! • **/ • public class Factorial2 { • public static long factorial(int n) { • if (n == 1) return 1; • else return n*factorial(n - 1); • } • }

N Factorial ( Recursive版 ) long factorial(int n) { if( n < 0) return -factorial(-n); if( n ==1 || n==0) return 1; return n * factorial(n-1); } /*告訴我 n-1 階乘, 我就給你 n 階乘*/ #include <stdio.h> int main( ){ printf("5! = %ld\n", factorial(5)); } C語言版本 N階乘就是N乘以N-1階乘

GCD -- 歐幾里得的最大公約數 • 輾轉相除法 (recursive概念!) GCD(m, n) = GCD(n, m%n) 但如果 n 是 0 則答案為 m long gcd(long m, long n) { if(n==0) return m; return gcd(n, m%n); } 如何寫成non-recursive version ?

GCD non-recursive version long gcd(long m, long n) { int r; /* remainder */ while(n != 0) { r = m%n; m = n; n = r; } return m; }

Recursive 也可以很有趣(1/3) Fibonacci Series: (費氏數列) 一開始有一對兔子小兔子隔一個月可以長大為成兔每對成兔每隔一個月可生出一對兔子假設兔子永遠不死, 問第 n 個月時有幾對兔子？ 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, . fib(n) 0 1 2 3 4 5 6 7 8 9 10 11 = n

Recursive 也可以很有趣(2/3) long fib(int n) { if( n < 0) return 0; if( n ==1 || n==0) return 1; return fib(n-1)+fib(n-2); } /*給我 n 我就告訴你第 n 個月時有幾對兔子 */ /*第n個月兔子數=第 n-1 個月兔子+第 n-2 個月兔子*/ /*但是, 最開始兩個月例外 ! */

Recursive 也可以很有趣(3/3) Fibonacci Series(費氏數列): 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, … 34/55 = 55/89 = 89/144 = 0.618 黃金分割比:) 神奇的費氏數列: 任何事物接近這些數字會有變化請看 . .可怕的巧合: !? 三重魔力 . . . 民國 34 年台灣光復, 民國89年變天國民黨從日本手上搶回台灣執政剛好55 年!

問題與思考 (Recursion) • 寫出 non-recursive版的Fibonacci function ? • 考慮小兔子隔兩個月可以長大為成兔 ? • 考慮成兔的懷孕期是兩個月 ? • 其他 Recursive 問題 • Recursive 版的 binary search • Quick sort • Hanoi tower problem • …

Factorial – cache ans in a Table • public class Factorial3 { • //create an array to cache values 0! Through 20! • static long[ ] table = new long[21]; • static {ans[0] = 1;} //factorial of 0 is 1 • static int last = 0; • public static long factorial(int n) { • if(n <= last) return ans[n]; • long tmp = ans[last]; int k = last; • while (k < n) { • tmp = (k+1) * tmp; /* (k+1) ! */ • if(last < 20) {++last; ans [last ] = tmp;} • ++k; • } /* while */ • return tmp; • } • } Cache 唸作 cash

Computing Primes (1/3) • Finding the largest prime number smaller than a specified integer: • Input integer m, find p m such that p is a prime and if there is prime p’ > p then p’ must be larger than m. • Finding all prime numbers that smaller than a specified integer? 1 4 6 8 9 10 12 14 15 16 18 20 2 3 5 7 11 13 17 19

2 2 3 3 4 4 Computing Primes (2/3) • Algorithm • main idea: find primes by eliminating multiples of the form k  j, where j is a prime smaller than square-root(m) and k is an integer such that k  j m. 2 i j  square-root(m) ... ... prime j i 2 2 3 2 i j j 4 2 i ... ... ...

