Monthly Archives: April 2010

Euler Problem 11

In the 20×20 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?

Looks like I need to develop a tricky algorithm.

Copy the strings, parse into 2D integer array. Then use 4 different calculation methods.

1) Up/Down

2) Left/Right

3) Diagonal Left to Right

4) Diagonal Right to Left

My first attempt failed because I was trying to put too much out-of-bounds error avoidance, which ended up missing the answer.

So instead I flipped my thinking and checked every cell for all eight possibilities and used a try{} block to ignore errors. It worked.

public class Euler11
{
    public static void main(String[] args)
    {
        System.out.print(“Problem 11:\n”);
        Euler11 e = new Euler11();
        System.out.print(“Answer = ” + e.Problem()+ “\n”);
    }
    public String Problem ()
    {
        String Grid = “08 02 22 97 38 15 00 40 00 75 04 05 07 78 52 12 50 77 91 08 “;
        Grid += “49 49 99 40 17 81 18 57 60 87 17 40 98 43 69 48 04 56 62 00 “;
        Grid += “81 49 31 73 55 79 14 29 93 71 40 67 53 88 30 03 49 13 36 65 “;
        Grid += “52 70 95 23 04 60 11 42 69 24 68 56 01 32 56 71 37 02 36 91 “;
        Grid += “22 31 16 71 51 67 63 89 41 92 36 54 22 40 40 28 66 33 13 80 “;
        Grid += “24 47 32 60 99 03 45 02 44 75 33 53 78 36 84 20 35 17 12 50 “;
        Grid += “32 98 81 28 64 23 67 10 26 38 40 67 59 54 70 66 18 38 64 70 “;
        Grid += “67 26 20 68 02 62 12 20 95 63 94 39 63 08 40 91 66 49 94 21 “;
        Grid += “24 55 58 05 66 73 99 26 97 17 78 78 96 83 14 88 34 89 63 72 “;
        Grid += “21 36 23 09 75 00 76 44 20 45 35 14 00 61 33 97 34 31 33 95 “;
        Grid += “78 17 53 28 22 75 31 67 15 94 03 80 04 62 16 14 09 53 56 92 “;
        Grid += “16 39 05 42 96 35 31 47 55 58 88 24 00 17 54 24 36 29 85 57 “;
        Grid += “86 56 00 48 35 71 89 07 05 44 44 37 44 60 21 58 51 54 17 58 “;
        Grid += “19 80 81 68 05 94 47 69 28 73 92 13 86 52 17 77 04 89 55 40 “;
        Grid += “04 52 08 83 97 35 99 16 07 97 57 32 16 26 26 79 33 27 98 66 “;
        Grid += “88 36 68 87 57 62 20 72 03 46 33 67 46 55 12 32 63 93 53 69 “;
        Grid += “04 42 16 73 38 25 39 11 24 94 72 18 08 46 29 32 40 62 76 36 “;
        Grid += “20 69 36 41 72 30 23 88 34 62 99 69 82 67 59 85 74 04 36 16 “;
        Grid += “20 73 35 29 78 31 90 01 74 31 49 71 48 86 81 16 23 57 05 54 “;
        Grid += “01 70 54 71 83 51 54 69 16 92 33 48 61 43 52 01 89 19 67 48”;
        String [] big = Grid.split(” “);
        int [] [] g = new int [21][21];
        int location = 0;
        for (int x = 1; x <= 20; x++)
        {
            for (int y = 1; y <= 20; y++)
            {
                g[x][y]= Integer.parseInt(big[location++]);
            }
        }
        int biggest = 0;
        for (int x = 1; x <= 20; x++)
        {
            for (int y = 1; y <= 20; y++)
            {
                int t[] = {0,0,0,0,0,0,0,0};
                try
                {
                    t[0]= g[x][y]*g[x+1][y]*g[x+2][y]*g[x+3][y];
                    t[1]= g[x][y]*g[x-1][y]*g[x-2][y]*g[x-3][y];
                    t[2]= g[x][y]*g[x][y+1]*g[x][y+2]*g[x][y+3];
                    t[3]= g[x][y]*g[x][y-1]*g[x][y-2]*g[x][y-3];
                    t[4]= g[x][y]*g[x+1][y+1]*g[x+2][y+2]*g[x+3][y+3];
                    t[5]= g[x][y]*g[x+1][y-1]*g[x+2][y-2]*g[x+3][y-3];
                    t[6]= g[x][y]*g[x-1][y-1]*g[x-2][y-2]*g[x-3][y-3];
                    t[7]= g[x][y]*g[x-1][y+1]*g[x-2][y+2]*g[x-3][y+3];
                }
                catch (java.lang.ArrayIndexOutOfBoundsException e)
                {
                }
                for (int i = 0; i <=7; i++)
                {
                    if (t[i] > biggest)
                    {
                        biggest = t[i];
                    }
                }
            }
        }
        return String.valueOf(biggest);
    }
}

Euler Problem 10

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

Find the sum of all the primes below two million.

