蒙特卡洛樹 2017 EC-Final L.SOS

最近看AlphaZero論文學習了蒙特卡洛樹，隱約記得很久以前EC-Final上有人說可以利用MTC打表，決定練練手。從構建到完成耗時兩天。

蒙特卡洛樹的學習可以參考：https://blog.csdn.net/ljyt2/article/details/78332802

題目可以參考：?https://vjudge.net/problem/1500546/origin

蒙特卡洛樹主要分為tree_travelsal， node_expand,? rollout，backpropogation 四個步驟. 在本例中分別對應choose，expand， rollout， update四個函數or代碼段， rollout調用wander函數來隨機選取局面。

Choose Node

引入UCB函數來評判

當UCB函數只根據平均值來選擇node的時候很容易陷入局部最優導致局面趨于平局，這是由于每一步只有少量的選擇能到達必勝態，隨機漫步難以發現這種必勝態，故需要使得樹發現更多的新節點。

UCB函數中UCB_C值越大，越注重探索， UCB_C值越小越注重原來的選擇。而C值大就意味著需要更多的迭代才能期望結果更接近最優。C值小就更可能陷入局部最優。這與過擬合非常相似。

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PS：不同于普通的DP算法和枚舉算法，蒙特卡洛樹的優點是可以控制迭代次數或者迭代時間，可以以此來獲得當前的最優解。

代碼不足之處：MCT沒有函數來清空節點內存， ans（）對最優節點的選取應該不利用UCB附加值（可以對其重寫），也許可以參照AlphaZero對MCT的變體來加速

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代碼如下：

/* author: InFiNiTeemo substract: This program is used for solving 2017 EC-Final Problem L with low effciency for subtle computer to table. many program segments can be optimized as well. It's a tutoral for greenhand to understand MTC */#include<bits/stdc++.h> #include<random> #include<windows.h> using namespace std;mt19937 generator((unsigned)time(NULL)); //#define DEBUG int sum = 0;class Node { private:const int INF = 1e8;const int choose[2] = { 'S', 'O' };const int UCT_C = 4;//evaluationdouble q, avg_q;int c;//transformationvector<Node*> son;//statevector<int> G;int size;Node* father;int turn; // turn & 1 enlarge the value, else ensmall the valuebool leaf;int end = -2; //-2 unjudged, -1 lose, 0 don't know/ draw, 1 win public:Node(int sz) :q(INF), c(0), size(sz), G(vector<int>(sz,0)), leaf(true), turn(0){}Node(vector<int> previous,int f_turn, int place, int kind) : c(0), size(previous.size()),G(previous), turn(f_turn+1), leaf(true){G[place] = kind;avg_q = turn & 1 ? INF : -INF;}void update(double value) {q += value;c++;avg_q = q / c;}double UCT() {if (c == 0) return avg_q;return avg_q + (turn&1?1:-1)* pow(UCT_C*log(sum*1.0)/c,0.5);}void add_edge(Node* son_nd) {son.emplace_back(son_nd);}bool is_leaf() {return leaf;}bool is_full() {return size - turn == 0;}//輸出只可能是1和-1int is_end() {if (end!=-2) return abs(end);else return abs(end = evaluate());} /*@The player on the offensive always choose the high score node, whereas the defensive choose the low score node, the score is based on the UCT()*/Node* choose_node() {try {if (turn - size == 0) {throw "Node.Choose Node: no left grid for node choosing.";}}catch(const char* msg){cerr << msg << endl;} #ifdef DEBUGcout << "--------CHOOSE_NODE---------" << endl;show();cout << "----------------------------" << endl;Sleep(300); #endif // DEBUGNode* t = NULL;for (auto candidate : son) { #ifdef DEBUGcout << "---------------------" << endl;candidate->show();cout << "-----------------------" << endl;Sleep(300); #endif // DEBUGif (t == NULL) t = candidate;else {//checkdouble delta = candidate->UCT() - t->UCT();int _turn = (turn & 1 ? -1 : 1);if (_turn*delta > 0) {t = candidate;}}}return t;}Node* wander() {try {if (turn - size == 0) {throw "Node.Wander Node: no left grid for wandering.";}}catch (const char* msg) {cerr << msg << endl;}int left = size - turn;int nxt = generator()%left+1, kind = generator()%2;for (int i = 0; i < size; i++) {if (G[i] == 0) {nxt--;}if (nxt == 0) return new Node(G, turn, i, choose[kind]);}}void expand_node() {try {if (turn - size == 0) {throw "Node.Expand_Node: no left grid for node expanding.";}}catch (const char* msg) {cerr << msg << endl;}leaf = false;for (int i = 0; i < size; i++) {if (G[i] == 0) {for (int j = 0; j < 2; j++) {Node* son = new Node(G, turn, i, choose[j]); #ifdef DEBUGcout << "-------EXPAND------------" << endl;son->show();cout << "-----------------------------" << endl;Sleep(300); #endifadd_edge(son);}}}}int visit_count() {return c;}int evaluate() {for (int i = 0; i < size - 2; i++) if (G[i] == 'S'&&G[i + 1] == 'O'&&G[i + 2] == 'S') {return (turn & 1 ? 1 : -1);}return 0;}void show() {cout << "Turn: " << turn << endl;for (auto x : G) {if (x == 0) x = '-';cout << setw(4) << char(x);}cout << endl;cout << "value: " <<UCT() << endl;} };class MTC_TREE { private:Node* root;//size of chessboard sizeint size;const int iter = 1e7;//debugqueue<Node*> Rollout_que;public:MTC_TREE(int sz=0):size(sz) {root = new Node(sz);root->expand_node();}void game() {//loopsum = 0;for (int i = 0; i < iter; i++) {single_game();sum++;}//ansans();}void single_game(){//init//chooseNode* p = root;queue<Node*> que;while (!p->is_leaf()) {p = p->choose_node();Node* q = p;que.push(q);}//rolloutNode* q = roll_out(p);//q->show();int x = q->evaluate();while(!Rollout_que.empty()) {Node* t = Rollout_que.front(); Rollout_que.pop();delete t;}//updatewhile (!que.empty()) {Node* t = que.front(); que.pop();if (t->visit_count() == 0 && t->is_leaf() && !t->is_full()) {t->expand_node();}t->update(x);}}Node* roll_out(Node* p) {while (!(p->is_full() || p->is_end())) {p = p->wander();Rollout_que.push(p);}return p;}void ans() {Node* p = root;while (!p->is_leaf()) {p = p->choose_node();}//cout << (p->is_leaf()?"true":"false") << endl;p->show();string result[] = { "B win", "Draw", "A win" };cout << result[p->evaluate()+1] << endl; } };int main() {for (int i = 3; i <= 100; i++) {MTC_TREE tree(i);cout << i << ": ";tree.game();}system("pause"); }

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