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Draw a square

Posted on 2007-03-17
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Last Modified: 2010-08-05
I have a function that is supposed to accept an odd number and draw a corresponding square with an '*' the square has two diagonals and all other spaces are left blank.Any ideas on how this can be done? I have started with a 2-d arrray

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Question by:wkellya
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by:Raymun
ID: 18742548
If you draw the output on paper and take note of the coordinates (your 2-d array indices), you'll see a pattern...
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by:marchent
marchent earned 100 total points
ID: 18742563
looks like homework :)
here is an easy process, you can implement this.

declare an array [100][100] and initialize with space.
then take an input odd N from user.
make all the columns of row[1] = '*' using loop
make all the columns of row[N] = '*' using loop
make all the rows of column[1] = '*' using loop
make all the rows of column[N] = '*' using loop

finaly print all NxN into screen from (1,1)

~marchent~
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Expert Comment

by:Raymun
ID: 18742604
marchent: I think wkellya wants to draw the diagonals in the square and not the perimeter

wkellya: You don't need an array to do this. If you do what I suggested above, you'll see a pattern and the *'s will occur in two conditions:

1) row == column
2) row == N - column - 1

where N is the input. Then, in your code, set up two for loops (one to traverse each row and the other to traverse each column) each up to N and finally output a "*" when either of the above two conditions are met, or " " otherwise.
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by:wkellya
ID: 18743224
I should have mentioned.The perimeter of the square has to be drawn and the diagonals. so I think what marchent was suggesting is on the right track a little more clarification would help though.
 

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by:marchent
marchent earned 100 total points
ID: 18743319
since the indexing from (1,1) so the loop for diagonal should be
    for(i=1;i<=N;i++)
        A[i][i] = A[i][N - i + 1] = '*';

the patterns is as below
doagonal -1
(1,1) : (2,2) : (3,3) : (4:4) ... (N:N)
diagonal -2
(1,N): (2,N-1) : (3,N-2)  ... (N:1)

~marchent~
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Author Comment

by:wkellya
ID: 18743337
Ok so at this point is it better to define the maximum square 19*19 fill that with zeros and  draw in the perimeter and the diagonals or fill the whole square with "*" and carve out the blank spots
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by:marchent
ID: 18743355
nope, fill with empty 19x19, and plot the *'s there using the loops.
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by:wkellya
ID: 18743392
And what about the perimeter? How do I loop to fill just the first and last columns and rows I've been trying to do both in one loop but it isn't working out
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by:marchent
ID: 18743728
for(i=1;i<=N;i++)
  A[1][i] = A[N][i] = '*';

first one is column 1 and last one is column N
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Expert Comment

by:Infinity08
ID: 18743731
>> I've been trying to do both in one loop but it isn't working out
Show us your attempt, please.
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Author Comment

by:wkellya
ID: 18743805
     for(int x=0;x<oddint;x++)
      {
            for(int y=0;y<oddint;y++)
            {
                  square[x][0]='*';
                  square[x][oddint]='*';
                  square[y][0]='*';
                  square[y][oddint]='*';
            
                  
            }
      }
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Expert Comment

by:Infinity08
ID: 18743828
It's a start, but it doesn't really do what you want it to do. To understand what it's doing, try following the loops, and write on a piece of paper what it's outputting.

The easiest way to draw a square with its diagonals, is to do it in 3 steps :

1) draw the top line
2) draw the middle part (one line at a time in a loop)
3) draw the bottom line

1) and 3) are easy. The harder part is 2).
In each iteration of the loop in 2), you can draw the first and the last position. For the diagonals, you'll have to find the pattern for the cross. Take a look at this cross, and try to find a way how you could draw that one line at a time :

    *     *
     *   *
      * *
       *
      * *
     *   *
    *     *
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Author Comment

by:wkellya
ID: 18743866
so whats wrong with what I attempted?
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Expert Comment

by:Infinity08
ID: 18743873
Take a look at what it does. I'll unroll the loops for oddint = 3 : First the innermost loop :

     for(int x=0;x<oddint;x++)
      {
                  square[x][0]='*';
                  square[x][oddint]='*';
                  square[0][0]='*';
                  square[0][oddint]='*';
                  square[x][0]='*';
                  square[x][oddint]='*';
                  square[1][0]='*';
                  square[1][oddint]='*';
                  square[x][0]='*';
                  square[x][oddint]='*';
                  square[2][0]='*';
                  square[2][oddint]='*';
      }

