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Added comments to the C code for all the problems solved so far.

C/p024.c

@@ -10,8 +10,6 @@
 #include <time.h>
 #include "projecteuler.h"
 
-void next_perm(int **perm, int n);
-void swap(int **vet, int i, int j);
 int compare(void *a, void *b);
 
 int main(int argc, char **argv)
@@ -43,7 +41,7 @@ int main(int argc, char **argv)
    {
       /* Function that generates permutations in lexicographic order.
        * Finish when the 1000000th is found.*/
-      next_perm(perm, 10);
+      next_permutation((void **)perm, 10, compare);
    }
 
    for(i = 0; i < 10; i++)
@@ -73,15 +71,6 @@ int main(int argc, char **argv)
    return 0;
 }
 
-void swap(int **vet, int i, int j)
-{
-   int *tmp;
-
-   tmp = vet[i];
-   vet[i] = vet[j];
-   vet[j] = tmp;
-}
-
 int compare(void *a, void *b)
 {
    int *n1, *n2;
@@ -91,38 +80,3 @@ int compare(void *a, void *b)
 
    return *n1 - *n2;
 }
-
-/* Implements SEPA (Simple, Efficient Permutation Algorithm)
- * to find the next permutation.*/
-void next_perm(int **perm, int n)
-{
-   int i, key;
-
-   /* Starting from the right of the array, for each pair of values
-    * if the left one is smaller than the right, that value is the key.*/
-   for(i = n - 2; i >= 0; i--)
-   {
-      if(compare((void *)perm[i], (void *)perm[i+1]) < 0)
-      {
-         key = i;
-         break;
-      }
-   }
-
-   /* If no left value is smaller than its right value, the
-    * array is in reverse order, i.e. it's the last permutation.*/
-   if(i == -1)
-   {
-      return;
-   }
-
-   /* Find the smallest value on the right of the key which is bigger than the key itself,
-    * considering that the values at the right of the key are in reverse order.*/
-   for(i = key + 1; i < n && compare((void *)perm[i], (void *)perm[key]) > 0; i++);
-
-   /* Swap the value found and the key.*/
-   swap(perm, key, i-1);
-   /* Sort the values at the right of the key. This is
-    * the next permutation.*/
-   insertion_sort((void **)perm, key+1, n-1, compare);
-}

C/p041.c

@@ -1,3 +1,8 @@
+/* We shall say that an n-digit number is pandigital if it makes use of all the digits 1 to n exactly once. For example, 2143 is a 4-digit pandigital 
+ * and is also prime.
+ *
+ * What is the largest n-digit pandigital prime that exists?*/
+
 #include <stdio.h>
 #include <stdlib.h>
 #include <math.h>
@@ -8,27 +13,34 @@ int count_digits(int n);
 
 int main(int argc, char **argv)
 {
-   int i, found = 0;
+   /* 8- and 9-digit pandigital numbers can't be prime, because 
+    * 1+2+...+8=36, which is divisible by 3, and 36+9=45 which is
+    * also divisible by 3, and therefore the whole number is divisible
+    * by 3. So we can start from the largest 7-digit pandigital number,
+    * until we find a prime.*/
+   int i = 7654321;
    double elapsed;
    struct timespec start, end;
 
    clock_gettime(CLOCK_MONOTONIC, &start);
 
-   for(i = 7654321; !found && i > 0; i -= 2)
+   while(i > 0)
    {
-
       if(is_pandigital(i, count_digits(i)) && is_prime(i))
       {
-         printf("Project Euler, Problem 41\n");
+         break;
-         printf("Answer: %d\n", i);
-         found = 1;
       }
+//       Skipping the even numbers.
+      i -= 2;
    }
 
    clock_gettime(CLOCK_MONOTONIC, &end);
 
    elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
 
+   printf("Project Euler, Problem 41\n");
+   printf("Answer: %d\n", i);
+
    printf("Elapsed time: %.9lf seconds\n", elapsed);
 
    return 0;

C/p042.c

@@ -1,3 +1,13 @@
+/* The nth term of the sequence of triangle numbers is given by, tn = ½n(n+1); so the first ten triangle numbers are:
+ *
+ * 1, 3, 6, 10, 15, 21, 28, 36, 45, 55, ...
+ *
+ * By converting each letter in a word to a number corresponding to its alphabetical position and adding these values we form a word value.
+ * For example, the word value for SKY is 19 + 11 + 25 = 55 = t10. If the word value is a triangle number then we shall call the word a triangle word.
+ *
+ * Using words.txt (right click and 'Save Link/Target As...'), a 16K text file containing nearly two-thousand common English words,
+ * how many are triangle words?*/
+
 #include <stdio.h>
 #include <stdlib.h>
 #include <string.h>
@@ -28,6 +38,7 @@ int main(int argc, char **argv)
    n = 1;
    len = strlen(line);
 
