C++ Diamond pattern output

[mimi7]

Hey you all. Does any of you know how to solve the code below to make a diamond pattern output? I just need help on the 'if' condition.Thanks in advance!


#include<iostream>
#include<cmath>
using namespace std;
int main()
{int x, y;
for (x=1;x<=5;x=x+1)
{
for (y=1;y<=5;y=y+1)
{
if(((abs(x-3)<y)&&(x+3>y)))
{
cout<<"*";
}
else
{
cout<<" ";
}
}
cout<<endl;
}
return 0;
}



Output (without the dashes):
--*
-***
*****
-***
--*

[Manga]

I did this...

int x, y, z;
 z=1;
 
 for (x=1;x<=5;x++)
 {
   for (y=1;y<=5;y++)
   {
     if(( (abs(z-3)<y) && (z+3>y) ))
     {
     cout<<"*";
     }
     else
     {
     cout<<" ";
     }
   }
   cout<<endl;
   if(x<3) {
	 z++;
   }
   else {
	 z--;
   }
 }

[mimi7]

Thank you very much for that, but Im sorry i forgot mention that i cannot replace or add more than 2 variables. My instructor said the rules are to not change the 'for' operations, to add more variables and to solve only the 'if' condition.

[Manga]

lol. It seems the teacher wants you to solve this in a very complicated way.
You could shove it up the teachers rear by doing this...

if((x==1 && y ==3) || (x==2 && y ==2) || (x==2 && y==3) || (x==2 && y==4))

Well, I think you see where I am going with this!!!

[keskiverto]

Code tags, please.

You almost get there:

if ( ( (abs(x-3) < y) && (x+3 > y) ) )
{
  cout << "*";
}

At least the "lower bound" looks reasonable:

if ( abs(x-3) < y )
{
  cout << "*";
}

What could be the "upper bound" for each row?

|x-3| : upper
  2   : y < 4
  1   : y < 5
  0   : y < 6
  1   : y < 5
  2   : y < 4

Do you notice a computable relation in that table?