Model: qchapp/MNLP_M2_document_encoder
Author: qchapp
Pipeline: sentence-similarity
Library: sentence-transformers
Tags: sentence-transformers, safetensors, bert, sentence-similarity, feature-extraction, generated_from_trainer, dataset_size:57499, loss:CosineSimilarityLoss, arxiv:1908.10084, base_model:sentence-transformers/all-MiniLM-L6-v2, base_model:finetune:sentence-transformers/all-MiniLM-L6-v2, text-embeddings-inference, endpoints_compatible, region:us
Base models (finetune): id: sentence-transformers/all-MiniLM-L6-v2
Card metadata: tags: sentence-transformers, sentence-similarity, feature-extraction, generated_from_trainer, dataset_size:57499, loss:CosineSimilarityLoss
base_model: sentence-transformers/all-MiniLM-L6-v2
widget: source_sentence: You are an AI-Coding assistant. User will you give you a task. Your goal is to complete the task as faithfully as you can.

Create a program that generates a unique and randomly generated color name for a given RGB value. The program should output the color name in title case.
sentences: To find the second smallest prime number in a given array of integers using recursion, we can use the following algorithm:

1. Initialize two variables `smallest` and `secondSmallest` to infinity and set `index` to -1.
2. Create a helper function `isPrime(num)` that takes an integer `num` as input and returns True if `num` is prime and False otherwise. This function can be implemented using a for loop to check if any number from 2 to the square root of `num` evenly divides `num`.
3. Create a recursive function `findSecondSmallestPrime(arr, currentSmallest, currentSecondSmallest, currentIndex)` that takes the array of integers `arr`, the current smallest prime number `currentSmallest`, the current second smallest prime number `currentSecondSmallest`, and the current index `currentIndex` as inputs.
4. In the base case, if `currentIndex` is equal to the length of `arr`, return `currentSecondSmallest` and `index`.
5. Otherwise, check if the current element `arr[currentIndex]` is less than or equal to `currentSmallest`.
   - If it is, call the recursive function with `arr`, `arr[currentIndex]`, `currentSmallest`, and `currentIndex + 1` as arguments.
   - If it is not, check if the current element `arr[currentIndex]` is a prime number using the `isPrime` helper function.
     - If it is, update `currentSecondSmallest` and `index` accordingly and call the recursive function with `arr`, `currentSmallest`, `currentSecondSmallest`, and `currentIndex + 1` as arguments.
     - If it is not, call the recursive function with `arr`, `currentSmallest`, `currentSecondSmallest`, and `currentIndex + 1` as arguments.
6. Finally, initialize `smallest` and `secondSmallest` to infinity and call the recursive function `findSecondSmallestPrime(arr, smallest, secondSmallest, 0)`.
7. Return the `secondSmallest` and `index` obtained from the recursive function.

Here is the implementation of the algorithm in Python:

```python
import math

def isPrime(num):
    if num < 2:
        return False
    for i in range(2, int(math.sqrt(num)) + 1):
        if num % i == 0:
            return False
    return True

def findSecondSmallestPrime(arr, currentSmallest, currentSecondSmallest, currentIndex):
    if currentIndex == len(arr):
        return currentSecondSmallest, currentIndex
    elif arr[currentIndex] <= currentSmallest:
        return findSecondSmallestPrime(arr, arr[currentIndex], currentSmallest, currentIndex + 1)
    elif isPrime(arr[currentIndex]):
        currentSecondSmallest = min(currentSecondSmallest, arr[currentIndex])
        return findSecondSmallestPrime(arr, currentSmallest, currentSecondSmallest, currentIndex + 1)
    else:
        return findSecondSmallestPrime(arr, currentSmallest, currentSecondSmallest, currentIndex + 1)

def findSecondSmallestPrimeNumber(arr):
    smallest = float('inf')
    secondSmallest = float('inf')
    index = -1
    secondSmallest, index = findSecondSmallestPrime(arr, smallest, secondSmallest, 0)
    return secondSmallest, index
```

