My goal is to implement the following method in parallel:
public static double[][] parallelAddMatrix(double[][] a, double[][] b), then test my program on randomly generated two lists of size 2000 x 2000. Finally I have to output the first 5 elements of matrix a and matrix b, and also the first five elements of the result matrix, which is what I'm having trouble with.
This is the part of my code where I create the first and second matrix.
public static void main(String[] args) {
int var1, var2;
final int matrices = 2000;
// creates first matrix
double[][] matrixA = new double[matrices][matrices];
for(var1 = 0; var1 < matrixA.length; var1++)
for (var2 = 0; var2 < matrixA[var1].length; var2++)
matrixA[var1][var2] = 1;
// creates second matrix
double[][] matrixB = new double[matrices][matrices];
for (var1 = 0; var1 < matrixB.length; var1++)
for (var2 = 0; var2 < matrixB[var1].length; var2++)
matrixB[var1][var2] = 1;
And then later created a function to create the result matrix...
public static double[][] parallelAddMatrix( double [][] a, double[][] b) {
//creates output matrix
double[][] resultMatrix = new double[a.length][a[0].length];
RecursiveAction task = new multiProcess(a, b, resultMatrix);
ForkJoinPool joinPool = new ForkJoinPool();
joinPool.invoke(task);
return resultMatrix;
}
How can I print out the first five elements for each of the three matrices?
I've tried stuff for the first and second matrix such as initializing var3, then under the "matrixA(orB)[var1][var2] = 1;", I put
for (var3 = 0; var3 < 5; var3++) {
System.out.println(var3);
}
and also tried
for (var3 = 0; var3 < 5; var3++) {
System.out.print(matrixA[var1][var2] + "");
}
System.out.println();
Please help on this, and please tell where it would be placed for each one (I might have trouble with brackets).
You'll need a nested for loop to iterate through the matrix, and a counter to see how many entries you've printed. Let's start with the easiest part: iterating over the matrix. I'll assume that the matrix is simply called matrix.
for (int i = 0; i < matrix.length; i++) {
for (int j = 0; j < matrix[i].length; j++) {
System.out.println(matrix[i][j]);
}
}
You probably already figured that out. Now we need a counter to count how many times we've printed out an entry from the matrix.
int num_printed = 0;
for (int i = 0; i < matrix.length; i++) {
for (int j = 0; j < matrix[i].length; j++) {
System.out.println(matrix[i][j]);
num_printed ++;
}
}
Ok. So now we need to stop once we've reached the end. We can't just use one break statement, because, we have two for loops.
int num_printed = 0;
for (int i = 0; i < matrix.length; i++) { // iterate over the rows
for (int j = 0; j < matrix[i].length; j++) { // iterate over the columns
if (num_printed == 5) { // if we've already printed five items, stop
break;
} else { // otherwise, print the next item
System.out.println(matrix[i][j]);
num_printed ++; // increment the counter
}
}
if (num_printed == 5) { // so that we don't go to the next row
break;
}
}
It's worth noting that you could create your own separate method, and only use a return statement:
public void print_five_elements() {
int num_printed = 0;
for (int i = 0; i < matrix.length; i++) { // iterate over the rows
for (int j = 0; j < matrix[i].length; j++) { // iterate over the columns
if (num_printed == 5) { // if we've already printed five items, stop
return;
} else { // otherwise, print the next item
System.out.println(matrix[i][j]);
num_printed ++; // increment the counter
}
}
}
}
More Specialized Approach
This approach allows you to use matrices that have less than five columns. However, since your matrix is 2000x2000, you could go for a much simpler approach. Use zero as the first index, and then just iterate up to five. Just keep in mind that this won't work if you have less than five columns:
public void print_five_elements_for_wide_matrix() {
for (int i = 0; i < 5; i++) {
System.out.println(matrix[0][i]);
}
}
Since the matrices are of size 2000 x 2000, you do not need nested loops to display first 5 elements from each of them.
int i;
//Display first 5 elements of matrixA
for(i=0; i<5; i++) {
System.out.print(matrixA[0][i] + " ");
}
System.out.println();
//Display first 5 elements of matrixB
for(i=0; i<5; i++) {
System.out.print(matrixB[0][i] + " ");
}
System.out.println();
double[][] result = parallelAddMatrix(matrixA, matrixB);
//Display first 5 elements of result
for(i=0; i<5; i++) {
System.out.print(result[0][i] + " ");
}
System.out.println();
Note that the above loops print the first 5 elements of the first row (i.e. row at index, 0) of each matrix. However, if you want to print the first element of the first 5 rows, just swap the indices e.g.
