OrderTest.javapublic class OrderTest { Get an arra.pdf

OrderTest.java
public class OrderTest {
/**
* Get an array of specified size and pass it to Order.order().
* Report the results.
*/
public static void main(String[] args) {
if (args.length != 1) {//1
System.out.println("Usage: java OrderTest sizeOfArray "
+ "tor tjava OrderTest arrayFile");
System.exit(1);
}
// create or read the int[]
int size = 0;
int[] array = new int[0];//5
try {
size = Integer.parseInt(args[0]);
array = ArrayOfInts.randomizedArray(size);
} catch (NumberFormatException nfe) {//8
try {
array = ArrayOfInts.arrayFromFile(args[0]);
size = array.length;
} catch (Exception e) {
System.err.println("unable to read array from " + args[0]);
System.exit(1);//14
}
}
System.out.println("before:");//15
for (int i = 0; i < array.length; i++) {//2 n
System.out.printf(((i+1) % 10 > 0) ? " %d" : " %d ", array[i]);//1
}
int myNum = Order.order(array); //this is the call we want to measure

System.out.println(" after:");//18
System.out.printf(((i+1) % 10 > 0) ? " %d" : " %d ", array[i]);
}
System.out.println(myNum);
}
}
ArrayOfInts.java
import java.io.BufferedWriter;
import java.io.File;
import java.io.FileNotFoundException;
import java.io.FileWriter;
import java.io.IOException;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Collections;
import java.util.InputMismatchException;
import java.util.Scanner;
public class ArrayOfInts {
/**
* Returns an array of consecutive ints from 1 to size.
*/
public static int[] orderedArray(int size) {
int[] a = new int[size];
for (int i = 0; i < size; i++) {
a[i] = i+1;
}
return a;
}
/**
* Returns a randomized array containing ints from 1 to size.

*/
public static int[] randomizedArray(int size) {
ArrayList aL = new ArrayList();
aL.add(i+1);
}
Collections.shuffle(aL);
a[i] = aL.get(i);
}
return a;
}
/**
* Writes an int[] to a plain-text file with ints separated by spaces.
* Useful for creating input files for repeatable tests.
*/
public static void arrayToFile(int[] array, String outfile) {
try {
FileWriter fw = new FileWriter(outfile);
BufferedWriter bw = new BufferedWriter(fw);
PrintWriter outFile = new PrintWriter(bw);
for (int i : array) {
outFile.print(i + " ");
}
outFile.close();
} catch (IOException e) {
System.err.println("Could not write to " + outfile + " " + e);
}
}
/**
* Read ints from a file and return them in an int[]
*/
public static int[] arrayFromFile(String infile) throws FileNotFoundException,

InputMismatchException {
Scanner scan = new Scanner(new File(infile));
while (scan.hasNext()) {
aL.add(scan.nextInt());
}
scan.close();
int[] a = new int[aL.size()];
for (int i = 0; i < a.length; i++) {
a[i] = aL.get(i);
}
return a;
}
}
Order.java
public class Order {
/**
* Take an int[] and reorganize it so they are in ascending order.
*/
public static int order(int[] array) {
int myNum = 0;
for (int next = 1; next < array.length; next++) {//2
//n
myNum++;
int val = array[next];
int index = next;
while (index > 0 && val < array[index - 1]) {//3
array[index] = array[index - 1];//n
index--;
myNum++;
}
array[index] = val;
}
return myNum;

}
}
BieberSortTest.java
public class BieberSortTest {
public static void main(String[] args){
if (args.length != 1) {
System.out.println("Usage: java BieberSortTest numberOfBiebz");
System.exit(1);
}
try {
int n = Integer.parseInt(args[0]);
BieberSort bs = new BieberSort(n);
bs.printBiebers();
System.out.println("---------");
bs.sort(); // this is the method to measure
bs.printBiebers();
System.out.println(bs.getLoop1()+" <==Loop1");
System.out.println(bs.getLoop2()+" <==Loop");
} catch (NumberFormatException nfe) {
System.out.println("number of Biebz must be a positive integer");
nfe.printStackTrace();
}
}
}
BiberSort.java
import java.util.Random;
public class BieberSort {
private Biebz[] theBiebz;

int counterLoop1=0,counterLoop2=0;
/**
* Inner class for the BieberSort
*/
public class Biebz implements Comparable{
private int b;
public Biebz(int b){
this.b = b;
}
public int compareTo(Biebz o) {
return this.b - o.b;
}
public String toString() {
return String.valueOf(this.b);
}
}
/**
* Construct a new BieberSort class with n fans
*/
public BieberSort(int n) {
this.theBiebz = new Biebz[n];
Random rdm = new Random(System.currentTimeMillis());
for(int i = 0; i< n;i++){
theBiebz[i] = new Biebz(rdm.nextInt(n));
}
}
public int getLoop1(){
return counterLoop1;
}
}
/**
* Prints all the Biebers that we have ...

