14 replies to this topic

[rawranator]

HI! new here. I posted this question in another forum but some guy got super angry for me even asking any questions so i figured i'll try another forums. I'll try to make this explanation short.

I have a programming class, one of the work assigned to us was to make a diamond using cout statments.

// diamond.cpp : Defines the entry point for the console application.
//

#include "stdafx.h"
#include <string>
#include <iostream>

using namespace std;


int _tmain(int argc, _TCHAR* argv[])
{
cout << " * " << endl;
cout << " *** " << endl;
cout << " ***** " << endl;
cout << " ******* " << endl;
cout << "*********" << endl;
cout << " ******* " << endl;
cout << " ***** " << endl;
cout << " *** " << endl;
cout << " * " << endl;

return 0;
}


so i made this.
for extra credit, he asked us to create a new Diamond class that uses Loops to create the diamond pattern using no more than two (2) cout statements.

i been trying to do this for weeks and this is what i've gotten.

int _tmain(int argc, _TCHAR* argv[])
{

for(int i = 0; i < 7; i++){
string Display;
cout << i << "*";

do{ cout << Display;

} while(false);


}

return 0;
}


I think i'm on the right track but some input would be nice. am i doing it right? am i doing it wrong? any tips as to what to do next? (kinda stumped)

this is c++, using visual c++ 2008 express edition to write/compile the code

[bobdark]

If I told you to draw a triangle, would you be able to implement it using only one statement including cout?

[rawranator]

bobdark said:

If I told you to draw a triangle, would you be able to implement it using only one statement including cout?

you mean something like ...?

cout
<< " * \n"
<<" *** \n"
<<"******\n"

[lintwurm]

What he means is that you will need two nested for-loops... one to count the rows and one for columns... I hope that makes sense...
for example...
your first part that prints out
*
**
***
****
*****

Will be something like this:

for(int j = 0; j <= 4 ; j++)
{
for(int i = 0; i <= j; i++)
{
cout << "*";
}
cout<<endl;
}
You can do the rest yourself... ^_^

Hope this helps... However if you need more help, you know where to find us...

[rawranator]

Thank you! now i need yo try and center this and loops it so the bottom repeats the same thing but backwards (if that made sense)

i'll post the code here once i figure it out, thanks again

[rawranator]

// Diamond.cpp : Defines the entry point for the console application.
//

#include "stdafx.h"
#include <iostream>
using namespace std;

class Diamond
{
private:
int m_nMaxStars;
public:
Diamond(int nMaxStars = 9)
{
if ((nMaxStars % 2) == 0)
nMaxStars += 1;

m_nMaxStars = nMaxStars;
}
virtual ~Diamond() {}

void Draw()
{
for (int i = 1; i <= m_nMaxStars; i += 2)
{
string strLine = "";
for (int j = 0; j < ((m_nMaxStars - i) / 2); j++)
strLine += " ";

for (int j = 0; j < i; j++)
strLine += "*";

DrawLine(strLine);
}

for (int i = (m_nMaxStars - 2); i > 0; i -= 2)
{
string strLine = "";
for (int j = 0; j < ((m_nMaxStars - i) / 2); j++)
strLine += " ";

for (int j = 0; j < i; j++)
strLine += "*";

DrawLine(strLine);

}
}
private:
void DrawLine(string strLine)
{
cout << strLine.c_str() << endl;
}
};

int _tmain(int argc, _TCHAR* argv[])
{
Diamond theDiamond;
theDiamond.Draw();

cin.get();
return 0;
}

[Zer033]

Another easy way to do it is:

static const int MAXROW = 11;
static const int MAXCOL = 11;

int diamond[MAXROW][MAXCOL] = {
{0,0,0,0,0,1,0,0,0,0,2},
{0,0,0,0,1,1,1,0,0,0,2},
{0,0,0,1,1,1,1,1,0,0,2},
{0,0,1,1,1,1,1,1,1,0,2},
{0,1,1,1,1,1,1,1,1,1,2},
{0,1,1,1,1,1,1,1,1,1,2},
{0,1,1,1,1,1,1,1,1,1,2},
{0,0,1,1,1,1,1,1,1,0,2},
{0,0,0,1,1,1,1,1,0,0,2},
{0,0,0,0,1,1,1,0,0,0,2},
{0,0,0,0,0,1,0,0,0,0,2},
};

for (int row = 0; row < MAXROW; row++)
{
    for (int col = 0; col < MAXCOL; col++)
    {
        if (diamond[row][col] == 0)
            cout << " ";
        if (diamond[row][col] == 1)
            cout << "*";
        if (diamond[row][col] == 2)
            cout << "\n";
    }
}

[l@mbd@]

I love these old school programming assignments!

