Started on 061, did a lot of stuff that was really really slow, now try to use bit masking

061/main.cpp

@@ -0,0 +1,135 @@
+/*
+Triangle, square, pentagonal, hexagonal, heptagonal, and octagonal numbers are all figurate (polygonal) numbers and are generated by the following formulae:
+
+Triangle	 	P3,n=n(n+1)/2	 	1, 3, 6, 10, 15, ...
+Square	 	P4,n=n2	 	1, 4, 9, 16, 25, ...
+Pentagonal	 	P5,n=n(3n−1)/2	 	1, 5, 12, 22, 35, ...
+Hexagonal	 	P6,n=n(2n−1)	 	1, 6, 15, 28, 45, ...
+Heptagonal	 	P7,n=n(5n−3)/2	 	1, 7, 18, 34, 55, ...
+Octagonal	 	P8,n=n(3n−2)	 	1, 8, 21, 40, 65, ...
+The ordered set of three 4-digit numbers: 8128, 2882, 8281, has three interesting properties.
+
+The set is cyclic, in that the last two digits of each number is the first two digits of the next number (including the last number with the first).
+Each polygonal type: triangle (P3,127=8128), square (P4,91=8281), and pentagonal (P5,44=2882), is represented by a different number in the set.
+This is the only set of 4-digit numbers with this property.
+Find the sum of the only ordered set of six cyclic 4-digit numbers for which each polygonal type: triangle, square, pentagonal, hexagonal, heptagonal, and octagonal, is represented by a different number in the set.
+*/
+
+#include <iostream>
+#include <chrono>
+#include <vector>
+#include <algorithm>
+#include <map>
+
+using namespace std;
+
+uint triangle(uint n){
+    return n * (n + 1) / 2;
+}
+
+uint square(uint n){
+    return n * n;
+}
+
+uint pentagonal(uint n){
+    return n * (3 * n - 1) / 2;
+}
+
+uint hexagonal(uint n){
+    return n * (2 * n - 1);
+}
+
+uint heptagonal(uint n){
+    return n * (5 * n - 3) / 2;
+}
+
+uint octogonal(uint n){
+    return n * (3 * n - 2);
+}
+
+bool validEnd(uint n){
+    return n % 100 >= 10;
+}
+
+void fillVector(std::vector<uint> & toFill, const uint bitmask, uint (*calc)(uint n)){
+    uint value = calc(0);
+
+    for(uint i = 0; value < 10000; ++i){
+        if(value >= 1000 && validEnd(value)){
+            toFill[value] |= bitmask;
+        }
+        value = calc(i + 1);
+    }
+}
+
+// Non-recursive version of Heap's algorithm for permutations
+// Source: https://en.wikipedia.org/wiki/Heap%27s_algorithm
+vector<vector<uint>> permutations(uint n, vector<uint> & A){
+    vector<vector<uint>> result;
+    vector<uint> c(n, 0);
+
+    result.push_back(A);
+
+    uint i = 1;
+    while(i < n){
+        if(c[i] < i){
+            if(i % 2 == 0){
+                swap(A[0], A[i]);
+            } else{
+                swap(A[c[i]], A[i]);
+            }
+            result.push_back(A);
+            c[i] += 1;
+            i = 1;
+        } else{
+            c[i] = 0;
+            i++;
+        }
+    }
+    return result;
+}
+
+bool checkCyclicness(const vector<uint> & v){
+    for(int i = 0; i < 5; ++i){
+        uint end = v[i] % 100;
+        uint start = v[i + 1] / 100;
+
+        if(end != start){
+            return false;
+        }
+    }
+    if(v[5] % 100 != v[0] / 100){
+        return false;
+    }
+
+    return true;
+}
+
+void search(vector<uint> & sequence, uint mask = 0){
+    
+}
+
+int main(){
+
+    cout << "Hello this is Patrick" << endl;
+    auto start = chrono::high_resolution_clock::now();
+
+    // creating big vector with the numeric values and their corresponding mask
+    uint finalMask = 0b111111000;
+
+    // Finally learned what exactly it means when 1 << n, which is nice
+    vector<uint> all(10000, 0);
+    fillVector(all, 1 << 3, triangle);
+    fillVector(all, 1 << 4, square);
+    fillVector(all, 1 << 5, pentagonal);
+    fillVector(all, 1 << 6, hexagonal);
+    fillVector(all, 1 << 7, heptagonal);
+    fillVector(all, 1 << 8, octogonal);
+    
+
+
+    cout << "Cycle not found" << endl;
+    auto duration = chrono::duration_cast<chrono::milliseconds>(chrono::high_resolution_clock::now() - start);
+    cout << (float)duration.count()/1000 << endl;
+    return 0;
+}