answer:Sure, here is an example of a Jupyter notebook cell that implements a simple Linear Regression model using Scikit-learn library.```python# Importing required librariesfrom sklearn.model_selection import train_test_split from sklearn.linear_model import LinearRegressionfrom sklearn import metricsimport matplotlib.pyplot as pltimport pandas as pdimport numpy as np# Load the dataseturl = "http://bit.ly/w-data"data = pd.read_csv(url)# Preprocessing the datasetX = data.iloc[:, :-1].values y = data.iloc[:, 1].values# Split the dataset into the training set and test setX_train, X_test, y_train, y_test = train_test_split(X, y, test_size=0.2, random_state=0)# Training the modelmodel = LinearRegression() model.fit(X_train, y_train)# Making predictionsy_pred = model.predict(X_test)# Comparing Actual output vs Predicted outputdf = pd.DataFrame({'Actual': y_test, 'Predicted': y_pred}) print(df)# Evaluating the modelprint('Mean Absolute Error:', metrics.mean_absolute_error(y_test, y_pred))print('Mean Squared Error:', metrics.mean_squared_error(y_test, y_pred))print('Root Mean Squared Error:', np.sqrt(metrics.mean_squared_error(y_test, y_pred)))```In this code, we are implementing a simple linear regression model to predict the score of a student based on the number of hours they study. The dataset used is a simple 2D dataset which is publicly available.

answer:In React, you can accomplish this by using the state and ref.Here's a simple example of how you can do this:```jsximport React, { useState, useRef } from 'react';function MyComponent() { const [selectedLink, setSelectedLink] = useState(null); const linkRef = useRef(); const handleClick = (event) => { event.preventDefault(); setSelectedLink(event.target.href); linkRef.current = event.target; // Now you have the selected link reference and url } return ( <p> This is a lexicaljs text with a <a href="https://lexical.dev/" onClick={handleClick} ref={linkRef}>link</a> that I want to get ref of... </p> );}export default MyComponent;```In this example, I'm using a click event handler to prevent the link from navigating, and I am storing the clicked link in the state of the component. Also, I'm using `useRef` hook to attach ref to the link.You can replace `handleClick` function with your custom function, and `selectedLink` and `linkRef.current` will be updated whenever a link is clicked.

answer:Here is the Python solution to the problem:The function calculates the rolling mean and median by iterating through the array. For each iteration, it calculates the current mean and median of the data up to the current element. It then checks if this pair (mean, median) has been seen before. If not, it adds it to the list of outputs. If it has been seen before, it continues to the next iteration. The median is calculated using the function numpy.median(), and the mean is calculated using numpy.mean().```Pythonfrom typing import List, Tupleimport numpy as npdef rolling_avg_median(numbers: List[int]) -> List[Tuple[float, float]]: means_medians = set() ans = [] for i in range(len(numbers)): sequence = numbers[:i+1] mean = np.mean(sequence) median = np.median(sequence) tuple_mean_median = (mean, median) if tuple_mean_median not in means_medians: ans.append(tuple_mean_median) means_medians.add(tuple_mean_median) return ans```You can test this code with the following unit test:```Pythondef test_rolling_avg_median(): assert rolling_avg_median([1, 2, 3, 4, 3, 5, 2]) == [(1.0, 1.0), (1.5, 1.5), (2.0, 2.0), (2.5, 2.5), (2.6666666666666665, 2.0), (2.8333333333333335, 2.5), (2.857142857142857, 3.0)] assert rolling_avg_median([1, 2, 2]) == [(1.0, 1.0), (1.5, 1.5), (1.6666666666666667, 2.0)] assert rolling_avg_median([4, 3, 5, 2]) == [(4.0, 4.0), (3.5, 3.5), (4.0, 4.0), (3.5, 3.5)]```

