Drawing mirrored arrows in C-CodePudding

I am having trouble with drawing this shape in C using asterisk:

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

I know how to make either the right or left arrow, but I don't know how to make them like this. This is the code that I have so far, which makes the left arrow.

#include <stdio.h>

int main() {
    int n,i,j;
    printf("N:");
    scanf("%d", &n);

    for(i=1; i<=n; i  ){
        for(j=i; j<n; j  ){
            printf(" ");
        }
        for(j=1; j<=i; j  ){
            printf("*");
        }
        printf("\n");
    }

    for(i=n; i>=1; i--){
        for(j=i; j<=n; j  ){
            printf(" ");
        }
        for(j=1; j<i; j  ){
            printf("*");
        }
        printf("\n");
    }

    return 0;
}

I would really appreciate the help.

CodePudding user response：

When trying to draw shapes in ASCII, you have to ask yourself what characters you are printing in what order. Draw it manually (with a keyboard, not on paper eh) if you have to.

In the case of your double arrows:

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

The sequence here starts with star, space x9, star, \n, star x2, space x7, star x2, \n, star x3, space x5, star x3, \n, ...

So here's one possible solution:

#include <stdio.h>

int main(void) {
  for (int i = 0; i < 5; i  ) {
    for (int j = 0; j < i   1; j  ) {
      putchar('*');
    }
    for (int j = 0; j < 9 - (i * 2); j  ) {
      putchar(' ');
    }
    for (int j = 0; j < i   1; j  ) {
      putchar('*');
    }
    putchar('\n');
  }
  for (int i = 4; i > 0; i--) {
    for (int j = 0; j < i; j  ) {
      putchar('*');
    }
    for (int j = 0; j < 11 - (i * 2); j  ) {
      putchar(' ');
    }
    for (int j = 0; j < i; j  ) {
      putchar('*');
    }
    putchar('\n');
  }
}

CodePudding user response：

You need to start by identifying the patterns in the figure.

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

We have a number of stars, a number of spaces, then a number of stars again.

1 *, 9 ␠, 1 *
2 *, 7 ␠, 2 *
3 *, 5 ␠, 3 *
4 *, 3 ␠, 4 *
5 *, 1 ␠, 5 *
4 *, 3 ␠, 4 *
3 *, 5 ␠, 3 *
2 *, 7 ␠, 2 *
1 *, 9 ␠, 1 *

We could also see it as a number of stars, an amount of padding, a space, an amount of padding, then a number of stars.

1 *, 4 ␠, 1 ␠, 4 ␠, 1 *
2 *, 3 ␠, 1 ␠, 3 ␠, 2 *
3 *, 2 ␠, 1 ␠, 2 ␠, 3 *
4 *, 1 ␠, 1 ␠, 1 ␠, 4 *
5 *, 0 ␠, 1 ␠, 0 ␠, 5 *
4 *, 1 ␠, 1 ␠, 1 ␠, 4 *
3 *, 2 ␠, 1 ␠, 2 ␠, 3 *
2 *, 3 ␠, 1 ␠, 3 ␠, 2 *
1 *, 4 ␠, 1 ␠, 4 ␠, 1 *

For the first 5 lines, we print

a number of stars equal to the line number
a number of spaces equal to 5 minus the line number
a single space
a number of spaces equal to 5 minus the line number,
a number of stars equal to the line number

If i is the line number, this simplifies to

i stars
( 5-i ) * 2 1 spaces
i stars

We can find a similar pattern for the last 4 lines. With this we can achieve the desired result using two loops.

Loop such that i goes from 1 to 5 inclusive,
1. Print i stars.
2. Print ( 5-i ) * 2 1 spaces.
3. Print i stars.
4. Print a line feed.
Loop such that i goes from 4 to 1 inclusive,
1. Print i stars.
2. Print ( 5-i ) * 2 1 spaces.
3. Print i stars.
4. Print a line feed.

We could even generalize this to a single loop.

Loop such that j goes from -4 to 4 inclusive,
1. Let i be 5 minus the absolute value of j.
2. Print i stars.
3. Print ( 5-i ) * 2 1 spaces.
4. Print i stars.
5. Print a line feed.

Finally, it's best if we avoid using 5 and derived value all over the place.

Loop such that j goes from -(n-1) to (n-1) inclusive,
1. Let i be n minus the absolute value of j.
2. Print i stars.
3. Print ( n-i ) * 2 1 spaces.
4. Print i stars.
5. Print a line feed.

#include <stdio.h>

void putcharx( char ch, size_t count ) {
   while ( count-- )
      putchar( ch );
}

int main( void ) {
   int n = 5;
   for ( int j = -(n-1); j <= (n-1);   j ) {
      int i = n - ( j >= 0 ? j : -j );
      putcharx( '*', i );
      putcharx( ' ', ( n-i ) * 2   1 );
      putcharx( '*', i );
      putchar( '\n' );
   }

   return 0;
}

Demo on Compiler Explorer

CodePudding user response：

Function drawing it for any (odd) number.

void drawArrow(unsigned x)
{
    if(x % 2)
        for(unsigned row = 0; row < x; row  )
        {
            for(unsigned col = 0; col <= x   1; col  )
            {
                if((row <= x / 2 || row == x / 2) && (col <= row || col > x - row)) putchar('*');
                else if((row > x / 2) && (col > row   1 || col < x - row)) putchar('*');
                else putchar(' ');

            }
            putchar('\n');
        }
}

int main(void)
{
    drawArrow(19);
}

https://godbolt.org/z/6WK55Kv1a

CodePudding user response：

When you enter 5

pattern will write this

1 * 4 blank 1blank 4 blank 1 * \n

2 * 3 blank 1blank 3 blank 2 * \n

3 * 2 blank 1blank 2 blank 3 * \n

4 * 1 blank 1blank 1 blank 4 * \n

5 * 0 blank 1blank 0 blank 5 *

You can use this code:

#include <stdio.h>

int n;
char ch;
printf("Enter the number of rows and any Character to print the pattern\n");
scanf("%d", &n);
printf("\n");

for (int i = 1; i <= n; i  )
{
    for (int j = 1; j <= i; j  )
    {
        printf("*");
    }

    int space = 2 * n - 2 * i;
    for (int j = 1; j <= space; j  )
    {
        printf(" ");
    }
    printf(" ");
    for (int k = 1; k <= i; k  )
    {
        printf("*");
    }
    printf("\n");
}

for (int i = n - 1; i >= 1; i--)
{
    for (int j = 1; j <= i; j  )
    {
        printf("*");
    }

    int space = 2 * n - 2 * i;
    for (int j = 1; j <= space; j  )
    {
        printf(" ");
    }
    printf(" ");
    for (int k = 1; k <= i; k  )
    {
        printf("*" );
    }
    printf("\n");
}
return 0;

}