document project euler till prob 12

project_euler/problem_1/sol1.c

@@ -1,14 +1,20 @@
-/*An Efficient code to print all the sum of all numbers that are multiples of 3
- * & 5 below N.*/
+/**
+ * \file
+ * \brief [Problem 1](https://projecteuler.net/problem=1) solution
+ *
+ * An Efficient code to print all the sum of all numbers that are multiples of 3
+ * & 5 below N.
+ */
 
 #include <stdio.h>
 
+/** Main function */
 int main()
 {
     int t;
     printf("Enter number of times you want to try");
     scanf("%d", &t);
-    while (t--)
+    while (t--) // while t > 0, decrement 't' before every iteration
     {
         unsigned long long N, p = 0, sum = 0;
         printf("Enter the value of N ");

project_euler/problem_1/sol2.c

@@ -1,14 +1,17 @@
-/*
-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 N.
-'''
-'''
-This solution is based on the pattern that the successive numbers in the series
-follow: 0+3,+2,+1,+3,+1,+2,+3.
-*/
+/**
+ * \file
+ * \brief [Problem 1](https://projecteuler.net/problem=1) solution
+ *
+ * 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 N.
+ *
+ * This solution is based on the pattern that the successive numbers in the
+ * series follow: 0+3,+2,+1,+3,+1,+2,+3.
+ */
 #include <stdio.h>
 
+/** Main function */
 int main()
 {
     int n = 0;

project_euler/problem_1/sol3.c

@@ -1,14 +1,16 @@
-/*
-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 N.
-'''
-'''
-This solution is based on the pattern that the successive numbers in the series
-follow: 0+3,+2,+1,+3,+1,+2,+3.
-*/
+/**
+ * \file
+ * \brief [Problem 1](https://projecteuler.net/problem=1) solution.
+ * This solution is based on the pattern that the successive numbers in the
+ * series follow: 0+3,+2,+1,+3,+1,+2,+3.
+ *
+ * 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 N.
+ */
 #include <stdio.h>
 
+/** Main function */
 int main()
 {
     int n = 0;
@@ -49,4 +51,5 @@ int main()
     }
 
     printf("%d\n", sum);
+    return 0;
 }

project_euler/problem_1/sol4.c

@@ -1,8 +1,14 @@
-/*An Efficient code to print all the sum of all numbers that are multiples of 3
- * & 5 below N.*/
+/**
+ * \file
+ * \brief [Problem 1](https://projecteuler.net/problem=1) solution
+ *
+ * An Efficient code to print all the sum of all numbers that are multiples of 3
+ * & 5 below N.
+ */
 
 #include <stdio.h>
 
+/** Main function */
 int main()
 {
     int t;

project_euler/problem_10/sol1.c

@@ -1,19 +1,25 @@
+/**
+ * \file
+ * \brief [Problem 10](https://projecteuler.net/problem=10) solution
+ */
 #include <math.h>
 #include <stdio.h>
 #include <stdlib.h>
 
-char is_prime(long n)
+/** Function to check if a number is prime */
+char is_prime(unsigned long n)
 {
-    for (long i = 2; i < sqrtl(n) + 1; i++)
+    for (unsigned long i = 2; i < sqrtl(n) + 1; i++)
         if (n % i == 0)
             return 0;
 
     return 1;
 }
 
-long long sum_of_primes(long N)
+/** Computes sum of prime numbers less than N */
+unsigned long long sum_of_primes(unsigned long N)
 {
-    long long sum = 2;
+    unsigned long long sum = 2;
 
     for (long i = 3; i < N; i += 2) /* skip even numbers */
         if (is_prime(i))
@@ -22,14 +28,15 @@ long long sum_of_primes(long N)
     return sum;
 }
 
+/** Main function */
 int main(int argc, char *argv[])
 {
-    long n = 100;
+    unsigned long n = 100;
 
     if (argc == 2)         /* if command line argument is provided */
         n = atol(argv[1]); /* use that as the upper limit */
 
-    printf("%ld: %lld\n", n, sum_of_primes(n));
+    printf("%ld: %llu\n", n, sum_of_primes(n));
 
     return 0;
 }

project_euler/problem_10/sol2.c

@@ -1,7 +1,12 @@
+/**
+ * \file
+ * \brief [Problem 10](https://projecteuler.net/problem=10) solution
+ */
 #include <math.h>
 #include <stdio.h>
 #include <stdlib.h>
 
