Stars rhombus using c++ - c++

Question

How can I output the below asterisks in rhombus shape in c++ using the least number of loops and variables. Maximum no. of stars = n (input taken from the user)
*
* *
* * * = n
* *
*
I tried doing it with just 2 loops but didn't succeed plus the program is too complicated to understand so didn't include it here.
So any algorithms you can think of?

Answer 1

That's not a pyramid, that's a rhombus :)
Anyway let's see the star number:
n=1 -> 1 star
n=2 -> 4 stars
n=3 -> 9 stars
The pattern looks like twice the sum from 1 to n minus n (because the middle row exists once not twice), so s(n) = 2 * (1+n)/2 * n - n = (1+n)*n-n = n*n hey it's the area of a square! Well how surprising. :-)
Now how to draw a n-sized rhombus:
there will be 2*n-1 rows
each row is w = 2*n-1 characters wide
each ith row (starting from i=0) has m = min(i+1, 2*n-1-i) stars, and m-1 spaces between them
so each row needs w - (m) - (m-1) spaces of space padding at sides, i.e. w/2 on left and on right.
Now go write it! :)

Answer 2

This is a common question for education I believe, and thus a Google search is very helpful.
Here is a ready made solution in Visual C++ (as some people learn better from code examples, the choice is yours)...
http://www.softwareandfinance.com/Visual_CPP/Loops_Diamond_Pattern.html
EDIT: Trying to reduce the for loops and keeping the spacing between stars, this is my best effort...
int i, j;
int n = 0;
std::cout << "Enter the maximum number of *:";
std::cin >> n;
std::cout << "\n\n";
for (i = 1; i <= n; i++)
{
std::cout.width(n - i);
std::cout.fill(' ');
std::cout << "";
for (j = 1; j <= i; j++)
std::cout << "* ";
std::cout.width(n - i);
std::cout.fill(' ');
std::cout << "";
std::cout << "\n";
}
for (i = n - 1; i >= 1; i--)
{
std::cout.width(n - i);
std::cout.fill(' ');
std::cout << "";
for (j = 1; j <= i; j++)
std::cout << "* ";
std::cout.width(n - i);
std::cout.fill(' ');
std::cout << "";
std::cout << "\n";
}
std::cout << "\n";
ONLY 2 LOOPS USED:
int i, j;
int n = 0, c = 0, inc = 1;
std::cout << "Enter the maximum number of *:";
std::cin >> n;
std::cout <<"\n\n";
for (i = 1; i <= (n * 2) - 1; i++)
{
c += inc;
if(i == n)
inc = -1;
std::cout.width(n - c);
std::cout.fill(' ');
std::cout << "";
for (j = 1; j <= c; j++)
std::cout << "* ";
std::cout.width(n - c);
std::cout.fill(' ');
std::cout << "";
std::cout <<"\n";
}
std::cout <<"\n";

Answer 3

Two loops? Surely you can get by with just one.. Here is a rough description of the alogirthm I came up with:
Loop through the (2*n-1)^2 square
Determine the amount of stars needed on the current row (i.e. distance to the row with n stars)
Determine the distance from the middle of the current index
If the absolute distance of the current index is smaller than the amount of stars on that row, it needs a star in odd/even indices depending if n is odd/even.
Add endlines end of rows
And here's a near IOCC-worthy quick improvisation of the algorithm that draws the rhombus with 2 variables, the loop counter and n=amount of stars in the middle row.
#include <cmath>
void print_rhombus(int n)
{
for (int i = 1; i <= ((2*n-1)*(2*n-1)); ++i) {
if ((abs((((2*n-1)+1)/2)-(i%(2*n-1) == 0?2*n-1:i%(2*n-1)))) - (n-abs((i%(2*n-1)==0?(i/(2*n-1)):(i/(2*n-1))+1) - n)) < 0 && (
(n%2==1 && ((n-abs((i%(2*n-1)==0?(i/(2*n-1)):(i/(2*n-1))+1) - n))%2==1 && (i%(2*n-1) == 0?2*n-1:i%(2*n-1))%2 == 1 || (n-abs((i%(2*n-1)==0?(i/(2*n-1)):(i/(2*n-1))+1) - n))%2==0 && (i%(2*n-1) == 0?2*n-1:i%(2*n-1))%2==0)) ||
(n%2==0 && ((n-abs((i%(2*n-1)==0?(i/(2*n-1)):(i/(2*n-1))+1) - n))%2==1 && (i%(2*n-1) == 0?2*n-1:i%(2*n-1))%2 == 0 || (n-abs((i%(2*n-1)==0?(i/(2*n-1)):(i/(2*n-1))+1) - n))%2==0 && (i%(2*n-1) == 0?2*n-1:i%(2*n-1))%2==1))
)) {
std::cout << "*";
} else {
std::cout << " ";
}
if (i%(2*n-1) == 0 && (n-abs((i%(2*n-1)==0?(i/(2*n-1)):(i/(2*n-1))+1) - n)) == n) {
std::cout << " = " << n << std::endl;
}
else if (i%(2*n-1) == 0) {
std::cout << std::endl;
}
}
}
It also prints the " = N", which I suppose may not have been the requirement. As you can see, sometimes a couple of extra descriptive variables would go a long ways. And if this were homework, you'd better be prepared to explain it..

Answer 4

Break the task down.
1) First write a program that outputs a square of asterisks. In the example above (n = 3, I think) a five by five square will contain all the asterisks you want to output. Two loops (one inside the other) is the way to do this.
2) Now you have all the asterisks you need, work out formulas for which asterisks should be skipped to get the diamond pattern you want. If you decide an asterisk should be skipped you output a space instead.
All programming is like this, you're faced with a complex problem, you break it down into smaller sub-problems.

Answer 5

Not filled rhombus like this one:
*
* *
* *
* *
* *
* *
*
Code c++:
#include <iostream>
#include <string>
using namespace std;
int main()
{
int w=7;
int h=7;
string fill("*");
string empty(" ");
for (int i = 0; i<w; i++)
{
for (int j = 0; j<h; j++)
{
if((j == h/2-i) or (j == h/2+i) or (j == -h/2+i) or (j+i == h+h/2-1) ) cout << fill;
else cout <<empty;
}
cout<<endl;
}
}