+   /* Count the words.*/
    for(i = 0; i < len; i++)
    {
       if(line[i] == ',')
@@ -42,6 +53,7 @@ int main(int argc, char **argv)
       return 1;
    }
 
+   /* Save the words in an array of strings.*/
    words[0] = strtok(line, ",\"");
 
    for(i = 1; i < n; i++)
@@ -49,6 +61,7 @@ int main(int argc, char **argv)
       words[i] = strtok(NULL, ",\"");
    }
 
+   /* For each word, calculate its value and check if it's a triangle number.*/
    for(i = 0; i < n; i++)
    {
       value = 0;

C/p044.c

@@ -1,3 +1,12 @@
+/* Pentagonal numbers are generated by the formula, Pn=n(3n−1)/2. The first ten pentagonal numbers are:
+ *
+ * 1, 5, 12, 22, 35, 51, 70, 92, 117, 145, ...
+ *
+ * It can be seen that P4 + P7 = 22 + 70 = 92 = P8. However, their difference, 70 − 22 = 48, is not pentagonal.
+ *
+ * Find the pair of pentagonal numbers, Pj and Pk, for which their sum and difference are pentagonal and D = |Pk − Pj| is minimised;
+ * what is the value of D?*/
+
 #include <stdio.h>
 #include <stdlib.h>
 #include <math.h>
@@ -14,6 +23,9 @@ int main(int argc, char **argv)
 
    n = 2;
 
+   /* Check all couples of pentagonal numbers until the right one
+    * is found. Use the function implemented in projecteuler.c to
+    * check if the sum and difference ot the two numbers is pentagonal.*/
    while(!found)
    {
       pn = n * (3 * n - 1) / 2;

C/p045.c

@@ -1,3 +1,13 @@
+/* Triangle, pentagonal, and hexagonal numbers are generated by the following formulae:
+ *
+ * Triangle	 	   T_n=n(n+1)/2	 	1, 3, 6, 10, 15, ...
+ * Pentagonal	 	P_n=n(3n−1)/2	1, 5, 12, 22, 35, ...
+ * Hexagonal	 	H_n=n(2n−1)	 	1, 6, 15, 28, 45, ...
+ *
+ * It can be verified that T_285 = P_165 = H_143 = 40755.
+ *
+ * Find the next triangle number that is also pentagonal and hexagonal.*/
+
 #include <stdio.h>
 #include <stdlib.h>
 #include <math.h>
@@ -18,6 +28,9 @@ int main(int argc, char **argv)
    while(!found)
    {
       i++;
+      /* Hexagonal numbers are also triangle numbers, so it's sufficient
+       * to generate hexagonal numbers (starting from H_144) and check if
+       * they're also pentagonal.*/
       n = i * (2 * i - 1);
       
       if(is_pentagonal(n))

C/p047.c

@@ -1,3 +1,16 @@
+/* The first two consecutive numbers to have two distinct prime factors are:
+ *
+ * 14 = 2 × 7
+ * 15 = 3 × 5
+ *
+ * The first three consecutive numbers to have three distinct prime factors are:
+ *
+ * 644 = 2² × 7 × 23
+ * 645 = 3 × 5 × 43
+ * 646 = 2 × 17 × 19.
+ *
+ * Find the first four consecutive integers to have four distinct prime factors each. What is the first of these numbers?*/
+
 #include <stdio.h>
 #include <stdlib.h>
 #include <math.h>
@@ -24,7 +37,9 @@ int main(int argc, char **argv)
       return 1;
    }
 
-   for(i = 645; !found && i < N - 3; i++)
+   /* Starting from 647, count the distinct prime factors of n, n+1, n+2 and n+3.
+    * If they all have 4, the solution is found.*/
+   for(i = 647; !found && i < N - 3; i++)
    {
       if(!primes[i] && !primes[i+1] && !primes[i+2] && !primes[i+3])
       {
@@ -54,6 +69,8 @@ int count_distinct_factors(int n)
 {
    int i, count=0;
 
+   /* Start checking if 2 is a prime factor of n. Then remove
+    * all 2s factore.*/
    if(n % 2 == 0)
    {
       count++;
@@ -64,6 +81,10 @@ int count_distinct_factors(int n)
       }while(n % 2 == 0);
    }
 
+   /* Check all odd numbers i, if they're prime and they're a factor
+    * of n, count them and then divide n for by i until all factors i
+    * are eliminated. Stop the loop when n=1, i.e. all factors have
+    * been found.*/
    for(i = 3; n > 1; i += 2)
    {
       if(primes[i] && n % i == 0)

C/p048.c

@@ -1,3 +1,7 @@
+/* The series, 1^1 + 2^2 + 3^3 + ... + 10^10 = 10405071317.
+ *
+ * Find the last ten digits of the series, 1^1 + 2^2 + 3^3 + ... + 1000^1000.*/
+
 #include <stdio.h>
 #include <stdlib.h>
 #include <string.h>
@@ -17,6 +21,7 @@ int main(int argc, char **argv)
    mpz_init_set_ui(sum, 0);
    mpz_init(power);
 
