art of programming yangjun chen dept. business computing university of winnipeg

Jan. 2004 1

Art of Programming

Yangjun Chen

Dept. Business ComputingUniversity of Winnipeg

Jan. 2004 2

Outline: Art of programmingOutline: Art of programming

• Computing factorial• Sorting numbers• Computing primes

Jan. 2004 3

Computing FactorialComputing Factorial

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

Jan. 2004 4

Computing FactorialsComputing Factorials

• A simple class to compute factorials:public class Factorial {

public static int factorial(int x) {int fact = 1;for (int i =2; i <= x; i ++) //loop

fact *= i; //shorthand for: fact=fact*i;

return fact;}

}public class ComputingFactorial {

public static void main(String arg[]) {int a = Factorial.factorial(Integer.parseInt(arg[0]));System.out.println(a);

}}

Jan. 2004 5


• Recursive Factorials

/*** 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 long factorial(long x) {if (x == 1) return 1;else return x*factorial(x - 1);

}}

Jan. 2004 6


• Caching factorials

public class Factorial3 {//create an array to cache values 0! Through 20!Static long[] table = new long[21];Static {table[0] = 1;} //factorial of 0 is 1//Remember the highest initialized value in the arraystatic int last = 0; public static long factorial(int x) {

while (last < x) {table [last + 1] = table[last]*(last + 1);last++;

}}

Jan. 2004 7

Sorting NumbersSorting 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

Jan. 2004 8


• A simple sorting algorithmmain idea:

AlgorithmInput: 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; goto step (2).

Jan. 2004 9


• A simple sorting algorithmmain idea:

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

Jan. 2004 10


• Sorting program:

import java.lang.*;public class SortNumber {

public static void sort(double[] nums) {for(int i = 0; i < nums.length - 1; 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;}}}

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public static void main (String[] args) {double[] nums = new double[10]; //Create an array to hold

numbersfor(int i = 0; i < nums.length; i++) //Generate random numbers

nums[i] = Math.random()*100;sort(nums); //Sort themfor (int j = 0; j < nums.length; j++) //Print them out

System.out.println(nums [j] );}

}

Jan. 2004 12

Sorting NumbersSorting Numbers• Quick sort

main idea:Algorithm quick_sort(from, center, to)Input: from - pointer to the starting position of array A

center - pointer to the middle position of array Ato - pointer to the end position of array A

Output: sorted array: A’1. Find the first element a = A(i) larger than or equal to A(center)

from A(from) to A(to);

2. Find the first element b = A(j) smaller than or equal to A(center) from

A(to) to A(from); 3. If i < j then exchange a and b;4. Repeat step from 1 to 3 until j <= i;5. If from < j then recursive call quick_sort(from,(from + j)/2, j);6. If i < to then recursive call quick_sort(i, (i+ to)/2, to);

Jan. 2004 13


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

center to

2299

66 44

Smaller than 5Smaller than 5 greater than 5greater than 5

i j

The center element is 5.

i = j = 5

• Quick sort

from

Jan. 2004 14


4th step: 4 5 6 10

5th step: 1 2 3 4

centerfrom to

33 77

tofrom center

i = 2

j = 2

8 92 1

Jan. 2004 15

6th step: 16th step: 1

The sequence contains only one element, no sorting.

7th step: 3 47th step: 3 4

i = j = 1

8th step: 8th step: 4 4

from tocenter

The center element is 4.

The sequence contains only one element, no sorting.

1 2 3 4 51 2 3 4 5

Jan. 2004 16

6 7 8 9 6 7 8 9 1010

... ...

Jan. 2004 17

18,

19,

18,

- quick sorting3, 4, 6, 1, 10, 9, 5, 20, 19, 18, 17, 2, 1, 14, 13, 12, 11, 8, 16, 15

3, 4, 6, 1, 10, 9, 5, 15, 19, 17, 2, 1, 14, 13, 12, 11, 8, 16, 20

18,16,3, 4, 6, 1, 10, 9, 5, 15, 17, 2, 1, 14, 13, 12, 11, 8, 20

3, 4, 6, 1, 10, 9, 5, 15, 16, 8, 17, 2, 1, 14, 13, 12, 11, 19, 20

i=17

j=16

3, 4, 6, 1, 10, 9, 5, 15, 16, 8, 17, 2, 1, 14, 13, 12, 11


ii jj

Jan. 2004 18


• Another Java program for the quick sort:

public class Sorter {public static void sort (int[] a, int from, int to) {

if ((a == null) || (a.length < 2)) return;int i = from, j = to;int center = a[(from + to)/2];do {

while ((i < to) && (a[i] < center)) i++;while ((j > from) && (a[j] > center)) j--;if (i < j) { int tmp =a[i]; a [i] = a[j]; a[j] = tmp;}i++; j--;}while (i <= j);

Jan. 2004 19


• Another Java program for the quick sort:

if (from < j) sort(a, from, j);if (i < to) sort(a, i, to);

}}

Jan. 2004 20

•Sorting by mergingMerging means the combination of two or more ordered sequence intoa single sequence. For example, can merge two sequences: 503, 703, 765and 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 765087 512 677 087

503 703 765512 677

087 503703 765512 677

Jan. 2004 21

•Merging algorithmAlgorithm 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 yj goto 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;

Jan. 2004 22

•Merge-sortingAlgorithm 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));

Jan. 2004 23

Computing PrimesComputing Primes

• 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 m.

than m.11 22 4433 55 7766 88 99 1212 1414 1515 1616 1818 20201010 1111 1313 1717 1919

Jan. 2004 24


• Algorithmmain 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.

...... ......primeprime22 ii jj square-root(m)

223344

...... ...... ......

22

33

44

22

33

44

22

22

22

ii

ii

ii

jj

jj

jj

Jan. 2004 25


Import java.lang.*;

public class Sieve {public static void main(String[] args) {

int max = 100; //Assign a default valuetry {max = Integer.parseInt(args[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; i < 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;

Jan. 2004 26


//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 - 1; !isprime[largest]; largest--); //empty loop bodySystem.out.println(“The largest prime less than or equal to “ + max + “is ”

+ largest); }}

Jan. 2004 27

Assignment #1Assignment #1(Assignment due:(Assignment due: Thu. Feb. 8, 2001)Thu. Feb. 8, 2001)

1. What does the "plateform independence" mean? How is it implemented in Java?

2. Summarize the basic concepts used in Java, such as class, method, variable,Constructor, ... (as many as possible)

3. Implement "Caching factorial" and compute 20! using your program.Trace 10 steps of the computation.

4. Implement "quick sort" and sort a sequence containing 20 integers:3, 4, 6, 1, 10, 9, 5, 20, 19, 18, 17, 2, 1, 14, 13, 12, 11, 8, 16, 15.Trace 10 steps of the computation.

5. Implement "Sieve" and find all the primes smaller than 200 using your program.Trace 10 steps of the computation.

art of programming yangjun chen dept. business computing university of winnipeg

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