Project Euler – Problem # 15 – Solved with Go

Starting in the top left corner in a 20 by 20 grid, how many routes are there to the bottom right corner?

Starting in the top left corner of a 2 x 2 grid, there are 6 routes (without backtracking) to the bottom right corner.

How many routes are there through a 20 x 20 grid?

**********

After doing a bit of research on this problem, it seems as though a simple Combinations formula will provide the answer. We know that any path will have 40 moves (20 right + 20 down), so for C(n,r) – n will equal 40 and r (the number of right moves) will equal 20 – C(40,20). Then we just plug the numbers into the formula – the ! means ‘factorial’. In order to solve this problem, I asked a question about creating a big.Int factorial in go on Stackoverflow.

                 n!
C(n,r) =   ———-
             r!*(n-r)!

The number of how many good routes we have can be found by finding how
many combinations of 20 R’s we can have in our 40 moves, so we want to
calculate:

                    40!
C(40,20) =   ———-
                20!*(40-20)!

Helpful Links:

One Possible Solution: Go – uses recursion for the factorial function

package main

import "fmt"
import "math/big"

func main() {

	n := big.NewInt(40)	// The total number of moves for any one path (right + down)
	r := big.NewInt(20)	// The total number of right moves for any one path

	t := big.NewInt(0).Sub(n, r)				// Part of Denominator
	x := big.NewInt(0).Mul(factorial(r), factorial(t))	// The complete Denominator
	Answer := big.NewInt(0).Div(factorial(n), x)		// Divide the Numerator by the Denominator

	fmt.Println(Answer)

}

func factorial(n *big.Int) (result *big.Int) {
	// This function finds the factorial of a number. For example: 5! = 5 * 4 * 3 * 2 * 1 = 120

	b := big.NewInt(0)
	c := big.NewInt(1)

	if n.Cmp(b) == -1 {
		result = big.NewInt(1)
	}
	if n.Cmp(b) == 0 {
		result = big.NewInt(1)
	} else {
		result = new(big.Int)
		result.Set(n)
		result.Mul(result, factorial(n.Sub(n, c)))
	}
	return
}
Advertisements

Project Euler – Problem # 15 – Solved with Java

Starting in the top left corner in a 20 by 20 grid, how many routes are there to the bottom right corner?

Starting in the top left corner of a 2 x 2 grid, there are 6 routes (without backtracking) to the bottom right corner.

How many routes are there through a 20 x 20 grid?

**********

After doing a bit of research on this problem, it seems as though a simple Combinations formula will provide the answer. We know that any path will have 40 moves (20 right + 20 down), so for C(n,r) – n will equal 40 and r (the number of right moves) will equal 20 – C(40,20). Then we just plug the numbers into the formula – the ! means ‘factorial’. I posted 2 versions of the solution: 1 without recursion – uses a for loop, and one with recursion – the factorial function calls itself.

                 n!
C(n,r) =   ———-
             r!*(n-r)!

The number of how many good routes we have can be found by finding how
many combinations of 20 R’s we can have in our 40 moves, so we want to
calculate:

                    40!
C(40,20) =   ———-
                20!*(40-20)!

Helpful Links:

One Possible Solution: Java – without recursion – (uses a for loop)

import java.math.BigInteger;

public class Problem_15 {
	public static void main(String args[]){
		int n = 40;		// The total number of moves for any one path (right + down)
		int r = 20;		// The total number of right moves for any one path
		
		System.out.println(factorial(n).divide(factorial(r).multiply(factorial(n - r))));
	}
	public static BigInteger factorial( int n1 )
    {
		BigInteger n = BigInteger.ONE;
		for (int i = 1; i <= n1; i ++) {
			n = n.multiply(BigInteger.valueOf(i));			
		}
		return n;
    }

}

One Possible Solution: Java – uses recursion (the factorial method calls itself)

import java.math.BigInteger;


public class Problem_15a {
	public static void main(String args[]){
		BigInteger n = BigInteger.valueOf(40);		// The total number of moves for any one path (right + down)
		BigInteger r = BigInteger.valueOf(20);		// The total number of right moves for any one path
		
		System.out.println(factorial(n).divide(factorial(r).multiply(factorial(n.subtract(r)))));
	}
	
	public static BigInteger factorial( BigInteger n )
	{
			if ( n.compareTo(BigInteger.ZERO) <= 0 ) { // base case
				return BigInteger.ONE;
			}
			else  {  // general case
				return ( n.multiply(factorial ( n.subtract(BigInteger.ONE)) ) );
			}
	}
}

Project Euler – Problem # 15 – Solved with Python

Starting in the top left corner in a 20 by 20 grid, how many routes are there to the bottom right corner?

Starting in the top left corner of a 2 x 2 grid, there are 6 routes (without backtracking) to the bottom right corner.

How many routes are there through a 20 x 20 grid?

**********

After doing a bit of research on this problem, it seems as though a simple Combinations formula will provide the answer. We know that any path will have 40 moves (20 right + 20 down), so for C(n,r) – n will equal 40 and r (the number of right moves) will equal 20 – C(40,20). Then we just plug the numbers into the formula – the ! means ‘factorial’ which is imported from the math class in python.

                 n!
C(n,r) =   ———-
             r!*(n-r)!

The number of how many good routes we have can be found by finding how
many combinations of 20 R’s we can have in our 40 moves, so we want to
calculate:

                    40!
C(40,20) =   ———-
                20!*(40-20)!

Helpful Links:

One Possible Solution: Python

# Python version = 2.7.2
# Platform = win32
# Elapsed Time: 20.9999084473 millisecs

from math import factorial
import time

def main():
    """Main Program"""
    start_time = time.time()

    n = 40      # The total number of moves for any one path (right + down)
    r = 20      # The total number of right moves for any one path

    print factorial(n) / (factorial(r) * factorial(n - r))
    print "Elapsed Time:", (time.time() - start_time) * 1000, "millisecs"

if __name__ == '__main__':
    main()