Project Euler – Problem # 11 – Solved with Go

What is the greatest product of four adjacent numbers on the same straight line in the 20 by 20 grid?

In the 2020 grid below, four numbers along a diagonal line have been marked in red.

08 02 22 97 38 15 00 40 00 75 04 05 07 78 52 12 50 77 91 08
49 49 99 40 17 81 18 57 60 87 17 40 98 43 69 48 04 56 62 00
81 49 31 73 55 79 14 29 93 71 40 67 53 88 30 03 49 13 36 65
52 70 95 23 04 60 11 42 69 24 68 56 01 32 56 71 37 02 36 91
22 31 16 71 51 67 63 89 41 92 36 54 22 40 40 28 66 33 13 80
24 47 32 60 99 03 45 02 44 75 33 53 78 36 84 20 35 17 12 50
32 98 81 28 64 23 67 10 26 38 40 67 59 54 70 66 18 38 64 70
67 26 20 68 02 62 12 20 95 63 94 39 63 08 40 91 66 49 94 21
24 55 58 05 66 73 99 26 97 17 78 78 96 83 14 88 34 89 63 72
21 36 23 09 75 00 76 44 20 45 35 14 00 61 33 97 34 31 33 95
78 17 53 28 22 75 31 67 15 94 03 80 04 62 16 14 09 53 56 92
16 39 05 42 96 35 31 47 55 58 88 24 00 17 54 24 36 29 85 57
86 56 00 48 35 71 89 07 05 44 44 37 44 60 21 58 51 54 17 58
19 80 81 68 05 94 47 69 28 73 92 13 86 52 17 77 04 89 55 40
04 52 08 83 97 35 99 16 07 97 57 32 16 26 26 79 33 27 98 66
88 36 68 87 57 62 20 72 03 46 33 67 46 55 12 32 63 93 53 69
04 42 16 73 38 25 39 11 24 94 72 18 08 46 29 32 40 62 76 36
20 69 36 41 72 30 23 88 34 62 99 69 82 67 59 85 74 04 36 16
20 73 35 29 78 31 90 01 74 31 49 71 48 86 81 16 23 57 05 54
01 70 54 71 83 51 54 69 16 92 33 48 61 43 52 01 89 19 67 48

The product of these numbers is 26 * 63 * 78 * 14 = 1788696.

What is the greatest product of four adjacent numbers in any direction (up, down, left, right, or diagonally) in the 20×20 grid?

One Possible Solution: Go

package main

import "fmt"

func main() {

	Grid := [20][20]int{
		{8, 02, 22, 97, 38, 15, 00, 40, 00, 75, 04, 05, 07, 78, 52, 12, 50, 77, 91, 8},
		{49, 49, 99, 40, 17, 81, 18, 57, 60, 87, 17, 40, 98, 43, 69, 48, 04, 56, 62, 00},
		{81, 49, 31, 73, 55, 79, 14, 29, 93, 71, 40, 67, 53, 88, 30, 03, 49, 13, 36, 65},
		{52, 70, 95, 23, 04, 60, 11, 42, 69, 24, 68, 56, 01, 32, 56, 71, 37, 02, 36, 91},
		{22, 31, 16, 71, 51, 67, 63, 89, 41, 92, 36, 54, 22, 40, 40, 28, 66, 33, 13, 80},
		{24, 47, 32, 60, 99, 03, 45, 02, 44, 75, 33, 53, 78, 36, 84, 20, 35, 17, 12, 50},
		{32, 98, 81, 28, 64, 23, 67, 10, 26, 38, 40, 67, 59, 54, 70, 66, 18, 38, 64, 70},
		{67, 26, 20, 68, 02, 62, 12, 20, 95, 63, 94, 39, 63, 8, 40, 91, 66, 49, 94, 21},
		{24, 55, 58, 05, 66, 73, 99, 26, 97, 17, 78, 78, 96, 83, 14, 88, 34, 89, 63, 72},
		{21, 36, 23, 9, 75, 00, 76, 44, 20, 45, 35, 14, 00, 61, 33, 97, 34, 31, 33, 95},
		{78, 17, 53, 28, 22, 75, 31, 67, 15, 94, 03, 80, 04, 62, 16, 14, 9, 53, 56, 92},
		{16, 39, 05, 42, 96, 35, 31, 47, 55, 58, 88, 24, 00, 17, 54, 24, 36, 29, 85, 57},
		{86, 56, 00, 48, 35, 71, 89, 07, 05, 44, 44, 37, 44, 60, 21, 58, 51, 54, 17, 58},
		{19, 80, 81, 68, 05, 94, 47, 69, 28, 73, 92, 13, 86, 52, 17, 77, 04, 89, 55, 40},
		{04, 52, 8, 83, 97, 35, 99, 16, 07, 97, 57, 32, 16, 26, 26, 79, 33, 27, 98, 66},
		{88, 36, 68, 87, 57, 62, 20, 72, 03, 46, 33, 67, 46, 55, 12, 32, 63, 93, 53, 69},
		{04, 42, 16, 73, 38, 25, 39, 11, 24, 94, 72, 18, 8, 46, 29, 32, 40, 62, 76, 36},
		{20, 69, 36, 41, 72, 30, 23, 88, 34, 62, 99, 69, 82, 67, 59, 85, 74, 04, 36, 16},
		{20, 73, 35, 29, 78, 31, 90, 01, 74, 31, 49, 71, 48, 86, 81, 16, 23, 57, 05, 54},
		{01, 70, 54, 71, 83, 51, 54, 69, 16, 92, 33, 48, 61, 43, 52, 01, 89, 19, 67, 48},
	}