You can now call the `findSecondSmallestPrimeNumber` function with an array of integers as an argument to find the second smallest prime number and its index in the array., Sure, here's a Python program that generates a unique and randomly generated color name for a given RGB value:
```python
import random
def generate_color_name(rgb):
    # Define a list of adjectives and nouns to use in color names
    adjectives = ['red', 'orange', 'yellow', 'green', 'blue', 'purple', 'pink', 'brown', 'gray']
    nouns = ['sky', 'ocean', 'leaf', 'sunset', 'fire', 'rose', 'chocolate', 'stone', 'smoke']
    
    # Convert the RGB value to a hexadecimal string
    hex_value = '#{:02x}{:02x}{:02x}'.format(*rgb)
    
    # Generate a random adjective and noun to use in the color name
    adjective = random.choice(adjectives)
    noun = random.choice(nouns)
    
    # Combine the adjective, noun, and hex value to create the color name
    color_name = '{} {} ({})'.format(adjective.title(), noun.title(), hex_value)
    
    return color_name
# Example usage
rgb_value = (255, 0, 128)
color_name = generate_color_name(rgb_value)
print(color_name)
```
This program defines a list of adjectives and nouns to use in color names, and generates a random combination of an adjective, noun, and hex value to create a unique color name. The `generate_color_name` function takes an RGB value as input, converts it to a hexadecimal string, and returns the color name in title case., Solution is here: 4
Explanation: The most commonly used term to designate the most intense degree of distraction and the complete absence of attention is "Mental Absence", option 4. While the other options like "Attentional perplexity", "Attention indifference", "Hyperprosexia", and "Aprosexia" are terms related to attention and functionality but they do not quite fit the description of the question., source_sentence: You are an AI-Coding assistant. User will you give you a task. Your goal is to complete the task as faithfully as you can.

Given a number n, count the total number of digit-only divisors of n that are also prime numbers. A digit-only divisor is defined as a divisor of n that only consists of the digits present in n. 

Constraints:
- 1 <= n <= 10^9

Example:
n = 123456

Explanation:
In the given test, we are counting the total number of digit-only divisors of 123456. The digit-only divisors of 123456 are 1, 2, 3, 4, 5, 6, 12, 123, 1234, 12345, and 123456. However, only 2 and 3 are prime numbers. Therefore, the output will be 2.
sentences: Reasoning:

Consider the following formulas:

$D(m,n) = \displaystyle\sum_{d\mid m}\sum_{k=1}^n\sigma_0(kd)$
$\sigma_0(kd) = \displaystyle\sum_{i\mid kd}1$

We can rewrite the function as:

$D(m,n) = \displaystyle\sum_{d\mid m}\sum_{k=1}^n\displaystyle\sum_{i\mid kd}1$

Which can be rewritten as:

$D(m,n) = \displaystyle\sum_{i\mid m}\displaystyle\sum_{d=1}^{n(i/m)}\sigma_0(id/m)$

This indicates that instead of having variable upper limits for each of the inner sums, we have constant upper limit for the second sum, which now depends on the first sum. 

The count of divisors for each number can be calculated using prime factorization. Let $n$ be a natural number and let its prime factorization be $n = p_1^{a_1}p_2^{a_2}...p_k^{a_k}$. The number of divisors of $n$ is $(a_1 + 1)(a_2 + 1)...(a_k + 1)$.

Parking this thought aside, define a new function $f(i)$ as $f(i) = \displaystyle\sum_{k=1}^i\sigma_0(k)$. Thus, we can rewrite the function as:

$D(m,n) = \displaystyle\sum_{i\mid m}f(n(i/m))$

$f(i)$ can be calculated by looping through 1 to i, and for each number calculating the number of divisors and adding that to f(i). Then to calculate D(m,n), we loop for i dividing m, calculate the integer value of n(i/m) and add f(value) to D.

