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question:# Rabin-Miller Primality Test Implementation Problem You are tasked with implementing the Rabin-Miller primality test. Your function should determine whether a given number is prime with high probability. The number tested might be very large, so efficiency is important. Function Signature ```python def is_probably_prime(n: int, k: int) -> bool: pass ``` Input 1. `n` (int): The number to be checked, where `n >= 2`. 2. `k` (int): The number of iterations for accuracy. Higher k provides higher conviction in the result. Output - Returns `True` if `n` is probably prime with an error chance of `4^-k`. - Returns `False` if `n` is definitely composite. Constraints - The function should handle large values of `n`. - Aim for a time complexity of O(k * log(n)^3). - The implementation should be space-efficient, using O(1) extra space. Examples 1. `is_probably_prime(17, 5)` should return `True` 2. `is_probably_prime(18, 5)` should return `False` 3. `is_probably_prime(2, 3)` should return `True` 4. `is_probably_prime(7919, 4)` should return `True` Requirements Ensure to handle the power-of-2 factorization correctly and manage random number generations consistently within the constraints.
answer:import random def miller_rabin_pass(a, s, d, n): Helper function for performing a single Miller-Rabin test. a_to_power = pow(a, d, n) if a_to_power == 1: return True for _ in range(s - 1): if a_to_power == n - 1: return True a_to_power = pow(a_to_power, 2, n) return a_to_power == n - 1 def is_probably_prime(n, k): Uses the Miller-Rabin primality test to determine whether n is a probable prime. if n in (2, 3): return True if n % 2 == 0 or n == 1: return False # Write n-1 as 2^s * d s = 0 d = n - 1 while d % 2 == 0: d //= 2 s += 1 for _ in range(k): a = random.randint(2, n - 2) if not miller_rabin_pass(a, s, d, n): return False return True
question:# Coding Challenge: Detecting the Start of a Cycle in a Linked List Background You are given a linked list that might contain a cycle. A cycle occurs if a node's `next` pointer points to one of the previous nodes in the list, causing a loop. Your task is to implement a function using Floyd's Cycle-Finding Algorithm to detect if a cycle exists in the linked list, and if it does, return the first node where the cycle begins. Objective Write a function: ```python def first_cyclic_node(head): :type head: Node :rtype: Node ``` - **Input**: A `head` of the singly linked list where each node is an instance of the `Node` class. - **Output**: The first node (instance of `Node` class) at the start of the cycle if a cycle is present; otherwise, return `None`. Constraints - Do not modify the linked list. - Aim for linear time complexity and constant space complexity. - The function should handle edge cases such as an empty list, a list with a single node, or a list where the cycle starts at the head. Example Below is an example of a linked list and how a cycle might appear: ``` 1 -> 2 -> 3 -> 4 -> 5 ^ | |_________| ``` For the list above, the cycle starts at the node with value `3`. Testing Your Solution You can use the following stub for functional testing purposes: ```python class Node: def __init__(self, x): self.val = x self.next = None def first_cyclic_node(head): # Implement your solution here pass # Example usage: # create linked list => 1 -> 2 -> 3 -> 4 -> 5 -> 3 (cycle starting at 3) head = Node(1) head.next = Node(2) node3 = Node(3) head.next.next = node3 head.next.next.next = Node(4) head.next.next.next.next = Node(5) head.next.next.next.next.next = node3 result_node = first_cyclic_node(head) if result_node: print(result_node.val) # Should output: 3 else: print("No cycle detected") ``` Additional Notes - Ensure your code runs efficiently for large linked lists. - Include comments to explain the key steps of your implementation.