And then completely :

                  square[0][0]='*';
                  square[0][oddint]='*';
                  square[0][0]='*';
                  square[0][oddint]='*';
                  square[0][0]='*';
                  square[0][oddint]='*';
                  square[1][0]='*';
                  square[1][oddint]='*';
                  square[0][0]='*';
                  square[0][oddint]='*';
                  square[2][0]='*';
                  square[2][oddint]='*';

                  square[1][0]='*';
                  square[1][oddint]='*';
                  square[0][0]='*';
                  square[0][oddint]='*';
                  square[1][0]='*';
                  square[1][oddint]='*';
                  square[1][0]='*';
                  square[1][oddint]='*';
                  square[1][0]='*';
                  square[1][oddint]='*';
                  square[2][0]='*';
                  square[2][oddint]='*';

                  square[2][0]='*';
                  square[2][oddint]='*';
                  square[0][0]='*';
                  square[0][oddint]='*';
                  square[2][0]='*';
                  square[2][oddint]='*';
                  square[1][0]='*';
                  square[1][oddint]='*';
                  square[2][0]='*';
                  square[2][oddint]='*';
                  square[2][0]='*';
                  square[2][oddint]='*';

Can you see what's not correct ?
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Author Comment

by:wkellya
ID: 18743887
Not exactly it is supposed to fill the perimeter i was going to do the diagonals in another loop
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Expert Comment

by:Infinity08
ID: 18743892
>> Not exactly it is supposed to fill the perimeter
And is it doing that ? Why not ? Take a look at the unrolled version of your loops in my last post to see wh yit's not working.
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Author Comment

by:wkellya
ID: 18743915
i'm not seeing it
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Expert Comment

by:Infinity08
ID: 18743957
>> i'm not seeing it

Well, then take a piece of paper and draw a 3 by 3 grid on that.
Then go over the unrolled loop I posted, and place mraks in the grid. Try to see what's going wrong.
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Author Comment

by:wkellya
ID: 18744071
So I should start with row 0 column 0?
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by:Infinity08
ID: 18744080
>> So I should start with row 0 column 0?
Yes, the first line :

                  square[0][0]='*';

marks row 0, column 0.

That's right.

Now do the same for the other lines. And you'll immediately see where the problem lies.
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Author Comment

by:wkellya
ID: 18744102
Its only assigning one '*'
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Expert Comment

by:Infinity08
ID: 18744110
>> Its only assigning one '*'
To be more exact, it's marking the same positions over and over again, and it's only marking the left and right sides of the square.

So, follow the 3 steps I listed earlier, and try to come up with new code to do them.
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Author Comment

by:wkellya
ID: 18744497
I'm just not getting this its just not working.Thanks for the help tho i'll split the points to be fair.
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Expert Comment

by:Infinity08
ID: 18744526
wkellya, you don't have to close this question until your problem is solved
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by:Infinity08
ID: 18744530
>> I'm just not getting this its just not working.
How far did you get. What do you have now ?
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Author Comment

by:wkellya
ID: 18744581
for(int x=0;x<oddint;x++)
      {
            for(int y=0;y<oddint;y++)
            {
                  square[0][y]='*';
                  square[oddint][y]='*';
            }
      }
for the top and bottom rows but i'm only getting the top
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Expert Comment

by:Infinity08
ID: 18744608
That's closer, but you don't need nested loops for this. Just this loop :

            for(int y=0;y<oddint;y++)
            {
                  square[0][y]='*';
                  square[oddint][y]='*';
            }

will achieve the same, namely to draw the left and right borders of the square.
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Author Comment

by:wkellya
ID: 18744652
so why isn't the bottom row being drawn with just this? won't I need to nest if I want to fill the left and right colomns?
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Expert Comment

by:Infinity08
ID: 18744675
>> so why isn't the bottom row being drawn with just this?
This doesn't "draw" anything - it's just filling in a matrix.

