### 计算机代写|蒙特卡洛树搜索代写Monte Carlo tree search代考|Math577

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• Foundations of Data Science 数据科学基础

## 计算机代写|蒙特卡洛树搜索代写Monte Carlo tree search代考|Monte Carlo Tree Search

Theorem 2. Under the assumptions of Theorem 1,
$$\left|\mathbb{E}\left[\bar{X}_n\right]-\mu^\right| \leq\left|\delta_n^\right|+O\left(\frac{K\left(C_p^2 \ln n+N_0\right)}{n}\right)$$
where $N_0=N_0(1 / 2)$.
To provide further intuition for the above, recall that $\delta_n^*$ is the distance between the optimal arm’s mean at time $t=n$, and the true optimal mean. Furthermore, the theorem is proven using the value $N_0(\epsilon)$ where $\epsilon=1 / 2$.

Furthermore, using Theorem 1, we can derive a lower bound on the number of times some arm $i$ will be played. This result follows:

Theorem 3. Under the assumptions of Theorem 3, there exists some positive constant $\rho$ such that for all arms $i$ and $n, T_i(n) \geq\lceil\rho \log (n)\rceil$.

We need the above result to prove that the optimal reward converges quickly to its true mean in the drifting multi armed bandit problem. We omit the proof of the result, but it is natural that it would require both the lower bound on $T_i(n)$ and the upper bound on $\mathbb{E}\left[T_i(n)\right]$. This quick convergence is one of the two most important properties for incorporating UCB into MCTS. If the drifting means converge slowly then we cannot hope to quickly find an optimal arm. The bound follows:
Theorem 4. Fix an arbitrary $\delta>0$ and let $\Delta_n=9 \sqrt{2 n \ln 2 / \delta}$. Let $n_0$ be such that
$$\sqrt{n_0} \geq O\left(K\left(C_p^2 \ln n_0+N_0(1 / 2)\right)\right)$$
Then for any $n \geq n_0$, under the assumptions of Theorem 1 the following bounds hold true:
\begin{aligned} & \mathbb{P}\left(n \bar{X}n \geq n \mathbb{E}\left[\bar{X}_n\right]+\Delta_n\right) \leq \delta \ & \mathbb{P}\left(n \bar{X}_n \leq n \mathbb{E}\left[\bar{X}_n\right]-\Delta_n\right) \leq \delta \end{aligned} Another extremely important result for using UCB in MCTS is that when the means can drift, UCB still finds the best arm when given infinite time. This result follows: Theorem 5. Under the assumptions of Theorem 1 it holds that $$\lim {t \rightarrow \infty} P\left(I_t \neq i^*\right)=0$$

For more detail on any of the theorems discussed in this section, please refer to [5]. In order to understand UCT, we must first relax the restraints on $K$-armed bandits to allow each arm’s mean $\mu_i$ to change over time t. While we can no longer rely on the assumption that the mean of each arm is fixed from $t=1$ onward, we can assume that the expected value of the empirical averages converge. We let $\bar{X}{i, n}=\frac{1}{n} \sum{t=1}^n X_{i, t}$ be the empirical average of arm $i$ at time $n$, and $\mu_{i, n}=\mathbb{E}\left[\tilde{X}{i, n}\right]$ be its expected value. Therefore, we now have a sequence of expected means for each arm $i$, namely $\mu{i, n}$. We assume that these expected means eventually converge to one final mean $\mu_i=\lim {n \rightarrow \infty} \mu{i, n}$. We further define a sequence of offsets for each arm as $\delta_{i, n}=\mu_{i, n}-\mu_i$. We also make the following assumptions about the reward sequence:

Assumption 1. Fix $1 \leq i \leq K$. Let $\left{\mathcal{F}{i, t}\right}_t$ be a fultration such that $\left{X{i, t}\right}_t$ is $\left{\mathcal{F}{i, t}\right}$-adapted and $X{i, t}$ is conditionally independent of $\mathcal{F}{i, t+1}, \mathcal{F}{i, t+2}, \ldots$ given $\mathcal{F}{i, t-1}{ }^1$. Then $0 \leq X{i, t} \leq 1$ and the limit of $\mu_{i, n}=\mathbb{E}\left[\bar{X}{i n}\right]$ exists. Further, we assume that there exist a constant $C_p>0$ and an integer $N_p$ such that for $n \geq N_p$. for any $\delta>0, \Delta_n(\delta)=C_p \sqrt{n \ln (1 / \delta)}$, the following bounds hold: \begin{aligned} & \mathbb{P}\left(n \bar{X}{i, n} \geq n \mathbb{E}\left[\bar{X}{i, n}\right]+\Delta_n(\delta)\right) \leq \delta \ & \mathbb{P}\left(n \bar{X}{i, n} \leq n \mathbb{E}\left[\bar{X}{i, n}\right]-\Delta_n(\delta)\right) \leq \delta \end{aligned} This assumption allows us to define a bias sequence $c{t, s}$ for time $t$ and sample size $s$ which satisfies Equation 3 and Equation 4. This sequence is as follows:
$$c_{t, s}=2 C_p \sqrt{\frac{\ln t}{s}}$$
We define $\Delta_i=\mu^-\mu_i$ to be the loss of arm $i$. Recall that since the expected mean of each arm converges, the mean offset $\delta_{i, t}$ converges to zero. Therefore, there exists a time $N_0(\epsilon)$ at which the uncertainty of the true mean rewards are guaranteed to be within a factor $\epsilon$ of their distance from the optimal mean, and the uncertainty of the optimal mean is guaranteed to be within the same factor $\epsilon$ of its distance to to closest suboptimal mean. Therefore, even though we still have some uncertainty as to what the true means really are, we have enough information to know which is probably the best, as $\mu_{N_0(c)}$ is closer to $\mu^$ than it is to any $\mu_{i, N_0(\epsilon)}$. More formally, $N_0(\epsilon): \mathbb{R} \rightarrow \mathbb{N}$ is a function which returns the minimum $t$ for which $2\left|\delta_{i, t}\right| \leq \epsilon \Delta_i$ for all arms $i$, and $2\left|\delta_{j^*, t}\right| \leq \epsilon \min _i \Delta_i$. Under Assumption 1, and using the preceding definitions, we can upper bound the expected number of times that a suboptimal arm will be played by UCB1 when the means are allowed to drift.

