From dcf5fa4241ce8d7dc3988b49dc53f8dc01973459 Mon Sep 17 00:00:00 2001
From: Filip Hlasek <fhlasek@gmail.com>
Date: Fri, 24 Jul 2020 19:59:49 -0700
Subject: [PATCH] fix, test: simplification and tests for
 number_of_positive_divisors (#975)

* fix, test: simplification and tests for number_of_positive_divisors

* Further refactor and documentation of number_of_positive_divisors

* Update the comments.

* Update the LaTeX documentation.
---
 math/number_of_positive_divisors.cpp | 92 +++++++++++++++++-----------
 1 file changed, 55 insertions(+), 37 deletions(-)

diff --git a/math/number_of_positive_divisors.cpp b/math/number_of_positive_divisors.cpp
index f157f7b41..9407c4dc8 100644
--- a/math/number_of_positive_divisors.cpp
+++ b/math/number_of_positive_divisors.cpp
@@ -1,74 +1,92 @@
 /**
  * @file
- * @brief C++ Program to calculate number of divisors
+ * @brief C++ Program to calculate the number of positive divisors
  *
- * This algorithm use the prime factorization approach.
- * Any number can be written in multiplication of its prime factors.
- * <br/>Let N = P1^E1 * P2^E2 ... Pk^Ek
- * <br/>Therefore. number-of-divisors(N) = (E1+1) * (E2+1) ... (Ek+1).
- * <br/>Where P1, P2 ... Pk are prime factors and E1, E2 ... Ek are exponents
-respectively.
+ * This algorithm uses the prime factorization approach.
+ * Any positive integer can be written as a product of its prime factors.
+ * <br/>Let \f$N = p_1^{e_1} \times p_2^{e_2} \times\cdots\times p_k^{e_k}\f$
+ * where \f$p_1,\, p_2,\, \dots,\, p_k\f$ are distinct prime factors of \f$N\f$ and
+ * \f$e_1,\, e_2,\, \dots,\, e_k\f$ are respective positive integer exponents.
+ * <br/>Each positive divisor of \f$N\f$ is in the form
+ * \f$p_1^{g_1}\times p_2^{g_2}\times\cdots\times p_k^{g_k}\f$
+ * where \f$0\le g_i\le e_i\f$ are integers for all \f$1\le i\le k\f$.
+ * <br/>Finally, there are \f$(e_1+1) \times (e_2+1)\times\cdots\times (e_k+1)\f$
+ * positive divisors of \f$N\f$ since we can choose every \f$g_i\f$
+ * independently.
  *
- * Example:-
- * <br/>N = 36
- * <br/>36 = (3^2 * 2^2)
- * <br/>number_of_positive_divisors(36) = (2+1) * (2+1) = 9.
+ * Example:
+ * <br/>\f$N = 36 = (3^2 \cdot 2^2)\f$
+ * <br/>\f$\mbox{number_of_positive_divisors}(36) = (2+1) \cdot (2+1) = 9\f$.
  * <br/>list of positive divisors of 36 = 1, 2, 3, 4, 6, 9, 12, 18, 36.
  *
- * Similarly if N is -36 at that time number of positive divisors remain same.
- *
- * Example:-
- * <br/>N = -36
- * <br/>-36 = -1 * (3^2 * 2^2)
- * <br/>number_of_positive_divisors(-36) = (2+1) * (2+1) = 9.
- * <br/>list of positive divisors of -36 = 1, 2, 3, 4, 6, 9, 12, 18, 36.
- *
+ * Similarly, for N = -36 the number of positive divisors remain same.
 **/
 
+#include <cassert>
 #include <iostream>
-#include <vector>
 
 /**
- * Algorithm
+ * Function to compute the number of positive divisors.
+ * @param n number to compute divisors for
+ * @returns number of positive divisors of n (or 1 if n = 0)
  */
 int number_of_positive_divisors(int n) {
-    std::vector<int> prime_exponent_count;
+    if (n < 0) {
+        n = -n; // take the absolute value of n
+    }
+
+    int number_of_divisors = 1;
+
     for (int i = 2; i * i <= n; i++) {
-        int prime_count = 0;
+        // This part is doing the prime factorization.
+        // Note that we cannot find a composite divisor of n unless we would
+        // already previously find the corresponding prime divisor and dvided
+        // n by that prime. Therefore, all the divisors found here will
+        // actually be primes.
+        // The loop terminates early when it is left with a number n which
+        // does not have a divisor smaller or equal to sqrt(n) - that means
+        // the remaining number is a prime itself.
+        int prime_exponent = 0;
         while (n % i == 0) {
-            prime_count += 1;
+            // Repeatedly divide n by the prime divisor n to compute
+            // the exponent (e_i in the algorithm description).
+            prime_exponent++;
             n /= i;
         }
-        if (prime_count != 0) {
-            prime_exponent_count.push_back(prime_count);
-        }
+        number_of_divisors *= prime_exponent + 1;
     }
     if (n > 1) {
-        prime_exponent_count.push_back(1);
+        // In case the remaining number n is a prime number itself
+        // (essentially p_k^1) the final answer is also multiplied by (e_k+1).
+        number_of_divisors *= 2;
     }
 
-    int divisors_count = 1;
+    return number_of_divisors;
+}
 
-    for (int i = 0; i < prime_exponent_count.size(); i++) {
-        divisors_count = divisors_count * (prime_exponent_count[i] + 1);
-    }
-
-    return divisors_count;
+/**
+ * Test implementations
+ */
+void tests() {
+    assert(number_of_positive_divisors(36) == 9);
+    assert(number_of_positive_divisors(-36) == 9);
+    assert(number_of_positive_divisors(1) == 1);
+    assert(number_of_positive_divisors(2011) == 2); // 2011 is a prime
+    assert(number_of_positive_divisors(756) == 24); // 756 = 2^2 * 3^3 * 7
 }
 
 /**
  * Main function
  */
 int main() {
+    tests();
     int n;
     std::cin >> n;
-    if (n < 0) {
-        n = -n;
-    }
     if (n == 0) {
         std::cout << "All non-zero numbers are divisors of 0 !" << std::endl;
     } else {
         std::cout << "Number of positive divisors is : ";
         std::cout << number_of_positive_divisors(n) << std::endl;
     }
+    return 0;
 }