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Added comments to the code of the first 5 problems in C.

C/p001.c

@@ -1,3 +1,7 @@
+/* If we list all the natural numbers below 10 that are multiples of 3 or 5, we get 3, 5, 6 and 9. The sum of these multiples is 23.
+ *
+ * Find the sum of all the multiples of 3 or 5 below 1000.*/
+
 #include <stdio.h>
 #include <stdlib.h>
 #include <time.h>
@@ -10,7 +14,9 @@ int main(int argc, char **argv)
 
    clock_gettime(CLOCK_MONOTONIC, &start);
 
-   for(i = 1; i < 1000; i++)
+   /* Simple brute-force approach: try every number between 3 and 999,
+    * check if it's a multiple of 3 or 5, if yes add it to the total.*/
+   for(i = 3; i < 1000; i++)
    {
       if (i % 3 == 0 || i % 5 == 0)
       {

C/p002.c

@@ -1,7 +1,15 @@
+/* Each new term in the Fibonacci sequence is generated by adding the previous two terms. By starting with 1 and 2, the first 10 terms will be:
+ *
+ * 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, ...
+ *
+ * By considering the terms in the Fibonacci sequence whose values do not exceed four million, find the sum of the even-valued terms.*/
+
 #include <stdio.h>
 #include <stdlib.h>
 #include <time.h>
 
+#define N 4000000
+
 int main(int argc, char **argv)
 {
    int fib0 = 1, fib1 = 2, fib2, sum = 2;
@@ -12,7 +20,9 @@ int main(int argc, char **argv)
 
    fib2 = fib0 + fib1;
 
-   while(fib2 <= 4000000)
+   /* Simple brute-force approach: generate every value in the Fibonacci
+    * sequence smaller than 4 million and if it's even add it to the total.*/
+   while(fib2 <= N)
    {
       if(fib2 % 2 == 0)
       {

C/p003.c

@@ -1,20 +1,25 @@
+/* The prime factors of 13195 are 5, 7, 13 and 29.
+ *
+ * What is the largest prime factor of the number 600851475143?*/
+
 #include <stdio.h>
 #include <stdlib.h>
 #include <math.h>
 #include <time.h>
 #include "projecteuler.h"
 
-int max_prime_factor(long int num);
+long int max_prime_factor(long int num);
 
 int main(int argc, char **argv)
 {
-   int res;
+   long int res;
    long int num = 600851475143;
    double elapsed;
    struct timespec start, end;
 
    clock_gettime(CLOCK_MONOTONIC, &start);
 
+   /* Use a function to calculate the largest prime factor.*/
    res = max_prime_factor(num);
 
    clock_gettime(CLOCK_MONOTONIC, &end);
@@ -22,41 +27,51 @@ 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 3\n");
-   printf("Answer: %d\n", res);
+   printf("Answer: %ld\n", res);
 
    printf("Elapsed time: %.9lf seconds\n", elapsed);
 
    return 0;
 }
 
-int max_prime_factor(long int num)
+/* Recursive approach: if num is prime, return num, otherwise
+ * recursively look for the largest prime factor of num divided
+ * by its prime factors until only the largest remains.*/
+long int max_prime_factor(long int num)
 {
-   int i, limit;
+   long int i;
 
+   /* Use function defined in projecteuler.c to check if a number is prime.*/
    if(is_prime(num))
    {
       return num;
    }
 
+   /* If num is even, find the largest prime factor of num/2.*/
    if(num % 2 == 0)
    {
       return max_prime_factor(num/2);
    }
    else
    {
-      limit = floor(sqrt(num));
-
-      for(i = 3; i <= limit; i += 2)
+      i = 3;
+      
+      while(1)
       {
+         /* If num is divisible by i and i is prime, find largest prime
+          * factor of num/i.*/
          if(num % i == 0)
          {
-            if(is_prime((long int)i))
+            if(is_prime(i))
             {
                return max_prime_factor(num/i);
             }
          }
+
+         i += 2;
       }
    }
 
+   /* Should never get here.*/
    return -1;
 }

C/p004.c

@@ -1,3 +1,7 @@
+/* A palindromic number reads the same both ways. The largest palindrome made from the product of two 2-digit numbers is 9009 = 91 × 99.
+ *
+ * Find the largest palindrome made from the product of two 3-digit numbers.*/
+
 #include <stdio.h>
 #include <stdlib.h>
 #include <time.h>
@@ -11,11 +15,15 @@ int main(int argc, char **argv)
 
    clock_gettime(CLOCK_MONOTONIC, &start);
 
+   /* Using a brute-force approach: generate every product of 3-digit numbers
+    * and check if it's palindrome. If the product found is greater than the
+    * current maximum, save the current product.*/
    for(i = 999; i >= 100; i--)
    {
       for(j = i; j >= 100; j--)
       {
          num = i * j;
+         /* Use the function defined in projecteuler.c to check if a number is palindrome.*/
          if(num > max && is_palindrome(num, 10))
          {
             max = num;

C/p005.c

@@ -1,3 +1,7 @@
+/* 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?*/
+
 #include <stdio.h>
 #include <stdlib.h>
 #include <time.h>
@@ -11,6 +15,7 @@ int main(int argc, char **argv)
 
    clock_gettime(CLOCK_MONOTONIC, &start);
 
