My study notes

optimize_basic_functions.cpp

#include <iostream>
#include <math.h>
using namespace std;

bool is_prime(int n) {
    if (n <= 1) return false;
    if (n <= 3) return true;
    if (n % 2 == 0 || n % 3 == 0) return false;
    for (int i = 5; i * i <= n; i += 6) 
        if (n % i == 0 || n % (i + 2) == 0) return false;
    return true;
}

bool is_fibo(int n) {
    int n1 = n * n * 5 - 4;
    int n2 = n * n * 5 + 4;
    float sqrt1 = sqrt(n1);
    float sqrt2 = sqrt(n2);
    return (int)sqrt1 == sqrt1 || (int)sqrt2 == sqrt2;
}

int get_fibo(int n) {
    double phi = (1 + sqrt(5)) / 2;
    return round(pow(phi, n) / sqrt(5));
}

// SAKAMOTO ALGORITHM to checks what day of the week it is
int day_of_week(int year, int month, int day) {
    int t[] = {0, 3, 2, 5, 0, 3, 5, 1, 4, 6, 2, 4};
    year -= month < 3;
    return (497 * year/400 + t[month - 1] + day) % 7;
}

optimize_basic_functions.js

function getWebName(url) {
	// http://example1.com/a/b?c=d => example1
	// http://www.example2.com/b?c=d => example2
	// https://ww.example3.com.vn => example3
	
	const hostnameParts = new URL(url).hostname.split('.');
	return hostnameParts[hostnameParts.length - 1].length === 2
		? hostnameParts[hostnameParts.length - 3]
		: hostnameParts[hostnameParts.length - 2];
}

// Check even and odd without `if else`
number = 3
["even", "odd"][number % 2] 

// Get intersection
const a = new Set([1,2,3]);
const b = new Set([4,3,2]);
const intersection = [...a].filter(x => b.has(x))
console.log(intersection) // [2, 3]

function getCookieField(name) {
   const cookie = document.cookie.split("; ").find(item => item.startsWith(`${name}=`));
   return cookie ? decodeURIComponent(cookie.split("=")[1]) : null;
}

(265 >>> 0).toString(2);
(_$=($,_=[]+[])=>$?_$($>>+!![],($&+!![])+_):_)(265);
/*
Đây ko phải là RegEx mà là hàm mũi tên (arrow function) với các tên hàm, tên biến và số (1) được thể hiện bằng các kí tự đặc biệt và sô 1 được thể hiện bằng biểu thức mảng như này +!![]
Đây là phiên bản dễ hiểu hơn một chút của đoạn mã:
(toBinary = (val, str = "") => val ? toBinary(val >> 1, (val & 1) + str):str)(265);
[]+[] chính là chuỗi trống "".
+!![] chính là số 1.
Dùng đệ quy để lấy từng bit và cộng dồn vào chuỗi str (ban đầu là trống ""). Điều kiện dừng là val bằng 0 (đoạn toán tử 2 ngôi chỗ val?... đấy).
Viết cho dễ nhìn và chú thích:
(
	toBinary = (val, str = "") => // gán toBinary cho hàm mũi tên với 2 tham số val và str (mặc định là "").
	val ? // nếu val khác 0...
	toBinary(val >> 1, (val & 1) + str) : // ... thì thực hiện đệ quy cho bit tiếp theo
	str // ...ngược lại kết thúc đệ quy và trả về giá trị
)(265); // gọi trực tiếp hàm toBinary
*/

parallel coordinates plot.py

# Compare hyperparameter search results
def plot_param_performace(clf, param_name, title):
    results = clf.search_cv.cv_results_
    plt.figure(figsize=(13, 5))
    plt.title(title)
    plt.xlabel(param_name)
    plt.ylabel("Score")
    plt.grid()

    ax = plt.gca()
    ax.set_ylim(0.96, 1)

    # Get the regular numpy array from the MaskedArray
    X_axis = np.array(results[f'param_{param_name}'].data, dtype=float)
    for scorer, color in zip(('test_score', 'train_score'), ('g', 'r')):
        for sample, style in (('mean', '-'), ('std', '--')):
            sample_score_mean = results[f'mean_{scorer}']
            sample_score_std = results[f'std_{scorer}']
            ax.fill_between(X_axis, sample_score_mean - sample_score_std,
                            sample_score_mean + sample_score_std,
                            alpha=0.1 if sample == 'mean' else 0, color=color)
            ax.plot(X_axis, sample_score_mean, style, color=color,
                    alpha=1 if sample == 'mean' else 0.7,
                    label=f'{scorer} ({sample})')

    plt.legend(loc="best")
    plt.tight_layout()
    plt.show()

plot_param_performace(rf_classifier, 'n_estimators', "Random Forest: Performance vs Number of Estimators")

