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question:# Challenging Markov Chain Simulation You are given a Markov Chain represented by a dictionary where keys are state names and values are dictionaries of probabilities of transitioning from the key state to other states. For instance: ```python my_chain = { 'A': {'A': 0.6, 'E': 0.4}, 'E': {'A': 0.7, 'E': 0.3} } ``` Your task is to write two functions based on the Markov Chain implementation provided: 1. `simulate_markov_chain(chain, start_state, steps)`: Given a Markov Chain `chain`, an initial state `start_state`, and number of steps to simulate `steps`, return the sequence of states visited. 2. `steady_state_probability(chain, threshold=1e-6)`: Compute the steady-state probabilities of each state. The steady-state probability distribution does not change after further applications of the transition probabilities. Function Descriptions and Constraints: - **simulate_markov_chain** - **Input**: - `chain` (dict): The Markov Chain. - `start_state` (str): The starting state. - `steps` (int): The number of steps to simulate. - **Output**: - List of states visited in the given number of steps, starting from `start_state`. - **steady_state_probability** - **Input**: - `chain` (dict): The Markov Chain. - `threshold` (float): Tolerance threshold below which to consider steady-state achieved. - **Output**: - Dictionary where keys are state names and values are their steady-state probabilities. Example Given the Markov Chain: ```python my_chain = { 'A': {'A': 0.6, 'E': 0.4}, 'E': {'A': 0.7, 'E': 0.3} } ``` **simulate_markov_chain** ``` simulate_markov_chain(my_chain, 'A', 5) ``` Could return a sequence like: `['A', 'A', 'E', 'A', 'E']` **steady_state_probability** ``` steady_state_probability(my_chain, threshold=1e-6) ``` Should return probabilities close to: `{'A': 0.6363, 'E': 0.3636}` Additional Constraints - Transition probabilities in the chain for each state must sum up to 1.0. - Assume there are no disconnected states in the provided Markov Chain.
answer:import numpy as np def simulate_markov_chain(chain, start_state, steps): Simulate the number of steps in a Markov Chain starting from a given state. Parameters: - chain (dict): The Markov Chain. - start_state (str): The starting state. - steps (int): The number of steps to simulate. Returns: - list of states visited in the given number of steps, starting from `start_state`. current_state = start_state state_sequence = [current_state] for _ in range(steps): next_state = np.random.choice(list(chain[current_state].keys()), 1, p=list(chain[current_state].values()))[0] state_sequence.append(next_state) current_state = next_state return state_sequence def steady_state_probability(chain, threshold=1e-6): Compute the steady-state probabilities of each state in a Markov Chain. Parameters: - chain (dict): The Markov Chain. - threshold (float): Tolerance threshold below which to consider steady-state achieved. Returns: - dictionary: Keys are state names and values are their steady-state probabilities. states = list(chain.keys()) n = len(states) # Transpose the matrix so we can multiply it on the left transition_matrix = np.array([[chain[i].get(j, 0) for j in states] for i in states]).T # Start with an equal probability distribution prob_dist = np.ones(n) / n while True: new_prob_dist = transition_matrix @ prob_dist if np.max(np.abs(new_prob_dist - prob_dist)) < threshold: break prob_dist = new_prob_dist return dict(zip(states, new_prob_dist))
question:**Problem Statement**: You are given a 2D board of characters and a list of strings representing words. Your task is to find all words in the list that can be formed by tracing a path in the board. Each word must be constructed from letters of sequentially adjacent cells, where "adjacent" cells are those horizontally or vertically neighboring. The same letter cell may not be used more than once for any word. **Function Signature**: ```python def find_words(board: List[List[str]], words: List[str]) -> List[str]: pass ``` **Input**: * `board`: A list of lists of characters, `board[i][j]` is a character in the board. * `words`: A list of strings representing the words to search for. **Output**: * Return a list of strings that are found in the board. **Constraints**: * The length of all words will sum to at most `15*10^4`. * The board will have a maximum of `100` rows and `100` columns. * Each word and each character of the board consists of lowercase English letters. * Words may overlap on the board as long as the same letter cell is not reused within each individual word search. **Example**: ```python board = [ ['o','a','a','n'], ['e','t','a','e'], ['i','h','k','r'], ['i','f','l','v'] ] words = ["oath", "pea", "eat", "rain"] # Example Output find_words(board, words) # Output: ["oath", "eat"] ``` **Directions**: 1. Construct a Trie (Prefix Tree) from the words list. 2. Implement the backtracking algorithm to search words from the board using the Trie. 3. Manage the visited status of cells to prevent reusing the same cell within a single word search. 4. Handle edge cases like empty board or words list gracefully. 5. Optimize for performance considering the large input constraints.