+/** Main function */
 int main(int argc, char *argv[])
 {
     long n = 100;

project_euler/problem_12/sol1.c

@@ -1,15 +1,21 @@
+/**
+ * \file
+ * \brief [Problem 11](https://projecteuler.net/problem=11) solution
+ */
 #include <math.h>
 #include <stdio.h>
 #include <stdlib.h>
 
+/**
+ * Get number of divisors of a given number
+ *
+ * If \f$x = a \times b\f$, then both \f$a\f$ and \f$b\f$ are divisors of
+ * \f$x\f$. Since multiplication is commutative, we only need to search till a
+ * maximum of \f$a=b = a^2\f$ i.e., till \f$\sqrt{x}\f$. At every integer till
+ * then, there are eaxctly 2 divisors and at \f$a=b\f$, there is only one
+ * divisor.
+ */
 long count_divisors(long long n)
-/*
-    If x = a * b, then both a and b are divisors of x.
-    Since multiplication is commutative, we only need to search
-    till a maximum of a=b = a^2 i.e., till sqrt(x).
-    At every integer till then, there are eaxctly 2 divisors
-    and at a=b, there is only one divisor.
-*/
 {
     long num_divisors = 0;
 
@@ -22,6 +28,7 @@ long count_divisors(long long n)
     return num_divisors;
 }
 
+/** Main function */
 int main(int argc, char **argv)
 {
     int MAX_DIVISORS = 500;
@@ -44,4 +51,4 @@ int main(int argc, char **argv)
            MAX_DIVISORS, triangle_number);
 
     return 0;
-}
+}

project_euler/problem_2/so1.c

@@ -1,13 +1,19 @@
-/*
-Problem:
-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 n,
-find the sum of the even-valued terms. e.g. for n=10, we have {2,8}, sum is 10.
-*/
+/**
+ * \file
+ * \brief [Problem 2](https://projecteuler.net/problem=2) solution
+ *
+ * Problem:
+ *
+ * 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
+ * n, find the sum of the even-valued terms. e.g. for n=10, we have {2,8}, sum
+ * is 10.
+ */
 #include <stdio.h>
 
+/** Main function */
 int main()
 {
     int n = 0;
@@ -27,4 +33,5 @@ int main()
     }
 
     printf("%d\n", sum);
+    return 0;
 }

project_euler/problem_3/sol1.c

@@ -1,13 +1,18 @@
-/*
-Problem:
-The prime factors of 13195 are 5,7,13 and 29. What is the largest prime factor
-of a given number N? e.g. for 10, largest prime factor = 5. For 17, largest
-prime factor = 17.
-*/
+/**
+ * \file
+ * \brief [Problem 3](https://projecteuler.net/problem=3) solution
+ *
+ * Problem:
+ *
+ * The prime factors of 13195 are 5,7,13 and 29. What is the largest prime
+ * factor of a given number N? e.g. for 10, largest prime factor = 5. For 17,
+ * largest prime factor = 17.
+ */
 #include <math.h>
 #include <stdio.h>
 
-int isprime(int no)
+/** Check if the given number is prime */
+char isprime(int no)
 {
     int sq;
 
@@ -30,6 +35,7 @@ int isprime(int no)
     return 1;
 }
 
+/** Main function */
 int main()
 {
     int maxNumber = 0;
@@ -69,4 +75,5 @@ int main()
             printf("%d\n", maxNumber);
         }
     }
+    return 0;
 }

project_euler/problem_3/sol2.c

@@ -1,11 +1,16 @@
-/*
-Problem:
-The prime factors of 13195 are 5,7,13 and 29. What is the largest prime factor
-of a given number N? e.g. for 10, largest prime factor = 5. For 17, largest
-prime factor = 17.
-*/
+/**
+ * \file
+ * \brief [Problem 3](https://projecteuler.net/problem=3) solution
+ *
+ * Problem:
+ *
+ * The prime factors of 13195 are 5,7,13 and 29. What is the largest prime
+ * factor of a given number N? e.g. for 10, largest prime factor = 5. For 17,
+ * largest prime factor = 17.
+ */
 #include <stdio.h>
 
+/** Main function */
 int main()
 {
     int n = 0;
@@ -24,4 +29,5 @@ int main()
     if (n > 1)
         prime = n;
     printf("%d\n", prime);
+    return 0;
 }

project_euler/problem_4/sol.c

@@ -1,5 +1,14 @@
+/**
+ * \file
+ * \brief [Problem 4](https://projecteuler.net/problem=4) solution
+ */
 #include <stdio.h>
 
+/** Check if number is palindromic
+ * \param[in] n number to check
+ * \returns 1 if palindromic
+ * \returns 0 if not palindromic
+ */
 int is_palindromic(unsigned int n)
 {
     unsigned int reversed = 0, t = n;
@@ -12,6 +21,7 @@ int is_palindromic(unsigned int n)
     return reversed == n;
 }
 