## 计算机代写|蒙特卡洛树搜索代写Monte Carlo tree search代考|UCT

UCT is simply a marriage of MCTS and UCB1. The main idea is to treat each internal node in MCTS as a $K$-armed bandit, where the arms are the actions available at that state. $\mathrm{A}$ separate instance of UCB1 is run on each internal state, modified to accommodate the drifting means present in this problem. The necessity of the drifting means generalization can be seen by realizing that UCT is essentially a tree of separate UCB1 instances. The means of the action rewards in each of these multi-armed bandit problems depend on the instances lower in the tree. As we explore the tree, we gain a clearer understanding of the true means, reducing noisiness introduced by a partial exploration deeper in the tree. Simply put, UCT is MCTS where we use UCB1 to select an action in the tree policy. See Algorithm 2 for pseudo code of the action selection.
Recall Theorems 4 and 5 from the previous section. These state that when means can drift in the multi armed bandit problem, UCB still finds the optimal solution quickly with high probability, and given enough time it always finds the optinal arn. When incorporating UCB into MCTS, we also get these properties in UCT, as stated in Theorem 6 .

Theorem 6. Consider algorithm UCT running on a game tree of depth $D$, branching factor $K$ with stochastic payoffs at the leaves. Assume that the payoffs lie in the interval [0,1]. Then the bias of the estimated expected payoff, $\bar{X}_n$, is $O\left(\left(K D \log (n)+K^D\right) / n\right)$. Further, the failure probability at the root converges to zero as the number of samples grows to infinity.

A detailed proof of the preceding theorem is beyond the scope of these lecture notes (see [5] for such a proof), however we will provide a sketch of the proof. We must induct on $D$ to prove the theorem. In the base case of $D=1$, UCT is reduced to a single instantiation of UCB. Therefore Theorems 4 and 5 lead directly to our desired result. In the inductive case, we assume the result holds for any tree of depth $D$ – 1 , and consider a tree of depth $D$. All of the children of the root node are trees of depth $D-1$ which, by induction, satisfy the theorem. Therefore, we consider only the single UCB instance at the root. Using the theorems from the previous section (particularly Theorems 2,4 , and 5 ), we can then show that the theorem holds at the root as well.

# 蒙特卡洛树搜索代考

## 计算机代写|蒙特卡洛树搜索代写Monte Carlo tree search代考|Monte Carlo Tree Search

Veft{\mathca|{F}{i, t}\right } } \text { -改编和 } X i , t \text { 有条件地独立于 } \mathcal { F } i , t + 1 , \mathcal { F } i , t + 2 , \ldots \text { 给予 } \mathcal { F } i , t – 1 ^ { 1 } \text { . 然后 } $0 \leq X i, t \leq 1$ 和极限 $\mu_{i, n}=\mathbb{E}[\bar{X} i n]$ 存在。此外，我们假设存在一个常数 $C_p>0$ 和一个整数 $N_p$ 这样 对于 $n \geq N_p$. 对于任何 $\delta>0, \Delta_n(\delta)=C_p \sqrt{n \ln (1 / \delta)}$ ，以下界限成立:
$$\mathbb{P}\left(n \bar{X} i, n \geq n \mathbb{E}[\bar{X} i, n]+\Delta_n(\delta)\right) \leq \delta \quad \mathbb{P}\left(n \bar{X} i, n \leq n \mathbb{E}[\bar{X} i, n]-\Delta_n(\delta)\right) \leq \delta$$

$$c_{t, s}=2 C_p \sqrt{\frac{\ln t}{s}}$$

## 计算机代写|蒙特卡洛树搜索代写Monte Carlo tree search代考|UCT

UCT 只是 MCTS 和 UCB1 的结合。主要思想是将 MCTS 中的每个内部节点视为一个 $K$-armed bandit， 其中 arms 是该状态下可用的动作。AUCB1 的单独实例在每个内部状态上运行，经过修改以适应此问题 中存在的漂移方法。通过认识到 UCT 本质上是一棵独立的 UCB1 实例树，可以看出漂移均值泛化的必要 性。在这些多臂老虎机问题中，行动奖励的方式取决于树中较低的实例。当我们探索这棵树时，我们对真 实均值有了更清晰的理解，减少了在树的更深处进行部分探索所引入的噪音。简单地说，UCT 是 MCTS， 我们使用 UCB1 在树策略中选择一个动作。有关动作选择的伪代码，请参见算法 2。

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