Develop Reference

Make a diagonal using zero for integer print in Java

i have a problem with my code. I'm making a program that should display like this:
0 2 3 4
5 0 7 8
9 10 0 11
12 13 14 0
Here's my code:
int rows = 4, count1=1, count2=4;
for(int i=1; i<=rows; i++){
for(int j=1; j<=rows; j++){
if(j==count1){
System.out.printf("0");
}else{
System.out.print(count1);
}
}
if(i<=rows){
count1++;
count2--;
}
System.out.printf("\n");
}
But the output shown is like this:
0 1 1 1
2 0 2 2
3 3 0 3
4 4 4 0
Can somebody tell me what is wrong with my code? Thank you

Use a counter (e.g. count in the code given below) initialized with 1 to print the values. Print 0 when i==j. Increase the counter whether you print the value of the counter or 0.
Do it as follows:
public class Main {
public static void main(String[] args) {
int rows = 4, count = 1;
for (int i = 1; i <= rows; i++) {
for (int j = 1; j <= rows; j++, count++) {
if (i == j) {
System.out.printf("%3d", 0);
} else {
System.out.printf("%3d", count);
}
}
System.out.printf("\n");
}
}
}
Output:
0 2 3 4
5 0 7 8
9 10 0 12
13 14 15 0

I can't see a need for two counters. Increment count1 on EVERY iteration of your inner loop. Replace the first condition with if (j == i).

Printing a squares triangle. How to mirror numbers?

So I've been working on this lab for a while now for my programming class and so far I think I'm on the right track.
However, I'm not quite sure how to mirror the numbers. So pretty much, my code is only printing the top half of the triangle. Anyway here is the actual assignment that was given to us:
Write a program using a Scanner that asks the user for a number n between 1 and 9 (inclusive). The program prints a triangle with n rows. The first row contains only the square of 1, and it is right-justified. The second row contains the square of 2 followed by the square of 1, and is right justified. Subsequent rows include the squares of 3, 2, and 1, and then 4, 3, 2 and 1, and so forth until n rows are printed.
Assuming the user enters 4, the program prints the following triangle to the console:
1
4 1
9 4 1
16 9 4 1
9 4 1
4 1
1
For full credit, each column should be 3 characters wide and the values should be right justified.
Now here is what I have written for my code so far:
import java.util.Scanner;
public class lab6 {
public static void main(String[] args) {
Scanner kybd = new Scanner(System.in);
System.out.println(
"Enter a number that is between 1 and 9 (inclusive): ");
// this is the value that the user will enter for # of rows
int rows = kybd.nextInt();
for (int i = rows; i > 0; i--) {
for (int j = rows; j > 0; j--)
System.out.print((rows - j + 1) < i ?
" " : String.format("%3d", j * j));
System.out.println();
}
}
}
And this is what that code PRINTS when I enter 4:
Enter a number that is between 1 and 9 (inclusive):
4
1
4 1
9 4 1
16 9 4 1
As you can see, I can only get the TOP half of the triangle to print out. I've been playing around trying to figure out how to mirror it but I can't seem to figure it out. I've looked on this website for help, and all over the Internet but I can't seem to do it.

Answer is:
public static void main(String... args) {
Scanner kybd = new Scanner(System.in);
System.out.println("Enter a number that is between 1 and 9 (inclusive): ");
int rows = kybd.nextInt(); // this is the value that the user will enter for # of rows
for (int i = -rows + 1; i < rows; i++) {
for (int j = -rows; j < 0; j++)
System.out.print(abs(i) > j + rows ? " " : String.format("%3d", j * j));
System.out.println();
}
}
Try think of this as how to find points(carthesians) that are betwean three linear functions(area of triangle that lied betwean):
y = 0 // in loops i is y and j is x
y = x + 4
y = -x -4
And here is example result for 4:
And 9:

