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# CSE510 Lecture 3
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# CSE510 Deep Reinforcement Learning (Lecture 3)
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## Introduction and Definition of MDPs
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content/CSE510/CSE510_L4.md
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# CSE510 Deep Reinforcement Learning (Lecture 4)
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Markov Decision Process (MDP) Part II
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## Recall from last lecture
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An Finite MDP is defined by:
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- A finite set of **states** $s \in S$
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- A finite set of **actions** $a \in A$
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- A **transition function** $T(s, a, s')$
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- Probability that a from s leads to $s'$, i.e.,
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$P(s'| s, a)$
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- Also called the model or the dynamics
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- A **reward function $R(s)$** ( Sometimes $R(s,a)$ or $R(s, a, s')$ )
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- A **start state**
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- Maybe a **terminal state**
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A model for sequential decision making problem under uncertainty
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### Optimal Policy and Bellman Optimality Equation
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The goal for a MDP is to compute or learn an optimal policy.
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- An **optimal policy** is one that achieves the highest value at any state
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$$
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\pi^* = \arg\max_\pi V^\pi(s)
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$$
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- We define the optimal value function using Bellman Optimality Equation
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$$
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V^*(s) = R(s) + \gamma \max_{a\in A} \sum_{s'\in S} P(s'|s,a) V^*(s')
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$$
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- The optimal policy is
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$$
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\pi^*(s) = \arg\max_{a\in A} \sum_{s'\in S} P(s'|s,a) V^*(s')
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$$
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### The Existence of the Optimal Policy
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Theorem: for any Markov Decision Process
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- There exists an optimal policy
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- There can be many optimal policies, but all optimal policies achieve the same optimal value function
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- There is always a deterministic optimal policy for any MDP
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## Solve MDP
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### Value Iteration
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Repeatedly update an estimate of the optimal value function according to Bellman Optimality Equation.
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1. Initialize an estimate for the value function arbitrarily
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$$
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\hat{V}(s) \gets 0, \forall s \in S
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$$
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2. Repeat, update:
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$$
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\hat{V}(s) \gets R(s) + \gamma \max_{a\in A} \sum_{s'\in S} P(s'|s,a) \hat{V}(s'), \forall s \in S
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$$
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<details>
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<summary>Example</summary>
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Suppose we have a robot that can move in a 2D grid. with the following dynamics:
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- with 80% probability, the robot moves in the direction of the action
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- with 10% probability, the robot moves in the direction of the action + 1 (wrap to left)
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- with 10% probability, the robot moves in the direction of the action - 1 (wrap to right)
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The gird ($V^0(s)$) is:
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|0|0|0|1|
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|0|*|0|-100|
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|0|0|0|0|
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If we fun the value iteration with $\gamma = 0.9$, we can update the value function as follows:
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$$
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V^1(s) = R(s) + \gamma \max_{a\in A} \sum_{s'\in S} P(s'|s,a) V^0(s')
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$$
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On point $(3,3)$, the best action is to move to the goal state, so:
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$$
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\begin{aligned}
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V^1((3,3)) &= R((3,3)) + \gamma \max_{a\in A} \sum_{s'\in S} P(s'|(3,3),\text{right}) V^0((3,4))
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&= 0+0.9 \times 0.8 \times 1 = 0.72
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\end{aligned}
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$$
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On point $(3,4)$, the best action is to move up so that you can stay in the grid with $90\%$ probability, so:
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$$
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\begin{aligned}
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V^1((3,4)) &= R((3,4)) + \gamma \max_{a\in A} \sum_{s'\in S} P(s'|(3,4),\text{up}) V^0((3,4))
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&= 1+0.9 \times (0.8+0.1) \times 1 = 1.81
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\end{aligned}
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$$
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On $t=1$, the value on grid is:
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|0|0|0.72|1.81|
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|0|*|0|-99.91|
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|0|0|0|0|
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</details>
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The general algorithm can be written as:
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```python
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# suppose we defined the grid as previous example
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grid = [
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[0, 0, 0, 1],
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[0, '*', 0, -100],
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[0, 0, 0, 0]
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]
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m,n = len(grid), len(grid[0])
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ACTIONS = {'up':(0,-1), 'down':(0,1), 'left':(-1,0), 'right':(1,0)}
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gamma = 0.9
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V = value_iteration(gamma, ACTIONS, grid)
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print(V)
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def get_reward(action, i, j):
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reward = 0
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reward += 0.8 * grid[i+action[0]][j+action[1]] if i+action[0] >= 0 and i+action[0] < m and j+action[1] >= 0 and j+action[1] < n and grid[i+action[0]][j+action[1]] != '*' else grid[i][j]
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reward += 0.1 * grid[i+action[0]][j+action[1]] if i+action[0] >= 0 and i+action[0] < m and j+action[1] >= 0 and j+action[1] < n and grid[i+action[0]][j+action[1]] != '*' else grid[i][j]
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reward += 0.1 * grid[i+action[0]][j+action[1]] if i+action[0] >= 0 and i+action[0] < m and j+action[1] >= 0 and j+action[1] < n and grid[i+action[0]][j+action[1]] != '*' else grid[i][j]
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return reward
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def value_iteration(gamma, ACTIONS, V):
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V_new=[[0]*m for _ in range(n)]
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while True:
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for i in range(m):
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for j in range(n):
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s = (i, j)
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V_new[i][j] = V[i][j] + gamma * max(get_reward(action, i, j) for action.values() in ACTIONS)
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if max(abs(V_new[i][j] - V[i][j]) for i in range(m) for j in range(n)) < 1e-6:
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break
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V = V_new
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return V
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```
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### Convergence of Value Iteration
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Theorem: Value Iteration converges to the optimal value function $\hat{V}\to V^*$ as $t\to\infty$.