+   /* Simply calculate the sum of the powers using the GMP Library.*/
    for(i = 1; i <= 1000; i++)
    {
       mpz_ui_pow_ui(power, i, i);

C/p049.c

@@ -1,3 +1,11 @@
+/* The arithmetic sequence, 1487, 4817, 8147, in which each of the terms increases by 3330, is unusual in two ways: (i) each of the three terms are prime,
+ * and, (ii) each of the 4-digit numbers are permutations of one another.
+ *
+ * There are no arithmetic sequences made up of three 1-, 2-, or 3-digit primes, exhibiting this property, but there is one other 4-digit
+ * increasing sequence.
+ *
+ * What 12-digit number do you form by concatenating the three terms in this sequence?*/
+
 #include <stdio.h>
 #include <stdlib.h>
 #include <math.h>
@@ -12,7 +20,7 @@ int *primes;
 
 int main(int argc, char **argv)
 {
-   int i, j, found = 0;
+   int i = 1489, j, found = 0;
    double elapsed;
    struct timespec start, end;
 
@@ -24,12 +32,18 @@ int main(int argc, char **argv)
       return 1;
    }
 
-   for(i = 1489; i < N && !found; i++)
+   /* Starting from i=1489 (bigger than the first number in the sequence given in the problem),
+    * check odd numbers. If they're prime, loop on even numbers j (odd+even=odd, odd+odd=even and
+    * we need odd numbers because we're looking for primes) up to 4254 (1489+2*4256=10001 which has
+    * 5 digits.*/
+   while(i < N)
    {
       if(primes[i])
       {
-         for(j = 1; j < 4255; j++)
+         for(j = 2; j < 4255; j += 2)
          {
+            /* If i, i+j and i+2*j are all primes and they have
+             * all the same digits, the result has been found.*/
             if(i + 2 * j < N && primes[i+j] && primes[i+2*j] &&
                   check_digits(i, i+j) && check_digits(i, i+2*j))
             {
@@ -38,6 +52,11 @@ int main(int argc, char **argv)
             }
          }
       }
+      if(found)
+      {
+         break;
+      }
+      i += 2;
    }
 
    free(primes);
@@ -47,7 +66,7 @@ int main(int argc, char **argv)
    elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
 
    printf("Project Euler, Problem 49\n");
-   printf("Answer: %d%d%d\n", i-1, i-1+j, i-1+2*j);
+   printf("Answer: %d%d%d\n", i, i+j, i+2*j);
 
    printf("Elapsed time: %.9lf seconds\n", elapsed);
 

C/p050.c

@@ -1,3 +1,13 @@
+/* The prime 41, can be written as the sum of six consecutive primes:
+ *
+ * 41 = 2 + 3 + 5 + 7 + 11 + 13
+ *
+ * This is the longest sum of consecutive primes that adds to a prime below one-hundred.
+ *
+ * The longest sum of consecutive primes below one-thousand that adds to a prime, contains 21 terms, and is equal to 953.
+ *
+ * Which prime, below one-million, can be written as the sum of the most consecutive primes?*/
+
 #include <stdio.h>
 #include <stdlib.h>
 #include <math.h>
@@ -21,14 +31,39 @@ int main(int argc, char **argv)
       return 1;
    }
 
-   for(i = 2; i < N; i++)
+   /* Starting from a prime i, add consecutive primes until the 
+    * sum exceeds the limit, every time the sum is also a prime
+    * save the value and the count if the count is larger than the
+    * current maximum. Repeat for all primes below N.
+    * A separate loop is used for i=2, so later only odd numbers are 
+    * checked for primality.*/
+   i = 2;
+   count = 1;
+   sum = i;
+
+   for(j = i + 1; j < N && sum < N; j++)
+   {
+      if(primes[j])
+      {
+         sum += j;
+         count++;
+
+         if(sum < N && primes[sum] && count > max)
+         {
+            max = count;
+            max_p = sum;
+         }
+      }
+   }
+
+   for(i = 3; i < N; i += 2)
    {
       if(primes[i])
       {
          count = 1;
          sum = i;
 
-         for(j = i + 1; j < N && sum < N; j++)
+         for(j = i + 2; j < N && sum < N; j += 2)
          {
             if(primes[j])
             {

Python/p041.py

@@ -15,14 +15,18 @@ def count_digits(n):
 def main():
     start = default_timer()
 
-    for i in range(7654321, 0, -2):
+    i = 7654321
+
+    while(i > 0):
         if is_pandigital(i, count_digits(i)) and is_prime(i):
-            print('Project Euler, Problem 41')
-            print('Answer: {}'.format(i))
             break
+        i = i - 2
     
     end = default_timer()
 
+    print('Project Euler, Problem 41')
+    print('Answer: {}'.format(i))
+    
     print('Elapsed time: {:.9f} seconds'.format(end - start))
 
 if __name__ == '__main__':