In the outer loop or stream you have to iterate from 1-n to n-1 (inclusive) and take absolute values for negative numbers. The rest is the same.
If n=6, then the triangle looks like this:
1
4 1
9 4 1
16 9 4 1
25 16 9 4 1
36 25 16 9 4 1
25 16 9 4 1
16 9 4 1
9 4 1
4 1
1
Try it online!
int n = 6;
IntStream.rangeClosed(1 - n, n - 1)
.map(Math::abs)
.peek(i -> IntStream.iterate(n, j -> j > 0, j -> j - 1)
// prepare an element
.mapToObj(j -> i > n - j ? " " : String.format("%3d", j * j))
// print out an element
.forEach(System.out::print))
// start new line
.forEach(i -> System.out.println());
See also: Output an ASCII diamond shape using loops

Another alternative :
public static void main(String args[]) {
Scanner kybd = new Scanner(System.in);
System.out.println("Enter a number that is between 1 and 9 (inclusive): ");
int rows = kybd.nextInt(); // this is the value that the user will enter for # of rows
int row = rows, increment = -1;
while (row <= rows){
for (int j = rows; j > 0; j--) {
System.out.print(rows - j + 1 < row ? " " : String.format("%3d", j * j));
}
System.out.println();
if(row == 1) {
increment = - increment;
}
row += increment;
}
}

The outer loop from 1-n to n-1 inclusive, and the inner decrementing loop from n to 0. The if condition is the absolute value of i should not be greater than n - j.
Try it online!
int n = 6;
for (int i = 1 - n; i <= n - 1; i++) {
for (int j = n; j > 0; j--)
if (Math.abs(i) > n - j)
System.out.print(" ");
else
System.out.printf("%3d", j * j);
System.out.println();
}
Output:
1
4 1
9 4 1
16 9 4 1
25 16 9 4 1
36 25 16 9 4 1
25 16 9 4 1
16 9 4 1
9 4 1
4 1
1
See also: Invert incrementing triangle pattern

print 2 arrays like matrix for my assignment

I want my output to be like this e.g. if the user inputs 3:
without using 2d array
1 2 3
1 1 2 3
2 1 4 6
3 3 6 9
My code so far
public void matrixmutilplication() {
String thenumberofmatrix = JOptionPane.showInputDialog(null, "Enter the number of column and rows ");
int i = Integer.parseInt(thenumberofmatrix);
int[] cloumnarray = new int[i];
int[] rowarray = new int[i];
for (int z = 0; z <= i - 1; z++) {
cloumnarray[z] = z + 1;
rowarray[z] = z + 1;
}
for (int j = 0; j < i; j++) {
System.out.println(cloumnarray[j] * rowarray[j]);
}
}
I tried different options and can't get this to work properly.