answer:class TrieNode: def __init__(self): self.children = {} self.is_end_of_word = False class Trie: def __init__(self): self.root = TrieNode() def insert(self, word): node = self.root for char in word: if char not in node.children: node.children[char] = TrieNode() node = node.children[char] node.is_end_of_word = True def search_prefix(self, prefix): node = self.root for char in prefix: if char not in node.children: return None node = node.children[char] return node def find_words(board, words): def backtrack(node, i, j, path, visited): if node.is_end_of_word: result.add(path) node.is_end_of_word = False # avoid duplicate entries if i < 0 or i >= len(board) or j < 0 or j >= len(board[0]) or (i, j) in visited: return char = board[i][j] if char not in node.children: return # mark the cell as visited visited.add((i, j)) next_node = node.children[char] # explore neighbor cells in the up, down, left, right direction for x, y in [(i-1, j), (i+1, j), (i, j-1), (i, j+1)]: backtrack(next_node, x, y, path + char, visited) # unmark the cell as visited visited.remove((i, j)) # create a trie of words trie = Trie() for word in words: trie.insert(word) result = set() # iterate each cell on the board for i in range(len(board)): for j in range(len(board[0])): backtrack(trie.root, i, j, '', set()) return list(result)
question:**Context**: When securing communication, it is sometimes necessary to validate that data received matches what was sent. In a simpler scenario, this can be exemplified by finding a single character that was added to a string after shuffling. **Problem**: You are given two strings `s` and `t`, where: - `s` is an original string. - `t` is a string created by shuffling the characters of `s` and adding one additional lowercase letter. Write a function `find_added_character(s: str, t: str) -> str` which finds and returns the letter that was added to the string `t`. **Input**: - `s`: A string of lowercase letters (1 <= len(s) <= 10^5). - `t`: A string of lowercase letters such that `t` contains all the characters of `s` plus one additional character (len(t) = len(s) + 1). **Output**: - Return a single character that was added to `t`. **Constraints**: - Both `s` and `t` consist of only lowercase letters. - The additional letter in `t` will appear exactly once. **Examples**: ```python assert find_added_character("abcd", "abecd") == 'e' assert find_added_character("", "y") == 'y' assert find_added_character("a", "aa") == 'a' assert find_added_character("xyz", "zxyz") == 'z' ``` **Performance Requirements**: - The time complexity should be O(n). - The space complexity should be O(1). **Notes**: - Utilize the XOR bitwise operation to solve this problem optimally. **Function Signature**: ```python def find_added_character(s: str, t: str) -> str: # TODO: Implement the function pass ```
answer:def find_added_character(s: str, t: str) -> str: Returns the single character that was added to string t. The function utilizes XOR to achieve O(n) time complexity and O(1) space complexity. result = 0 for char in s: result ^= ord(char) for char in t: result ^= ord(char) return chr(result)
question:# Context: The Fibonacci sequence is a series of numbers where each number is the sum of the two preceding ones, usually starting with 0 and 1. While it is simple to define the nth Fibonacci number recursively, this method can be highly inefficient for large n. Alternative methods utilize dynamic programming or iterative methods to overcome these inefficiencies. # Task: Write a Python function to compute the nth Fibonacci number in an efficient manner. You must use an iterative approach to achieve optimal time and space complexity. # Requirements: 1. **Function Name**: `fib_iter_optimized` 2. **Input**: - An integer `n` (0 <= n <= 10^6), representing the position in the Fibonacci sequence. 3. **Output**: - An integer, the nth Fibonacci number. 4. **Constraints**: - The solution should have a time complexity O(n). - The solution should use O(1) additional space for variables apart from the input and output. # Example Usage: ```python assert fib_iter_optimized(0) == 0 assert fib_iter_optimized(1) == 1 assert fib_iter_optimized(10) == 55 assert fib_iter_optimized(50) == 12586269025 ``` # Notes: * Ensure that your function handles the edge cases correctly like n=0 and n=1. * Pay attention to performance. The function should execute within a reasonable time frame for very large n (up to 10^6).
answer:def fib_iter_optimized(n): Computes the nth Fibonacci number using an iterative approach. Args: n : int The position in the Fibonacci sequence (0 <= n <= 10^6). Returns: int : the nth Fibonacci number. if n == 0: return 0 elif n == 1: return 1 a, b = 0, 1 for _ in range(2, n + 1): a, b = b, a + b return b