+/** Main function */
 int main(void)
 {
     unsigned int i, j, max = 0;

project_euler/problem_401/sol1.c

@@ -1,3 +1,9 @@
+/**
+ * \file
+ * \brief [Problem 401](https://projecteuler.net/problem=401) solution
+ *
+ * Sum of squares of divisors
+ */
 #include <stdint.h>
 #include <stdio.h>
 #include <stdlib.h>
@@ -8,9 +14,17 @@
 #include <omp.h>
 #endif
 
-#define MOD (uint64_t)1e9
-#define MAX_L 5000
+#define MOD (uint64_t)1e9 /**< modulo limit */
+#define MAX_L 5000        /**< chunk size of array allocation */
 
+/**
+ * Check if a number is present in given array
+ * \param[in] N number to check
+ * \param[in] D array to check
+ * \param[in] L length of array
+ * \returns 1 if present
+ * \returns 0 if absent
+ */
 char is_in(uint64_t N, uint64_t *D, uint64_t L)
 {
     uint64_t i;
@@ -20,6 +34,12 @@ char is_in(uint64_t N, uint64_t *D, uint64_t L)
     return 0;
 }
 
+/**
+ * Get all integer divisors of a number
+ * \param[in] N number to find divisors for
+ * \param[out] D array to store divisors in
+ * \returns number of divisors found
+ */
 uint64_t get_divisors(uint64_t N, uint64_t *D)
 {
     uint64_t q, r;
@@ -31,43 +51,49 @@ uint64_t get_divisors(uint64_t N, uint64_t *D)
         return 1;
     }
 
+    // search till sqrt(N)
+    // because after this, the pair of divisors will repeat themselves
     for (i = 1; i * i <= N + 1; i++)
     {
-        r = N % i;
+        r = N % i; // get reminder
 
+        // reminder = 0 if 'i' is a divisor of 'N'
         if (r == 0)
         {
             q = N / i;
-            if (!is_in(i, D, num))
+            if (!is_in(i, D, num)) // if divisor was already stored
             {
                 D[num] = i;
                 num++;
             }
-            if (!is_in(q, D, num))
+            if (!is_in(q, D, num)) // if divisor was already stored
             {
                 D[num] = q;
                 num++;
             }
         }
-        if (num == MAX_L)
+
+        if (num == MAX_L) // limit of array reached, allocate more space
             D = (uint64_t *)realloc(D, MAX_L * sizeof(uint64_t) << 1);
     }
     return num;
 }
 
 /**
- * sum of squares of all integer factors
- **/
+ * compute sum of squares of all integer factors of a number
+ * \param[in] N
+ * \returns sum of squares
+ */
 uint64_t sigma2(uint64_t N)
 {
-    uint64_t sum = 0, DD, L;
+    uint64_t sum = 0, L;
     int64_t i;
     uint64_t *D = (uint64_t *)malloc(MAX_L * sizeof(uint64_t));
 
     L = get_divisors(N, D);
     for (i = 1; i < L; i++)
     {
-        DD = (D[i] * D[i]) % MOD;
+        uint64_t DD = (D[i] * D[i]) % MOD;
         sum += DD;
     }
 
@@ -78,13 +104,14 @@ uint64_t sigma2(uint64_t N)
 /**
  * sum of squares of factors of numbers
  * from 1 thru N
- **/
+ */
 uint64_t sigma(uint64_t N)
 {
     uint64_t s, sum = 0;
     int64_t i;
 
 #ifdef _OPENMP
+// parallelize on threads
 #pragma omp parallel for reduction(+ : sum)
 #endif
     for (i = 0; i <= N; i++)
@@ -95,15 +122,17 @@ uint64_t sigma(uint64_t N)
     return sum % MOD;
 }
 
+/** Main function */
 int main(int argc, char **argv)
 {
     uint64_t N = 1000;
 
     if (argc == 2)
         N = strtoll(argv[1], NULL, 10);
-    else if (argc > 2)
+    else if (N > 2)
     {
         fprintf(stderr, "Wrong number of input arguments!\n");
+        printf("Usage:\t ./sol1.c [N=1000]");
         return -1;
     }
 

project_euler/problem_5/sol.c

@@ -1,5 +1,13 @@
+/**
+ * \file
+ * \brief [Problem 5](https://projecteuler.net/problem=5) solution
+ */
 #include <stdio.h>
 