	Product := 0
	largest := 0

	// Check horizontally
	for i := 0; i < 20; i++ {
		for j := 0; j < 17; j++ {
			Product = Grid[i][j] * Grid[i][j+1] * Grid[i][j+2] * Grid[i][j+3]
			if Product > largest {
				largest = Product
			}
		}
	}

	// Check vertically
	for i := 0; i < 17; i++ {
		for j := 0; j < 20; j++ {
			Product = Grid[i][j] * Grid[i+1][j] * Grid[i+2][j] * Grid[i+3][j]
			if Product > largest {
				largest = Product
			}
		}
	}

	// Check diagonally right
	for i := 0; i < 17; i++ {
		for j := 0; j < 17; j++ {
			Product = Grid[i][j] * Grid[i+1][j+1] * Grid[i+2][j+2] * Grid[i+3][j+3]
			if Product > largest {
				largest = Product
			}
		}
	}

	// Check diagonally left
	for i := 0; i < 17; i++ {
		for j := 3; j < 20; j++ {
			Product = Grid[i][j] * Grid[i+1][j-1] * Grid[i+2][j-2] * Grid[i+3][j-3]
			if Product > largest {
				largest = Product
			}
		}
	}

	fmt.Println(largest)
}

Project Euler – Problem # 11 – Solved with Java

What is the greatest product of four adjacent numbers on the same straight line in the 20 by 20 grid?

In the 2020 grid below, four numbers along a diagonal line have been marked in red.

08 02 22 97 38 15 00 40 00 75 04 05 07 78 52 12 50 77 91 08
49 49 99 40 17 81 18 57 60 87 17 40 98 43 69 48 04 56 62 00
81 49 31 73 55 79 14 29 93 71 40 67 53 88 30 03 49 13 36 65
52 70 95 23 04 60 11 42 69 24 68 56 01 32 56 71 37 02 36 91
22 31 16 71 51 67 63 89 41 92 36 54 22 40 40 28 66 33 13 80
24 47 32 60 99 03 45 02 44 75 33 53 78 36 84 20 35 17 12 50
32 98 81 28 64 23 67 10 26 38 40 67 59 54 70 66 18 38 64 70
67 26 20 68 02 62 12 20 95 63 94 39 63 08 40 91 66 49 94 21
24 55 58 05 66 73 99 26 97 17 78 78 96 83 14 88 34 89 63 72
21 36 23 09 75 00 76 44 20 45 35 14 00 61 33 97 34 31 33 95
78 17 53 28 22 75 31 67 15 94 03 80 04 62 16 14 09 53 56 92
16 39 05 42 96 35 31 47 55 58 88 24 00 17 54 24 36 29 85 57
86 56 00 48 35 71 89 07 05 44 44 37 44 60 21 58 51 54 17 58
19 80 81 68 05 94 47 69 28 73 92 13 86 52 17 77 04 89 55 40
04 52 08 83 97 35 99 16 07 97 57 32 16 26 26 79 33 27 98 66
88 36 68 87 57 62 20 72 03 46 33 67 46 55 12 32 63 93 53 69
04 42 16 73 38 25 39 11 24 94 72 18 08 46 29 32 40 62 76 36
20 69 36 41 72 30 23 88 34 62 99 69 82 67 59 85 74 04 36 16
20 73 35 29 78 31 90 01 74 31 49 71 48 86 81 16 23 57 05 54
01 70 54 71 83 51 54 69 16 92 33 48 61 43 52 01 89 19 67 48

The product of these numbers is 26 * 63 * 78 * 14 = 1788696.

