Nested loops and numbers pyramid

If it is allowed to "translate" your question...

Actually we have a string "1234567890".
We need to type in each line a string that contains N characters from position i plus the same N characters but in the opposite direction.
In each line we need to increase N and i.
If this is right, I attached a c-file (main.c) that does it. I compiled in VS 2005.

* code removed - assignment *
evilrix EE ZA

                                              

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pgnatyuk, there's no point in posting full code solutions for what is very likely an assignment, but is certainly intended for learning. It's not allowed by this site's rules in fact (and you know that). So, please don't do that.

Smanyx, I suggest to ignore pgnatyuk's reply, and to try to do this yourself with our help. See the hint in my previous post for example.

Sorry.
Now I think it is really a homework. I hope, I didn't understand the question and my code is wrong.

pgnatyuk:

The output required has to be a isosceles triangle with leading space characters!
And I need to use just nested loops, not advanced concepts such as pointers...
Thanks

Smanyx

Smanyx, please ignore my post with the code.

Does my hint help you further, Smanyx ?

A good way to start is to try to generate the left half of the pyramid only, ie. :

----------1
---------23
--------345
-------4567
------56789
-----678901
----7890123
---89012345
--901234567
-0123456789

And one that works, you can add the right half.

Infinity08:

I am kind of struggling with the idea of implementing the two loops within the k loop..
I understand that from the top of the triangle to its bottom, that central axis has the following numbers: 1,3,5,7,9.
With this code I get the above output generated twice:
int main()
{
 int i, j, row;
for (i = 1, i <=2, i++)
{
for (j =1, j <=9, j+=2)
{
printf("%d", j);
printf("\n");
row++;
}
}
system("pause");
return 0;

}
This code gives me the following output:
1
3
5
7
9
1
3
5
7
9
Which is kind of the spine of the triangle.
From here, I would like to understand  how I can go about, implementing for each row, the leading space characters and the other digits, counting up and down on each side of the axis and apparently beginning and stopping at the row number.

I don't know how to continue with this idea.

I think (I can be wrong), you need to think that you have 2 loops where the first handles the position and the count of the characters and the inner loop creates the line to be printed.

Your line is '1234567890'

so, for the first iteration you take the first character - index 0 in the given string and you need to type only 1 character.
For the second iteration you take secon and third characters from the given string - they have indexes 1 and 2.
Then 3 characters - 2, 3 and 4.
And so on.
so the main loop is simple. Right?

Seems like I have to stop here, because people will say that I'm doing your homework. Sorry. Bye.

First of all it's not a homework.

I may be wrong with my last post. I was just a little confused...
I better come back to the first idea then...
I'm still thinking. I will post if I make any progress.

Ok, one step at a time.

You have the outer loop (from your original code) that goes from 1 to 10 :

        for(i=1; i<=10; i++)

Each iteration of this loop corresponds to one row in the pyramid, starting with the top-most.

If we look at the left half of each row, we see that each row starts with the current row number (1 for the first row, 2 for the second row, etc.). And conveniently, we have that row number available in the variable i. The next number on each row is one more than the previous, etc. basically counting, starting from i.

With this information, can you add a second loop inside the above loop that would print the left half of the pyramid ?

No worries evilrix.
I understand.
Thanks.

Hi there,

So far, I have managed to come up with just what seems to be a 'third' of the solution. This is my code:

#include<stdio.h>
#include<stdlib.h>

int main()

{     int i,j,k;

     for(i=0; i<10; i++) //manages the pyramid steps(new lines)
     {
          for (j = 1; j < 10-i; j++)//manages leading space characters
          {
              printf(" ");
               
              }
          for (k=i+1; k<= 2*i+1; k++)//manages what should be written in one line
          {
              printf("%d", k);
          }
         printf("\n");
    }

      system("pause");
    return 0;

}

The output looks partially right, but whenever numbers get greater then 10, how can I, looking at this code, reset that value to 0 and so on...

>> The output looks partially right, but whenever numbers get greater then 10, how can I, looking at this code, reset that value to 0 and so on...

There is a very easy way using the modulo operator (%). The result of (value % 10) is a value between 0 and 9 (inclusive).

Not quite there yet !

Look at what I have, what am I doing wrong?

#include<stdio.h>
#include<stdlib.h>

int main()

{     int i,j,k,m;

     for(i=0; i<10; i++) //manages the pyramid steps(new lines)
     {
          for (j = 1; j < 10-i; j++)//manages leading space characters
          {
              printf(" ");
               
              }
          for (k=i+1; k<= 2*i+1; k++)//manages what should be written in one line
          {
              if (k >= 10)
                    {
                     m = k % 10;
                     printf("%d", m);
                    }
                    else
                    {
                    printf("%d", k);
                    }
          }
              for (k=i+1; k<= 2*i-1; k++)
              {
                    if (k >= 10)
                    {
                     m = k % 10;
                     printf("%d", m);
                    }
                    else
                    {
                    printf("%d", k);
                    }
              }
         printf("\n");
    }

    system("pause");
    return 0;

}

>>               if (k >= 10)
>>                     {
>>                      m = k % 10;
>>                      printf("%d", m);
>>                     }
>>                     else
>>                     {
>>                     printf("%d", k);
>>                     }

You don't have to check whether k is larger than 10. You can simply do :

        printf("%d", k % 10);



>> Look at what I have, what am I doing wrong?

So, the left half of the pyramid is ok. The right half is almost ok, but you're skipping one value. For example, on the third row, you have 3453 instead of 34543.

So you need to adjust the limits to start with the correct value

>> So you need to adjust the limits to start with the correct value

And you'll need to count down instead of up too. The right half of the pyramid has decreasing values.

I am really struggling with the condition the loop for the right half of the pyramid.
I have exausted all possible avenues( I mean from what I believe)... I am now just groping around in the dark. It's may not be rational what I am doing now but, is this any where close?

for (s = k-1; s >= 2*i-1; s--)

k being the last value not written (that stopped the first incrementing loop)
I really need more hints ...

>> for (s = k-1; s >= 2*i-1; s--)

Well, you basically have to do the same thing as the first loop, but in reverse order. So the stop condtion of the second loop is the same as the start condition of the first loop.

I can feel that I'm getting closer but still not quite there yet.
What should I change to this loop ? Is it leading in the right direction?

for (k=2*1+1; k>= i+1; k++)

Bingo !!!!

I got it. I finally got it. This is such a relief !
Infinity08:, thank you so much for your guidance. I really, truly appreciate it.

Here is the final code:

#include<stdio.h>
#include<stdlib.h>
 
int main()
 
{     int i,j,k;
 
     for(i=0; i<10; i++) //manages the pyramid steps(new lines)
     {
          for (j = 1; j < 10-i; j++)//manages leading space characters
          {
              printf(" ");
               
		  }
          for (k=i+1; k<= 2*i+1; k++)//left part of the pyramid, increments
          {
              
			  printf("%d", k % 10); 
			  
          }
			  
		  for (k=2*i; k>= i+1; k--)//right part of the pyramid, decrements
		  {
			  		
			  printf("%d", k % 10 ); 
			
		  }
         printf("\n");
    }
 
	system("pause");
    return 0;
 
}
                                              

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