Can you show the complete code you're using ?
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Author Comment

by:wkellya
ID: 18744704
that is it other than the initialization  zeros and the print out

for( int i=0;i<oddint;i++)
      {
            for(int j=0;j<oddint;j++)
            
                  printf("%c  " ,square[i][j]);
                    
            
     }
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by:Infinity08
ID: 18744722
>> that is it other than the initialization  zeros and the print out
Can you show the code completely, please ?

You can already start by adding a newline in the output loops :

      for( int i=0;i<oddint;i++)
      {
            for(int j=0;j<oddint;j++) {
                  printf("%c  " ,square[i][j]);
            }
            printf("\n");
     }
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Expert Comment

by:Infinity08
ID: 18744726
btw, is there a reason that you're using a matrix ? Why not just directly output ?
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Author Comment

by:wkellya
ID: 18745026
isn't a 2d array easier?
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Expert Comment

by:Infinity08
ID: 18745038
>> isn't a 2d array easier?
Well, it's an extra step. If you want to use it, fine - it's your choice.

Any progress ? How about the complete code ?
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Author Comment

by:wkellya
ID: 18745080
no progress i just  just took a break complete function
void drawSquare(int oddint)
{
      
      int square[row_size][col_size];
      
      for(int i=0;i<row_size;i++)
      {
            for(int j=0;j<col_size;j++)
            {
                  square[i][j]=0;
          }
    }

      for(int x=0;x<oddint;x++)
      {
            for(int y=0;y<oddint;y++)
            {
                  square[0][y]='*';
                  square[oddint-1][y]='*';
                           
            }
      }

      for( int i=0;i<oddint;i++)
      {
            for(int j=0;j<oddint;j++)
            
                  printf("%c " ,square[i][j]);
             
            
        }
             
}
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Author Comment

by:wkellya
ID: 18745112
i upped the points does anyone want to jump in here and tell me where i'm going wrong?
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Expert Comment

by:Infinity08
ID: 18745145
>>                   square[i][j]=0;
For initializing, it's better to use a space like this :

                  square[i][j] = ' ';


And about the main part where you fill the matrix :

      for(int x=0;x<oddint;x++)
      {
            for(int y=0;y<oddint;y++)
            {
                  square[0][y]='*';
                  square[oddint-1][y]='*';
                           
            }
      }

As I said earlier, there is no reason to have a nested loop here. I repeat : the following does exactly the same :

            for(int y=0;y<oddint;y++)
            {
                  square[0][y]='*';
                  square[oddint-1][y]='*';
                           
            }

It just draws the left and right side of the square.

Did you try with the 3 steps I suggested earlier ?

Just fyi : we are not allowed to do the work for you, since this is most likely homework, but we will help you find the solution yourself.
Please take note of the suggestions made, make some modifications to your code, and post it again if you have further questions/problems.

If something is not clear, then please tell us so, clearly indicating what it is that is not clear. Just saying that it still doesn't work, does not help us understand what problem you are having.
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Author Comment

by:wkellya
ID: 18745192
the second loop is for the bottom line. Are u saying i can't draw the perimeter using the two existing loops?How many loop will i need?One for each line?

ps:i don't want anyone to tell me anything of course i wouldn't learn anything if they did i just need some pointers on what is wrong with my code since I'm not seeing it ur way.I saw ur three steps  and this is the approach i was always using but the problem lies in trying to implement them.
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Expert Comment

by:Infinity08
ID: 18745204
>> and this is the approach i was always using but the problem lies in trying to implement them.

Ok, then explain me your approach. The code that you showed is not enough for me to understand what you're trying to do.
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Author Comment

by:wkellya
ID: 18745245
     for(int x=0;x<oddint;x++)
      {
            for(int y=0;y<oddint;y++)
            {
                  square[0][y]='*';//Fill in all the columns in row 0
                  square[oddint-1][y]='*';//Fill in all the column in the last row
                  square[x][0]='*'; //Fill in all the rows in column 0 for left line
                  square[x][oddint-1]='*';//Fill in all the rows in  last column for the right line

            }
      }
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Expert Comment

by:Raymun
ID: 18745365
>>  isn't a 2d array easier?
If you don't use an array, then you'd have to output the image in a single pass. This is a little more difficult because you'd have to draw the diagonals and the perimeter at the same time. With an array, you can construct the image in it with multiple passes and finally one last pass to output. Using an array, one strategy might be to first construct the perimeter and then make a second pass to construct the diagonals.