Use a prime number generator (from earlier) and an accumulator.

Update: My first attempt failed. Brute force generator from earlier problem took 33 minutes to generate the sum, but the number was off. So I’ll need to refine the algorithm. Found this on Wikipedia:

To find all the prime numbers less than or equal to a given integer n by Eratosthenes’ method:

  1. Create a list of consecutive integers from two to n: (2, 3, 4, …, n),
  2. Initially, let p equal 2, the first prime number,
  3. While enumerating all multiples of p starting from p2, strike them off from the original list,
  4. Find the first number remaining on the list after p (it’s the next prime); let p equal this number,
  5. Repeat steps 3 and 4 until p2 is greater than n.
  6. All the remaining numbers in the list are prime.

Works like a dream and fast!

public class Euler10
{
    public static void main(String[] args)
    {
        System.out.print(“Problem 10:\n”);
        Euler10 e = new Euler10();
        System.out.print(“Answer = ” + e.Problem()+ “\n”);
    }
    public String Problem ()
    {
        long accumulator=0;
        int n = 2000000;
        boolean [] isPrime = new boolean [n+1];
        for (int i = 2; i < n; i++)
        {
            isPrime[i] = true;
        }
        for (int p = 2; p*p < n; p++)
        {
            if (isPrime[p] == true)
            {
                for (int m = p*p; m<n; m=m+p)
                {
                    if (m % p == 0)
                    {
                        isPrime[m] = false;
                    }
                }
            }
        }
        for (int i = 2; i < n; i++)
        {
            if (isPrime[i] == true)
            {
                System.out.print(i+”\n”); //remove to speed up
                accumulator += i;
            }
        }
        return String.valueOf(accumulator);
    }
}

Euler Problem 9

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

a^(2) + b^(2) = c^(2)

For example, 3^(2) + 4^(2) = 9 + 16 = 25 = 5^(2).

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

Algebra time. I had to go back to the formal definition for ALL triplets. Its easy to confuse formulae that only create some, not all.

Just created nested loops to run through k, n & m and calculate:

a = k * (Math.pow(m,2) –  Math.pow(n,2));
b = k * (2 * m * n);
c = k * (Math.pow(m,2) + Math.pow(n,2));

then test each to see if the sum is 1000.

public class Euler9
{
    public static void main(String[] args)
    {
        System.out.print(“Problem 9:\n”);
        Euler9 e = new Euler9();
        System.out.print(“Answer = ” + e.Problem()+ “\n”);
    }
    public String Problem ()
    {
        double a = 0;
        double b = 0;
        double c = 0;
        double product = 0;
        for (int k = 1; k <=100; k++)
        {
            for (int n = 1; n <= 100; n++)
            {
                for (int m = 1; m <= 100; m++)
                {
                    if (m > n)
                    {
                        a = k * (Math.pow(m,2) –  Math.pow(n,2));
                        b = k * (2 * m * n);
                        c = k * (Math.pow(m,2) + Math.pow(n,2));
                        if (a+b+c == 1000 && a > 0 && b > 0 && c > 0)
                        {
                            product = a * b * c;
                            System.out.print(“k:” + k +” n:” + n+ ” m:” + m + ” a:” + a +” b:” + b+ ” c:” + c + “\n”);
                        }
                    }
                }
            }
        }
        return String.format(“%.0f”, product);
    }
}

Euler Problem 8

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

Whoa. Got to check if Java can handle 1000 digits. If not then we’ve got to use arrays or Strings.