What is the greatest product of four adjacent numbers in any direction (up, down, left, right, or diagonally) in the 20×20 grid?

One Possible Solution: Java

public class Problem_11 {
	public static void main(String [] args){
		int Grid [][] = {
		{8,02,22,97,38,15,00,40,00,75,04,05,07,78,52,12,50,77,91,8},
		{49,49,99,40,17,81,18,57,60,87,17,40,98,43,69,48,04,56,62,00},
		{81,49,31,73,55,79,14,29,93,71,40,67,53,88,30,03,49,13,36,65},
		{52,70,95,23,04,60,11,42,69,24,68,56,01,32,56,71,37,02,36,91},
		{22,31,16,71,51,67,63,89,41,92,36,54,22,40,40,28,66,33,13,80},
		{24,47,32,60,99,03,45,02,44,75,33,53,78,36,84,20,35,17,12,50},
		{32,98,81,28,64,23,67,10,26,38,40,67,59,54,70,66,18,38,64,70},
		{67,26,20,68,02,62,12,20,95,63,94,39,63,8,40,91,66,49,94,21},
		{24,55,58,05,66,73,99,26,97,17,78,78,96,83,14,88,34,89,63,72},
		{21,36,23,9,75,00,76,44,20,45,35,14,00,61,33,97,34,31,33,95},
		{78,17,53,28,22,75,31,67,15,94,03,80,04,62,16,14,9,53,56,92},
		{16,39,05,42,96,35,31,47,55,58,88,24,00,17,54,24,36,29,85,57},
		{86,56,00,48,35,71,89,07,05,44,44,37,44,60,21,58,51,54,17,58},
		{19,80,81,68,05,94,47,69,28,73,92,13,86,52,17,77,04,89,55,40},
		{04,52,8,83,97,35,99,16,07,97,57,32,16,26,26,79,33,27,98,66},
		{88,36,68,87,57,62,20,72,03,46,33,67,46,55,12,32,63,93,53,69},
		{04,42,16,73,38,25,39,11,24,94,72,18,8,46,29,32,40,62,76,36},
		{20,69,36,41,72,30,23,88,34,62,99,69,82,67,59,85,74,04,36,16},
		{20,73,35,29,78,31,90,01,74,31,49,71,48,86,81,16,23,57,05,54},
		{01,70,54,71,83,51,54,69,16,92,33,48,61,43,52,01,89,19,67,48}
		};
		
		int Product = 0;
		int largest = 0;
		
		// Check horizontally
		for(int i = 0; i < 20; i++){
			for(int j = 0; j < 17; j++){
				Product = Grid[i][j] * Grid[i][j + 1] * Grid[i][j + 2] * Grid[i][j + 3];
				if(Product > largest){
					largest = Product;
				}
			}	
		}
		
		// Check vertically
		for(int i = 0; i < 17; i ++){
			for(int j = 0; j < 20; j++){
				Product = Grid[i][j] * Grid[i + 1][j] * Grid[i + 2][j] * Grid[i + 3][j];
				if(Product > largest){
					largest = Product;
				}
			}
		}
		
		// Check diagonally right
		for(int i = 0; i < 17; i++){
			for(int j = 0; j < 17; j++){
				Product = Grid[i][j] * Grid[i + 1][j + 1] * Grid[i + 2][j + 2] * Grid[i + 3][i + 3];
				if(Product > largest){
					largest = Product;
				}
			}
		}
		
		// Check diagonally left
		for(int i = 0; i < 17; i ++){
			for(int j = 3; j < 20; j ++){
				Product = Grid[i][j] * Grid[i + 1][j - 1] * Grid[i + 2][j  - 2] * Grid[i + 3][j - 3];
				if(Product > largest){
					largest = Product;
				}
			}
		}
		System.out.format("Max Product = %d", + largest);
	}
}

Project Euler – Problem # 10 – Solved with Go

Calculate the sum of all the primes below two million.

