Daniel Pedigo

Challenging Websites

admin — Sat, 03 Jan 2009 01:09:46 +0000

As most of you can tell, I’ve been pretty fascinated with Project Euler for quite some time (I have a lot more challenges done than I’ve actually posted). However, these last few weeks I’ve been neglecting Project Euler and keeping myself busy with a few other “challenge”-type websites that I’ve found. If any of you are interested in having hours and hours disappear in the blink of an eye, keep reading. Seriously, I’m not joking.

First up is an amazingly clever riddle website titled Notpron. Think of this as a no holds barred, anything goes type of riddle. The first few levels are quite easy, as to give you a grasp on how it works, but they become much harder as you progress. I’ve spent countless hours on Notpron over the last few weeks only to progress to level 18. There are 140 levels available. According to the website, only 3 people have been “certified” as having completed all of the levels. I definitely recommend everyone gives this website a try.

My next recommendation is a website called Bright Shadows. Bright shadows has many different categories of challenges available in hopes that there will be a category suitable to you. As most of these challenges require programming experience, I would not recommend this website to anyone that is not at least a little bit familiar with programming concepts. Some of the categories include: programming, Cryptography, Steganography, logic, and my favorite, cracking. Regardless, I still give this website an A+ for finally convincing me to whip out a disassembler and “go at it”.

My last recommendation (not because the others are any better) for challenge websites is Project Euler. This website is based around challenging mathematical/computer programming problems. Even someone like me, who would rather be doing things that are not math related, should still find this website to be a lot of fun.

Even though I’ve recommended a few websites to check out, there are still many more out there that are a lot of fun. A few runner-ups include:

If any one knows of any other websites similar to these worth mentioning, leave a comment and I’ll add it to the list.

Project Euler: Problem #11

admin — Fri, 28 Nov 2008 23:05:25 +0000

I’ve been so preoccupied as of late that I’ve completely neglected the website! I’ll try to be a little better in the future. So anyways, I’ve finally got around to posting the answer to Problem 11. It’s not at all a hard problem, in fact, I think the hardest part was filling that pesky multi-dimensional array with the values!

Problem #11:
Description: What is the greatest product of four adjacent numbers in any direction (up, down, left, right, or diagonally) in the 20×20 grid?

#define LENGTH 20
#define HEIGHT 20
 
int ProblemEleven()
{
  int result = 0;
 
  int g[][LENGTH] = {
    {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},
    {20,69,36,41,72,30,23,88,34,62,99,69,82,67,59,85,74,40,36,16},
    {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}
  };
 
  // first case : left <-> right
  for ( int i = 0; i < HEIGHT; i++ )
  {
    for ( int j = 0; j < (LENGTH - 3); j++ )
    {
      int temp = g[i][j] * g[i][j+1] * g[i][j+2] * g[i][j+3];
      if ( temp > result )
      {
        result = temp;
      }
    }
  }
 
  // second case : up <-> down
  for ( int i = 0; i < LENGTH; i++ )
  {
    for ( int j = 0; j < (HEIGHT - 3); j++ )
    {
      int temp = g[j][i] * g[j+1][i] * g[j+2][i] * g[j+3][i];
      if ( temp > result )
      {
        result = temp;
      }
    }
  }
 
  // third case : diagonal -> right
  for ( int i = 0; i < (LENGTH - 3); i++ )
  {
    for ( int j = 0; j < (HEIGHT - 3); j++ )
    {
      int temp = g[j][i] * g[j+1][i+1] * g[j+2][i+2] * g[j+3][i+3];
      if ( temp > result )
      {
        result = temp;
      }
    }
  }
 
  // fourth case : diagonal -> left
  for ( int i = LENGTH; i > 3; i-- )
  {
    for ( int j = 0; j < (HEIGHT - 3); j++ )
    {
      int temp = g[j][i] * g[j+1][i-1] * g[j+2][i-2] * g[j+3][i-3];
      if ( temp > result )
      {
        result = temp;
      }
    }
  }
 
  return result;
}

Project Euler: Problem #10

admin — Thu, 20 Nov 2008 06:53:11 +0000

Okay, slightly off topic, but I’ve got to throw this out there. Never, ever, ever write something that people will see when it’s late at night and you’re tired. If you’re me, you will not only incorrectly express your feelings due to your terrible writing skills, but you’ll also make numerous mistakes misrepresenting the facts. That’s what happened bashing on the LC-3 assembly language the other night. To be honest, I’m not sure why I picked on LC-3. I’m going to avoid late night writing altogether. Oh, and Stephen, I wasn’t complaining about the number registers–that’s just silly.