In terms of complexity, the calculation of f(i) can be up to n, and the calculation of D(m,n) can be up to m, so this method is efficient for large inputs.

Now, to implement this logic in python, we iterate from 1 to sqrt(n) and then from sqrt(n) to 1, calculate the factor for the i-th term, add f(i*term) to the result for numbers with 1 factor, and add 2*f(i*term) to the result for numbers with 2 factors. This is done to overcome the double counting problem that occurs due to the initial looping from 1 to sqrt(n).

Finally, for very large number such as 200! and 10^12, calculating f(i) will be time consuming. So we use memoization to store the calculated values of f(i). Lastly, we calculate the values modulo (10^9 +7) after every addition and multiplication operation to prevent overflow of integer values. 

Python code snippet:

```python
MAX = 10**12
num_dict = {i: i-1 for i in range(1, 200+1)}
f = [0] * (MAX + 1)
mod = 10**9 + 7

#Calculate n!
def calc_fact(n):
    fact=1 
    for i in range(1,n+1): 
        fact=(fact*i)%mod
    return fact

#Calculate f(i)
for i in range(1, MAX + 1):
    for j in range(2 * i, MAX +1, i):
        f[j] += 1

for i in range(2, MAX + 1):
    f[i] += f[i-1]

N = calc_fact(200)
result = 0

#Calculate D(m,n)
i = 1

while i * i <= N:
    term = N // i
    result=((result + num_dict[term] * (f[min(i, MAX // term) - 1])%mod)%mod)
    i += 1

for term in range(N // (i-1), 0, -1):
    lo = N // (term + 1)
    hi = min(N // term, MAX // term) 
    if hi >= lo:
        count = (hi - lo)
        val = (f[hi] - f[lo] + mod) % mod
        result=(result + count * val % mod) % mod

print(result)
```
I would like to mention that, this python script will take more than expected time to run because it involves lots of calculation with large numbers. Therefore, if you are thinking to execute this script then make sure you have good configurations to support this much of high level computation., The options are:

- Sentence A: "He made green paint with blue and yellow paint."
- Sentence B: "He made green paint with red and yellow paint."

To determine which sentence doesn't make sense, we need to understand how colors are mixed. Primary colors are colors that cannot be mixed from other colors, and include blue, red, and yellow. When mixed together, primary colors create secondary colors, such as green, which is made by mixing blue and yellow.

Therefore, Sentence A makes sense as green paint can be made by mixing blue and yellow paint, which are primary colors. Meanwhile, Sentence B does not make sense as red and yellow paint cannot be mixed to create green paint, as red and yellow are not primary colors.

Confidence: 95%, To solve this problem, we can iterate through each digit-only divisor of n and check if it is a prime number.

Here's the step-by-step algorithm:

1. Initialize a variable `count` to 0 to keep track of the number of digit-only divisors that are prime.
2. Iterate through each digit-only divisor of n.
  - Start a loop from 1 to the square root of n.
  - Check if i is a divisor of n using the modulus operator (n % i == 0).
  - If i is a divisor, convert i to a string and check if all its digits are present in n.
  - If i passes both conditions, increment `count` by 1.
  - Also, check if n divided by i is different from i. If so, convert n/i to a string and check if all its digits are present in n.
  - If n/i passes both conditions, increment `count` by 1.
3. Return the value of `count`.