public static void matrixmutilplication() {
String thenumberofmatrix = JOptionPane.showInputDialog(null, "Enter the number of column and rows ");
int i = Integer.parseInt(thenumberofmatrix);
for (int a = 0; a <= i; a++) {
for (int b = 0; b <= i; b++) {
// top corner, don't print nothing
if (a == 0 && b == 0) System.out.print("\t");
// top row 0-1, 0-2, 0-3 etc... just 1,2,3...
else if (a == 0) {
System.out.print(b + "\t");
// last line, print extra line break
if (b == i)
System.out.print("\n");
}
// first column 1-0, 2-0, 3-0... just a + space (tabulator)
else if (b == 0) System.out.print(a + "\t");
// any other cases, are candidates to multiply and give result
else System.out.print(a*b + "\t");
}
//look this is out of scope of nested loops, so,
// in each a iteration, print line break :)
System.out.print("\n");
}
}
public static void main(String[] args) throws Exception {
matrixmutilplication();
}
OUTPUT (3)
1 2 3
1 1 2 3
2 2 4 6
3 3 6 9
OUTPUT (5)
1 2 3 4 5
1 1 2 3 4 5
2 2 4 6 8 10
3 3 6 9 12 15
4 4 8 12 16 20
5 5 10 15 20 25
But problem (for me) is the numbers are not padded in the natural order, so, to achieve your goal, exactly as in your demo, will need a bit of padding like this
public static void matrixmutilplication() {
String thenumberofmatrix = JOptionPane.showInputDialog(null, "Enter the number of column and rows ");
int i = Integer.parseInt(thenumberofmatrix);
for (int a = 0; a <= i; a++) {
for (int b = 0; b <= i; b++) {
if (a == 0 && b == 0) System.out.print("\t");
else if (a == 0) {
System.out.print(String.format("%3s", b));
if (b == i)
System.out.print("\n");
}
else if (b == 0) System.out.print(a + "\t");
else System.out.print(String.format("%3s", a*b));
}
System.out.print("\n");
}
}
public static void main(String[] args) throws Exception {
matrixmutilplication();
}
OUTPUT (7)
1 2 3 4 5 6 7
1 1 2 3 4 5 6 7
2 2 4 6 8 10 12 14
3 3 6 9 12 15 18 21
4 4 8 12 16 20 24 28
5 5 10 15 20 25 30 35
6 6 12 18 24 30 36 42
7 7 14 21 28 35 42 49
What looks quite good :)

So this should be pretty simple.
public void matrixmutilplication() {
String thenumberofmatrix = JOptionPane.showInputDialog(null, "Enter the number of column and rows ");
int i = Integer.parseInt(thenumberofmatrix);
for (int a = 0; a < i; a++) {
for (int b = 0; b < i; b++) {
System.out.print(a*b + "\t");
}
System.out.print("\n");
}
}

Whenever you're working with a matrix involving two arrays (especially if you're trying to a solve a problem that deals with patterns), you want to have a nested for loop like so:
for(int row = 0; row < numSelected; row++) {
for(int col = 0; col < numSelected; col++) {
...
}
}
That way, each cell in the matrix will be covered. Now using that, you can try multiplying the row index and the col index and storing that to the correct cell.

Java Looping Arrays

How do I make this loop properly? it right now So it loops but it does not loop properly. It does this
Here are the numbers:
15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 [1]
How many positions do you want to shift?: 2
2 1 15 14 13 12 11 10 9 8 7 6 5 4 3 [3]
How many positions do you want to shift?: 4
the [] are where its suppose to ask me for my input instead of me just putting in a input
its suppose to run like this:
re are the numbers:
15 14 13 12 11 10 9 8 7 6 5 4 3 2 1
How many positions do you want to shift?: 1
2 1 15 14 13 12 11 10 9 8 7 6 5 4 3
How many positions do you want to shift?: 4
System.out.println("Here are the numbers:");
for (i=0; i<numberArray.length; i++) {
System.out.print(numberArray[i] + " ");
}
while (x != input.nextInt()){
System.out.printf("How many positions do you want to shift?: ");
int shiftTimes=input.nextInt();
for( i = 0; i < shiftTimes; ++i)
shift.Shifter(numberArray);
for(j = 0; j < numberArray.length; j++)
System.out.printf(numberArray[j]+" ");
}
}
}
Also How Do I make it exit the program when I enter in a invalid number and how do I get get it to read a negative value and get it to shift left
Edit: heres my shifter code
public static void Shifter(int[] list)
{
int i;
if (list.length < 2) return;
int last = list[list.length - 1];
for(i = list.length - 1; i > 0; i--) {
list[i] = list[i - 1];
}
list[0] = last;
}

