Two solutions for Goldbach's conjecture and their complexity.

puzzles/interviews/training/goldbachs_conjecture.cpp

@@ -3,68 +3,94 @@
 #include <iostream>
 #include <cmath>
 #include <algorithm>
+#include <chrono>
 
-using namespace std;
+/*
+ Given an even number ( greater than 2 ), return two prime numbers whose 
+ sum will be equal to given number.
 
-static std::vector<int> GetPrimes(const size_t n) {
-    std::vector<int> primes;
-    std::vector<int> lp(n + 1, 0);
+NOTE A solution will always exist. read Goldbach’s conjecture
 
-for (auto i = 2u; i <= n; ++i) {
-    if (lp[i] == 0) {
-        lp[i] = i;
-        primes.push_back(i);
+Example:
+
+
+Input : 4
+Output: 2 + 2 = 4
+
+If there are more than one solutions possible, return the lexicographically 
+smaller solution.
+*/
+
+/*
+ *  My understanding is:
+ *  The first solution time complexity is O(n^2/log_n) + O(n) memory complexity.
+ *  The second solution id O(n^2) complexity + O(0) memory complexity.
+ */
+#define SOLUTION 1
+
+#if SOLUTION == 1
+
+std::vector<size_t> primes;
+std::unordered_set<size_t> primes_set;
+
+void init_primes(const size_t n)
+{
+
+    std::vector<bool> prime_table(n/2+1, true); // Only odd numbers.
+    primes.push_back(2);
+
+    for (auto p = 3; p <= n; p+=2) {
+        if ( prime_table[p/2] ) {
+            primes.push_back(p);
+            for ( int j = p*3; j <= n; j+=p*2 ) {
+                prime_table[j/2] = false;
             }
-    for (auto j = 0u; j < primes.size() && primes[j] <= lp[i] && i * primes[j] <= n; ++j)
-        lp[i * primes[j]] = primes[j];
         }
-return primes;
-
     }
 
-inline bool is_prime( int n, const vector<int>& primes ){
-    int sqrt_n = floor(sqrt(double(n)));
-    for ( int i = 0; i < primes.size() && primes[i] <= sqrt_n; ++i ){
-        if ( n % primes[i] == 0 ) {
+    primes_set.insert(primes.begin(), primes.end());
+}
+
+std::vector<int> primesum(int A)
+{
+    for ( int p : primes ){
+        if ( primes_set.find(A-p) != primes_set.end() ) {
+            return {p, A-p};
+        }
+    }
+
+    return {}; 
+}
+
+#elif SOLUTION == 2
+
+bool is_prime(int a)
+{
+    if ( a <2 )
+        return false;
+
+    int end = std::floor(std::sqrt(a));
+    for ( int i = 2; i <= end; ++i ){
+        if ( a % i == 0) {
             return false;
         }
     }
     return true;
 }
 
-vector<int> primesum(int A) {
+std::vector<int> primesum(int A)
+{
 
-    vector<int> primes;
-
-#if 0
     for ( int i = 2; i < A; ++i) {
-        if ( is_prime(i,primes) ){
-            primes.push_back(i);
+        if ( is_prime(i) && is_prime(A-i) ){
+            return {i,A-i};
         }
     }
-#endif
-
-    primes = GetPrimes(A);
-
-#if 1
-    unordered_set<int> primes_set(primes.begin(),primes.end());
-    for ( int i = 0; i < primes.size(); ++i ){
-        if ( primes_set.find(A-primes[i]) != primes_set.end() ) {
-            return {primes[i], A-primes[i]};
-        }
-    }
-#else
-    for ( int i = 0; i < primes.size(); ++i ) {
-        auto it = std::lower_bound(primes.begin(), primes.end(), A - primes[i]);
-        if ( it != primes.end() && *it == A-primes[i] ) {
-            return {primes[i], A-primes[i]};
-        }
-    }
-#endif
-
     return {};
 }
 
+#endif
+
 void solve_and_print(int A){
     auto r = primesum(A);
     std::cout << r[0] << ' ' << r[1] << std::endl;
@@ -72,8 +98,23 @@ void solve_and_print(int A){
 
 int main()
 {
+    auto begin = std::chrono::high_resolution_clock::now();
+
+#if SOLUTION == 1
+    init_primes(16777218);
+#endif
+
     solve_and_print(4);
     solve_and_print(10);
 
     solve_and_print(16777214);
+    solve_and_print(16777218);
+
+    for ( int i = 4; i < 1000000; i+=2 ){
+        primesum(i);
+    }
+
+    auto end = std::chrono::high_resolution_clock::now();
+    std::chrono::duration<double> seconds = end - begin;
+    std::cout << "Time: " << seconds.count() << std::endl;
 }