System.out.println(matrixA[i][0] + " ");
Try this:
Think of the first set of brackets as the row and the second set as the column.
for (int row = 0; row < 5; row++) {
for (int col = 0; col < 5; col++) {
System.out.print(matrixA[row][col] + " ");
}
System.out.println();
}
Since "multi-dimensional" arrays are really arrays of arrays you can do it like this if you wanted to print out the whole matrix
for (double[] row : matrixA) {
System.out.println(Arrays.toString(row));
}
Because of this, each row can be a different length. So you may have to get the length to print them out like you first wanted to.
for (int row = 0; row < matrixA.length; row++) {
for (int col = 0; col < matrixA[row].length; col++) {
System.out.print(matrixA[row][col] + " " );
}
}
Rows of different length of a "2D" array are known as ragged-arrays.
Related
I have a loop to loop through 0-200 and if the number matches the number in the list. I will put it inside the freq[][]. However, I'm having problem into putting the numbers I found into the freq[][] considering that it needs to be in the size of [10][20].
public static void example(List<Integer> numbers, List<Integer> elements, int[][] list){
int index = 0;
int[][] freq = new int[10][20];
for (int i = 0; i < 200; i++){
for (int x = 0; x < list.length; x++){
for (int y = 0; y < list[x].length; y++){
if (list[x][y] == i){
freq[][index] = i;
}
}
}
}
}
Keep it as
if (list[x][y]== i){
freq[x][y] = i;
}
else {
freq[x][y] = 0; // if not matched
}
So that freq will be a two dimensional array with 10rows and 20 columns .
-Element at 5th index position in the list will be in 0*5th position in the 10*20 array.
-Element at 199 th index position in the list will be in 9*19th position in the 10*20 array.
first, you make a loop to read 200 numbers inside this loop you want to loop 2d array to compare its elements by every number and make condition if a number exist in list but it in freq[][] this code put every number Achieves the condition in freq array and otherwise but 0
for (int i = 0; i <= 200; i++) {
for (int j = 0; j < list.length; j++) {
for (int k = 0; k < list[j].length; k++) {
if(list[j][k]==i)
freq[j][k]=i;
}
}
}
if you use `
else
freq[j][k]=0;
`
that mean it start putting in array the number or 0, and finally, you get an array don't match you want, so let if condition only without else I test it and it work for me
I was trying to do a 2D array program to demonstrate a TRANSPOSE but I am getting error .. here is my code.
import java.util.Scanner;
/* To demonstrate TRANSPOSE USING 2-D array */
public class Array_2ddd {
public static void main(String args[]) {
Scanner s1 = new Scanner(System.in);
int i, j;
int myArray1[][] = new int[9][9];
int myArray2[][] = new int[9][9];
for (i = 0; i < 9; i++) {
for (j = 0; j < 9; j++) {
System.out.println("Enter array from 1 to 9");
myArray1[i][j] = s1.nextInt();
System.out.print("your array is" + myArray2[i][j]);
}
}
// Transposing now...
for (i = 0; i < 9; i++) {
for (j = 0; j < 9; j++) {
myArray2[i][j] = myArray1[j][i];
}
}
// After transposing
for (i = 0; i < 9; i++) {
for (j = 0; j < 9; j++) {
System.out.print("Your array is as follow" + myArray2[i][j]);
}
}
}
}
EDIT: My error during runtime (Solved)
EDIT 2: Solved
EDIT 3: The loop is in infinity ..it keeps on asking for values fromt the user even when i wrote i<9 and j<9..it still keeps on asking for values till infinity..
There are several errors in your code, also it is recommend that the dimensions of the array is to be declared as a final int, so your code works for all matrix sizes and that debugging is easier. In your original code, the errors are:
At the input step, you are printing one element of myArray[2] before you perform the transpose. That means, you are getting your array is0.
In the section commented "After transposing", you are outputting your array wrong. Namely, for each entry, you call System.out.print("Your array is as follow" + myArray2[i][j]);, and that you forgot to add a new line after each row (when inner loop is finished).
"..it keeps on asking for values fromt the user even when i wrote i<9 and j<9..it still keeps on asking for values till infinity.." There are 81 entries for the 9-by-9 case and you did not output which i,j index to be applied. You probably mistaken an infinite loop with a long but terminating loop.
Your transpose step is good.