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<details>
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<summary>Proof</summary>
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For any estimate of the value function $\hat{V}$, we define the Bellman backup operator $\operatorname{B}:\mathbb{R}^{|S|}\to \mathbb{R}^{|S|}$ by
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$$
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\operatorname{B}(\hat{V}(s)) = R(s) + \gamma \max_{a\in A} \sum_{s'\in S} P(s'|s,a) \hat{V}(s')
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$$
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Note that $\operatorname{B}(V^*) = V^*$.
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Since $\|\max_{x\in X}f(x)-\max_{x\in X}g(x)\|\leq \max_{x\in X}\|f(x)-g(x)\|$, for any value function $V_1$ and $V_2$, we have
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$$
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\begin{aligned}
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|\operatorname{B}(V_1(s))-\operatorname{B}(V_2(s))|&= \gamma \left|\max_{a\in A} \sum_{s'\in S} P(s'|s,a) V_1(s')-\max_{a\in A} \sum_{s'\in S} P(s'|s,a) V_2(s')\right|\\
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&\leq \gamma \max_{a\in A} \left|\sum_{s'\in S} P(s'|s,a) V_1(s')-\sum_{s'\in S} P(s'|s,a) V_2(s')\right|\\
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&\leq \gamma \max_{a\in A} \sum_{s'\in S} P(s'|s,a) |V_1(s')-V_2(s')|\\
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&\leq \gamma \max_{s\in S}|V_1-V_2|
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\end{aligned}
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$$
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</details>
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Assume $0\leq \gamma < 1$, and reward $R(s)$ is bounded by $R_{\max}$.
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Then
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$$
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V^*(s)\leq \sum_{t=0}^\infty \gamma^t R_{\max} = \frac{R_{\max}}{1-\gamma}
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$$
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Let $V^k$ be the value function after $k$ iterations of Value Iteration.
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$$
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\max_{s\in S}|V^k(s)-V^*(s)|\leq \frac{R_{\max}}{1-\gamma}\gamma^k
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$$
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#### Stopping condition
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We can construct the optimal policy arbitrarily close to the optimal value function.
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If $\|V^k-V^{k+1}\|<\epsilon$, then $\|V^k-V^*\|\leq \epsilon\frac{\gamma}{1-\gamma}$.
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So we can select small $\epsilon$ to stop the iteration.
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### Greedy Policy
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Given a $V^k$ that is close to the optimal value $V^*$, the greedy policy is:
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$$
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\pi_{g}(s) = \arg\max_{a\in A} \sum_{s'\in S} T(s',a,s') V^k(s')
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$$
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Here $T(s',a,s')$ is the transition function between state $s'$ and $s$ with action $a$.
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This selects the action looks best if we assume that we get value $V^k$ in one step.
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#### Value of a greedy policy
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If we define $V_g$ to be the value function of the greedy policy, then
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This is not necessarily optimal, but it is a good approximation.
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In homework, we will prove that if $\|V^k-V^*\|<\lambda$, then $\|V_g-V^*\|\leq 2\lambda\frac{\gamma}{1-\gamma}$.
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So we can set stopping condition so that $V_g$ has desired accuracy to $V^*$.
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There is a finite $\epsilon$ such that greedy policy is $\epsilon$-optimal.