There’s something about Project Euler and prime numbers that just go hand in hand it seems. If you’re ever in doubt in regards with how to solve a problem you’ll probably need to generate some prime numbers. No, seriously. So I’ve been talking about the day I would need to create a prime number sieve and that day has finally come. I decided to go ahead and implement a version of the Sieve of Eratosthenes for this and all future problems requiring me to calculate a list of prime numbers. This thing is insanely fast compared to my previous code examples for generating prime numbers; you can generate the first two million primes in under 2 seconds. If you were feeling (slightly) more adventurous, you could combine this sieve with Wheel Factorization to further optimize the code.

Problem #9:
Description: Find the sum of all the primes below two million.

#include 
 
int* SieveOfEratosthenes( int size )
{
  int* primes = new int[size];
 
  // populate the array with numbers
  for ( int i = 0; i <= size; i++ )
  {
    primes[i] = i;
  }
 
  // one isn't prime... sorry.
  primes[1] = 0;
 
  for ( int i = 2; i <= sqrt(size); i++ )
  {
    if ( primes[i] != 0 )
    {
      for ( int j = (i * i); j <= size; j++ )
      {
        if ( primes[j] != 0 )
        {
          if ( (primes[j] % primes[i]) == 0 )
          {
            primes[j] = 0;
          }
        }
      }
    }
  }
 
  return primes;
}
 
INT64 ProblemTen()
{
  INT64 result = 0;
  int* primes = SieveOfEratosthenes(2000000);
 
  for ( int i = 0; i <= 2000000; i++ )
  {
    if ( primes[i] != 0 )
    {
      result += primes[i];
    }
  }
 
  return result;
}

Project Euler: Problem #9

admin — Mon, 17 Nov 2008 20:12:22 +0000

Ahh, one of the most well-known theorems in math it seems: The Pythagorean theorem. This is a nice and easy problem to solve using a good ol’ brute force algorithm.

Problem #9
Description: Find the only Pythagorean triplet, {a, b, c}, for which a + b + c = 1000.

int ProblemNine()
{
  for ( int a = 1; a <= 1000; a++ )
  {
    for ( int b = 1; b <= (1000 - a); b++ )
    {
      int c = 1000 - (a + b);
      if ( (a * a) + (b * b) == (c * c) )
      {
        return (a * b * c);
      }
    }
  }
}

Project Euler: Problem #8

admin — Fri, 14 Nov 2008 18:39:58 +0000

Due to the size of the number provided in the problem, and the fact that I’ve yet to find a decent big integer library to use, I ended up treating this number as a character array and just converting groups of five into numbers as I need them. Not really an efficient method of doing this, but it still completes extremely quickly.

If one wanted to speed up the algorithm some, they could create temporary integer variables to hold the four values that will be used again in the next iteration of the for loop, thus saving some time by not needing to recalculate those values.

Problem #8
Description: Find the greatest product of five consecutive digits in the 1000-digit number.

#include 
#include 
 
int ProblemEight()
{
  char* n = "73167176531330624919225119674426574742355349194934 \
             96983520312774506326239578318016984801869478851843 \
             85861560789112949495459501737958331952853208805511 \
             12540698747158523863050715693290963295227443043557 \
             66896648950445244523161731856403098711121722383113 \
             62229893423380308135336276614282806444486645238749 \
             30358907296290491560440772390713810515859307960866 \
             70172427121883998797908792274921901699720888093776 \
             65727333001053367881220235421809751254540594752243 \
             52584907711670556013604839586446706324415722155397 \
             53697817977846174064955149290862569321978468622482 \
             83972241375657056057490261407972968652414535100474 \
             82166370484403199890008895243450658541227588666881 \
             16427171479924442928230863465674813919123162824586 \
             17866458359124566529476545682848912883142607690042 \
             24219022671055626321111109370544217506941658960408 \
             07198403850962455444362981230987879927244284909188 \
             84580156166097919133875499200524063689912560717606 \
             05886116467109405077541002256983155200055935729725 \
             71636269561882670428252483600823257530420752963450";
 
  int product = 0;
 
  for ( int i = 0; i <= (strlen(n) - 5); i++ )
  {
    char n1 = n[i];
    char n2 = n[i+1];
    char n3 = n[i+2];
    char n4 = n[i+3];
    char n5 = n[i+4];
 
    int temp = atoi(&n1) * atoi(&n2) * atoi(&n3)
                         * atoi(&n4) * atoi(&n5);
 
    if ( temp > product )
    {
      product = temp;
    }
  }
 
  return product;
}

Website Updates

admin — Fri, 14 Nov 2008 02:50:07 +0000

I decided that instead of solving and posting a Euler Project problem tonight, I would spend some time updating a few areas of my blog that have been neglected as of late. I have re-posted my game of Tic Tac Toe and updated the project page that links to it. I have also begun the lengthy process of adding metadata and optimizing the web pages for search engines. I’ll hopefully have that completed by this weekend.

