interviews/training->training

puzzles/training/goldbachs_conjecture.cpp

@@ -0,0 +1,120 @@
+#include <vector>
+#include <unordered_set>
+#include <iostream>
+#include <cmath>
+#include <algorithm>
+#include <chrono>
+
+/*
+ Given an even number ( greater than 2 ), return two prime numbers whose 
+ sum will be equal to given number.
+
+NOTE A solution will always exist. read Goldbach’s conjecture
+
+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;
+            }
+        }
+    }
+
+    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;
+}
+
+std::vector<int> primesum(int A)
+{
+
+    for ( int i = 2; i < A; ++i) {
+        if ( is_prime(i) && is_prime(A-i) ){
+            return {i,A-i};
+        }
+    }
+    return {};
+}
+
+#endif
+
+void solve_and_print(int A){
+    auto r = primesum(A);
+    std::cout << r[0] << ' ' << r[1] << std::endl;
+}
+
+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;
+}