Bellman Equation

When the RF problem is stochastic, there isn't a sequence of rewards that you see for sure -> what we're interested in is not maximizing the return (because that's a random number) but maximizing the average value of the sum of discounted rewards. In cases where both the state and action space are discrete we can estimate the action-value function iteratively by using the Bellman equation:

$$ Q_{i+1}(s,a) = R + \gamma \max_{a'}Q_i(s',a') $$

This iterative method converges to the optimal action-value function $Q^*(s,a)$ as $i\to\infty$. This means that the agent just needs to gradually explore the state-action space and keep updating the estimate of $Q(s,a)$ until it converges to the optimal action-value function $Q^*(s,a)$:

• However, in cases where the state space is continuous it becomes practically impossible to explore the entire state-action space. Consequently, this also makes it practically impossible to gradually estimate $Q(s,a)$ until it converges to $Q^*(s,a)$.

In the Deep $Q$-Learning, we solve this problem by using a neural network to estimate the action-value function $Q(s,a)\approx Q^*(s,a)$. It can be trained by adjusting its weights at each iteration to minimize the mean-squared error in the Bellman equation:

• Using neural networks in RF to estimate action-value functions has proven to be highly unstable -> Use a Target Network (soft update) and Experience Replay storing the agent's states, actions, rewards the agent receives in a memory buffer and then samples a random mini-batch to generate uncorrelated experiences for training agent.
• Towards the end of training, the agent will lean towards selecting the action that it believes (based on past experiences) will maximize $Q(s,a)$ -> We will set the minimum 𝜖 value = 0.01 (not 0) because we always want to keep a little bit of exploration during training.

Deep reinforcement learning

rl_formalism.mp4	In the standard “agent-environment loop” formalism, an agent interacts with the environment in discrete time-steps t=0,1,2,... At each $t$, the agent uses a policy $\pi$ to select an action $A_t$ based on its observation of the environment's state $S_t$. The agent receives a numerical reward $R_t$ and on the next time step, moves to a new state $S_{t+1}$.

def print_number_of_trainable_model_parameters(model):
    trainable_model_params = 0
    all_model_params = 0
    for _, param in model.named_parameters():
        all_model_params += param.numel()
        if param.requires_grad:
            trainable_model_params += param.numel()
    return f"trainable model parameters: {trainable_model_params}\nall model parameters: {all_model_params}\npercentage of trainable model parameters: {100 * trainable_model_params / all_model_params:.2f}%"

print(print_number_of_trainable_model_parameters(original_model))

Calculate $\frac{\partial{L}}{\partial{w}}$ without chain rule, where L is loss function of a simple Logistic Regression model with binary cross-entropy:

I know this is not practical way but I just want to strengthen my understanding by looking for a pure substitution method. Please correct me if I was doing something wrong 😅

$$\eqalign{ L &= -[y * ln(\frac{1}{1 + e^{-(wx+b)}}) + (1 - y) * ln(1 - \frac{1}{1 + e^{-(wx+b)}})] \\ &= -[y * ln(\frac{1}{1 + e^{-(wx+b)}}) + (1 - y) * ln(\frac{e^{-(wx+b)}}{1 + e^{-(wx+b)}})] \\ }$$

Now, we used the fact that $ln(\frac{1}{1 + e^{-z}}) = -ln(1 + e^{-z})$ and $ln(\frac{e^{-z}}{1 + e^{-z}}) = -z - ln(1 + e^{-z})$, where $z = wx+b$:

$$\eqalign{ L &= -[-y * ln(1 + e^{-(wx+b)}) + (1 - y) * (-(wx+b) - ln(1 + e^{-(wx+b)}))] \\ &= -[-y * ln(1 + e^{-(wx+b)}) - (1 - y) * ln(1 + e^{-(wx+b)}) - (1 - y) * (wx+b)] \\ &= ln(1 + e^{-(wx+b)}) + (1 - y) * wx + (1 - y) * b] \\ }$$

Now, find $\frac{\partial{L}}{\partial{w}}$:

$$\eqalign{ \frac{\partial{L}}{\partial{w}} &= \frac{-x * e^{-(wx+b)}}{1 + e^{-(wx+b)}} + (1 - y) * x] \\ &= \frac{-x * e^{-(wx+b)} + (1 - y) * x * (1 + e^{-(wx+b)})}{1 + e^{-(wx+b)}} \\ &= \frac{-x * e^{-(wx+b)} + (1 - y) * x + (1 - y) * x * e^{-(wx+b)}}{1 + e^{-(wx+b)}} \\ &= \frac{-xy * e^{-(wx+b)} + x - xy}{1 + e^{-(wx+b)}} \\ &= \frac{-xy * (1 + e^{-(wx+b)}) + x}{1 + e^{-(wx+b)}} \\ &= -xy + \frac{x}{1 + e^{-(wx+b)}} \\ &= (\frac{1}{1 + e^{-(wx+b)}} - y) * x \\ &= (sigmoid(z) - y) * x }$$