Nope. Strings it is. Simple count & product check.

public class Euler8
{
    public static void main(String[] args)
    {
        System.out.print(“Problem 8:\n”);
        Euler8 e = new Euler8();
        System.out.print(“Answer = ” + e.Problem8()+ “\n”);
    }
    public String Problem8 ()
    {
        String huge = “73167176531330624919225119674426574742355349194934”;
        huge += “96983520312774506326239578318016984801869478851843”;
        huge += “85861560789112949495459501737958331952853208805511”;
        huge += “12540698747158523863050715693290963295227443043557”;
        huge += “66896648950445244523161731856403098711121722383113”;
        huge += “62229893423380308135336276614282806444486645238749”;
        huge += “30358907296290491560440772390713810515859307960866”;
        huge += “70172427121883998797908792274921901699720888093776”;
        huge += “65727333001053367881220235421809751254540594752243”;
        huge += “52584907711670556013604839586446706324415722155397”;
        huge += “53697817977846174064955149290862569321978468622482”;
        huge += “83972241375657056057490261407972968652414535100474”;
        huge += “82166370484403199890008895243450658541227588666881”;
        huge += “16427171479924442928230863465674813919123162824586”;
        huge += “17866458359124566529476545682848912883142607690042”;
        huge += “24219022671055626321111109370544217506941658960408”;
        huge += “07198403850962455444362981230987879927244284909188”;
        huge += “84580156166097919133875499200524063689912560717606”;
        huge += “05886116467109405077541002256983155200055935729725”;
        huge += “71636269561882670428252483600823257530420752963450”;
        int biggest=0;
        for (int i=0; i<=999-5; i++)
        {
            int product = Integer.parseInt(String.valueOf(huge.charAt(i)));
            product = product * Integer.parseInt(String.valueOf(huge.charAt(i+1)));
            product = product * Integer.parseInt(String.valueOf(huge.charAt(i+2)));
            product = product * Integer.parseInt(String.valueOf(huge.charAt(i+3)));
            product = product * Integer.parseInt(String.valueOf(huge.charAt(i+4)));
            if (product > biggest)
            {
                biggest = product;
            }
        }
        return String.valueOf(biggest);
    }
}

Euler Problem 7

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

What is the 10001^(st) prime number?

Just a simple prime number generator.

public class Euler7
{
    public static void main(String[] args)
    {
        System.out.print(“Problem 7:\n”);
        Euler7 e = new Euler7();
        System.out.print(“Answer = ” + e.Problem7()+ “\n”);
    }
    public String Problem7 ()
    {
        int x = 2;
        int y = 0;
        int c = 1;
        int answer = 0;
        while (c <= 10001)
        {
            if (x % 2 != 0 || x == 2)
            {
                for (y = 2; y <= x/2; y++)
                {
                    if (x % y == 0)
                    {
                        break;
                    }
                }
                if (y > x/2)
                {
                    System.out.print(c + “: ” + x + “\n”);
                    answer = x;
                    c++;
                }
            }
            x++;
        }
        return String.valueOf(answer);
    }
}

Euler Problem 6

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

1^(2) + 2^(2) + … + 10^(2) = 385

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

(1 + 2 + … + 10)^(2) = 55^(2) = 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.

First write the script to verify the first 10, then just add a 0 when it works.

public class Euler6
{
    public static void main(String[] args)
    {
        System.out.print(“Problem 6:\n”);
        Euler6 e = new Euler6();
        System.out.print(“Difference = ” + e.Problem6()+ “\n”);
    }
    public String Problem6 ()
    {
        int sumOfSquares = 0;
        int squareOfSums = 0;
        for (int i = 1; i <= 100; i++)
        {
            sumOfSquares += i*i;
            squareOfSums += i;
        }
        squareOfSums = squareOfSums*squareOfSums;
        return String.valueOf(squareOfSums – sumOfSquares);
    }
}

Euler Problem 5

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

I’m thinking brute force. there’s probably something simpler using factorials, but this works albeit slowly. Took about 90 seconds on my loaded system.

public class Euler5
{
    public static void main(String[] args)
    {
        System.out.print(“Problem 5:\n”);
        Euler5 e = new Euler5();
        System.out.print(“Smallest number = ” + e.Problem5()+ “\n”);
    }
    public String Problem5 ()
    {
        int first = 0;
        for (int i = 20; i <= 300000000; i++)
        {
            boolean test = true;
            for (int n = 1; n <= 20; n++)
            {
                if (i % n != 0)
                {
                    test = false;
                }
            }
            if (test == true)
            {
                first = i;
                break;
            }
        }
        return String.valueOf(first);
    }
}