Computing Primes (3/3) • Finding all prime numbers that smaller than a specified integer? 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

Example: Find Primes (1/2) • Import java.lang.*; • public class Sieve { • public static void main(String[] args) { • int max = 100; //Assign a default value • try {max = Integer.parseInt(arg[0]);} • catch (Exception e) {} //Silently ignore exceptions. • //Create an array that specifies whether each number is prime or not. • Boolean[] isprime = new boolean[max+1]; • //Assume that all numbers are primes, until proven otherwise. • for (int i = 0; <= max; i++) isprime[i] = true; • //We know that that 0 and 1 are not prime. Make a note of it. • isprime[0] = isprime[1] = false;

Example: Find Primes (2/2) • //To compute all primes less than max, we need to rule out multiples of all • //integers less than the square root of max. • int n = (int) Math.ceil(Math.sqrt(max)); • for (int i = 0; i <= n; i++) { • if (isprime[i]) { int k = 2; • for (int j = k*i; j <= max; j = (k ++)*i) • isprime[j] = false; } • } • int largest; • for (largest = max; !sprime[largest]; largest--); //empty loop body • System.out.println("The largest prime less than or equal to " + max + " is " • + largest); • } • }

Sorting Numbers • Sorting: • Input n numbers, sort them such that the numbers are ordered increasingly. • 3 9 1 6 5 4 8 2 10 7 • 1 2 3 4 5 6 7 8 9 10

Selection Sort (1/4) • A simple sorting algorithm • Purpose: • Input: an array A containing n integers. • Output: sorted array. • 1. i := 2; • 2. Find the least element a from A(i) to A(n); • 3. If a is less than A(i - 1), exchange A(i - 1) and a; • 4. i := i + 1; if (i <= n) goto step (2).

Selection Sort (2/4) • 1st step: 3 9 1 6 5 4 8 2 10 7 • 2nd step: 1 9 3 6 5 4 8 2 10 7 • 1 2 3 6 5 4 8 9 10 7 • … ... swap swap

Selection Sort (3/4) • Sorting program: • public class SelSort { • public static void sort(double[ ] nums) { • for(int i = 0; i < nums.length; i++) { • int min = i +1; • for (int j = i+2; j < nums.length; j++) { • if (nums[j] < nums[min]) min = j; • } • if (nums[i] > nums[min]) { • double tmp; • tmp = nums[i]; nums[i] = nums[min]; nums[min] = tmp; • } • } • } • }

Selection Sort (4/4) • In separate file: • public class SortNumber { • public static void main (String[ ] args) { • double[ ] nums = new double[10]; //Create an array to hold numbers • for(int i = 0; i < nums.length; i++) //Generate random numbers • nums[i] = Math.random( )*100; • SelSort.sort(nums); //Sort them • for (int j = 0; j < nums.length; j++) //Print them out • System.out.println(nums [j] ); • } • } Math.random( ) : see next slide

Insertion Sort • public class InsSort { • public static void sort(double[ ] nums) { • for(int i = 1; i < nums.length; i++) { • double tmp = nums[i]; /* we want to put this in proper position */ • k = i – 1; • while ((k>=0) && nums[k] > tmp ) { /* smaller than me */ nums[k+1] = nums[k]); /* move it down */ • --k; • } • nums[k+1] = tmp; /* copy it into correct position */ • } • } • }

Test the Insertion Sort • In separate file as previous example (Selection sort): • public class SortNumber { • public static void main (String[ ] args) { • double[ ] nums = new double[10]; //Create an array to hold numbers • for(int i = 0; i < nums.length; i++) //Generate random numbers • nums[i] = Math.random( )*100; • InsSort.sort(nums); //Sort them using insertion sort • for (int j = 0; j < nums.length; j++) //Print them out • System.out.println(nums [j] ); • } • }

Bubble Sort • public class BubSort { • public static void sort(double[ ] nums) { • for(int i = nums.length-2; i >=0; --i) { • int flag = 0; /* Assume no exchange */ • for(k=0; k <= i; ++k) { /* walk through all every pass */ • if ( nums[k] > nums[k+1] ) { /* incorrect order */ • double tmp=nums[k]; nums[k]=nums[k+1]; nums[k+1]=tmp; • ++flag; • } • } • if(flag == 0) break; /* no need to do next pass */ • } • } • }

Quick Sort (1/6) • Algorithm quick_sort(array A, from, to) • Input: from - pointer to the starting position of array A • to - pointer to the end position of array A • Output: sorted array: A’ • 1. Choose any one element as the pivot; • 2. Find the first element a = A[i] larger than or equal to pivot from • A[from] to A[to]; • 3. Find the first element b = A[j] smaller than or equal to pivot from • A[to] to A[from]; • 4. If i < j then exchange a and b; • 5. Repeat step from 2 to 4 until j <= i; • 6. If from < j then recursive call quick_sort(A, from, j); • 7. If i < to then recursive call quick_sort(A, i, to);

