This article intends to show the different ways to print a Pascal’s triangle and also format it so that it looks like an isosceles triangle.
Understanding how a Pascal’s triangle is built would be easier by considering the following figure which shows the first six rows of the Pascal’s triangle.
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
Each row begins and ends with the number one. The remaining numbers are obtained by summing the two numbers that lie directly above that number on the previous row and the number that follows it. So, in order to find the numbers in the nth row of the triangle, we need the values of the (n-1) the row. Let’s say that we have computed the fourth row – 1 3 3 1. Now, the fifth row has five elements. The first and the last element would be one. The remaining elements would be (1+3), (3+3), (3+1) = 4, 6, 4. So, the complete fifth row would be 1 4 6 4 1.
Pascal’s triangle using loops
We will first see how to generate a (unformatted ) Pascal’s triangle using simple looping statements. We maintain two int arrays, named currentRow and previousRow. Initially, previousRow would be initialised with { 1 } – the contents of the first row of the Pascal’s triangle. After that, we have a loop whose loop counter, i runs from 2 to n where n is the number of rows that we wish to display. At any iteration, I represents the row number that we are printing. The currentRow will be initialised with an array of i (the loop counter) elements with the first and the last element of the array set to ‘1’. Next, we have an inner loop whose task is to compute the elements of currentRow. To do so, the loop counter, j runs from 0 to i-3. The numbers in the array previousRow at indices j and j+1 are added and the result is stored at the index j+1 of currentRow. When the inner loop exits, currentRow would be populated. The values stored in currentRow are printed and previousRow is assigned with currentRow so that it can be used in the next iteration of the outer loop.
Given below is a complete program which takes an input n from the user and prints the first n lines of the Pascal’s triangle.
import java.util.Scanner;
Public class PascalTriangle {
Public static void print(int n) {
int[] previousRow;
int[] currentRow = {1};
printArray(currentRow);
previousRow = currentRow;
for (int i = 2; i <= n; i++) {
currentRow = new int[i];
currentRow[0] = 1;
currentRow[i – 1] = 1;
for (int j = 0; j <= i – 3; j++) {
currentRow[j + 1] = previousRow[j] + previousRow[j + 1];
}
printArray(currentRow);
previousRow = currentRow;
}
}
Public static void printArray(int[] array) {
for (int i = 0; i < array.length; i++) {
System.out.print(array[i] + ” “);
}
System.out.println();
}
Public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
System.out.print(“Enter the row number upto which Pascal’s triangle has to be printed: “);
int row = scanner.nextInt();
print(row);
}
}
Here is a sample execution of the above program.
Enter the row number up to which Pascal’s triangle has to be printed: 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
Pascal’s triangle as binomial coefficients
The Pascal’s triangle can also be visualised as the binomial coefficients in the expansion of (x+y)n where n is the row of the Pascal’s triangle, with the rows labelled starting from n=0. Using this observation, one can calculate the values in the Pascal’s triangle by the direct application of the nCk formulae. Now, the rows would be labelled from n=0 and the elements within a row would be labelled from k=0. Using this numbering scheme, the element in row n and at position k can be calculated as
nCk = n!/(k! * (n-k)! )
Given below is the program which uses this method to print the Pascal’s triangle.
import java.util.Scanner;
Public class PascalTriangle {
Public static void print(int row) {
for (int n = 0; n < row; n++) {
for (int k = 0; k <= n; k++) {
System.out.print(nCk(n, k) + ” “);
}
System.out.println();
}
}
Public static int nCk(int n, int k) {
int numerator = fact(n);
int denominator = fact(k) * fact(n – k);
int result = (int) (numerator / denominator);
return result;
}
Public static int fact(int num) {
int result = 1;
for (int i = 1; i <= num; i++) {
result = result * i;
}
return result;
}
Public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
System.out.print(“Enter the row number upto which Pascal’s triangle has to be printed: “);
int row = scanner.nextInt();
print(row);
}
}
Generating Pascal’s triangle using recursion
Since calculating the value of a particular position in the triangle uses previously calculated values, this problem can also be solved using recursion. The number at position row i and column j would be represented as pascal(i,j) with I and j starting from 0. There would be two base conditionswhich are related to the first and last elements, which are always one. These two conditions can be expressed as
pascal ( i, 0 ) = 1
pascal ( i, i ) = 1
Following is the recursive formula used to calculate the remaining elements :
pascal ( i, j ) = pascal ( i – 1 , j -1 ) + pascal ( i – 1 , j )
Given below is the program which uses the recursive definition of the Pascal’s triangle
import java.util.Scanner;
Public class PascalTriangle {
Public static void print(int n) {
for (int i = 0; i < n; i++) {
for (int j = 0; j <= i; j++) {
System.out.print(pascal(i, j) + ” “);
}
System.out.println();
}
}
Public static int pascal(int i, int j) {
if (j == 0) {
return 1;
} else if (j == i) {
return 1;
} else {
return pascal(i – 1, j – 1) + pascal(i – 1, j);
}
}
Public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
System.out.print(“Enter the row number upto which Pascal’s triangle has to be printed: “);
int row = scanner.nextInt();
print(row);
}
}