Here is a refined version of your code which allows you to input array (in reading order, or more technically, row-major order), create a transposed array. You can copy and compare your current code with this code directly to test it.
public static void main(String args[]) {
final int m = 9; // Rows
final int n = 9; // Columns
Scanner s1 = new Scanner(System.in);
int i, j;
int myArray1[][] = new int[m][n]; // Original array, m rows n cols
int myArray2[][] = new int[n][m]; // Transposed array, n rows m cols
// Input
for (i = 0; i < m; i++) {
for (j = 0; j < n; j++) {
// Should be only prompt.
// Improved to show which entry will be affected.
System.out.printf("[%d][%d]" + "Enter array from 1 to 9\n", i, j);
myArray1[i][j] = s1.nextInt();
}
}
// Transposing now (watch for the ordering of m, n in loops)...
for (i = 0; i < n; i++)
{
for (j = 0; j < m; j++)
{
myArray2[i][j] = myArray1[j][i];
}
}
// After transposing, output
System.out.print("Your array is:\n");
for (i = 0; i < m; i++) {
for (j = 0; j < n; j++) {
System.out.print(myArray1[i][j] + " ");
}
System.out.println(); // New line after row is finished
}
System.out.print("Your transposed array is:\n");
for (i = 0; i < n; i++) {
for (j = 0; j < m; j++) {
System.out.print(myArray2[i][j] + " ");
}
System.out.println();
}
s1.close();
}
For an array with three rows (m = 3) and four columns (n = 4), I inputted the numbers from 0 to 9, and then 0, 1, 2. As expected, the output should be:
Your array is:
0 1 2 3
4 5 6 7
8 9 0 1
Your transposed array is:
0 4 8
1 5 9
2 6 0
3 7 1
You define your matrix as 9x9
int myArray1[][] = new int[9][9];
But actually you want to insert 10x10 items:
for (i = 0; i <= 9; i++)
{
for (j = 0; j <= 9; j++)
So either:
Redefine your arrays to store 10x10 items
int myArray1[][] = new int[10][10];
Only read and store 9x9 items in your defined array
for (i = 0; i < 9; i++) {
for (j = 0; j < 9; j++)
You haven't close your first outer for loop i.e in line 17 and change your array size to 10,as you wanted take 10 input (for 0 to 9 = 10 values).
Want to write the diagonal of an 2-dimensional array (n*n Matrix) into an one-dimensional array.
1 2 3
4 5 6 => 1 5 9
7 8 9
public int[] getDiagonalFromArray(int[][] two_d_array){
int[] diagonal_array = new int[two_d_array[0].length];
int k=0;
for (int i = 0; i < two_d_array[0].length; i++) {
for (int j = 0; j < two_d_array[1].length; j++) {
for (int l = 0; l < two_d_array[0].length; l++) {
diagonal_array[k]=two_d_array[i][j];} //HERE SHOULD BE THE ERROR... HOW DO I CYCLE THROUGH THE 1dim "diagonal_array"?
}
}
return diagonal_array;
}
This method delivers wrong values.
This method of mine works, but just Prints the diagonale, instead of putting it into an 1dim array.
public void getDiagonal(int[][] two_d_array){
//int[] diagonal_array = new int[two_d_array[0].length];
for (int i = 0; i < two_d_array[0].length; i++) {
for (int j = 0; j < two_d_array[1].length; j++) {
if (i==j) System.out.print(two_d_array[i][j]+" ");
}
}
}
Where is the logical difference? I tried the if-clause on the first method, but it raises the "outofbound"-Exception.
Thanks in advance.
Why do you need more than one loop?
for (int i = 0; i < two_d_array[0].length; i++) {
diagonal_array[i]=two_d_array[i][i];
}
Seems to be enough to me.
If your matrix has the same width and height, this is a solution:
public int[] getDiagonal(int[][] two_d_array){
int[] diagonal_array = new int[two_d_array.length];
for (int i = 0; i < two_d_array.length; i++) {
diagonal_array[i] = two_d_array[i][i];
}
return diagonal_array;
Here, I consider principal diagonal elements to be the set of elements , where n & m are the number of rows and the number of columns (per row?) respectively.
Thus, the number of diagonal elements is never greater than min(numOfRows, numOfColumns).