>> int square[row_size][col_size];
Do you mean char array? You can also initialize the array without using for loops:
char square[row_size][col_size] = { ' ' };

>> Are u saying i can't draw the perimeter using the two existing loops?
He's saying the operations

    square[0][y]='*';
    square[oddint-1][y]='*';

don't depend on the outer loop (there is no x) so it is not needed. However, it is needed in order to draw the left and right sides, so including it as you did above is good.

>> square[x][0]='*'; //Fill in all the rows in column 0 for left line
>> square[x][oddint-1]='*';//Fill in all the rows in  last column for the right line
This code is correct but placed in a bad spot. It is being executed on every iteration of the inner loop. Although the output will be correct, the code redundantly writes to the same array indices (because x does not change).
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Author Comment

by:wkellya
ID: 18745669
If it is correct why is the output wrong?
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Expert Comment

by:Raymun
ID: 18745932
You need to print a new line character ( '\n' ) after printing each row.
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Expert Comment

by:Infinity08
ID: 18746441
>> If it is correct why is the output wrong?

I've already told you that, wkellya. See my earlier posts ...


>> Ok, then explain me your approach. The code that you showed is not enough for me to understand what you're trying to do.

You still didn't explain me your approach ... can you do that please. If you're not happy with the 3-step approach I suggested, then at least tell us yours, so we can help you implement it !
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Expert Comment

by:Infinity08
ID: 18746490
>> If you don't use an array, then you'd have to output the image in a single pass. This is a little more difficult

And for the record, I don't agree with this for 3 reasons :

1) You have to use a 2d array, which makes the whole code more complicated.
2) Using my 3 steps to do it in a single pass is extremely easy to implement, so you don't need to be distracted by a 2d array.
3) I'm not sure that the assignment allows the use of a 2d array. It kind of defeats the purpose of the exercise.
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Author Comment

by:wkellya
ID: 18747274
     for(int x=0;x<oddint;x++)
      {
            for(int y=0;y<oddint;y++)
            {
                  
                    square[0][y]='*';//Fill in all the columns in row 0
                          square[oddint-1][y]='*';//Fill in all the column in the last row
                          square[x][0]='*'; //Fill in all the rows in column 0 for left line
                          square[x][oddint-1]='*';//Fill in all the rows in  last column for the right line
                           
            }
      }

In this approach I'm trying to draw the perimeter. Once that is done correctly. I will implement a for loop to pass in the diagonals. Do you understand what I am saying? I understand in your step 2) you want me to draw the cross one line at a time but i have attempted that and it didn't work so so to want to try it this way.
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Author Comment

by:wkellya
ID: 18747430
for(int i=0;i<=oddint;i++)

      {
            square[i][i] = square[i][oddint - i + 1] = '*';
      }

This works for the first diagonal but only up to 5 beyond that it kind of overlaps
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Accepted Solution

by:
Infinity08 earned 100 total points
ID: 18747536
You don't need the nested loop as I said earlier. Try just this :

            for(int y=0;y<oddint;y++)
            {
                    square[0][y]='*';//Fill in all the columns in row 0
                    square[oddint-1][y]='*';//Fill in all the column in the last row
                    square[y][0]='*'; //Fill in all the rows in column 0 for left line
                    square[y][oddint-1]='*';//Fill in all the rows in  last column for the right line
            }

This will set the whole square (all 4 sides).


And for the cross, what you posted in your last code is very good :

            for(int i=0;i<oddint;i++)
            {
                  square[i][i] = square[i][oddint - i - 1] = '*';
            }

I made only a minor modification, namely the upper limit for i in the for loop, and the column index.


That's it - you've got working code !!

Now, you could realise that you don't need two loops, and that you can just combine both loops into one loop.
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Expert Comment

by:Infinity08
ID: 18747546
Note that I did only minor modifications to your code - you did most of the work yourself !! That's the way it should be :)
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Author Comment

by:wkellya
ID: 18747858
Yes thats correct. thanks a bundle everyone!
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