The sum of the primes below 10 is 2 + 3 + 5 + 7 = 17.

Find the sum of all the primes below two million.

**********

When running this program on http://play.golang.org/, I get the error message [process took too long]; however, when running it on my local machine I get the correct answer.

One Possible Solution: Go

package main

import "fmt"
import "math"

var Sum int64 = 0

func main() {
	
	counter := 1
	for counter < 2000000 {
		if isPrime(counter) {
			Sum += int64(counter)
		}
		counter += 2
	}
	fmt.Println(Sum + 2)

}

func Sqrt(x float64) float64 {
	x = math.Sqrt(x)
	return x
}

func isPrime(n int) (isP bool) {
	if n == 1 {
		return false
	}
	if n == 2 {
		return true
	}
	k := int(Sqrt(float64(n)))
	for i := 2; i <= k; i++ {
		if n%i == 0 {
			return false
		}

	}
	return true

}

Project Euler – Problem # 10 – Solved with Java & Python

Calculate the sum of all the primes below two million.

The sum of the primes below 10 is 2 + 3 + 5 + 7 = 17.

Find the sum of all the primes below two million.

One Possible Solution: Java

public class Problem_10 {
	public static void main(String [] args){
		// create Problem_10 object
		Problem_10 prob_10Obj = new Problem_10();
		
		// answer will be a long
		long Sum = 0;
		
		// start the counter at 1 and increase by 2 (all odd numbers)
		int counter = 1;
		
		// while counter is less than two million
		while (counter < 2000000) {
			
			// checks to see if counter is a prime number - if true - add to Sum
			if (prob_10Obj.isPrime(counter)) {
				Sum += counter;
			}
			// increase counter by 2 -> 1, 3, 5, 7, 9, 11, etc.
			counter += 2;
		}
		// add 2 to the Sum - our counter does not count the first prime number -> 2.
		System.out.println(Sum + 2); 
	}

	public boolean isPrime(int n)
	    {
		if (n == 1){
			return false;
		}
	    int k = (int) Math.sqrt(n);
	        for (int i = 2; i <= k; i++)
	        {
	            if (n % i == 0)
	                return false;
	        }
	        return true;
	    }

}

One Possible Solution: Python

# Python version = 2.7.2
# Platform = win32

def primes(n):
    """Prime number generator up to n - (generates a list)"""
    ## {{{ http://code.activestate.com/recipes/366178/ (r5)
    if n == 2: return [2]
    elif n < 2: return []
    s = range(3, n + 1, 2)
    mroot = n ** 0.5
    half = (n + 1) / 2 - 1
    i = 0
    m = 3
    while m <= mroot:
        if s[i]:
            j = (m * m - 3) / 2
            s[j] = 0
            while j < half:
                s[j] = 0
                j += m
        i = i + 1
        m = 2 * i + 3
    return [2]+[x for x in s if x]

   ## end of http://code.activestate.com/recipes/366178/ }}}

def main():
    """Main Program"""
    print "Sum of all the primes below two million = %d" % sum(primes(2000000))

if __name__ == '__main__':
    main()

Project Euler – Problem # 9 – Solved with Java & Python

Find the only Pythagorean triplet, {a, b, c}, for which a + b + c = 1000.

A Pythagorean triplet is a set of three natural numbers, a b c, for which,

a² + b² = c²

For example, 3² + 4² = 9 + 16 = 25 = 5².

There exists exactly one Pythagorean triplet for which a + b + c = 1000.
Find the product abc.

**********

The wikipedia page on Pythagorean triple is helpful in solving this problem. Particularly the part about
Euclid’s formula: a = m² – n², b = 2mn, c = m² + n².