And so, you can always try:
public int[] getDiagonalFromArray(int[][] 2DArray){
int[] diagonalArray = new int[Math.min(2DArray.length, 2DArray[0].length]);
int k=0;
for (int i = 0; i < 2DArray.length && k < diagonalArray.l length; ++i) {
for (int j = 0; j < 2DArray[i].length && k < diagonalArray.l length; ++j) {
if (i == j) {
diagonalArray[k++]=2DArray[i][j];
}
}
}
return diagonalArray;
}
Threw in some bounds checks for good measure.
Your input matrix must at be at least rectangular (square makes most sense), otherwise, the code will behave unreliably.
This is the same as #Andreas' answer, but I sacrifice performance and brevity here for the sake of understanding.
The following method uses some nested loops. I have a 2D array which I have taken row 1 and made a 1D array called keys. I am now iterating across the 2D array and comparing each value from keys to each item in the 2D array starting with row number 2.
So if I have the following 2D array:
1 1
1 0
1 0
upon the first iteration row 1 index 1 will compare and find that the first item in each following row is equal to the first item of row 1 (keys). However, when the next iteration begins since keys has another 1 in position 2 it will again be compared to the 1 in row 2, and this is the type of situation that I want to discriminate. The goal is to find common elements so I can't count the same element twice. I'm trying to convert my array collections to a 2D arraylist but was thinking it might not be necessary. What should be returned from above would be a 1D array with the values {1, null}. Here is the method
public Comparable[] findCommonElements(Comparable[][] collections){
Comparable[] keys = new Comparable[collections[0].length];
for(int row = 0; row < 1; row++){
for(int column = 0; column < collections[row].length; column++){
keys[column] = collections[row][column];
}
}
int comparison = 0;
Comparable[] result = new Comparable[keys.length];
for(int i = 0; i < keys.length; i++){
int found = 0;
for(int r = 1; r < collections.length; r++){
for(int c = 0; c < collections[r].length; c++){
comparison += 1;
if(keys[i].compareTo(collections[r][c]) == 0){
found += 1;
break;
}
}
if (found == collections.length - 1){
result[i] = keys[i];
}
}
}
for(int i = 0; i < result.length; i++){
System.out.print(result[i] + "\t");
}
System.out.println();
System.out.println("Comparisons made: " + comparison);
return result;
}
It sounds like you are just trying to find out if the entire column has alll of the same values, in which case the third nested loop is unnecesary because you just want to loop through each row but for one specific column at a time, so if you were to change it to:
for(int i = 0; i < keys.length; i++){
int found = 0;
for(int r = 1; r < collections.length; r++){
comparison += 1;
if(keys[i].compareTo(collections[r][i]) == 0){
found += 1;
}
if (found == collections.length - 1){
result[i] = keys[i];
}
}
}
That should give you the result you are looking for.
If you are actually doing it differently where you are checking that a certain number exists in every row, but you only want to check it once, then you can store what has already been checked and continue if that is repeated, as follows:
List<Comparable> checked = new ArrayList<Comparable>();
outer:
for(int i = 0; i < keys.length; i++)
{
for(Comparable c : list)
{
if(keys[i].compareTo(c) == 0) continue outer;
}
list.add(keys[i];
}
My goal is to
Create a 5x5 array and fill it with random integers in the range of 1 to 25.
Print this original array
Process the array, counting the number of odds, evens, and summing all of the elements in the array.
Print the total odds, evens, and the sum.
Im not sure how to do it and my teacher is very confused and cannot help me. I was wondering if i could get some guidance.
Here is my code:
public class Processing {
public static void main(String[] args) {
Random Random = new Random();
int[][] Processing = new int[5][5];
for (int x = 0; x < 5; x++) {
int number = Random.nextInt(25);
Processing[x] = number;
}
for (int i = 0; i < 5; i++) {
Processing[i] = new int[10];
}
}
}
Please follow naming conventions for your variables. See here: http://en.wikipedia.org/wiki/Naming_convention_(programming)#Java
Anyways, you have to nest your loops as follows:
for(int i = 0; i < 5; i ++) {
for(int j = 0; j < 5; j++) {
yourArray[i][j] = random.nextInt(25);
}
}
i is the row number and j is the column number, so this would assign values to each element in a row, then move on to the next row.
I'm guessing this is homework so I won't just give away the answer to your other questions, but to set you on the right track, here's how you would print the elements. Again, use two nested loops:
for(int i = 0; i < 5; i ++) {
for(int j = 0; j < 5; j++) {
// print elements in one row in a single line
System.out.print(yourArray[i][j] + " ");
}
System.out.println(); //return to the next line to print next row.
}