I need to make a triforce containing 3 lines of number for each triangle

Question

Each line is the same number and in all three triangles the number is the same.

Input:

• The first line contains an integer n that represents the number of data sets to follow.
• Each data set will consist of 1 integer m that represents the number that the Triforce of Courage contains throughout.

Output:

Print out a triforce, using this template (replacing 0 with the integer specified):

     0
    000
   00000
  0     0
 000   000
00000 00000

Assumptions: The number to be replaced will be:0<=m<=9

Sample Input: 1 2

Sample Output:

     2
    222
   22222
  2     2
 222   222
22222 22222

The following is my code thus far:

import java.lang.Math;
import java.util.Scanner;
import java.io.*;
import java.util.*;
import java.io.FileNotFoundException;

public class TriforceOfCourage {
     public static void main(String[] args) throws FileNotFoundException {
        Scanner scan=new Scanner(new File("num.dat"));
        int n = scan.nextInt();
        int count = 0;
        while(count<n) {
            for (int i=0; i<3; i++) {
                for (int k=0; k<3-i; k++) {
                    System.out.print(" ");
                }
                for (int j=0; j<i*2+1; j++) {
                    System.out.print(n);
                }
                System.out.println("");
            }
            break;
        }
    }
}

Answer 1

There are many possible solutions to this problem. In general, there are many good strategies you would typically use to solve this:

1. Look for a pattern. Can you find a pattern that can be easily expressed through logic and algebraic operations?
2. Can the problem be broken down into smaller, more manageable parts? Sometimes what appears to be a complex problem can be separated into smaller, very simple problems.
3. When in doubt, work things out on paper.

Let's take your example. At a glance, personally, I would divide the output into two halves; the top, consisting of a single triangle, and the bottom, consisting of a two triangles side by side. Further, let's focus on just being able to draw a triangle at all. Let's say we want to produce this, a triangle at an arbitrary location:

            11
  012345678901
 0       x
 1      xxx
 2     xxxxx

We notice the triangle is centered on column 7, and we suspect it will help us if we can draw a triangle in any column, so let's let center be the center column of our triangle. A lot of these types of strategies involve coming up with a way to parameterize the problem. For this approach we want to find the answer to this question:

• Given row, column, and center, should we draw a character at that location?

Let's take a simple algebraic approach first, one row at a time, and see if we notice any patterns:

• row == 0: Here we output only when column == center.
• row == 1: Here we output when column >= center - 1 && column <= center + 1.
• row == 2: Here we output when column >= center - 2 && column <= center + 2.

Notice a pattern? Think about it for a second. Recognizing that row == 0 isn't actually a special case, the pattern is:

• Output when column >= center - row && column <= center + row.

Great! Now we can output a triangle very easily:

int center = 7; // From our example.

for (int row = 0; row < 3; ++ row) {
    for (int column = 0; column < 11; ++ column) {
        if (column >= center - row && column <= center + row)
            System.out.print("x"); // Replace with whatever character to print.
        else
            System.out.print(" ");
    }
    System.out.println(); // Line break after each row, of course.
}

But what about two triangles, for the bottom half? The simplest of course would be to do the exact same as above, but since we have two triangles, we have two centers (say, centerL and centerR), and can simply add a second if block in our bottom-half loop -- same logic for both centers, all in one loop. I'll leave this as an exercise to you.

Now, like I said, there are many possible solutions. Choose the one that makes the most sense for you and is easiest for you to get your head around. In fact, as a learning exercise, I would suggest trying to implement this program with at least three different algorithms. For example:

• Loop over all 6 rows (instead of top and bottom half) and put all 3 triangles in the same loop.
• Create a 2D array and draw the triangles into it, then output the contents.
• Try to implement the above without using if at all (e.g. loops to output from center - row to center + row, and separate loops for the borders) - this is similar to your current approach.
• Try to create a method that can draw triangles of any height, not just 3.
• Try to create a method that can draw any "triangle of triangles", e.g. 3, 4, 5 rows of triangles.

And of course,

Answer 2

As promised, i will be posting my solution. Thanks to MR. Jason i was able to solve this problem and now i need to make some slight modifications to where it reads in the numbers from a file but that i can comprehend with easily.

 // By Rexhep Rexhepi
 // 11/12/14
 // BIG THANKS TO MR. Jason!
 import java.lang.Math;
 import java.util.Scanner;
 import java.io.*;
 import java.util.*;
 import java.io.FileNotFoundException;

 public class TriforceOfCourage
 {

     public static void main(String[] args) throws FileNotFoundException
         {

            Scanner scan=new Scanner(new File("num.dat"));
            int n = scan.nextInt();
            int count = 0;
            int center = 5;
            int centerL = 2;
            int centerR = 8;

             for (int row = 0; row < 3; ++ row)
                 {
        for (int column = 0; column < 11; ++ column)
                {
            if (column >= center - row && column <= center + row)
                System.out.print("x"); // Replace with whatever character to print.
            else
                System.out.print(" ");
                }
        System.out.println(); // Line break after each row, of course.
            }

        for(int rowB = 0; rowB < 3; rowB++)
        {
        for (int columnL= 0;columnL < 5; columnL++)
        {
            if (columnL>= centerL - rowB && columnL <= centerL + rowB)
                System.out.print("x");
                else
                    System.out.print(" ");
        }
        for (int columnR = 5; columnR<11; columnR++)
        {
            if (columnR>= centerR-rowB && columnR <= centerR+rowB)
                System.out.print("x");
                else
                    System.out.print(" ");
        }
        System.out.println();
        }

} }

Answer 3

In the below code triforce(...) will print the required output.

public void triforce(int n, int m) {
    triangleWithSpaces(2, 1, n, m);
    triangleWithSpaces(2, 2, n, m);
}

public void triangleWithSpaces(int s, int t, int n, int m) {
    for (int i = 0; i <= n; i++) {
        for (int k = 0; k < t; k++) {
            for (int j = 0; j < ((n * 2 + 1) / 2 + 1) * (s - t); j++) {
                System.out.print(" ");
            }
            for (int j = n - i; j > 0; j--) {
                System.out.print(" ");
            }
            for (int j = 0; j < i * 2 + 1; j++) {
                System.out.print(m);
            }
            for (int j = n - i; j > 0; j--) {
                System.out.print(" ");
            }
            System.out.print(" ");
        }
        System.out.println();
    }
}

Sample run :

triforce(2, 0);

     0  
    000 
   00000
  0     0  
 000   000 
00000 00000

triforce(3, 9);

       9    
      999   
     99999  
    9999999 
   9       9    
  999     999   
 99999   99999  
9999999 9999999