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### Problem of Value Iteration and Policy Iteration
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- It is slow $O(|S|^2|A|)$
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- The max action at each state rarely changes
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- The policy converges before the value function
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### Policy Iteration
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Interleaving polity evaluation and policy improvement.
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1. Initialize a random policy $\hat{\pi}$
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2. Compute the value function $V^{\pi}$
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3. Update the policy $\pi$ to be greedy policy with respect to $V^{\pi}$
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$$
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\pi(s)\gets \arg\max_{a\in A} \sum_{s'\in S} P(s'|s,a) V^{\pi}(s')
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$$
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4. Repeat until convergence
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### Exact Policy Evaluation by Linear Solver
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Let $V^{\pi}\in \mathbb{R}^{|S|}$ be a vector of values for each state, $r\in \mathbb{R}^{|S|}$ be a vector of rewards for each state.
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Let $P^{\pi}\in \mathbb{R}^{|S|\times |S|}$ be a transition matrix for the policy $\pi$.
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$$
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P^{\pi}_{ij} = P(s_{t+1}=i|s_t=j,a_t=\pi(s_t))
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$$
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The Bellman equation for the policy can be written in vector form as:
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$$
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\begin{aligned}
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V^{\pi} &= r + \gamma P^{\pi} V^{\pi} \\
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(I-\gamma P^{\pi})V^{\pi} &= r \\
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V^{\pi} &= (I-\gamma P^{\pi})^{-1} r
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\end{aligned}
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$$
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- Proof involves showing that each iteration is also a contraction and monotonically improve the policy
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- Convergence to the exact optimal policy
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- The number of policies is finite
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In real world, policy iteration is usually faster than value iteration.
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#### Policy Iteration Complexity
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- Each iteration runs in polynomial time in the number of states and actions
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- There are at most |A|n policies and PI never repeats a policy
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- So at most an exponential number of iterations
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- Not a very good complexity bound
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- Empirically O(n) iterations are required
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- Challenge: try to generate an MDP that requires more than that n iterations
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### Generalized Policy Iteration
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- Generalized Policy Iteration (GPI): any interleaving of policy evaluation and policy improvement
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- independent of their granularity and other details of the two processes
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### Summary
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#### Policy Iteration vs Value Iteration
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- **PI has two loops**: inner loop (evaluate $V^{\pi}$)
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and outer loop (improve $\pi$)
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- **VI has one loop**: repeatedly apply
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$V^{k+1}(s) = \max_{a\in A} [r(s,a) + \gamma \sum_{s'\in S} P(s'|s,a) V^k(s')]$
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- **Trade-offs**:
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- PI converges in few outer steps if you can evaluate quickly/accurately;
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- VI avoids expensive exact evaluation, doing cheaper but many Bellman optimality updates.
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- **Modified Policy Iteration**: partial evaluation + improvement.
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- **Modified Policy Iteration**: partial evaluation + improvement.
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# CSE5313 Coding and information theory for data science (Lecture 4)
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Algebra over finite fields
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$\mathbb{N}$ is the set of natural numbers.
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by adding additive inverse, we can define the set of integers $\mathbb{Z}$.
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by adding multiplicative inverse, we can define the set of rational numbers $\mathbb{Q}$.
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by adding limits, we can define the set of real numbers $\mathbb{R}$.
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by adding polynomial roots, we can define the set of complex numbers $\mathbb{C}$.
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Another two extensions $\mathbb{H}$ for quaternions and $\mathbb{O}$ for octonions.
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## Few theorems about extension of fields
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- As long as it is irreducible, the choice of $f(x)$ does not matter.
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- If $f_1(x), f_2(x)$ are irreducible of the same degree, then $\mathbb{Z}_p[x] \mod f_1(x) \cong \mathbb{Z}_p[x] \mod f_2(x)$.
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- Over every $\mathbb{Z}_p$ ($p$ prime), there exists an irreducible polynomial of every degree.
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- All finite fields of the same size are isomorphic.
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- All finite fields are of size $p^d$ for prime $p$ and integer $d$.
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Corollary: This is effectively the only way to construct finite fields.
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## Common notations
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$\mathbb{F}_q$ is the finite field with $q$ elements. where $q$ is a prime power.
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$\mathbb{F}_q^t$ is a vector field of dimension $t$ over $\mathbb{F}_q$. (no notion of multiplication)
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$\mathbb{F}_{q^t}$ is the finite field with $q^t$ elements. (as extension field of $\mathbb{F}_q$)
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$\mathbb{F}_{q^t}\Rightarrow \mathbb{F}_q^t$. Every extension field of $\mathbb{F}_q$ is a vector field over $\mathbb{F}_q$.