*/
public void printBiebers(){
for(Biebz bz:theBiebz){
System.out.println(bz);
}
}
/**
* Sorts a collection of Bieber objects
*/
public void sort(){
for(int i =0;ii;j--){
counterLoop2++;
if(theBiebz[i].compareTo(theBiebz[j]) > 0 ){
swap(i,j);
}
}
}
}
private void swap(int i, int j){
Biebz b =theBiebz[i];
theBiebz[i] = theBiebz[j];
theBiebz[j]=b;
}
}
FindTest.java
public class FindTest {
/**
* Get an array of specified size and find the indexes of all elements
* in the array using Find.find().
*/
System.out.println("Usage: java FindTest sizeOfArray "
+ "tor tjava FindTest arrayFile");

System.exit(1);
}
int size = 0;
int counter = 0;
int[] array = new int[0];
try {
try {
System.exit(1);
}
}
// find the index of every element
double average=0;
for (int i = 1; i <= array.length; i++) {
//this is the method of interest, but you'll need to average
//the number of statements needed to find each element for
//meaningful results
int index = Find.find(array, i);
average = Find.getAverage(array);
System.out.printf("%d found at index %d ", i, index);
}
System.out.println("The average is: "+average);
System.out.println(size);
}
}

Find.java
public class Find {
/**
* Return index where value is found in array or -1 if not found.
*/
public static int find(int[] array, int value) {
for (int i = 0; i < array.length; i++) {
if (array[i] == value) {
return i;
}
}
return -1;
}
public static double getAverage(int[] myArray){
int sum=0;
double average=0;
for (int i = 0; i < myArray.length; i++) {
sum=sum+myArray[i];
}
average=sum/myArray.length;
return average;
}
}
ConsecutivePairs.java
public class ConsecutivePairs {
private int n;
private long pairsCount;
private long numStatements = 0; //looking for the dominant section as n gets large

/**
* Generates all combinations of numberOfBits and counts all matched bit pairs.
*/
public ConsecutivePairs(int numberOfBits) throws IllegalArgumentException {
if (numberOfBits < 1) {
throw new IllegalArgumentException("number of bits must be positive");
}
n = numberOfBits;
pairsCount = 0;
int[] bit = new int[n];
for (int i = 0; i < bit.length; i++) {
bit[i] = 0;
}
//generate all combinations of n "bits" by mimicking a binary counter
while(positivesCount(bit) < n) {
numStatements++; //are these the most important statements?
//count "bits" that match their neighbor
for(int i = 0; i < bit.length - 1; i++) {
//numStatements++; //are these the most important statements?
if(bit[i] == bit[i+1]) {
pairsCount++;
}
}
//add 1
int b = 0;
while (b < bit.length && bit[b] == 1) {
bit[b] = 0;
b++;

}
if (b < bit.length) {
bit[b] = 1;
}
}
//still need to count last state where all bits are 1s
pairsCount++;
}
}
}
private int positivesCount(int[] intsArray) {
int positives = 0;
for(int i = 0; i < intsArray.length; i++) {
if(intsArray[i] > 0) {
positives++;
}
}
return positives;
}
public long getPairsCount() {
return pairsCount;
}

public int getNumberOfBits() {
return n;
}
public long getNumStatements() {
return numStatements;
}
}
ConsecutivePairsTest.java
public class ConsecutivePairsTest {
System.out.println("Usage: java ConsecutivePairsTest numberOfBits");
System.exit(1);
}
try {
ConsecutivePairs pairs = new ConsecutivePairs(n); //this is the call to measure
System.out.printf("for n = %d, there are %d matched bit pairs ",
n, pairs.getPairsCount());
System.out.printf("approximately %d statements were executed to calculate this
answer",
pairs.getNumStatements());
} catch (Exception nfe) {
System.out.println("number of bits must be a positive integer");
System.exit(1);
}
}
}
analysis.odt
Haseeb Nain