Choose 5 as pivot Quick Sort (2/6) • Quick sort • main idea: • 1st step: 3 1 6 5 4 8 10 7 • 2nd step: 3 2 1 5 8 9 10 7 • 3rd step: 3 2 1 4 5 6 8 9 10 7 from to 9 2 j i 6 4 Smaller than any integer right to 5 greater than any integer left to 5

pivot Quick Sort (3/6) to from • Quick sort • 4th step: 2 4 5 6 10 9 • 5th step: 1 2 3 4 5 pivot from to 1 3 8 7 6th step: 5 6 7 8 10 9 7th step: 7 8 10 9 8th step: 9 10

Quick Sort (4/6) • public class QuickSorter { • public static void sort (int[ ] a, int from, int to) { • if ((a == null) || (a.length < 2)) return; • int i = from, j = to; • int pivot = a[(from + to)/2]; • do { • while ((i < to) && (a[i] < pivot)) i++; • while ((j > from) && (a[j] >= pivot)) j--; • if (i < j) { int tmp =a[i]; a [i] = a[j]; a[j] = tmp;} • i++; j--; • }while (i <= j); • if (from < j) sort(a, from, j); • if (i < to) sort(a, i, to); } • }

Quick Sort (5/6) 3, 4, 6, 1, 10, 9, 5, 20, 19, 14, 12, 2, 15, 21, 13, 18, 17, 8, 16, 1 1, 12, 2, 15, 21, 13, 18, 17, 8, 16, 3, 4, 6, 1, 10, 9, 5, 20, 19, 14 j i 8, 14 1, 12, 2, 15, 21, 13, 18, 17, ,16, 19, 3, 4, 6, 1, 10, 9, 5, 20, i j 3, 4, 6, 1, 10, 9, 5, 8, 13 , 1, 12, 2,15, 21, 19, 18, 17, 20, 16, 14 i j 3, 4, 6, 1, 10, 9, 5, 8, 13, 1, 12, 2, 14, 21, 19, 18, 17, 20, 16, 15 14 i 3, 4, 6, 1, 10, 9, 5, 8, 13, 1, 12, 2 j

Quick Sort (6/6) • public class BQSorter { • public static void sort (int[ ] a, int from, int to) { • if ((a == null) || (a.length < 2) || from >= to) return; • int k = (from + to)/2; int tmp =a[to]; a [to] = a[k]; a[k] = tmp; • int pivot = a[to]; int i = from, j = to-1; • while(i < j ) { • while ((i < j) && (a[i] < pivot)) i++; • while ((i < j) && (a[j] >= pivot)) j--; • if (i < j) { tmp =a[i]; a [i] = a[j]; a[j] = tmp;} • }; • tmp =a[i]; a [i] = a[to]; a[to] = tmp; • if (from < i-1) sort(a, from, i-1); • if (i < to) sort(a, i+1, to); } • }

Merge Sort (1/3) • Merging means the combination of two or more ordered sequence into • a single sequence. For example, can merge two sequences: 503, 703, 765 • and 087, 512, 677 to obtain a sequence: 087, 503, 512, 677, 703, 765. • A simple way to accomplish this is to compare the two smallest items, • output the smallest, and then repeat the same process. 503 703 765 087 512 677 503 703 765 512 677 087 703 765 512 677 087 503

Merge Sort (2/3) • Algorithm Merge(s1, s2) • Input: two sequences: s1 - x1  x2 ...  xm and s2 - y1  y2 ...  yn • Output: a sorted sequence: z1  z2 ...  zm+n. • 1.[initialize] i := 1, j := 1, k := 1; • 2.[find smaller] if xi yjgoto step 3, otherwise goto step 5; • 3.[output xi] zk.:= xi, k := k+1, i := i+1. If i  m, goto step 2; • 4.[transmit yj ...  yn] zk, ..., zm+n := yj, ..., yn. Terminate the algorithm; • 5.[output yj] zk.:= yj, k := k+1, j := j+1. If j  n, goto step 2; • 6.[transmit xi ...  xm] zk, ..., zm+n := xi, ..., xm. Terminate the algorithm;

Merge Sort (3/3) • Algorithm Merge-sorting(s) • Input: a sequences s = < x1, ..., xm> • Output: a sorted sequence. • 1. If |s| = 1, then return s; • 2. k := m/2; • 3. s1 := Merge-sorting(x1, ..., xk); • 4. s2 := Merge-sorting(xk+1, ..., xm); • 5. return(Merge(s1, s2));

Time complexity of Mergesort • Takes roughly n·log2n comparisons. • Without the shortcut, there is no best or worst case. • With the optional shortcut, the best case is when the array is already sorted: takes only (n-1) comparisons.