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$\mathbb{F}_q^t\nRightarrow \mathbb{F}_{q^t}$. Additional structure is required.
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> [!IMPORTANT]
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>
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> From now on, we will use $+,\cdot$ to denote the modulo operations $\oplus,\odot$.
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### Few exercises
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Let $F$ be a field and $\alpha(x)\in F[x]$. Prove that for $\beta\in F$, show $\alpha(\beta)=0\iff (x-\beta)\mid \alpha(x)$.
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<details>
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<summary>Proof</summary>
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$\implies$:
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Suppose $x-\beta\mid \alpha(x)$, then $\alpha(x)=(x-\beta)q(x)$ for some $q(x)\in F[x]$.
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Thus, $\alpha(\beta)=(\beta-\beta)q(\beta)=0$.
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$\impliedby$:
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We proceed by induction over $\deg(\alpha(x))$.
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Base case: $\deg(\alpha(x))=1$, then $\alpha(x)=\alpha_0+\alpha_1x$ for some $\alpha_0,\alpha_1\in F$.
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Then $\alpha(\beta)=\alpha_0+\alpha_1\beta=0\iff \alpha_1\beta$. So $\alpha_0=-\alpha_1\beta$.
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So $\alpha(x)=\alpha_1 x-\alpha_1\beta=\alpha_1 (x-\beta)$.
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Inductive step: Suppose $\deg(\alpha(x))>1$, and the condition holds for all polynomials $r(x)$ with $\deg(r(x))<\deg(\alpha(x))$.
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Then $a(x)=(x-\beta)q(x)+r(x)$ for some $q(x),r(x)\in F[x]$ with $\deg(r(x))<1$ (By euclid's division algorithm).
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By our inductive hypothesis, $r(x)=0\iff (x-\beta)\mid r(x)$.
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So $\alpha(x)=(x-\beta)q(x)+r(x)=0\iff (x-\beta)\mid \alpha(x)$.
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</details>
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Let $F$ be a field and $a(x)\in F[x]$. Prove that for $\beta\in F$, show $a(\beta)=0\iff f(x)\mid a(x)$.
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<details>
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<summary>Solution</summary>
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$9=3^2$
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We extend $\mathbb{Z}_3=\mathbb{F}_3$ to $\mathbb{F}_{3^2}$.
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We extend this by using an irreducible polynomial of degree 2.
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Need to find an irreducible polynomial $f(x)=x^2+\alpha\in \mathbb{F}_3[x]$.
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> [!TIP]
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>
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> In $\mathbb{F}_3$, $\forall x\in \mathbb{F}_3$, $x^2\neq 2$. So $f(x)=x^2+1$ has no root in $\mathbb{F}_3$.
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Consider $f(x)=x^2-2=x^2+1$.
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Suppose for contradiction that $f(x)$ is reducible, then $f(x)=a(x)b(x)$ for some $a(x),b(x)\in \mathbb{F}_3[x]$. And $a(x),b(x)$ are both of degree 1.
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So $f(x)=(\alpha_1 x-a_2)(\beta_1 x-\beta_2)$ for some $\alpha_1,\alpha_2,\beta_1,\beta_2\in \mathbb{F}_3$, and $\alpha_1\beta_1\neq 0$.
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So $f(\frac{a_2}{a_1})=0$
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This contradicts the fact that $f(x)$ has no root in $\mathbb{F}_3$.
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</details>
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## Summary from last lecture
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A recipe for constructing a field $\mathbb{F}$ with $p^t$ elements ($p$ prime).
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- Construct $\mathbb{Z}_p$.
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- Find an irreducible polynomial $f(x)$ of degree $t$ (always exists).
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- Let $\mathbb{F} = \mathbb{Z}_p[x] \mod f(x)$.
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- The elements: polynomials in $\mathbb{Z}_p[x]$ of degree at most $t-1$.
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- Addition and multiplication $\mod f(x)$.
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Facts:
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- Choice of $f(x)$ does not matter.
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- Always end up with isomorphic $\mathbb{F}$ (identical up to renaming of elements).
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- All finite fields are of size $p^t$ for prime $p$ and some $t$.
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- The above recipe is unique.
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- The above recipe is unique.