Computer Science 221
Mason Vail
FindSort:
Part 1 Section 1
Because this sort relies on finding a specific value at the within the array. And in doing so finds
multiple values at multiple points within the array it must be at least O(n). This is simply because
the amount of recursion is directly related to the size of the array.
Part 1 Section 2
Table 1 FindSort Data
FindSort
Input Average NumStatements
1000 500 500000
10000 5000 50000000
20000 10000 200000000
30000 15000 450000000
40000 20000 800000000
50000 25000 1250000000
Figure 1: FindSort Average
Figure 2: FindSort Statments
Part 1 Section 3
This outcome did not support my hypothesis, I believed that the array simply relied on the
length of the array, n, this is not the case. While the average was constant, the actual amount of
time the statements were run followed more closely to a O(n2).
Order Sort:
Part 2 Section 1
Because this sort relies on the value of a point in an array, and then compares the two arrays,
and then switches the position of the array based on its size in comparison, this sort will be
greater in magnitude than the previous FindSort. The order.java class seems to make the
comparison, so it must always run twice for each loop, the loop is the size of the array, so it will
repeat n times. When putting these together I arrive at a O(n2).
Part 2 Section 2
Table 2: OrderSort Data
OrderSort
Input NumStatements

1000 247753
10000 25256933
20000 1.00E+08
30000 2.24E+08
40000 3.98E+08
50000 6.23E+08
Figure 3: OrderSort Statements
Part 2 Section 3
After plotting the number of statements in relation to the number of inputs, I noticed that the
graph looked similar to the quadratic of the second power. This confirms my hypothesis of
O(n2).
BeiberSort:
Part 3 Section 1
BeiberSort is similar to Consecutive Sort, as it compares two values and then orders them from
lowest to greatest. Because it also follows though with two for loops, I believe this will also be
O(n2).
Part 3 Section 2
Table 3: BeiberSort Data
BeiberSort
Input Count
1000 499500
10000 49995000
20000 199990000
30000 449985000
40000 799980000
50000 1249975000
Figure 4: BeiberSort Statements
Part 3 Section 3
The information exracted from the amount of statements and the amount of inputs provides a
graph with a plot equivalent to a quadratic of power 2. This further confirms my hypothesis that
O(n2)
ConsecutiveSort:
Part 4 Section 1
Because the consecutive sort merges concepts from the previous sorting methods, I believe its
number of statements run must be larger then anything before. The consecutive sort will find

values, then compare the values and report values where two values are equivalent and
consecutive. The use of a while loop will also contribute to the amount of statements run,
because it directly relates to the size of “n”. In the end I believe this to be an exponential
increase, so O(2n).
Part 4 Section 2
Table 4: ConsecutiveSort Data
ConsecutiveSort
Input NumStatements
5 31
10 1023
15 32767
20 1048575
25 33554431
Figure 5: ConsecutiveSort Statements
Part 4 Section 3
The information provided by the input and the number of statements further confirms my
hypothesis. The slope was minimal at first by begins to ramp up significantly after 20. On my
excel sheet the data went negative for the value between 15 and 20, I believe this is simply an
error on the behalf of Excel.
Solution
OrderTest.java
public class OrderTest {
/**
* Get an array of specified size and pass it to Order.order().
* Report the results.
*/
if (args.length != 1) {//1
System.out.println("Usage: java OrderTest sizeOfArray "
+ "tor tjava OrderTest arrayFile");
System.exit(1);
}

int size = 0;
int[] array = new int[0];//5
try {
} catch (NumberFormatException nfe) {//8
try {
System.exit(1);//14
}
}
System.out.println("before:");//15
System.out.printf(((i+1) % 10 > 0) ? " %d" : " %d ", array[i]);//1
}
int myNum = Order.order(array); //this is the call we want to measure
System.out.println(" after:");//18
System.out.printf(((i+1) % 10 > 0) ? " %d" : " %d ", array[i]);
}
System.out.println(myNum);
}
}
ArrayOfInts.java
import java.io.BufferedWriter;

import java.io.File;
import java.io.FileNotFoundException;
import java.io.FileWriter;
import java.io.IOException;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Collections;
import java.util.InputMismatchException;
import java.util.Scanner;
public class ArrayOfInts {
/**
* Returns an array of consecutive ints from 1 to size.
*/
public static int[] orderedArray(int size) {
a[i] = i+1;
}
return a;
}
/**
* Returns a randomized array containing ints from 1 to size.
*/
public static int[] randomizedArray(int size) {
aL.add(i+1);
}
Collections.shuffle(aL);
a[i] = aL.get(i);
}
return a;
}