Here's the implementation in Python:

```python
import math

def is_prime(num):
    if num < 2:
        return False
    for i in range(2, int(math.sqrt(num)) + 1):
        if num % i == 0:
            return False
    return True

def count_digit_only_prime_divisors(n):
    count = 0
    for i in range(1, int(math.sqrt(n)) + 1):
        if n % i == 0:
            str_i = str(i)
            if all(digit in str(n) for digit in str_i):
                if is_prime(i):
                    count += 1
            if i != n // i:
                str_divisor = str(n // i)
                if all(digit in str(n) for digit in str_divisor):
                    if is_prime(n // i):
                        count += 1
    return count

# Test the function with the example input
n = 123456
result = count_digit_only_prime_divisors(n)
print(result)  # Output: 2
```

The time complexity of this algorithm is O(sqrt(n) * d), where d is the number of digits in n., source_sentence: You are an AI-Coding assistant. User will you give you a task. Your goal is to complete the task as faithfully as you can.

You are tasked with creating a Python class that manages a storage pool for volumes. The class should provide methods to add, remove, and retrieve volumes, as well as to check for orphaned volumes. Additionally, the class should be tested using the `unittest` framework to ensure its functionality.

Your task is to implement the `StoragePool` class with the following methods:
- `add_volume(volume_name)`: Adds a volume with the given name to the storage pool.
- `remove_volume(volume_name)`: Removes the volume with the given name from the storage pool.
- `get_volume_list()`: Returns a list of all volumes in the storage pool.
- `get_orphaned_volumes()`: Returns a list of orphaned volumes, i.e., volumes that are not associated with any host.

You should also write unit tests for the `StoragePool` class using the `unittest` framework. The tests should cover the functionality of the class, including adding and removing volumes, retrieving the volume list, and checking for orphaned volumes.

Your task is to complete the implementation of the `StoragePool` class and write unit tests to ensure its correctness.
sentences: ```python
import unittest
from unittest.mock import Mock

class StoragePool:
    def __init__(self):
        self.volumes = {}
        self.hosts = {}

    def add_volume(self, volume_name):
        self.volumes[volume_name] = None

    def remove_volume(self, volume_name):
        if volume_name in self.volumes:
            del self.volumes[volume_name]

    def get_volume_list(self):
        return list(self.volumes.keys())

    def associate_volume_with_host(self, volume_name, host_name):
        if volume_name in self.volumes:
            self.volumes[volume_name] = host_name
            self.hosts[host_name] = volume_name

    def get_orphaned_volumes(self):
        return [volume for volume, host in self.volumes.items() if host is None]

class TestStoragePool(unittest.TestCase):
    def setUp(self):
        self.storage_pool = StoragePool()

    def test_add_volume(self):
        self.storage_pool.add_volume('volume1')
        self.assertIn('volume1', self.storage_pool.get_volume_list())

    def test_remove_volume(self):
        self.storage_pool.add_volume('volume2')
        self.storage_pool.remove_volume('volume2')
        self.assertNotIn('volume2', self.storage_pool.get_volume_list())

    def test_get_orphaned_volumes(self):
        self.storage_pool.add_volume('volume3')
        self.storage_pool.add_volume('volume4')
        self.storage_pool.associate_volume_with_host('volume3', 'host1')
        orphaned_volumes = self.storage_pool.get_orphaned_volumes()
        self.assertIn('volume4', orphaned_volumes)
        self.assertNotIn('volume3', orphaned_volumes)