Few minor nits with your program...

1. You don't need a destructor. And certainly not a virtual one at that.

2. The idea is to follow the boundaries of the star. To walk you through my example:

a. Suppose we have a maximum of 9 stars, and the inner k loop is the horizontal printing loop. Initially, we want to print a star when its index is between 4 and 6 (stars/2 and stars/2 + 2)

b. Those boundaries increase and decrease respectively as we loop vertically (the j loop).

c. We want to do this twice (i loop). The first pass we decrease the lower boundary, and increase the upper boundary.

d. On the second i loop pass, we increase the lower boundary, and decrease the upper.

For any ODD number of stars...

#include <iostream>

class Diamond
{
    int stars;
    
    public:
        
        Diamond(int stars);
        void print();
};

Diamond::Diamond(int stars)
: stars(stars) {}

void Diamond::print()
{
    int min = stars/2;
    int max = min + 2;
    
    for(int i = 0; i < 2; i++)
    {
        for(int j = 0; j < stars/2 + i; j++)
        {
            for(int k = 1; k < stars + 1; k++)
            {
                if( k < max && k > min )
                {
                    std::cout << "*";
                }
                else std::cout << " ";
            }
            
            std::cout << "\n";
            if( i == 0 ) min--; else min++;
            if( i == 0 ) max++; else max--;
        }
    }
}

int main()
{
    Diamond diamond(21);
    diamond.print();
    return 0;
}

Edited by l@mbd@, 06 June 2010 - 10:30 AM.
none of your business

[rawranator]

Thanks guys!

[JCoder]

Blah. Classes? Arrays? Nested loops? Variables? Sooo complicated and error prone!

I love these old school programming assignments. If the students only knew how simple and short the answers could be, if they used the right tool for the job.

def repeat(n: Int)(code: => Unit) = 
  1 to n foreach {_ => code}

def diamond(n: Int) = 
  for (i <- -n to n) {
    val m = math.abs(i) 
    repeat(m)(print(" "))
    repeat((n - m) * 2 + 1)(print("*"))
    println 
  }

Results:

scala> diamond(5)
     *
    ***
   *****
  *******
 *********
***********
 *********
  *******
   *****
    ***
     *

Edited by JCoder, 15 June 2010 - 11:17 PM.

[Lance]

#include <stdio.h>

// draw a diamond on the console
//
// first thing we need to know is the half height on the diamond - hh
// to be able to place our diamond freely, there is an optional displacement - d
void diamond (unsigned hh, unsigned d=0, char c='*');

// A diamond consists of lines. Each line is n spaces follow by m stars
// where n and m is dependent on the index of the line
//
void line(unsigned n, unsigned m, char c)
{
    while(n--!=0)
        printf(" ");
    while(m--!=0)
        printf("%c",c);
    printf("\n");
}

void diamond(unsigned hh, unsigned d, char c)
{
    unsigned i;
    for(i=0; i<hh; ++i)
        line(d+hh-i,2*i+1, c);
    while(--i!=0)
        line(d+hh-i+1,2*i-1, c);
}


int main(int argc, char* argv[])
{
    diamond(6);

    diamond(7,10);


    diamond(5, 3, '$');

    return 0;
}

      *
     ***
    *****
   *******
  *********
 ***********
  *********
   *******
    *****
     ***
      *
                 *
                ***
               *****
              *******
             *********
            ***********
           *************
            ***********
             *********
              *******
               *****
                ***
                 *
        $
       $$$
      $$$$$
     $$$$$$$
    $$$$$$$$$
     $$$$$$$
      $$$$$
       $$$
        $

[JCoder]

@Lance: the best C version so far. :) However, still over 50% longer than mine and uses more than one statement per line.
If you used abs, you could make your version shorter.