answer:class Node: def __init__(self, x): self.val = x self.next = None def first_cyclic_node(head): Detects the start of a cycle in a linked list using Floyd's Cycle-Finding Algorithm. Returns the first node where the cycle begins, or None if there is no cycle. if not head or not head.next: return None slow = head fast = head # Phase 1: Determine if there is a cycle while fast and fast.next: slow = slow.next fast = fast.next.next if slow == fast: break if not fast or not fast.next: return None # Phase 2: Find the start of the cycle slow = head while slow != fast: slow = slow.next fast = fast.next return slow
question:# Ternary Search Algorithm Implementation and Edge Case Handling Problem Statement Implement the ternary search algorithm to find a given key in a sorted array. Your function should return the index of the key if found, and -1 if the key is not present in the array. Additionally, you need to handle edge cases and ensure the calculations for `mid1` and `mid2` are correct. Function Signature ```python def ternary_search(left: int, right: int, key: int, arr: list[int]) -> int: pass ``` Input - `left` (int): The left boundary of the array to search (inclusive). - `right` (int): The right boundary of the array to search (inclusive). - `key` (int): The value to search for within the array. - `arr` (list of int): A list of integers sorted in non-decreasing order. Output - Return the index of the `key` within the array if found. - Return `-1` if the `key` is not found in the array or if the input is invalid. Constraints - You must ensure the function handles all edge cases correctly, including: - The array being empty. - The `left` and `right` range being invalid. - The key is not present within the specified range. - Aim for an efficient solution with a time complexity of O(log3(N)). Examples Consider the following examples: **Example 1:** ```python arr = [1, 2, 3, 4, 5, 6] left = 0 right = 5 key = 4 ternary_search(left, right, key, arr) => 3 ``` **Example 2:** ```python arr = [1, 2, 3, 4, 5, 6] left = 0 right = 5 key = 7 ternary_search(left, right, key, arr) => -1 ``` **Example 3:** ```python arr = [10, 20, 30, 40, 50, 60] left = 1 right = 4 key = 25 ternary_search(left, right, key, arr) => -1 ``` Evaluation Criteria - Correctness: The function should return the correct index if the key is present and -1 if it is not. - Efficiency: The function should run within the expected time complexity. - Edge Case Handling: The function should correctly handle all specified edge cases.
answer:def ternary_search(left: int, right: int, key: int, arr: list[int]) -> int: Perform a ternary search on a sorted array to find a key. :param left: The left boundary of the search range (inclusive). :param right: The right boundary of the search range (inclusive). :param key: The value to search for. :param arr: The sorted list of integers. :return: The index of the key if found, otherwise -1. if left > right or not arr: return -1 while left <= right: # Calculate the first and second mid points mid1 = left + (right - left) // 3 mid2 = right - (right - left) // 3 # Check if key is at any mid if arr[mid1] == key: return mid1 if arr[mid2] == key: return mid2 # Reduce the search range to the appropriate third if key < arr[mid1]: right = mid1 - 1 elif key > arr[mid2]: left = mid2 + 1 else: left = mid1 + 1 right = mid2 - 1 return -1
question:# Scenario: You are working on a symbolic mathematical solver that needs to handle polynomials for various algebraic computations. You are tasked to extend the existing `Polynomial` class by implementing a method to find the derivative of a polynomial with respect to a given variable. # Task: Implement a method called `derivative` within the `Polynomial` class to compute the derivative of the polynomial with respect to a specified variable index. # Function Signature: ```python def derivative(self, var_index: int) -> Polynomial: Compute the derivative of the polynomial with respect to the given variable index. ``` # Input: - `var_index` (int): The index of the variable with respect to which the derivative is to be computed. # Output: - Returns a `Polynomial` representing the derivative of the original polynomial. # Constraints: - The `Polynomial` consists of `Monomial` instances that contain rational coefficients and an arbitrary number of variables. - The variable index provided will always be valid within the set of variables in the polynomial. # Notes: - The derivative of a polynomial ( P(x) = c cdot x^n ) with respect to ( x ) is ( P'(x) = n cdot c cdot x^{n-1} ). # Example: ```python p = Polynomial([ Monomial({1: 3}, 4), # Represents 4*(a_1)^3 Monomial({2: 1}, 5), # Represents 5*(a_2) Monomial({1: 2, 2: 1}, 2) # Represents 2*(a_1)^2*(a_2) ]) print(p.derivative(1)) # Expected output: 12*(a_1)^2 + 4*(a_1)*(a_2) print(p.derivative(2)) # Expected output: 5 + 2*(a_1)^2 ``` # Implementation: Extend the Polynomial class by adding the `derivative` method.
answer:from collections import defaultdict class Monomial: def __init__(self, exponents, coefficient): exponents: dict where keys are variable indices and values are their respective exponents. coefficient: the coefficient of the monomial. self.exponents = exponents self.coefficient = coefficient def __repr__(self): return f'Monomial({self.exponents}, {self.coefficient})' def derivative(self, var_index): if var_index in self.exponents and self.exponents[var_index] > 0: new_exponents = self.exponents.copy() exponent = new_exponents[var_index] new_exponents[var_index] -= 1 if new_exponents[var_index] == 0: new_exponents.pop(var_index) new_coefficient = self.coefficient * exponent return Monomial(new_exponents, new_coefficient) else: return None class Polynomial: def __init__(self, monomials): monomials: a list of Monomial instances. self.monomials = monomials def derivative(self, var_index): derivative_monomials = [] for monomial in self.monomials: derivative_monomial = monomial.derivative(var_index) if derivative_monomial is not None: derivative_monomials.append(derivative_monomial) return Polynomial(derivative_monomials) def __repr__(self): return f'Polynomial({self.monomials})'