/**
* Writes an int[] to a plain-text file with ints separated by spaces.
* Useful for creating input files for repeatable tests.
*/
public static void arrayToFile(int[] array, String outfile) {
try {
FileWriter fw = new FileWriter(outfile);
BufferedWriter bw = new BufferedWriter(fw);
PrintWriter outFile = new PrintWriter(bw);
for (int i : array) {
outFile.print(i + " ");
}
outFile.close();
} catch (IOException e) {
System.err.println("Could not write to " + outfile + " " + e);
}
}
/**
* Read ints from a file and return them in an int[]
*/
public static int[] arrayFromFile(String infile) throws FileNotFoundException,
InputMismatchException {
Scanner scan = new Scanner(new File(infile));
while (scan.hasNext()) {
aL.add(scan.nextInt());
}
scan.close();
int[] a = new int[aL.size()];
for (int i = 0; i < a.length; i++) {
a[i] = aL.get(i);
}
return a;
}

}
Order.java
public class Order {
/**
* Take an int[] and reorganize it so they are in ascending order.
*/
public static int order(int[] array) {
int myNum = 0;
for (int next = 1; next < array.length; next++) {//2
//n
myNum++;
int val = array[next];
int index = next;
while (index > 0 && val < array[index - 1]) {//3
array[index] = array[index - 1];//n
index--;
myNum++;
}
array[index] = val;
}
return myNum;
}
}
BieberSortTest.java
public class BieberSortTest {
public static void main(String[] args){
System.out.println("Usage: java BieberSortTest numberOfBiebz");
System.exit(1);
}

try {
BieberSort bs = new BieberSort(n);
bs.printBiebers();
System.out.println("---------");
bs.sort(); // this is the method to measure
bs.printBiebers();
System.out.println(bs.getLoop1()+" <==Loop1");
System.out.println(bs.getLoop2()+" <==Loop");
System.out.println("number of Biebz must be a positive integer");
}
}
}
BiberSort.java
import java.util.Random;
public class BieberSort {
private Biebz[] theBiebz;
int counterLoop1=0,counterLoop2=0;
/**
* Inner class for the BieberSort
*/
public class Biebz implements Comparable{
private int b;
public Biebz(int b){
this.b = b;
}
public int compareTo(Biebz o) {
return this.b - o.b;
}

public String toString() {
return String.valueOf(this.b);
}
}
/**
* Construct a new BieberSort class with n fans
*/
public BieberSort(int n) {
this.theBiebz = new Biebz[n];
Random rdm = new Random(System.currentTimeMillis());
for(int i = 0; i< n;i++){
theBiebz[i] = new Biebz(rdm.nextInt(n));
}
}
}
}
/**
* Prints all the Biebers that we have ...
*/
public void printBiebers(){
for(Biebz bz:theBiebz){
System.out.println(bz);
}
}
/**
* Sorts a collection of Bieber objects
*/
public void sort(){
for(int i =0;ii;j--){
counterLoop2++;
if(theBiebz[i].compareTo(theBiebz[j]) > 0 ){

swap(i,j);
}
}
}
}
private void swap(int i, int j){
Biebz b =theBiebz[i];
theBiebz[i] = theBiebz[j];
theBiebz[j]=b;
}
}
FindTest.java
public class FindTest {
/**
* Get an array of specified size and find the indexes of all elements
* in the array using Find.find().
*/
System.out.println("Usage: java FindTest sizeOfArray "
+ "tor tjava FindTest arrayFile");
System.exit(1);
}
int size = 0;
int counter = 0;
int[] array = new int[0];
try {
try {

System.exit(1);
}
}
// find the index of every element
double average=0;
for (int i = 1; i <= array.length; i++) {
//this is the method of interest, but you'll need to average
//the number of statements needed to find each element for
//meaningful results
int index = Find.find(array, i);
average = Find.getAverage(array);
System.out.printf("%d found at index %d ", i, index);
}
System.out.println("The average is: "+average);
System.out.println(size);
}
}
Find.java
public class Find {
/**
* Return index where value is found in array or -1 if not found.
*/
public static int find(int[] array, int value) {
for (int i = 0; i < array.length; i++) {
if (array[i] == value) {
return i;
}
}
return -1;

}
public static double getAverage(int[] myArray){
int sum=0;
double average=0;
for (int i = 0; i < myArray.length; i++) {
sum=sum+myArray[i];
}
average=sum/myArray.length;
return average;
}
}
ConsecutivePairs.java
public class ConsecutivePairs {
private int n;
private long pairsCount;
private long numStatements = 0; //looking for the dominant section as n gets large
/**
* Generates all combinations of numberOfBits and counts all matched bit pairs.
*/
public ConsecutivePairs(int numberOfBits) throws IllegalArgumentException {
if (numberOfBits < 1) {
throw new IllegalArgumentException("number of bits must be positive");
}
n = numberOfBits;
pairsCount = 0;
int[] bit = new int[n];
for (int i = 0; i < bit.length; i++) {
bit[i] = 0;