if __name__ == '__main__':
    unittest.main()
```, Some common food and beverage sources that can aid in preventing dehydration are:

1. Water: Water is the best source of hydration. Drinking plenty of water throughout the day can help prevent dehydration.

2. Fruits: Fruits such as watermelon, pineapple, oranges, strawberries, and grapefruit have high water content, and their juicy texture helps to rehydrate your body.

3. Vegetables: Vegetables such as cucumber, lettuce, celery, and tomatoes have a high water content and can be a good source of hydration.

4. Sports drinks: Sports drinks like Gatorade or Powerade are designed to replenish electrolytes and hydrate your body. However, be mindful of the added sugars in these drinks, which may not be necessary for everyone.

5. Soup: Soups like chicken noodle or vegetable soup can help to rehydrate your body and provide nutrition simultaneously.

Remember, it's essential to stay hydrated throughout the day, especially during physical activity and in hot weather. Staying hydrated can help prevent dehydration and its associated risks., Of course! I'd be happy to help. Based on the input you provided, here is the sum of all the numerical elements in the list A between the positions 2 and 2 (including positions 2 and 2):

The list A is: ['L', '8647', 't', '2523', 'Z', 't', 'U', '5739', '4801', 'K', 'P', '1619', 's', '1785']

There are no numerical elements in the list between positions 2 and 2, so the sum is 0.

I hope that helps! Let me know if you have any other questions., source_sentence: Using only the digits in the current year (2021), can you calculate the predicted closing stock price of [Apple Inc.] for the last trading day of the year? Hint: Start by finding the sum of the digits in the current month and day, then multiply by the number of days left in the year. Divide that number by the sum of the digits in the current year, and then multiply by the stock's closing price on the last trading day of the previous year. Good luck!
sentences: It's not possible to accurately predict a stock's closing price using only numerical operations on the current year and other random numbers because stock prices depend on numerous factors in the real market. These factors include supply and demand, economic indicators, market sentiment, company earnings, and other variables that cannot be determined by the digits in the current year. Additionally, stock market volatility adds uncertainty to any attempt to predict future prices solely based on the past.

However, as an AI language model, I can provide valuable information and insights on stock market trends, technical analysis, and historical price movements. To make an informed decision about investing in a stock, it's crucial to analyze the company's financials, market position, and management team, as well as staying abreast of industry shifts and global economic trends. It's also recommended to seek guidance from a certified financial advisor before making any investment decisions., It sounds like you're looking to improve the reliability and handling of your webhooks, particularly in the case where your customers' servers go down and come back up again. Here are a few suggestions on how you can improve this situation:

1. Implement a progressive backoff strategy: Instead of immediately sending a large number of webhooks to a customer's server the moment it comes back online, you can implement a progressive backoff strategy. This involves gradually increasing the interval between retries, giving the customer server time to catch up and process the webhooks before sending more.
2. Limit the number of retries: Instead of retrying indefinitely, you can limit the number of times you'll try to deliver a webhook. After the maximum number of retries is reached, you can drop the webhook and move on to the next one. This can help prevent overloading the customer server with a large backlog of webhooks.
3. Provide a way for customers to pause/resume webhook delivery: Providing a way for customers to pause/resume webhook delivery can give them more control over their servers and help prevent overloading. This can be done through an API or a user interface.
4. Monitor the status of customer servers: Continuously monitoring the status of customer servers can help you detect when they go down and come back up. This can allow you to adjust your webhook delivery strategy accordingly, such as implementing a progressive backoff or limiting the number of retries.
5. Consider using a message queue: Implementing a message queue between your system and the customer server can help buffer the webhooks and prevent overloading the customer server. The queue can hold the webhooks until the customer server is available to process them.

By implementing one or more of these strategies, you can improve the reliability and handling of your webhooks, particularly in the case where your customers' servers go down and come back up again., Here is an example of a for loop that prints the first 10 powers of a number x, without using the exponentiation operator or any built-in function:

```python
def calculate_power(base, exponent):
    result = 1
    for _ in range(exponent):
        result *= base
    return result

x = 2  # Example number
for i in range(10):
    print(calculate_power(x, i))
```

In this code, we define a function `calculate_power()` that takes a `base` and an `exponent` as input, and calculates the power of `base` raised to the `exponent` using a for loop. The function initializes a variable `result` to 1, and then iterates `exponent` times, multiplying `base` to `result` in each iteration. Finally, the function returns the calculated power.

Outside the function, we set `x` to the desired number, and then use a for loop to print the first 10 powers of `x`. Inside the loop, we call the `calculate_power()` function with `x` as the base and the current value of `i` as the exponent. The result is then printed to the console., source_sentence: You are an AI-Coding assistant. User will you give you a task. Your goal is to complete the task as faithfully as you can.

Organize two distinct compilations, one constituted of numeric integers and the other of distinct English language words. The numeric composition should be portrayed in a sequentially increasing mode, whereas the English language words composition should be sequenced in lexicographical order.
numeric_compilation = [2, 4, 6, 1]
word_compilation = ['zanzibar', 'banana', 'apple']
sentences: The given Python code can be corrected to:

```python
import os
import shutil

def delete_directory(path):
    if os.path.exists(path) and os.path.isdir(path):
        shutil.rmtree(path)
    else:
        print('The given path does not exist or is not a directory')
	
delete_directory('/test-directory')
```

Key modifications in the corrected code:

1. **Check if the directory exists:** It is always a good practice to check whether the directory you are trying to delete exists or not using `os.path.exists(path) and os.path.isdir(path)`. This will help avoid `FileNotFoundError` if the directory does not exist.

2. **Erroneous use of `os.rmdir()`:** The original code attempts to use `os.rmdir()` incorrectly. Specifically, it is trying to remove a file using `os.rmdir(file)`, which is problematic because `os.rmdir()` can only remove empty directories and not files or non-empty directories. Instead of `os.rmdir(file)`, the corrected code utilizes `shutil.rmtree(path)`, which deletes a directory and all its contents.

3. **Use of `shutil.rmtree(path)`:** The `shutil` module provides a higher-level operation `rmtree(path)` in Python, which recursively deletes an entire directory tree, meaning it deletes the directory and its contents, including its subdirectories and their contents as well.

The `os` module in Python provides a variety of operations for interacting with the operating system, including file I/O operations, process management, and directory management. Misusing these functions could lead to significant issues:

1. **Security Risk:** Improper usage or unintentional handling of `os` functions, such as deleting a file or directory or executing a system command, can lead to accidental deletion or modification of important system files and directories leading to system instability or expose system vulnerabilities to attackers.

2. **Stability Risk:** A wrong system call can affect the stability of the system, causing system crashes or slowing down the overall performance.

To do operations on the file system, it is important to validate all inputs, ensure that the necessary permissions are granted, and handle exceptions properly. Aim to follow the principle of least privilege—meaning the code should only have the permissions it needs and no more—to reduce potential damage if something goes wrong., To efficiently calculate the sum of all the elements in a given list, we can use a simple iterative approach that iterates through each element in the list and adds it to a running sum. This solution has a time complexity of O(n) since it needs to iterate through all the elements in the list.

Here's an example implementation in Python:

```python
def calculate_sum(lst):
    total_sum = 0

    for num in lst:
        total_sum += num

    return total_sum
```

This implementation has a space complexity of O(1) because it only uses a constant amount of extra space to store the running sum.

Even for large input lists with thousands or millions of elements, this solution should be efficient as it only requires a single pass through the list. The memory usage is minimized as it only uses a constant amount of extra space regardless of the size of the input list., You can solve this problem by using the python built-in `sort()` method which sorts a list in a ascending manner. 

Here is the python program which can solve the above problem:

```python
#initial integer list
numeric_compilation = [2, 4, 6, 1]

# Sort the integer list in ascending order
numeric_compilation.sort()

print('Numeric Compilation sorted in ascending order:', numeric_compilation)

#initial word list
word_compilation = ['zanzibar', 'banana', 'apple']

# Sort the word list in lexicographical order
word_compilation.sort()

print('Word Compilation sorted in lexicographical order:', word_compilation)
```

When you run above python program, it will sort the integer list in ascending order and word list in lexicographical order and output the sorted lists:

Numeric Compilation sorted in ascending order: [1, 2, 4, 6]
Word Compilation sorted in lexicographical order: ['apple', 'banana', 'zanzibar']
pipeline_tag: sentence-similarity
library_name: sentence-transformers
Likes: 0 | All-time downloads: 50
Created: 2025-05-24 20:10:59