+/** Compute [Greatest Common Divisor
+ * (GCD)](https://en.wikipedia.org/wiki/Greatest_common_divisor) of two numbers
+ * using Euclids algorithm
+ */
 unsigned long gcd(unsigned long a, unsigned long b)
 {
     unsigned long r;
@@ -17,12 +25,16 @@ unsigned long gcd(unsigned long a, unsigned long b)
     return b;
 }
 
+/** Compute [Least Common Multiple
+ * (LCM)](https://en.wikipedia.org/wiki/Least_common_multiple) of two numbers
+ */
 unsigned long lcm(unsigned long a, unsigned long b)
 {
     unsigned long long p = (unsigned long long)a * b;
     return p / gcd(a, b);
 }
 
+/** Main function */
 int main(void)
 {
     unsigned long ans = 1;

project_euler/problem_6/sol.c

@@ -1,5 +1,10 @@
+/**
+ * \file
+ * \brief [Problem 6](https://projecteuler.net/problem=6) solution
+ */
 #include <stdio.h>
 
+/** Main function */
 int main(void)
 {
     unsigned s1 = 0, s2 = 0, i;

project_euler/problem_7/sol.c

@@ -1,6 +1,11 @@
+/**
+ * \file
+ * \brief [Problem 7](https://projecteuler.net/problem=7) solution
+ */
 #include <stdio.h>
 #include <stdlib.h>
 
+/** Main function */
 int main(void)
 {
     char *sieve;
@@ -9,7 +14,7 @@ int main(void)
     size_t n = 1000000;
     const unsigned target = 10001;
 
-    sieve = calloc(n, sizeof *sieve);
+    sieve = (char *)calloc(n, sizeof(char));
     for (i = 2; i < n; i++)
     {
         if (!sieve[i])

project_euler/problem_8/sol1.c

@@ -1,6 +1,17 @@
+/**
+ * \file
+ * \brief [Problem 8](https://projecteuler.net/problem=8) solution
+ */
 #include <stdio.h>
 #include <stdlib.h>
 
+/** Compute the product of two numbers in a file
+ *
+ * \param[in] fp pointer to file that is already open
+ * \param[in] start_pos line number of the first numer
+ * \param[in] num_digits number of digits on the line to multiply
+ * \returns expected product
+ */
 long long int get_product(FILE *fp, long start_pos, int num_digits)
 {
     char ch = ' '; /* temporary variable to store character read from file */
@@ -46,6 +57,7 @@ long long int get_product(FILE *fp, long start_pos, int num_digits)
     return prod;
 }
 
+/** Main function */
 int main(int argc, char *argv[])
 {
     int position = 0;

project_euler/problem_8/sol2.c

@@ -1,7 +1,12 @@
+/**
+ * \file
+ * \brief [Problem 8](https://projecteuler.net/problem=8) solution
+ */
 #include <stdio.h>
 #include <stdlib.h>
 #include <string.h> /* for memmove */
 
+/** Main function */
 int main(int argc, char *argv[])
 {
     int position = 0, num_bad_chars = 0;

project_euler/problem_9/sol1.c

@@ -1,5 +1,11 @@
+/**
+ * \file
+ * \brief [Problem 9](https://projecteuler.net/problem=9) solution - A naive
+ * implementation
+ */
 #include <stdio.h>
 
+/** Main function */
 int main(void)
 {
     for (int a = 1; a < 300; a++)

project_euler/problem_9/sol2.c

@@ -1,19 +1,24 @@
+/**
+ * \file
+ * \brief [Problem 9](https://projecteuler.net/problem=9) solution
+ *
+ Problem Statement:
+    A Pythagorean triplet is a set of three natural numbers, \f$a < b < c\f$,
+ for which, \f$a^2 + b^2 = c^2\f$. For example, \f$3^2 + 4^2 = 9 + 16 = 25 =
+ 5^2\f$. There exists exactly one Pythagorean triplet for which \f$a + b + c =
+ 1000\f$. Find the product abc.
+
+
+    Given \f$a^2 + b^2 = c^2\f$ and \f$a+b+c = n\f$, we can write:
+    \f{eqnarray*}{
+        b &=& \frac{n^2 - 2an}{2n - 2a}\\
+        c &=& n - a - b
+    \f}
+ */
 #include <stdio.h>
 #include <stdlib.h>
 
-/**
- Problem Statement:
-    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.
-
-
-    Given a^2 + b^2 = c^2 and a+b+c = n, we can write:
-        b = (n^2 - 2*a*n) / (2*n - 2*a)
-        c = n - a - b
- **/
-
+/** Main function */
 int main(void)
 {
     int N = 1000;