Java Asterisk Rectangle

Question

For my CS class, my instructor has given us the assignment of creating a rectangle made of asterisks with diagonal lines drawn through it in Java.

He also told us to write it in as few bytes as possible. I've gotten it down to 190 bytes, but I need to find a few to simplify this code even more to decrease bytes. Can anyone help me with this?

This code is functional:

interface d{static void main(String[]a){for(int z=0,w=new Byte(a[0]),h=new Byte(a[1]);z<h*w;){int y=z/w,x=z++%w;System.out.print((x>w-2)?"*\n":(y%(h-1)*x*((y-x)%3)==0)?"*":" ");}}}

Input is 10 10.

Output:

**********
**  *  * *
* *  *  **
*  *  *  *
**  *  * *
* *  *  **
*  *  *  *
**  *  * *
* *  *  ** 

Answer 1

Logically, there should be an asterisk (*) every time i == j and i+j==w-1(for diagonals), i == 0 and j == 0 (for top line and left side) and j == w-1 and i==h-1 (for right side and bottom line).

class d {
    public static void main(String[] a) {
        for(int i=0,w=new Byte(a[0]),h=new Byte(a[1]);i<h;i++) {
            for(int j=0;j<w;j++) {
                System.out.print(i==0 || j==0 || i==h-1 || i+j==w-1 || j==w-1 || i==j ? "*":" ");
            }
            System.out.println();
        }
    }
}

Answer 2

The code you provided could be reduced by doing this:

• replace * with 42 and " " by ' ' (taking advantage of the charcode for *)
• move ((y - x) % 3) to the beginning of the things to be multiplied and remove the surrounding parenthesis
• remove surrounding parenthesis from (x > w - 2) and from ((y - x) % 3 * y % (h - 1) * x == 0)

The resulting code would be:

interface r{static void main(String[]a){for(int z=0,w=new Byte(a[0]),h=new Byte(a[1]);z<h*w;){int y=z/w,x=z++%w;System.out.print(x>w-2?"*\n":(y-x)%3*y%(h-1)*x==0?42:' ');}}}

Note: the last line is missing in the example output of the question! The output of the sample code is different.

Answer 3

178 174 chars

interface T{static void main(String[]a){for(int w=new Byte(a[0]),h=new Byte(a[1]),j,i=0;i<h;i++)for(j=0;j<w;System.out.print((i-j)%3*i*j!=0&i<h-1&++j<w?" ":j<w?"*":"*\n"));}}

Try it online!

Readable:

interface T{
    static void main(String[]a){
        for(int w=new Byte(a[0]),h=new Byte(a[1]),j,i=0;i<h;i++)
            for(j=0;j<w;System.out.print(
                    (i-j)%3*i*j!=0&i<h-1&++j<w?" ":j<w?"*":"*\n"));
    }
}

---

Output for field: 30 10:

******************************
**  *  *  *  *  *  *  *  *  **
* *  *  *  *  *  *  *  *  *  *
*  *  *  *  *  *  *  *  *  * *
**  *  *  *  *  *  *  *  *  **
* *  *  *  *  *  *  *  *  *  *
*  *  *  *  *  *  *  *  *  * *
**  *  *  *  *  *  *  *  *  **
* *  *  *  *  *  *  *  *  *  *
******************************

Answer 4

I don't actually have Java on my computer so I can't test this but I think it should work for 174 bytes and almost definitely could be golfed more

class d{static void main(String[]a){for(int i=0,w=new Byte(a[0]),h=new Byte(a[1]);i<h;i++)for(int j=0;j<w;j++)System.out.print(j==w-1?"*\n":i%h-1==0||!j||i+j%2==0?"*":" ");}}

Whitespace added for clarity:

class d {
    static void string main(String[] a) {
        for(int i=0,w=new Byte(a[0]),h=new Byte(a[1]);i<h;i++)
            for(int j=0;j<w;j++)
                System.out.print(j==w-1 ? "*\n" : i % h-1 == 0 || !j || i+j % 2 == 0 ? "*":" ");
    }
}

print *\n for the last char in each line, * for all of the first and last lines and the first column, and * for any time that the sum of the row and column is even, otherwise prints