Thread: A diamond problem

Hello everybody, I need some help here

How am I gonna display a diamond shape as shown below by only using an output statements that prints a single asterisk(*) and using nested for loop?

*
***
*****
*******
*****
***
*

you must be in Dr. Li class

look for patterns

it isn't that hard

becoming a good programmer does require you to be a good problem solver, and looking for patterns is certainly the key to solving this problem. However, problem solving doesn't teach you the syntax of the language. Personally, I feel there should be a problem solving course separate from syntax instructions, but I am only a lowly peon, so what does my opinion matter. Therefore, I will share my approach to solving this problem, although there are several different possible approach's to use.

Notice that there is symetry around the middle line of the diamond as well as around the middle column of the diamond. If you substitute the char 'a' for spaces it is easier to see this:

aa*aa a*a
..........*a ***
***** a*a
..........*a
aa*aa

Notice also the that number of rows equals the number of columns and that to make a diamond you need an odd number of rows and columns (unless you can figure out how to do half spaces, but that is beyond me). Notice that the number of lines where the number of * increases is one half the maximum number of * plus one (if you use integer math). Notice that in each line the number of a's decreases by one before the * startand by one after the * end, and that the number of * increases or decreases by 2 each line. Note that the number of a's before the * start in the top section of the diamond is the number of columns minus the * divided by 2. Note that if you say line 1 is row 0 then the number of * in the top part of the diamond is one plus the quantity row number times two. And you can figure out similar relationships for the lower part of the diamond.

Based on these observations, I would try coding for the top part of the diamond separately from the bottom. I would use a loop to print out the a's one at a time and a separate loop to print out the * one a at time. I would use the row/line information and maximum number of * per line to do the calculations necessary to determine how many a's and how many * to print on each line.

At this point I would try to put all this into a coherent set of instructions in english (pseudocode) before writing actual program code.

Given differences in opinion as to how much information/help to give, etc. ; it's probably best to let you work on it from here, posting code/pseudocode/error statements/etc. with specific questions, etc. as you go along.

Well, this isn't exactly what you're looking for, but it should help you to figure out what you need.

#include <iostream.h>
#include <math.h>

void diamond(int num)
{
for (int lines = 1; lines <= (num); lines++)
{
for (int spaces = 0; spaces < (abs(num - lines)); spaces++)
{
cout<<" ";
}

for (int stars = 0; stars < (lines + lines - 1); stars++)
{
if (stars % 2 == 0)
cout<<"+";
else
cout<<"*";
}
cout<<endl;
}

int temp = 0;

for (int bottom_lines = (num-1); bottom_lines >= 0; bottom_lines--)
{
for (int bottom_spaces = 0; bottom_spaces <= temp; bottom_spaces++)
{
cout<<" ";
}

for (int bottom_stars = (bottom_lines + bottom_lines - 1); bottom_stars > 0; bottom_stars--)
{
if (bottom_stars % 2 == 0)
cout<< "*";
else
cout<<"+";
}

cout<<endl;
temp++;
}


}

int main()
{
int number;

cout << "Please enter the number for the drawing" << endl << " of the diamond." << endl;
cin >> number;

if ((number < 3) || (number > 10))
{
cout << "Sorry, incorrect number. Try again." << endl;
return 0;
}

diamond(number);

return 0;
}