One other thing I’ve done tonight is recreate my PAD file for the MHTML Converter, which in turn should start driving a lot of traffic to this website again. Believe it or not, it is quite the popular little utility. I’m sure I’ll begin to see the website traffic increase again very shortly.

Project Euler: Problem #7

admin — Thu, 13 Nov 2008 05:06:15 +0000

Work has been rather busy as of late and as a result my lunch breaks have been shorter than normal, which means that I don’t have as much free time to do these little problems. Luckily, I had enough time to complete this problem during lunch and some time tonight to post the problem to my blog. So, to avoid further delay, here is problem 7 in all of it’s glory. Oh, and I’ll get around to make a prime number generator later…

Problem #7
Description: Find the 10001st prime.

int ProblemSeven()
{
  int total = 0;
  int prime = 0;
  int curr = 2;
 
  while ( total < 10001 )
  {
    bool isPrime = true;
 
    // check to see if it's really prime
    for ( int j = 2; j <= (curr / 2); j++ )
    {
      if ( curr % j == 0 )
      {
        isPrime = false;
        break;
      }
    }
 
    if ( isPrime )
    {
      total++;
      prime = curr;
    }
 
    curr++;
  }
 
  return prime;
}

Project Euler: Problem #6

admin — Tue, 11 Nov 2008 17:05:35 +0000

Not really much to say about this problem as it was very straightforward. Reading the code should be more than enough to figure it out. I guess we’ll just have to wait for a more exciting problem.

Problem #6
Description: Find the difference between the sum of the squares of the first one hundred natural numbers and the square of the sum.

int ProblemSix()
{
  int sumSquares = 0;
  int sumNumbers = 0;
 
  for ( int i = 1; i <= 100; i++ )
  {
    sumSquares += (i * i);
    sumNumbers += i;
  }
 
  sumNumbers *= sumNumbers;
 
  return (sumNumbers - sumSquares);
}

Project Euler: Problem #5

admin — Fri, 07 Nov 2008 15:01:44 +0000

Here is actually the problem that originally turned me onto Project Euler. The idea here is to find the smallest number that can be divided evenly by the numbers one through twenty. Now, the beginner might think to check each number against the numbers one through twenty until they found the smallest number. While this works, it’s not quite as efficient as it could be. For instance, what do the numbers four, eight and sixteen have in common? What about seven and fourteen? They all have common factors.

Problem #5
Description: What is the smallest number divisible by each of the numbers 1 to 20?

int ProblemFive()
{
  int n = 1;
 
  while ( true ) 
  {
    if ( n % 19 == 0 && n % 18 == 0 &&
         n % 17 == 0 && n % 16 == 0 &&
         n % 15 == 0 && n % 14 == 0 &&
         n % 13 == 0 && n % 11 == 0 )
    {
      return n;
    }
 
    n++;
  }
}

Project Euler: Problem #4

admin — Thu, 06 Nov 2008 22:38:57 +0000

While the code below is not the most elegant, it’s definitely functional and to be fair, was written in maybe 10 minutes. Regardless, I’m really looking forward to some of these upcoming problems! Some of them are looking HARD.

Problem #4
Description: Find the largest palindrome made from the product of two 3-digit numbers.

#include 
#include 
 
int ProblemFour()
{
  int result = 0;
  char* palindrome = new char[6];
 
  for ( int i = 999; i >= 100; i-- )
  {
    for ( int j = 999; i >= 100; j-- )
    {
      int temp = i * j;
      itoa(temp, palindrome, 10);
 
      // Our palindrome will be of the even variety...
      int length = strlen(palindrome);
      if ( length % 2 == 0 )
      {
        char* pFront = palindrome;
        char* pEnd = palindrome + length - 1;
 
        // check to see if it is a palindrome
        bool isPalindrome = true;
        while ( pEnd > pFront )
        {
          if ( *pFront != *pEnd )
          {
            isPalindrome = false;
            break;
          }
 
          pFront++;
          pEnd--;
        }
 
        if ( isPalindrome && temp > result )
        {
          result = temp;
        }
      }
    }
  }
 
  delete palindrome;
  return result;
}