answer:Yes, you are correct. The Observer pattern is already effectively implemented in the .NET framework through the use of events and delegates.In the Observer pattern, an object (the 'subject') maintains a list of objects observing it (the 'observers'), and notifies them automatically of any state changes. Events in .NET essentially adhere to this pattern. An event is declared in a 'publisher' class and other 'subscriber' classes attach their methods (event handlers) to this event. When the event is triggered, all attached methods are automatically called, hence notifying them of the state change.So, while you can implement the Observer pattern itself, in .NET it's often more idiomatic to use events.

answer:From your statement, you're asking for permutations between the elements of multiple lists. Therefore, we could use the itertools.product() function in Python to generate the Cartesian product of input iterables which is equal to the nested for-loops.Please try the following script.```pythonimport itertoolsdef unique_permutations(*lists): # Get the Cartesian product of the lists result = list(itertools.product(*lists)) # Return unique permutations by converting list of tuples to list of lists return [list(p) for p in result]# Testinglist1 = [1, 2, 3]list2 = [4, 5, 3]print(unique_permutations(list1, list2))```This script will generate all permutations (with repetition) of the elements of the input lists.Note: The time complexity of getting the Cartesian product is linear with the total number of output tuples, so it's consistent with your request for a time complexity of no worse than O(n*m!). Consider, however, the complexity may raise when converting tuples to lists to get unique permutations as required in your problem.

answer:Sorry, as an AI model, I don't have the capability to develop browser add-ons or extensions. However, I can suggest outline of how to create the add-on you described.Here's a basic example of how this Google Chrome Extension might look in terms of programming, using JavaScript and CSS.JavaScript (contentscript.js):```javascriptvar links = document.getElementsByTagName("a");for (var i = 0; i < links.length; i++) { links[i].className += " emeraldGreen";}```This script would add the class "emeraldGreen" to all hyperlinks when the page loads.CSS (styles.css):```cssa.emeraldGreen { color: #50C878 !important;}```This style would change the color of each link with the "emeraldGreen" class to emerald green.Note `#50C878` is Hex color code for Emerald Green.To implement this as an extension you need below files -manifest.json:```json{ "manifest_version": 2, "name": "Hyperlink Color Changer", "version": "1.0", "permissions": [ "activeTab" ], "content_scripts": [ { "matches": ["<all_urls>"], "css": ["styles.css"], "js": ["contentscript.js"] } ]}```This manifest file is crucial for the browser to know what to do with the extension. The section "content_scripts" is where it is indicated that both the javascript and CSS files should be injected into every webpage that the user visits.This is a very simple illustration and depending individual project requirements, it might needs to be adjusted or expanded. For example, you may wish to add options allowing users to enable/disable the extension or choose when and where it should work. Always keep best practices and security in mind when developing extensions.

answer:Unfortunately, as a text-based AI model, I can't run Python code directly. Nevertheless, I can provide you with a general outline of how you may approach this problem using the Python programming language.You can use Natural Language Processing (NLP) libraries like NLTK and TextBlob, or more advanced machine learning libraries like Scikit-Learn, Tensorflow, and PyTorch to create sentiment analysis models. You can also use data visualization libraries like Matplotlib and Seaborn to create a sentiment intensity graph.Here's a skeleton Python script that shows how you can approach this:```pythonimport pandas as pdfrom sklearn.model_selection import train_test_splitfrom sklearn.feature_extraction.text import CountVectorizerfrom sklearn.linear_model import LogisticRegressionfrom sklearn.metrics import accuracy_scoreimport matplotlib.pyplot as plt# Load your dataset# data should be a pandas dataframe where the texts are in 'text' column # and the sentiments in 'sentiment' column. Sentiments should be labeled.data = pd.read_csv('data.csv')# Splitting the data into training and testingX_train, X_test, y_train, y_test = train_test_split( data['text'], data['sentiment'], test_size=0.2, random_state=42)# Create the CountVectorizervectorizer = CountVectorizer()# Transform the training dataX_train_vectorized = vectorizer.fit_transform(X_train)# Train the modelmodel = LogisticRegression().fit(X_train_vectorized, y_train)# Predict sentiments for test narratives predictions = model.predict(vectorizer.transform(X_test))# Check accuracyprint('Accuracy: ', accuracy_score(y_test, predictions))# Plotting the sentimentsplt.hist(predictions, color = 'blue', edgecolor = 'black')plt.title('Sentiment Intensity')plt.xlabel('Sentiment'); plt.ylabel('Frequency')plt.show()```Please replace 'data.csv' with your actual dataset. In this script, a simple Logistic Regression model is used for sentiment analysis. However, this basic model might not perform optimally for your dataset, and you might consider more complex models or pre-trained models, such as those available in the NLTK, TextBlob or Transformer-based models like BERT, for better performance.Also, note that in this example, labels for sentiments (joy, sadness, anger, etc.) should already be provided in the data. If your sentences do not have labels, you'll need to use unsupervised learning techniques which are quite different and often more complex.

answer:You can use `numpy.linalg.eig` function in the NumPy package to calculate the eigenvalues and the associated eigenvectors of a square matrix.Here is how to do it in code:```pythonimport numpy as npA = np.array([[5, 1], [-3, 8]])eig_values, eig_vectors = np.linalg.eig(A)print("The eigenvalues are: ", eig_values)print("The associated eigenvectors are: ", eig_vectors)```You can copy this code to your Python editor, run it, and you will get the eigenvalues and the associated eigenvectors of the matrix. The output format will be two arrays, the first array representing the eigenvalues and the second array showing the associated eigenvectors, each in a new line.Note that due to the nature of numerical calculations, these might not be exact - but they will be very close to the real values. The output will be in complex form if the eigenvalues or eigenvectors are complex numbers.Disclaimer: The output will depend on the Numpy's algorithm and it might be variant on different machines due to round off error. Hence, the result is expected to be close to the exact solution rather than exactly the same.

answer:Here is a Python solution using depth first search traversal with postorder.```# Definition for a binary tree node.# class TreeNode:# def __init__(self, x):# self.val = x# self.left = None# self.right = Noneclass Solution: def delNodes(self, root: TreeNode, to_delete: List[int]) -> List[TreeNode]: to_delete_set = set(to_delete) res = [] def helper(root, is_root): if not root: return None root_deleted = root.val in to_delete_set if is_root and not root_deleted: res.append(root) root.left = helper(root.left, root_deleted) root.right = helper(root.right, root_deleted) return None if root_deleted else root helper(root, True) return res```This solution uses a helper function to traverse the tree by postorder, which ensures that we traverse the bottom of the tree first. If a node is a root node and it's not deleted, it's added to the result list. If a node is deleted, its children would become the root node of the forest. Hence, we can call the helper function for its children nodes with 'is_root' set as True. If a node is not deleted, its children would still be its children nodes, so 'is_root' is set as False.We check every node if it's in 'to_delete_set' and mark it as 'root_deleted'. If root is deleted, we return None to its parent (this node is deleted, so it can't be a child node any more), else we return the node itself (this node is retained, so it's still the child node of its parent).In a nutshell, we use DFS to traverse all nodes and meanwhile deleting those required ones as we traverse the tree.The space complexity of the solution is O(N) where N is the total number of nodes. The time complexity is also O(N) as we're visiting each node once.