C Program to print Pascal Triangle

Question:

Algorithm:

Explanation:

Pascal's Triangle can be printed without using an array because every row can be generated directly from the previous value in the same row. The first value of every row is always 1, so the variable c is set to 1 before printing a new row. The outer loop controls the row number. When i is 0, the program prints only one value. When i is 1, it prints two values, and so on. Therefore the inner loop runs from 0 to i, giving exactly the required number of terms in each row.

The important part of the program is the formula c = c * (i - j) / (j + 1). This formula gives the next binomial coefficient in the current row. For example, in row 4, the values are 1, 4, 6, 4 and 1. Starting with 1, the formula calculates the next value using only the current value, the row number and the column number. Since the multiplication is done before division, the result remains correct for the integer values in Pascal's Triangle.

The variable space is used only for formatting. It prints leading blanks before each row so the output appears in a triangular shape. After each row, space is decreased, making the next row begin slightly earlier. The logic of the triangle does not depend on these spaces; they simply make the pattern easier to read on the screen.

C Program:

/*
 * Program to print Pascal Triangle
 * @author : www.guideforschool.com
 * @Program Type : C Program
 */

#include <stdio.h>

int main(void)
{
    int i, j, n, k, space;
    long c;

    printf("\nEnter no. of lines : ");
    scanf("%d", &n);

    printf("\n");
    space = n;

    for(i = 0; i < n; i++)
    {
        c = 1;

        // Print leading spaces to shape the triangle.
        for(k = space; k >= 0; k--)
        {
            printf("   ");
        }

        space--;

        // Print values of the current row.
        for(j = 0; j <= i; j++)
        {
            printf("%6ld", c);
            c = c * (i - j) / (j + 1);
        }

        printf("\n");
    }

    return 0;
}

Output:

Enter no. of lines : 7

                             1
                          1     1
                       1     2     1
                    1     3     3     1
                 1     4     6     4     1
              1     5    10    10     5     1
           1     6    15    20    15     6     1