How do I create this structure using only a for loop

Question

Yes, this is homework, but I cant figure out how to do it using only for loops.

enter image description here

I was able to reproduce the structure with a for loop and if else statements but my instructor told me that won't fly.

public class Problem3 {
public static void main(String[] args) {
    /** Code under here */
        int box = 22;
        for(int i=22; i>0; i--)
        {
         if(i==18)
             System.out.println("\\\\"+"!!!!!!!!!!!!!!!!!!"+"//");
         else if(i == 14) 
             System.out.println("\\\\\\\\"+"!!!!!!!!!!!!!!"+"////");
         else if (i== 10)
             System.out.println("\\\\\\\\\\\\"+"!!!!!!!!!!"+"//////");
         else if (i == 6)
             System.out.println("\\\\\\\\\\\\\\\\"+"!!!!!!"+"////////");
         else if (i == 2)
             System.out.println("\\\\\\\\\\\\\\\\\\\\"+"!!"+"//////////");
         else if(i == 22)
             System.out.println("!!!!!!!!!!!!!!!!!!!!!!");
        }


   System.out.println("");
}    
}

Use a nested loop to construct the string. – SLaks Oct 21 '13 at 15:20 — SLaks, Oct 21 '13 at 15:20
@SLaks ... I know I need to do that. My question is how. – Christopher Baldwin Oct 21 '13 at 15:44 — Christopher Baldwin, Oct 21 '13 at 15:44
Oh boy am I sending this one around to my pupils. – MLProgrammer-CiM Oct 21 '13 at 16:10 — MLProgrammer-CiM, Oct 21 '13 at 16:10

score 4 · Accepted Answer · answered Oct 21 '13 at 15:23

4

Some hints:

The lines are always symetric
The number of \ at the beginning of a line can be calculated from the line number (starting at 0)
The number of characters on a line is constant, so the number of ! can also be calculated if you know how many \ there are.

Based on these observations, you should be able to produce a code with only 2 levels of nested loops (I didn't say 2 loops, just 2 levels).

answered Oct 21 '13 at 15:23

Kaidjin

1,433
1
12
18

I just wanted to say thanks. I can't upvote your post because im too new. – Christopher Baldwin Oct 21 '13 at 15:35
You're welcome. You can probably accept the answer though, to let everyone know that you found a solution to your problem. – Kaidjin Oct 21 '13 at 15:37
I'm still trying to figure it out though :P Don't worry, I'll give you your just rewards. From the code 'given' below by Muhammad, I got the idea to just build each part separately. Would that work? Or will the box not form correctly. – Christopher Baldwin Oct 21 '13 at 15:41
Yes, you have to build each part of the line separately. The trick is to figure out the rules which defines the lines in function of the number of the line. – Kaidjin Oct 21 '13 at 15:57

score 1 · Answer 2 · edited May 23 '17 at 12:21

1

There are 22 characters in a row, and row n has 2*(n-1) \ characters, 2*(n-1) / characters and 22 - 4*(n-1) ! characters.

You should first build the string you want to print out (for each line). That will involve repeating the characters (for which this question provides a few ideas)

edited May 23 '17 at 12:21

Community

1
1

answered Oct 21 '13 at 15:26

SheetJS

22,470
12
65
75

OldCurmudgeon · Answer 3 · 2013-10-21T15:51:10.840

0

Notice that in each row i (if you number them from 0) there are 2 * i \ characters at the start, 2 * i '/' characters at the end and the rest of the row is filled with '!' characters.

So your code will look something like:

for ( int row = 0; row < 6; row++ ) {
  int slashes = 2 * row;
  for ( int slash = 0; slash < slashes; slash++ ) {
    // Print a `\`
  }
  // Work out how many '!'s you need and print them.
  // ... you try this bit
  for ( int slash = 0; slash < slashes; slash++ ) {
    // Print a `/`
  }
}

edited Oct 21 '13 at 15:51

answered Oct 21 '13 at 15:24

OldCurmudgeon

64,482
16
119
213

I see that, but how do I get nested for loops to print the different parts of the triangle? Actually @Muhammad-S 's answer that's being mercilessly downvoted is starting to make sense when I combine it with your explanation. Should I build each part separately – Christopher Baldwin Oct 21 '13 at 15:39

score 0 · Answer 4 · answered Oct 21 '13 at 15:25

Notice that each time, there are two more slashes/backslashes on each side (and thus four less !s in the middle). You can make a for loop that increases the number of slashes, and calculates the number of !s to print to fill out the length to 22, and then print out all necessary characters for the current line.

score 0 · Answer 5 · answered Oct 21 '13 at 16:40

Is this good enough? I'm learning java too. Feedback appreciated.

    public class Problem3 {
public static void main(String[] args) {
    /** Code under here */
    int box = 22;
    int line = 0;
    String char_1 = "w";
    String char_2 = "q";
    String char_3 = "e";
    String string = "";
    for (int i = 0; i < 6; i++) {
        for (int j = 0; j < line * 2; j++) {
            System.out.print(char_1);
        }
        for (int k = box; k > 0; k --) {
            System.out.print(char_2);
        }
        for (int l = 0; l < line * 2; l++){
            System.out.print(char_3);
        }
        System.out.println(string);
        line++;
        box = box - 4;
    }

}

}

But I'm Not A Wrapper Class · Answer 6 · 2013-10-21T15:38:04.923

The number_of_lines is the number of lines inside the expected output (which is 6 in this case). The number_of_characters is the number of characters each line should have (which is 22 in this case).

We know that there is a pattern. The left side has \, the middle has !, and the right side has /. This means we need one outer loop to loop through the lines and three inner loops for each of these sections.

How do we get know how many of each characters we need?

The first line has 0 \, 22 !, and 0 /

The second line has 2 \, 18 !, and 2 /

The third line has 4 \, 14 !, and 4 /

The fourth line has 6 \, 10 !, and 6 /

The fifth line has 6 \, 10 !, and 6 /

The sixth line has 8 \, 6 !, and 8 /

The seventh line has 10 \, 2 !, and 10 /

There is a pattern. Each line has the \ and / count increment by 2 and the ! count decrement by 4.

Number of \ equals two times the current line number (assuming the first line number is 0). Same goes with /. The number of ! is the total number of characters expected per line minus four times the current line number (assuming the first line number is 0).

public void printCoolStuff(int number_of_lines, int number_of_characters){

    for(int i = 0; i < number_of_lines; i ++){

        String left= "";
        for(int j = 0; j < (2*i); j++){
            left= left+ "\"
        }

        String middle = "";
        for(int j = 0; j < (number_of_characters - 4*i); j++){
            middle = middle + "!"
        }

        String right= "";
        for(int j = 0; j < (2*i); j++){
            right = right + "/"
        }

        System.out.println(left+middle+right);
    }
}

Great! Now OP gets (probably) a working code without learning a thing! — Pankaj, Oct 21 '13 at 15:28
While I appreciate the answer.. I'm trying to get a grasp on using java. Thanks though, man! — Christopher Baldwin, Oct 21 '13 at 15:36
Im pretty sure what you did was build each part and then print them together. Will the compiler really fill in the blanks of the box like that? — Christopher Baldwin, Oct 21 '13 at 15:40
That last statement I print "left+middle+right". That should print it all out together (per line). — But I'm Not A Wrapper Class, Oct 21 '13 at 15:42
Thanks for the edit man! If I could upvote to dig you out of the hole, I would. :3 — Christopher Baldwin, Oct 21 '13 at 15:42

How do I create this structure using only a for loop

6 Answers6