Helpful link:

Pythagorean triple – http://en.wikipedia.org/wiki/Pythagorean_triple

One Possible Solution: Java

public class Problem_09 {
	public static void main(String [] args){
		for (int n = 1; n < 500; n++) {
			for (int m = (n + 1); m < 500; m ++) {
				int a = (m * m) - (n * n);
				int b = 2 * m * n;
				int c = (m * m) + (n * n);
				if (a + b + c == 1000) {
					int product = a * b * c;
					System.out.format("a = %d" + "b = %d" + "c = %d" + newline, a, b, c);
					System.out.format("Product = %d" + newline, product);
				}
				
			}	
		}	
	}
	public static String newline = System.getProperty("line.separator");
}

One Possible Solution: Python

import math
    
def main():
    """Main program"""
    for n in xrange(1, 500):
        for m in xrange(n + 1, 500):
            a = int(math.pow(m, 2) - math.pow(n, 2))
            b = 2 * m * n
            c = int(math.pow(m, 2) + math.pow(n, 2))
            if a + b + c == 1000:
                product = a * b * c
                print "a = %d, b = %d, c = %d" % (a, b, c)
    print "Product = %d" % (product)

if __name__ == '__main__':
    main()

Project Euler – Problem # 9 – Solved with Go

Find the only Pythagorean triplet, {a, b, c}, for which a + b + c = 1000.

A Pythagorean triplet is a set of three natural numbers, a b c, for which,

a² + b² = c²

For example, 3² + 4² = 9 + 16 = 25 = 5².

There exists exactly one Pythagorean triplet for which a + b + c = 1000.
Find the product abc.

**********

The wikipedia page on Pythagorean triple is helpful in solving this problem. Particularly the part about
Euclid’s formula: a = m² – n², b = 2mn, c = m² + n².

Helpful link:

Pythagorean triple – http://en.wikipedia.org/wiki/Pythagorean_triple

One Possible Solution: Go

package main

import "fmt"
import "math"

func main() {
	for n := 1; n < 500; n++ {
		for m := (n + 1); m < 500; m++ {
			a := int(math.Pow(float64(m), 2) - math.Pow(float64(n), 2))
			b := 2 * m * n
			c := int(math.Pow(float64(m), 2) + math.Pow(float64(n), 2))
			if a+b+c == 1000 {
				product := a * b * c
				fmt.Println(a, b, c)
				fmt.Println(product)
			}
		}

	}
}

Project Euler – Problem # 8 – Solved with Go

Find the greatest product of five consecutive digits in the 1000-digit number.

73167176531330624919225119674426574742355349194934
96983520312774506326239578318016984801869478851843
85861560789112949495459501737958331952853208805511
12540698747158523863050715693290963295227443043557
66896648950445244523161731856403098711121722383113
62229893423380308135336276614282806444486645238749
30358907296290491560440772390713810515859307960866
70172427121883998797908792274921901699720888093776
65727333001053367881220235421809751254540594752243
52584907711670556013604839586446706324415722155397
53697817977846174064955149290862569321978468622482
83972241375657056057490261407972968652414535100474
82166370484403199890008895243450658541227588666881
16427171479924442928230863465674813919123162824586
17866458359124566529476545682848912883142607690042
24219022671055626321111109370544217506941658960408
07198403850962455444362981230987879927244284909188
84580156166097919133875499200524063689912560717606
05886116467109405077541002256983155200055935729725
71636269561882670428252483600823257530420752963450

One Possible Solution: Go

package main

import "fmt"
import "strconv"

const fileIn = "73167176531330624919225119674426574742355349194934" +
	"96983520312774506326239578318016984801869478851843" +
	"85861560789112949495459501737958331952853208805511" +
	"12540698747158523863050715693290963295227443043557" +
	"66896648950445244523161731856403098711121722383113" +
	"62229893423380308135336276614282806444486645238749" +
	"30358907296290491560440772390713810515859307960866" +
	"70172427121883998797908792274921901699720888093776" +
	"65727333001053367881220235421809751254540594752243" +
	"52584907711670556013604839586446706324415722155397" +
	"53697817977846174064955149290862569321978468622482" +
	"83972241375657056057490261407972968652414535100474" +
	"82166370484403199890008895243450658541227588666881" +
	"16427171479924442928230863465674813919123162824586" +
	"17866458359124566529476545682848912883142607690042" +
	"24219022671055626321111109370544217506941658960408" +
	"07198403850962455444362981230987879927244284909188" +
	"84580156166097919133875499200524063689912560717606" +
	"05886116467109405077541002256983155200055935729725" +
	"71636269561882670428252483600823257530420752963450"