Binary Search (1/5) • Works for an array of numbers or objects that can be compared and are arranged in ascending (or descending) order. • A very fast method: only 20 comparisons are needed for an array of 1,000,000 elements; (30 comparisons can handle 1,000,000,000 elements; etc.)

Binary Search (2/5) • Main idea: “divide and conquer”: • compare the target value to the middle element; • if equal, all set • if smaller, apply binary search to the left half; • if larger, apply binary search to the right half. NJ v ... MI MN MO MS MT NC ND ... NJ ... MI MN MO MS MT NC ND ...

Binary Search (3/5) • Recursive implementation: public int binarySearch (int arr [ ], int value, int left, int right) { int middle = (left + right) / 2; if (value == arr [middle] ) return middle; else if ( value < arr[middle] && left < middle ) return binarySearch (arr, value, left, middle - 1); else if ( value > arr[middle] && middle < right ) return binarySearch (arr, value, middle + 1, right ); else return -1; // Not found }

Binary Search (4/5) • Iterative implementation: public int binarySearch (int arr [ ], int value, int left, int right) { while (left <= right) { int middle = (left + right) / 2; if ( value == arr [middle] ) return middle; else if ( value < arr[middle] ) right = middle - 1; else /* if ( value > arr[middle] ) */ left = middle + 1; } return -1; // Not found }

Binary Search (5/5) • The average number of comparisons is roughly log2n • The worst-case is log2 (n + 1) rounded up to an integer (e.g. 2 for n = 3, 3 for n = 7, 4 for n = 15, etc.)

PRNG • Pseudo Random Number Generator • java.lang.Math.random( ) • A real (double) number: Uniform random number [0, 1) • java.util.Random • A utility class in java.util package • Has many miscellaneous random # generating functions • java.lang.Math.random( ) actually uses nextDouble( ) in this class • nextInt( ) returns a number like int rand( ) in C language • nextGaussian( ) returns a # in standard Normal distribution

java.util.Random (1/2) • Benchmarks uses the java.util.Random class — a more controlled way to generate random numbers. • Constructors: • If we set the same “seed,” we get the same“random” sequence. Default “seed”: System.currentTimeMillis() Random generator = new Random(); Random generator2 = new Random(seed); long seed;

java.util.Random (2/2) • Methods: All 232 possible int values are produced with (approximately) equal probability int k = generator.nextInt (); int k = generator.nextInt (n); double x = generator.nextDouble (); 0 k < n 0 x < 1

Normal Distribution Random number

java.util.Arrays • Provides static methods for dealing with arrays. • Works for arrays of numbers, Strings, (and “comparable”Objects). • Methods: int pos = Arrays.binarySearch (arr, target); Arrays.sort (arr); Arrays.sort (arr, from, to); Arrays.fill (arr, value); // fills arr with a given value Arrays.fill (arr, from, to, value);

java.util.ListInterface • The List library interface describes a list of objects in abstract terms • In a list, objects are arranged in sequence obj0, obj1, ..., objn-1 • In Java, a list holds references to objects • A list can contain duplicate objects (both obj1.equals(obj2) and obj1 == obj2 )

List Methods (a Subset) • int size(); • boolean isEmpty (); • boolean add (Object obj); • void add (int i, Object obj); • Object set(int i, Object obj); • Object get(int i); • Object remove(int i); • boolean contains(Object obj); • int indexOf(Object obj); returns true inserts obj as the i-th value; i must be from 0 to size() i must be from 0 to size() -1 use equals to compare objects

java.util.ArrayList (1/6) • Implements List using an array • Keeps track of the list capacity (the length of the allocated array) and list size (the number of elements currently in the list) capacity size "Cat" "Hat" "Bat"

Programming in Java