+   /* Function define in projecteuler.c to find the least common multiple of multiple numbers.*/
    res = lcmm(n, 20);
 
    clock_gettime(CLOCK_MONOTONIC, &end);

C/projecteuler.c

@@ -7,20 +7,29 @@ int partition(void **array, int l, int r, int (*cmp)(void *lv, void *rv));
 
 int is_prime(long int num)
 {
-   int i, limit;
+   long int i, limit;
 
    if(num <= 3)
    {
+      /* If num is 2 or 3 then it's prime.*/
       return num == 2 || num == 3;
    }
 
+   /* If num is divisible by 2 or 3 then it's not prime.*/
    if(num % 2 == 0 || num % 3 == 0)
    {
       return 0;
    }
 
+   /* Any number can have only one prime factor greater than its
+    * square root. If we reach the square root and we haven't found
+    * any smaller prime factors, then the number is prime.*/
    limit = floor(sqrt(num));
 
+   /* Every prime other than 2 and 3 is in the form 6k+1 or 6k-1.
+    * If I check all those value no prime factors of the number 
+    * will be missed. If a factor is found, the number is not prime
+    * and the function returns 0.*/
    for(i = 5; i <= limit; i += 6)
    {
       if(num % i == 0 || num % (i + 2) == 0)
@@ -29,11 +38,42 @@ int is_prime(long int num)
       }
    }
 
+   /* If no factor is found up to the square root of num, num is prime.*/
    return 1;
 }
 
+int is_palindrome(int num, int base)
+{
+   int reverse = 0, tmp;
+
+   tmp = num;
+
+   /* Start with reverse=0, get the rightmost digit of the number using 
+    * modulo operation (num modulo base), add it to reverse. Remove the
+    * rightmost digit dividing num by the base, shift the reverse left
+    * multiplying by the base, repeat until all digits have been inserted
+    * in reverse order.*/
+   while(tmp > 0)
+   {
+      reverse *= base;
+      reverse += tmp % base;
+      tmp /= base;
+   }
+
+   /* If the reversed number is equal to the original one, then it's palindrome.*/
+   if(num == reverse)
+   {
+      return 1;
+   }
+
+   return 0;
+}
+
 long int gcd(long int a, long int b)
 {
+   /* Euclid's algorithm for the greatest common divisor:
+    * gcd(a, 0) = a
+    * gcd(a, b) = gcd(b, a modulo b)*/
    if(b == 0)
    {
       return a;
@@ -42,16 +82,19 @@ long int gcd(long int a, long int b)
    return gcd(b, a%b);
 }
 
+/* Least common multiple algorithm using the greatest common divisor.*/
 long int lcm(long int a, long int b)
 {
    return a * b / gcd(a, b);
 }
 
+/* Recursive function to calculate the least common multiple of more than 2 numbers.*/
 long int lcmm(long int *values, int n)
 {
    int i;
    long int value;
 
+   /* If there are only two numbers, use the lcm function to calculate the lcm.*/
    if(n == 2)
    {
       return lcm(values[0], values[1]);
@@ -65,6 +108,7 @@ long int lcmm(long int *values, int n)
          values[i] = values[i+1];
       }
 
+      /* Recursively calculate lcm(a, b, c, ..., n) = lcm(a, lcm(b, c, ..., n)).*/
       return lcm(value, lcmm(values, n-1));
    }
 }
@@ -207,27 +251,6 @@ void quick_sort(void **array, int l, int r, int (*cmp)(void *lv, void *rv))
    quick_sort(array, i+1, r, cmp);
 }
 
-int is_palindrome(int num, int base)
-{
-   int reverse = 0, tmp;
-
-   tmp = num;
-
-   while(tmp > 0)
-   {
-      reverse *= base;
-      reverse += tmp % base;
-      tmp /= base;
-   }
-
-   if(num == reverse)
-   {
-      return 1;
-   }
-
-   return 0;
-}
-
 int is_pandigital(int value, int n)
 {
    int *digits;

C/projecteuler.h

@@ -2,14 +2,14 @@
 #define _PROJECTEULER_INCLUDED
 
 int is_prime(long int);
-long int lcmm(long int *values, int n);
-long int lcm(long int a, long int b);
+int is_palindrome(int num, int base);
 long int gcd(long int a, long int b);
+long int lcm(long int a, long int b);
+long int lcmm(long int *values, int n);
 int *sieve(int n);
 int count_divisors(int n);
 void insertion_sort(void **array, int l, int r, int (*cmp)(void *lv, void *rv));
 void quick_sort(void **array, int l, int r, int (*cmp)(void *lv, void *rv));
-int is_palindrome(int num, int base);
 int is_pandigital(int value, int n);
 int is_pentagonal(long int n);
 

Python/projecteuler.py

@@ -1,6 +1,6 @@
 #!/usr/bin/python3
 
-from math import sqrt, floor
+from math import sqrt, floor, gcd
 
 from numpy import ndarray, zeros
 
@@ -19,12 +19,6 @@ def is_prime(num):
 
     return True
 
-def gcd(a, b):
-    if b == 0:
-        return a
-
-    return gcd(b, a%b)
-
 def lcm(a, b):
     return a * b // gcd(a, b)