answer:# Required Librariesimport boto3from botocore.exceptions import NoCredentialsError, PartialCredentialsError, SSLError, BotoCoreError, EndpointConnectionErrordef get_from_dynamodb(table_name):# Initialize an empty response dictionary response = {}try:# Attempt to create a dynamodb resource object using Amazon's boto3 SDK dynamodb = boto3.resource('dynamodb')# Reference the specified DynamoDB table table = dynamodb.Table(table_name)# Attempt to retrieve all items from the specified table response = table.scan()except NoCredentialsError: print('No AWS Credentials provided')except PartialCredentialsError: print('Incomplete AWS Credentials provided')except SSLError: print('SSL Connection could not be established')except BotoCoreError as e: print('AWS Service Error: ', e)except EndpointConnectionError: print('Endpoint Connection Error: Check your AWS config.')except Exception as e: # General Exception print('Unexpected error: ', e)# return the response dictionary (contains items from dynamodb table if successful)return response# Usagetablename = 'mytable'data = get_from_dynamodb(tablename)

answer:Here is a simple Python program to solve this task:```python# creating a hash table (dictionary) to store student datastudent_dict = {}# adding students to the dictionarystudent_dict[1] = ('John', 24)student_dict[2] = ('Tom', 26)student_dict[3] = ('Harry', 20)# printing the dictionaryprint(student_dict)```In the above snippet, we first create an empty dictionary, then we add student data to it. Each key in the dictionary is a unique student ID, and the value is a tuple with two elements: the student's name and age.When you run this code, it will display:```python{1: ('John', 24), 2: ('Tom', 26), 3: ('Harry', 20)}```This is just a basic implementation. Based on your needs, you may want to add more functionality, such as methods to add, delete and update students data, to retrieve a student data by ID, etc.

answer:The task requires to find the 100,000,000th squarefree Fibonacci number and provide its last 16 digits precede by a comma and the number in scientific notation.The solution involves generating Fibonacci numbers, checking each number for squarefreeness, and counting how many squarefree numbers have been generated until reaching the 100,000,000th one.Python's sympy library can be used to generate Fibonacci number and to check an arbitrary size number for squarefreeness. Because of the size of a Fibonacci number, division remainders, and mod-pow functions can't handle the number, so an alternative method, like the pseudoprime method, should be employed.However, computing the 100,000,000th Fibonacci number that is squarefree may take an unfeasibly long time with Python due to the limitations of computational power and time.Here is a Python function for generating Fibonacci numbers and checking if they are squarefree:```pythonfrom sympy import is_squarefree, fibonaccidef fib_squarefree(n): i = count = 0 while count < n: i += 1 fib = fibonacci(i) if is_squarefree(fib): count += 1 return fib# To print the 100,000,000th Fibonacci number that is squarefree# print(fib_squarefree(100_000_000))```This code works for smaller inputs, but finding the 100,000,000th squarefree Fibonacci number is still beyond current computing capabilities.

answer:Here is a simple Python implementation of an iterative addition-based multiplication algorithm:```pythondef multiply(x, y): result = 0 for i in range(y): result += x # This is equivalent to x*y return result```This code works by adding up `x` a total of `y` times. However, this approach isn't very optimized, since it takes a time complexity of `O(Y)`, i.e., it loops `Y` times.A more optimized approach would be to use the concept of Bit Manipulation and Bit Shifting. Bit shifting allows us to multiply or divide a number by 2. Below is a Python implementation:```pythondef multiply_optimized(x, y): result = 0 while(y > 0): # If second number becomes odd, add the first number to result if((y & 1) == 1): result = result + x # Double the first number and halve the second number x = x << 1 y = y >> 1 return result```In this algorithm, we initialize result as 0. In the while loop, we consider y to be an array of bits. We iterate through each bit, and if the bit is set (i.e., 1), we add the corresponding power of 2 multiple of x to the result.During this, x is doubled (with << operation), and y is halved (with >> operation) in each iteration.This solution has a faster runtime as it takes a time complexity of `O(log Y)`, since it halves `Y` in each step. It also does not use the multiplication or division operators.

answer:# We can use the label function in scipy.ndimage which will assign a unique id to each # connected cluster of cells in the image. # We apply this function to a binary version of the image where all values < threshold are True# and all values >= threshold are False. The unique labels (id's) are returned in a numpy array of # same shape as the image. # We then use the np.unique function to count the unique labels in the resulting array. # The number of unique labels - 1 (as one label is given to the background noise) is the number of # regions where the values are below the specified threshold.labeled_img, num_labels = ndimage.label(img < threshold)result = num_labels&lt;/code&gt;END SOLUTION

answer:```cpp#include<iostream>#include<map>using namespace std;class PhoneBook { private: map<string, string> phoneBook; public: // Method to add a phone number void addPhoneNumber(string name, string phoneNumber) { phoneBook[name] = phoneNumber; } // Method to remove a phone number void removePhoneNumber(string name) { phoneBook.erase(name); } // Method to modify an existing phone number void modifyPhoneNumber(string name, string newPhoneNumber) { phoneBook[name] = newPhoneNumber; } // Method to display a phone number void getPhoneNumber(string name) { if (phoneBook.find(name) == phoneBook.end()) { cout << "Name not found in phone book" << endl; } else { cout << "Phone number of " << name << " is " << phoneBook[name] << endl; } } // Method to display all phone numbers void getAllPhoneNumbers() { cout << "Phone book:" << endl; for (auto& x: phoneBook) { cout << "Name: " << x.first << ", Phone Number: " << x.second << endl; } } }; ```Here, map is being used to associate each name with a phone number. A class called `PhoneBook` is defined that provides methods for adding, modifying and removing a phone number from a phone book. Adding, modifying and removing phone numbers makes use of map's insert, update and erase operations respectively. Please note:The `getPhoneNumber` and `getAllPhoneNumbers` methods are provided to display phone numbers for testing/debugging.