This should work for right shift. It should work with inputs larger then array length as well.
for (int i = shiftTimes%numberArray.length; i > 0; i--) {
System.out.print(numberArray[numberArray.length - i] + " ");
}
for (int i = 0; i < numberArray.length - shiftTimes%numberArray.length; i++) {
System.out.print(numberArray[i] + " ");
}
Reversing this logic should produce a left shift approach.
An invalid input would be the length of the array (because the result will be the same) or 0 because that doesn't do anything:
if (shiftTimes == numberArray.length || shiftTimes == 0) {
// present error to user
}
UPDATE: Putting the logic in your function. Also updated the invalid input check.
public static void Shifter(int[] list, int input)
{
for (int i = input%list.length; i > 0; i--) {
System.out.print(list[list.length - i] + " ");
}
for (int i = 0; i < list.length - input%list.length; i++) {
System.out.print(list[i] + " ");
}
}
The function call would be:
Shifter(numberArray, shiftTimes);

Need help in pattern printing

I need to print the following pattern and i almost did with the coding part.
1
1 2 1
1 2 4 2 1
1 2 4 8 4 2 1
1 2 4 8 16 8 4 2 1
1 2 4 8 16 32 16 8 4 2 1
1 2 4 8 16 32 64 32 16 8 4 2 1
1 2 4 8 16 32 64 128 64 32 16 8 4 2 1
Following is the program I tried
public class MyPattern {
public static void main(String[] args) {
for (int i = 0; i <= 7; i++) {
for (int j = 1; j <= 7 - i; j++) {
System.out.print(" ");
}
for (int j = 0; j <= i; j++) {
int n = (int) Math.pow(2.0D, j);
if (n > 100) {
System.out.print(" " + n);
} else if (n > 10) {
System.out.print(" " + n);
} else {
System.out.print(" " + n);
}
}
for (int j = i - 1; j >= 0; j--) {
int n = (int) Math.pow(2.0D, j);
if (n > 100) {
System.out.print(" " + n);
} else if (n > 10) {
System.out.print(" " + n);
} else {
System.out.print(" " + n);
}
}
System.out.print('\n');
}
}
}
When running the program I got the following output
1
1 2 1
1 2 4 2 1
1 2 4 8 4 2 1
1 2 4 8 16 8 4 2 1
1 2 4 8 16 32 16 8 4 2 1
1 2 4 8 16 32 64 32 16 8 4 2 1
1 2 4 8 16 32 64 128 64 32 16 8 4 2 1
But I need the output aligned to left (as specified first). Please help.

Well it's clearly caused by this part of code:
for (int j = 1; j <= 7 - i; j++) {
System.out.print(" ");
}
Have you tried running it without it?

if (n > 100) {
System.out.print(" " + n);
} else if (n > 10) {
System.out.print(" " + n);
} else {
System.out.print(" " + n);
}
Could also just be, as it does not matter what n is - it will all just do the same.
System.out.print(" " + n);

Comment the line:
//System.out.print(" ");
In the first for loop.

I hope this code helps you understand a few things.
// Make it ready for the loop, no point calling Math.pow() every loop - expensive
import static java.lang.Math.pow;
public class MyPattern {
public void showTree(int treeDepth) {
// Create local method fields, we try to avoid doing this in loops
int depth = treeDepth;
String result = "", sysOutput = "";
// Look the depth of the tree
for( int rowPosition = 0 ; rowPosition < depth ; rowPosition++ ) {
// Reset the row result each time
result = "";
// Build up to the centre (Handle the unique centre value here)
for( int columnPosition = 0 ; columnPosition <= rowPosition ; columnPosition++ )
result += (int) pow(2, columnPosition) + " ";
// Build up from after the centre (reason we -1 from the rowPosition)
for ( int columnPosition = rowPosition - 1 ; columnPosition >= 0 ; columnPosition-- )
result += (int) pow(2, columnPosition) + " ";
// Add the row result to the main output string
sysOutput += result.trim() + "\n";
}
// Output only once, much more efficient
System.out.print( sysOutput );
}
// Good practice to put the main method at the end of the methods
public static void main(String[] args) {
// Good practice to Create Object of itself
MyPattern test = new MyPattern();
// Call method on object (very clear this way)
test.showTree(5);
}
}