def calculate_values():
    value1 = 10
    value2 = 20
    return value1, value2

result1, result2 = calculate_values()
print("Result 1:", result1)
print("Result 2:", result2)

from typing import Optional

from datetime import datetime

def convert_timestamp_string_to_epoch(timestamp: str) -> Optional[int]:
    epoch_time = None
    time_obj = datetime.strptime(timestamp, "%Y-%m-%d %H:%M:%S.%f")
    epoch_time = int((time_obj - datetime(1970, 1, 1)).total_seconds() * 1000)
    return epoch_time


print(int(convert_timestamp_string_to_epoch("2025-08-01 13:11:47.171")))
#above outputs 1754053907171.0

#how to I remove the .0 ?

import re
_proposal_regex = r'(?:(?:(IKE|ESP):)?[\w/]+(?:/NO_EXT_SEQ)?(?:, ?(IKE|ESP):[\w/]+(?:/NO_EXT_SEQ)?)*)?'
_proposals_re = rf'(?P<proposals>{_proposal_regex}|)'
pattern = rf'received proposals: {_proposals_re}'
match = re.match(pattern, 'received proposals: ')
print(match.group('proposals') if match else "No match")  # Prints "No match"

import itertools
import string
import time

def guess_password(real):
    chars = string.ascii_lowercase + string.ascii_uppercase + string.digits + string.punctuation
    attempts = 0
    for password_length in range(1, 9):
        for guess in itertools.product(chars, repeat=password_length):
            startTime = time.time()
            attempts += 1
            guess = ''.join(guess)
            if guess == real:
                return 'password is {}. found in {} guesses.'.format(guess, attempts)
            loopTime = (time.time() - startTime);
            print(guess, attempts, loopTime)

print("\nIt will take A REALLY LONG TIME to crack a long password. Try this out with a 3 or 4 letter password and see how this program works.\n")
val = input("Enter a password you want to crack that is 9 characters or below: ")
print(guess_password(val.lower()))

def generate_floyds_triangle(num_rows):
    triangle = []
    number = 1
    for row in range(num_rows):
        current_row = []
        for _ in range(row + 1):
            current_row.append(number)
            number += 1
        triangle.append(current_row)
    return triangle


def display_floyds_triangle(triangle):
    for row in triangle:
        for number in row:
            print(number, end=" ")
        print()


# Prompt the user for the number of rows
num_rows = int(input("Enter the number of rows for Floyd's Triangle: "))

# Generate Floyd's Triangle
floyds_triangle = generate_floyds_triangle(num_rows)

# Display Floyd's Triangle
display_floyds_triangle(floyds_triangle)

#You are given a two-digit integer n. Return the sum of its digits.
#Example
#For n = 29 the output should be solution (n) = 11
def solution(n):
    return (n//10 + n%10)