v_pattern_with_explanation_v1.c

/*

    v_pattern_with_explanation_v1.c

    Task:
    https://web.facebook.com/photo.php?fbid=2607667832670872&set=p.2607667832670872&type=3&theater
    https://web.facebook.com/groups/c.programing/permalink/1577186559106855/?comment_id=1577227462436098

-*-------*-
--*-----*--
---*---*---
----*-*----
-----*-----

I will try to explain one way of solving this task.

Observing the V pattern, we notice the following:
The pattern consists of n rows.
In each line, before the first * there are n1b space characters.
In each line, there are n2b space characters between the first * and the second *.
Each line ends with *, except last one.
After the second * there are n3b space characters (this is not necessary).

We will write a function that has the following arguments:
n        - number of rows in the pattern,
ch       - the character by which the pattern is drawn,
space_ch - a character that draws a space.

For all n pattern rows,
first we draw n1b space_ch characters (-) before first ch (*),
then we draw the first ch (*),
then we draw n2b space_ch characters (-),
then we draw second ch (*) (except in the last row),
and finally we draw n2b blank characters (-) (this is not necessary).

For that we need three for loops.

The first for loop is for rows and that is why we call its iterator r.
From the picture pattern we see that there are as many rows r as the argument n.
So in the first for loop r goes from 0 to n for all n rows.

How to calculate n1b ?
n1b must depend on r.
From the picture of pattern it is obvious that there are r space_ch characters
in every row r.
n1b = r

Next is the most tricky part of the code !!!
How to calculate n2b ?
n2b must depend on n and r.
From the picture of pattern we see that:

n   r   n2b

5   0   7
5   1   5
5   2   3
5   3   1
5   4   0

So from n and r we have to come up with a formula to get n2b.
This is where your creativity comes to the fore.
By observing, thinking and using the method of trial and error,
we finally come to a formula:
n2b = 2*(n-r)-3

With this formula, in last row, case n=5 and r=4, result is -1 (not 0)
but even that satisfies the condition in for loop
for(c=0; c<2*(n-r)-3; c++) .


    You can find all my C programs at Dragan Milicev's pastebin:

    https://pastebin.com/u/dmilicev

*/

#include <stdio.h>

// Draw V pattern, version 1 with n1b and n2b which are not necessary
void draw_v_pattern_v1( int n, char ch, char space_ch )
{
    int r,c;                        // represent row and column for loops iterators
    int n1b;                        // n1b is number of space_ch characters (-) before first ch (*)
    int n2b;                        // n2b is number of space_ch characters (-) between two (*) in row

    for(r=0; r<n; r++)              // for all n pattern rows
    {
        n1b=r;
        for(c=0; c<n1b; c++)        // first we draw n1b space characters (-) before first ch (*)
            putchar(space_ch);      // character for space

        putchar(ch);                // then we draw the first ch (*)

        n2b = 2*(n-r)-3;            // this is the most tricky part of the code !!!
        for(c=0; c<n2b; c++)        // then we draw n2b space characters (-) between two (*) in row
            putchar(space_ch);      // character for space

        if( r < n-1 )               // then in all rows except the last row
            putchar(ch);            // we draw the second ch (*)

        putchar('\n');              // then we go to new pattern row
    }
}

// Draw V pattern, version 2, without n1b and n2b
void draw_v_pattern_v2( int n, char ch, char space_ch )
{
    int r,c;                        // for loops iterators

    for(r=0; r<n; r++)              // for all n pattern rows
    {
        for(c=0; c<r; c++)          // first we draw n1b space characters (-) before first ch (*)
            putchar(space_ch);      // character for space

        putchar(ch);                // then we draw the first ch (*)
                                    // next line is the most tricky part of the code !!!
        for(c=0; c<2*(n-r)-3; c++)  // then we draw n2b space characters (-) between two (*) in row
            putchar(space_ch);      // character for space

        if( r < n-1 )               // then in all rows except the last row
            putchar(ch);            // we draw the second ch (*)

        putchar('\n');              // then we go to new pattern row
    }
}

// Draw V pattern, version 3, code from Patrick Benedict Lirio
// https://web.facebook.com/pbenedict.lirio
// https://web.facebook.com/photo.php?fbid=1602215383269843&set=p.1602215383269843&type=3&theater
void draw_v_pattern_v3( int n, char ch, char space_ch )
{
    int r,c;                        // for loops iterators
    int x=1;                        // x coordinate for print
    int y=2*n-1;                    // y coordinate for print

    for(r=0; r<n; r++)              // for n rows of pattern
    {
        for(c=0; c<2*n+1; c++)      // for all characters in one row r
            if( c==x || c==y )      // decide witch character to put
                putchar(ch);        // character for drawing
            else
                putchar(space_ch);  // character for space
        x++;                        // increase x coordinate
        y--;                        // decrease y coordinate
        putchar('\n');              // new pattern row
    }
}

// Draw V pattern, version 4, code from Keshav Sahu
// https://web.facebook.com/profile.php?id=100042709719960
// https://pastebin.com/f5r47eQ6
void draw_v_pattern_v4( int n, char ch, char space_ch )
{
    int r,c;                        // for loops iterators

    for(r=0; r<n; r++){             // for n rows of pattern
        for(c=0; c<2*n-1; c++){     // for all characters in one row r
            if(c == r || c == 2*n-r-2)  // decide witch character to put
                putchar(ch);        // character for drawing
            else
                putchar(space_ch);  // character for space
        }
        putchar('\n');              // new pattern row
    }
}

// Draw V pattern, version 5, code from Keshav Sahu
// https://web.facebook.com/profile.php?id=100042709719960
// https://pastebin.com/f5r47eQ6
void draw_v_pattern_v5( int n, char ch, char space_ch )
{
    int r,c;                        // for loops iterators

    for(r=0; r<n; r++){             // for n rows of pattern
        for(c=0; c<2*n; c++){       // for all characters in one row r
            if(c == r || c == 2*n-r-1)  // decide witch character to put
                putchar(ch);        // character for drawing
            else
                putchar(space_ch);  // character for space
        }
        putchar('\n');              // new pattern row
    }
}


int main(void)
{
    int n=5;                        // the number of rows that make up the pattern
    char ch='*';                    // the character with which we draw the pattern
    char space_ch='-';              // the character with which we draw a space

    printf("\n draw_v_pattern_v1() \n\n");
    draw_v_pattern_v1(n, ch, space_ch);
    printf("\n");

    printf("\n draw_v_pattern_v2() \n\n");
    draw_v_pattern_v2(n, ch, space_ch);
    printf("\n");

    printf("\n draw_v_pattern_v3() \n\n");
    draw_v_pattern_v3(n, ch, space_ch);
    printf("\n");

    printf("\n draw_v_pattern_v4() \n\n");
    draw_v_pattern_v4(n, ch, space_ch);
    printf("\n");

    printf("\n draw_v_pattern_v5() \n\n");
    draw_v_pattern_v5(n, ch, space_ch);
    printf("\n");

    return 0;

} // main()