answer:Below I have written a Python program that takes four input parameters and constructs a matrix. After constructing the matrix, it then iterates through the matrix in a spiraling, vortex-shaped pattern starting from the top left. Here I have assumed that the fourth parameter is for determining the size of the matrix (4x4).Also, it is important to note that the problem lacks specific details like what kind of manipulations needs to be done on numerical value etc. So I'm assuming that we don't need to perform any additional operations. This solution will just put the values inside the matrix and then print them in spiraling order:```Python# Function to create matrix with 4 parametersdef create_matrix(a, b, c, d): return [[a, b, c, d] for _ in range(4)]# Function to print matrix in a spiral orderdef spiral_order(matrix): result = [] while matrix: result += matrix.pop(0) if matrix and matrix[0]: for row in matrix: result.append(row.pop()) if matrix: result += matrix.pop()[::-1] if matrix and matrix[0]: for row in matrix[::-1]: result.append(row.pop(0)) return result# Example parametersparams = [1, 2, 3, 4]matrix = create_matrix(*params)print("Original Matrix:")for row in matrix: print(row)print("nMatrix in Spiral Order:")print(spiral_order(matrix))```This program will create a 4x4 matrix with the provided parameters and then print the values in a spiraling pattern starting from the top left.

answer:Solving the entire problem in single code block isn't practically achievable as it requires multiple steps, several libraries and in-depth cryptographic understanding. However, I'll try to give a detailed approach on how to solve it. The programming language I'll use here is Python. Step 1 : RSA Key Generation Python has a library “Cryptography” which provides cryptographic recipes and primitives. from cryptography.hazmat.primitives import serialization as crypto_serialization from cryptography.hazmat.primitives.asymmetric import rsa from cryptography.hazmat.backends import default_backend as crypto_default_backend def generate_RSA_key(): key = rsa.generate_private_key(backend=crypto_default_backend(), public_exponent=65537, key_size=2048) private_key = key.private_bytes(crypto_serialization.Encoding.PEM, crypto_serialization.PrivateFormat.PKCS8, crypto_serialization.NoEncryption()) public_key = key.public_key().public_bytes(crypto_serialization.Encoding.OpenSSH, crypto_serialization.PublicFormat.OpenSSH) return private_key.decode('utf-8'), public_key.decode('utf-8')Step 2: Encoding the keys in Base64 import base64 private_key_encoded = base64.b64encode(private_key) public_key_encoded = base64.b64encode(public_key)Step 3: Implement Diffie Hellman for Secure Key Distribution Python supports PyCryptoDome library which can be used for Diffie Hellman key exchange. from Crypto.Protocol.KDF import HKDF from Crypto.Random import get_random_bytes from Crypto.Cipher import AES from Crypto.Protocol.KDF import HKDF from Crypto.PublicKey import ECC # Generate a key pair with curve 'P-256' key1 = ECC.generate(curve='P-256') # Encrypt a message using the ECC key secret = get_random_bytes(16) cipher_aes = AES.new(secret, AES.MODE_EAX) ecc_key = key1.export_key(format='DER', use_pkcs8=True, protection="pbkdf2withhmac-sha1anddesede", passphrase=b'my password') # Securely store the ECC key with open("my_ecc_key.pem", "wb") as f: f.write(ecc_key) You can repeat the steps for the second user and exchange the keys.Step 4: Managing Key Lifecycle Key lifecycle management includes creating, using, storing, archiving, and deleting keys. The two most prominent aspects are key rotation and key destruction.Key Rotation - Key rotation can be done by scheduling a script for generating and storing keys as shown in step 1 and step 2. Key Destruction - Key Destruction/Expiration can be done by simply deleting the file where key is stored after a certain period of time.Disclaimer: Security of keys isn't covered in this solution. Storing the keys securely, protecting them from unauthorized access is crucial to security and this solution provides only rudimentary key generation and sharing.

answer:Here is a solution using Python programming language:```pythondef lucas_series(n): if n == 0: return 2 if n == 1: return 1 else: return lucas_series(n - 1) + lucas_series(n - 2)lucas_numbers = [lucas_series(i) for i in range(15)]print(lucas_numbers)```This script implements a simple recursive function `lucas_series(n)` to calculate the Lucas numbers, then uses list comprehension to generate the first 15 numbers in the series.However, considering efficiency, the above solution is not quite efficient because of the repeated calculations in the recursive approach. Here is a more efficient way to compute Lucas numbers:```pythonlucas_numbers = [2, 1] # starting values# calculate the restfor _ in range(13): # we already have first two, so we need next 13 to make 15 lucas_numbers.append(sum(lucas_numbers[-2:]))print(lucas_numbers)```This script calculates each number as the sum of the last two numbers, avoiding the inefficiencies of the recursive approach.