Print the shape of an X on screen - c++

Question

I want to print an X on screen like this:
* *
* *
* *
*
* *
* *
* *
I tried with this code:
int main(){
bool back = false;
for (int i = 0; i < 7; ++i) {
if (i == 4)
back = true;
if (!back){
for (int j = 0; j < i; ++j) {
cout << " ";
}
} else{
for (int j = 7-i-1; j > 0; --j) {
cout << " ";
}
}
cout << "*" << endl;
}
}
The result is missing the right half:
*
*
*
*
*
*
*
The problem is that I can't figure out how to print the spaces between the stars and the stars that follow them.

Answer 1

A more educational approach to solving this problem requires 2 loops.
The first for loop controls the height of the output, i.e. the number of lines printed. Each iteration prints a single line and ends it with a std::endl.
The second is a nested for loop, which controls the width and prints characters horizontally, i.e. it prints asterisk(s) and spaces for that line. Each iteration prints either a space or an asterisk.
This illustration might help to understand the values of the variables when x_size = 5:
(width)
0 1 2 3 4
(height) ---------------------
0 | * | | | | * | asterisk_pos = 0, end_pos = 4, inc = 1
---------------------
1 | | * | | * | | asterisk_pos = 1, end_pos = 3, inc = 1
---------------------
2 | | | * | | | asterisk_pos = 2, end_pos = 2, inc = 1
---------------------
3 | | * | | * | | asterisk_pos = 1, end_pos = 3, inc = -1
---------------------
4 | * | | | | * | asterisk_pos = 0, end_pos = 4, inc = -1
---------------------
Source code:
int main()
{
int x_size = 7; // size of the drawing
int asterisk_pos = 0; // initial position of the asterisk
int inc = 1; // amount of increment added to asterisk_pos after an entire line has been printed
// height is the line number
for (int height = 0; height < x_size; height++)
{
// width is the column position of the character that needs to be printed for a given line
for (int width = 0; width < x_size; width++)
{
int end_pos = (x_size - width) - 1; // the position of the 2nd asterisk on the line
if (asterisk_pos == width || asterisk_pos == end_pos)
cout << "*";
else
cout << " ";
}
// print a new line character
cout << std::endl;
/* when the middle of x_size is reached,
* it's time to decrease the position of the asterisk!
*/
asterisk_pos += inc;
if (asterisk_pos > (x_size/2)-1)
inc *= -1;
}
return 0;
}
Output with x_size = 7:
* *
* *
* *
*
* *
* *
* *
Output with x_size = 3:
* *
*
* *

Answer 2

Observe the sequence in each line. Look at the first part you have:
0 space, 1 *, 5 spaces, 1 *, 0 space
1 space, 1 *, 3 spaces, 1 *, 1 space
2 spaces, 1 *, 1 space, 1 *, 2 spaces
Then for line i: i spaces followed by 1 * followed by 5-2 i spaces, followed by 1 *, followed by i spaces
Then the following should work:
for (int line=0; line<3; line++) {
for (int n=0; n<line; n++) cout << ' ';
cout << '*';
for (int n=0; n<5-2*line; n++) cout << ' ';
cout << '*';
for (int n=0; n<line; n++) cout << ' ';
cout << endl;
}
The middle line 3 is obvious, and the following is the reverse of the first part.
Another way is to observe sequence of positions of *: (0,6) (1,5) (2,4) (3,3) (4,2) (5,1) (6,0), thus:
for (int line=0; line<7; line++) {
int pos1 = line;
int pos2 = 6-line;
for (int n=0; n<7; n++) {
if (n==pos1 || n==pos2) cout << '*';
else cout << ' ';
}
cout << endl;
}
You can then obviously remove pos1 and pos2...

Answer 3

spaces between in upper part are decreasing by 2 and start with line - 2
spaces between in down part are incensing by 2
here how I solve your problem
void printSpaces(int count)
{
for (int i = 0; i < count; ++i) {
cout << " ";
}
}
int main()
{
int lines = 7;
int spaceBefore = 0;
int spaceBetween = lines - 2;
bool backword = false;
for (int i = 0; i < lines; ++i)
{
printSpaces(spaceBefore);
cout << "*";
if (spaceBetween > 0)
{
printSpaces(spaceBetween);
cout << "*";
}
else
{
backword = true;
}
cout << "\n";
spaceBefore = backword ? spaceBefore-1 : spaceBefore+1;
spaceBetween = backword ? spaceBetween+2 : spaceBetween-2;
}
return 0;
}

Answer 4

The pattern consists of two equations: x = y and x + y = 4
Just loop through the axes and plot the points that fall on any of the lines.
( y )
0 1 2 3 4
( x ) ---------------------
0 | * | | | | * |
---------------------
1 | | * | | * | |
---------------------
2 | | | * | | |
---------------------
3 | | * | | * | |
---------------------
4 | * | | | | * |
---------------------
Two Equations
x = y
x + y = 4
#include <iostream>
int main() {
int num_lines = 7;
auto on_line1 = [](int x, int y) {
return x == y;
};
auto on_line2 = [num_lines](int x, int y) {
return (x + y) == (num_lines - 1);
};
for(int x = 0; x < num_lines; x++) { // Simple looping
for(int y = 0; y < num_lines; y++) { // through the axes
if(on_line1(x, y) or on_line2(x, y)) { // If on any of the line
std::cout << '*'; // Then plot it
} else {
std::cout << ' '; // Else leave it
}
}
std::cout << '\n';
}
return 0;
}
PS: I copied the ascii table from the other answer.

Answer 5

If you're not required to loop you can create a string and print it.
#include <iostream>
#include <string>
int main(int argc, char * argv[]){
std::string myX("* *\n * * \n * * \n * \n * * \n * * \n* *\n");
std::cout << myX;
return 0;
}