}
//generate all combinations of n "bits" by mimicking a binary counter
while(positivesCount(bit) < n) {
numStatements++; //are these the most important statements?
pairsCount++;
}
}
//add 1
int b = 0;
while (b < bit.length && bit[b] == 1) {
bit[b] = 0;
b++;
}
if (b < bit.length) {
bit[b] = 1;
}
}
//still need to count last state where all bits are 1s

pairsCount++;
}
}
}
private int positivesCount(int[] intsArray) {
int positives = 0;
for(int i = 0; i < intsArray.length; i++) {
if(intsArray[i] > 0) {
positives++;
}
}
return positives;
}
public long getPairsCount() {
return pairsCount;
}
public int getNumberOfBits() {
return n;
}
public long getNumStatements() {
return numStatements;
}
}
ConsecutivePairsTest.java
public class ConsecutivePairsTest {

System.out.println("Usage: java ConsecutivePairsTest numberOfBits");
System.exit(1);
}
try {
ConsecutivePairs pairs = new ConsecutivePairs(n); //this is the call to measure
System.out.printf("for n = %d, there are %d matched bit pairs ",
n, pairs.getPairsCount());
System.out.printf("approximately %d statements were executed to calculate this
answer",
pairs.getNumStatements());
} catch (Exception nfe) {
System.out.println("number of bits must be a positive integer");
System.exit(1);
}
}
}
analysis.odt
Haseeb Nain
Computer Science 221
Mason Vail
FindSort:
Part 1 Section 1
Because this sort relies on finding a specific value at the within the array. And in doing so finds
multiple values at multiple points within the array it must be at least O(n). This is simply because
the amount of recursion is directly related to the size of the array.
Part 1 Section 2
Table 1 FindSort Data
FindSort
Input Average NumStatements
1000 500 500000

10000 5000 50000000
20000 10000 200000000
30000 15000 450000000
40000 20000 800000000
50000 25000 1250000000
Figure 1: FindSort Average
Figure 2: FindSort Statments
Part 1 Section 3
This outcome did not support my hypothesis, I believed that the array simply relied on the
length of the array, n, this is not the case. While the average was constant, the actual amount of
time the statements were run followed more closely to a O(n2).
Order Sort:
Part 2 Section 1
Because this sort relies on the value of a point in an array, and then compares the two arrays,
and then switches the position of the array based on its size in comparison, this sort will be
greater in magnitude than the previous FindSort. The order.java class seems to make the
comparison, so it must always run twice for each loop, the loop is the size of the array, so it will
repeat n times. When putting these together I arrive at a O(n2).
Part 2 Section 2
Table 2: OrderSort Data
OrderSort
Input NumStatements
1000 247753
10000 25256933
20000 1.00E+08
30000 2.24E+08
40000 3.98E+08
50000 6.23E+08
Figure 3: OrderSort Statements
Part 2 Section 3
After plotting the number of statements in relation to the number of inputs, I noticed that the
graph looked similar to the quadratic of the second power. This confirms my hypothesis of
O(n2).
BeiberSort:
Part 3 Section 1

BeiberSort is similar to Consecutive Sort, as it compares two values and then orders them from
lowest to greatest. Because it also follows though with two for loops, I believe this will also be
O(n2).
Part 3 Section 2
Table 3: BeiberSort Data
BeiberSort
Input Count
1000 499500
10000 49995000
20000 199990000
30000 449985000
40000 799980000
50000 1249975000
Figure 4: BeiberSort Statements
Part 3 Section 3
The information exracted from the amount of statements and the amount of inputs provides a
graph with a plot equivalent to a quadratic of power 2. This further confirms my hypothesis that
O(n2)
ConsecutiveSort:
Part 4 Section 1
Because the consecutive sort merges concepts from the previous sorting methods, I believe its
number of statements run must be larger then anything before. The consecutive sort will find
values, then compare the values and report values where two values are equivalent and
consecutive. The use of a while loop will also contribute to the amount of statements run,
because it directly relates to the size of “n”. In the end I believe this to be an exponential
increase, so O(2n).
Part 4 Section 2
Table 4: ConsecutiveSort Data
ConsecutiveSort
Input NumStatements
5 31
10 1023
15 32767
20 1048575

25 33554431
Figure 5: ConsecutiveSort Statements
Part 4 Section 3
The information provided by the input and the number of statements further confirms my
hypothesis. The slope was minimal at first by begins to ramp up significantly after 20. On my
excel sheet the data went negative for the value between 15 and 20, I believe this is simply an
error on the behalf of Excel.

OrderTest.javapublic class OrderTest { Get an arra.pdf

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