Java nested loops

Question

Program Description:

Write a program to print 21 rows of X's in the shape of a large X as illustrated below. Be sure so the two rows intersect at the "11" row.

Here is what I want as an output:

Sample output image

Here is what I have so far.

public class Program168h {

    public static void main (String [] args)  {
        String d= "X";
        for (int a = 1; a < 23; a++) {
            for (int b = a; b >= 1; b--) {   
                System.out.print(" ");
            }
            System.out.print(d);
            for (int x = a; x < 22; x++) {
                System.out.print("  ");
            }
            System.out.print(d);
            System.out.println();
        }
    }
}

This only produces the first half of the X, I do not know how to produce the lower half.

How do i move on to getting the output? I'm currently stuck. Sorry for being unclear. — Basam, Oct 16 '15 at 15:06
"How do i move on to getting the output?" - Can you see no output? What are you running this in? — Chris, Oct 16 '15 at 15:08
I'm running this in BlueJ and When i run this all I get is half of the X, — Basam, Oct 16 '15 at 15:09
Add (by using [edit]) your actual question (about BlueJ and the half of the picture) to your post. — PM 77-1, Oct 16 '15 at 15:14
So you are asking how to finish your X to get the lower half? — Chris, Oct 16 '15 at 15:14

score 1 · Accepted Answer · edited Oct 16 '15 at 16:30

1

Try this one:

int xSize = 21;
int ySize = 21;
String sign = "X";

for (int i = 0; i < xSize; ++i) {
    for (int j = 0; j < ySize; ++j) {
        if (i == j) {
            System.out.print(sign);
        } else if (i == ySize - j - 1) {
            System.out.print(sign);
        } else {
            System.out.print(" ");
        }

    }
    System.out.println();
}

explanation: The first operate on Xaxis coordinates, second for operates on Yaxis. Our task is to cover diagonal. Covering first diagonal is where coordinateX == coordinateY. In code is if(i==j). These are points (1,1), (2,2)...... Second diagonal are points where (x,y)= (20,1),(19,2),(18,3) .... This situation covers second if(i == ySize - j - 1) .

edited Oct 16 '15 at 16:30

brimborium

9,362
9
48
76

answered Oct 16 '15 at 15:26

Edwin Jarosiński

71
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2

Since you provided solution to this homework can you also add explanation how you created it so OP could learn from it? What is the purpose of `if (i == j)` and `if (i == ySize - j - 1)`? – Pshemo Oct 16 '15 at 15:38
its positions of first diagonal where X's occur. positions (1,1) (2,2) (3,3) .... . Next 'if' is second diagonal, positions (20,20) (19,19) ..... – Edwin Jarosiński Oct 16 '15 at 15:58
It is not me who may need this explanation but OP. Edit your answer and add that this informations there. – Pshemo Oct 16 '15 at 16:00

score 0 · Answer 2 · answered Oct 16 '15 at 15:44

You can try that:

public class ProductX {
        public static void main(String[] args) {
        for (int i = 0; i <= 10; i++) {
                for (int j = 0; j < 10; j++) {
                System.out.print(" ");
                if (i == j) {
                    System.out.print("X");
                }
                if(j == 9-i){
                    System.out.print("X");
                }
            }
        System.out.println();}

    }
}

Sharath Bhaskara · Answer 3 · 2015-10-16T18:24:53.460

Though the above solutions work perfectly, I tried to experiment by not using nested for and the soultion is as below. This will have higher performance than using nested for which has a complexity of O(n2) when compared to O(n) in this.

     public void testXFormation() {
        final int countOfLines = 21;
        int countOfSpaceBefore = 0;
        int countOfSpacesAfter = countOfLines -2 ;// 2 characters
        boolean halfReached = false;
        for (int index = 0; index < countOfLines; index++) {
            printSpaces(countOfSpaceBefore); // print required no. of spaces
            System.out.print("x"); // print first x
            printSpaces(countOfSpacesAfter); // print required no. of spaces after x
            if (index != (countOfLines / 2))// Avoid printing double, in the middle
                System.out.print("x");
            System.out.println(""); // move to next line

            /* Once you reach half matrix we need to reverse the logic */
            if (index >= (countOfLines - 1) / 2) {
                halfReached = true;
            }

            /* Reversing the logic for the spaces to be printed */
            if (halfReached) {
                countOfSpaceBefore--;
                countOfSpacesAfter += 2;
            } else {
                countOfSpaceBefore++;
                countOfSpacesAfter -= 2;
            }
        }

    }

    private void printSpaces(int count) {
        for (int i = 0; i < count; i++)
            System.out.print(" ");
    }

Java nested loops

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