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||||
|
||||
## Algebra over finite fields
|
||||
|
||||
### Groups
|
||||
|
||||
A group is a set $G$ with an operation $\cdot$ that satisfies the following axioms:
|
||||
|
||||
1. Closure: $\forall a,b\in G, a\cdot b\in G$.
|
||||
2. Associativity: $\forall a,b,c\in G, (a\cdot b)\cdot c=a\cdot (b\cdot c)$.
|
||||
3. Identity: $\exists e\in G, \forall a\in G, a\cdot e=e\cdot a=a$.
|
||||
4. Inverses: $\forall a\in G, \exists a^{-1}\in G, a\cdot a^{-1}=a^{-1}\cdot a=e$.
|
||||
|
||||
> [!IMPORTANT]
|
||||
>
|
||||
> 1. Operator $\cdot$ is not necessarily commutative. (if so then it's called an abelian group)
|
||||
> 2. May not be finite/infinite.
|
||||
> 3. May represented in power notation $a^n=a^{n-1}\cdot a$.
|
||||
|
||||
#### Examples of groups
|
||||
|
||||
$(\mathbb{Z},+)$ is an abelian group.
|
||||
$(\mathbb{Q},+)$ is an abelian group.
|
||||
|
||||
$(\{\mathbb{R}\setminus\{0\},\cdot\})$ is an abelian group.
|
||||
|
||||
Let $\mathbb{Z}_n^*=\{x\in \mathbb{Z}_n:gcd(x,n)=1\}$.
|
||||
|
||||
Then $(\mathbb{Z}_n^*,\cdot)$ is a group. If $n$ is prime, then $\mathbb{Z}_n^*$ only need to remove $0$.
|
||||
|
||||
#### Order of element in group
|
||||
|
||||
Let $(G,\cdot)$ be a **finite** group.
|
||||
|
||||
Then there exists $k\in\mathbb{N}$ such that $a^k=e$.
|
||||
|
||||
<details>
|
||||
<summary>Proof</summary>
|
||||
|
||||
Consider the sequence $a,a^2,a^3,\cdots$.
|
||||
|
||||
Since $G$ is finite, there exists $i,j\in\mathbb{N}$ such that $a^i=a^j$.
|
||||
|
||||
Then $a^i=a^j\iff a^{i-j}=e$.
|
||||
|
||||
So there exists $k=i-j\in\mathbb{N}$ such that $a^k=e$.
|
||||
|
||||
</details>
|
||||
|
||||
The order of an element $a\in G$ (denoted as $\mathcal{O}(a)$) is the smallest positive integer $n$ such that $a^n=e$.
|
||||
|
||||
Examples:
|
||||
|
||||
order of $5$ in $(\mathbb{Z}_6,+)$ is $6$.
|
||||
|
||||
If $a^l=e$, then $\mathcal{O}(a)\mid l$.
|
||||
|
||||
<details>
|
||||
<summary>Proof</summary>
|
||||
|
||||
Let $\mathcal{O}(a)=m$, then if $a^l=e$, then by definition of order, $l\geq m$. So $\exists q,r\in\mathbb{N}$ such that $l=qm+r$ and $r<m$.
|
||||
|
||||
So $a^l=a^{qm+r}=a^r=e$. This contradicts the definition of order. That $l$ is the smallest positive integer such that $a^l=e$.
|
||||
|
||||
</details>
|
||||
|
||||
### Cyclic groups
|
||||
|
||||
A group $G$ is called cyclic if there exists $a\in G$ such that $\mathcal{O}(a)=|G|$.
|
||||
|
||||
$G=\{a^i|i\in\mathbb{Z}\}$
|
||||
|
||||
one element is a generator of the group.
|
||||
|
||||
#### Exercises for cyclic groups
|
||||
|
||||
Is $(\mathbb{Z}_n,+)$ cyclic? If so, find a generator.
|
||||
|
||||
<details>
|
||||
<summary>Solution</summary>
|
||||
|
||||
Yes, the generator is $\{a|a\in \mathbb{Z}_n,gcd(a,n)=1\}$.
|
||||
|
||||
</details>
|
||||
@@ -6,4 +6,5 @@ export default {
|
||||
CSE5313_L1: "CSE5313 Coding and information theory for data science (Lecture 1)",
|
||||
CSE5313_L2: "CSE5313 Coding and information theory for data science (Lecture 2)",
|
||||
CSE5313_L3: "CSE5313 Coding and information theory for data science (Lecture 3)",
|
||||
CSE5313_L4: "CSE5313 Coding and information theory for data science (Lecture 4)",
|
||||
}
|
||||
Reference in New Issue
Block a user