func main() {
	largest := 0
	for i := 0; i < (len(fileIn) - 4); i++ {
		fiveDigits := fileIn[i:(i + 5)]

		s1 := fiveDigits[0:1]
		s2 := fiveDigits[1:2]
		s3 := fiveDigits[2:3]
		s4 := fiveDigits[3:4]
		s5 := fiveDigits[4:5]

		x1, _ := strconv.Atoi(s1)
		x2, _ := strconv.Atoi(s2)
		x3, _ := strconv.Atoi(s3)
		x4, _ := strconv.Atoi(s4)
		x5, _ := strconv.Atoi(s5)

		product := x1 * x2 * x3 * x4 * x5

		if product > largest {
			largest = product
		}

	}
	fmt.Println("Greatest product is: ", largest)
}

Project Euler – Problem # 7 – Solved with Go

Find the 10001st prime.

By listing the first six prime numbers: 2, 3, 5, 7, 11, and 13, we can see that the 6th prime is 13.

What is the 10 001st prime number?

One Possible Solution: Go

package main

import "fmt"
import "math"

func Sqrt(x float64) float64 {
	sR := math.Sqrt(x)
	return sR
}

func isPrime(n int) (isP bool) {
	if n < 2 {
		return false
	}
	if n == 2 {
		return true
	}
	k := int(Sqrt(float64(n)))
	for i := 3; i <= k; i += 2 {
		if n%i == 0 {
			return false
		}
	}
	return true
}

func main() {
	Switch := 1	// switch for the for loop
	number := 1	// start with 1 and increment by 2 (i.e - 1, 3, 5, 7, 9, etc.)

	// start with 1 since we will not count 2, which is a prime
	primes := 1

	for Switch >= 1 {
		if isPrime(number) {	// check for primality if true, add 1 to prime counter
			primes += 1
		}
		if primes == 10001 {	// if prime counter == 10001, print that prime number
			fmt.Println(number)
			Switch = 0
		}
		number += 2
	}
}

Project Euler – Problem # 6 – Solved with Go

What is the difference between the sum of the squares and the square of the sums?

The sum of the squares of the first ten natural numbers is,

1² + 2² + … + 10² = 385

The square of the sum of the first ten natural numbers is,

(1 + 2 + … + 10)² = 55² = 3025

Hence the difference between the sum of the squares of the first ten natural numbers and the square of the sum is 3025 – 385 = 2640.

Find the difference between the sum of the squares of the first one hundred natural numbers and the square of the sum.

One Possible Solution: Go

package main

import "fmt"
import "math"

var (
	sum_of_squares	float64	= 0
	square_of_sum	float64	= 0
	Sum		float64	= 0
)

func main() {
	for i := 1; i < 101; i++ {
		square := math.Pow(float64(i), 2)
		sum_of_squares += square
		Sum += float64(i)
	}
	square_of_sum = math.Pow(float64(Sum), 2)
	fmt.Println("sum_of_squares = ", int(sum_of_squares))
	fmt.Println("square_of_sum = ", int(square_of_sum))
	Difference := square_of_sum - sum_of_squares
	fmt.Println("Difference = ", int(Difference))

}

Project Euler – Problem # 5 – Solved with the Go programming language

What is the smallest number divisible by each of the numbers 1 to 20?

2520 is the smallest number that can be divided by each of the numbers from 1 to 10 without any remainder.

What is the smallest positive number that is evenly divisible by all of the numbers from 1 to 20?

One Possible Solution: Go

package main

import (
	"fmt"
)

func evenlyDivisible(n int) (eD int) {
	/*Checks if number (n) is evenly divisible by the numbers 1 -20.*/

	counter := 0
	for i := 1; i < 21; i++ {
		if n%i == 0 {
			counter += 1
			if counter == 20 {
				return n
			}

		} else {
			return 1
		}
	}
	return 1
}
func main() {
	/*Main program - since we know that 2520 is evenly divisible by
	the first 10 integers, we know that the number in question is a
	multiple of 2520. So we start with 2520 and increment by that much
	until we find the smallest number that is evenly divisible by the
	first 20 integers (1-20)*/

	n := 2520
	Switch := 1
	for Switch >= 1 {
		smallest := evenlyDivisible(n)
		if smallest != 1 {
			fmt.Println(smallest)
			Switch = 0

		} else {
			n += 2520
		}
	}
}