# question3.py
from itertools import product
V='∀'
E='∃'

def tt(f,n) :
  xss=product((0,1),repeat=n)
  print('function:',f.__name__)
  for xs in xss : print(*xs,':',int(f(*xs)))
  print('')

# this is the logic for part A (p\/q\/r) /\ (p\/q\/~r) /\ (p\/~q\/r) /\ (p\/~q\/~r) /\ (~p\/q\/r) /\ (~p\/q\/~r) /\ (~p\/~q\/r) /\ (~p\/~q\/~r)

def parta(p,q,r) :
  
  a=(p or q or r) and (p or q or not r) and (p or not q or r)and (p or not q or  not r)
  b=(not p or q or r ) and (not p or q or not r) and (not p or not q or r) and (not p or not q or not r)
  c= a and b
  return c

def partb(p,q,r) :
  a=(p or q and r) and (p or not q or not r) and (p or not q or not  r)and (p or q or  not r)
  b=(not p or q or r ) and (not p or q or not r) and (not p or not q or r) and (not p or not q or not r)
  c= a and b
  return c

print("part A:")
tt(parta,3)

print("part B:")
tt(partb,3)

# Function to check Armstrong number
def is_armstrong_number(number):
    # Convert number to string to iterate over its digits
    num_str = str(number)
    
    # Calculate the sum of the cubes of each digit
    digit_sum = sum(int(digit) ** len(num_str) for digit in num_str)
    
    # Compare the sum with the original number
    if digit_sum == number:
        return True
    else:
        return False

# Prompt user for a number
number = int(input("Enter a number: "))

# Check if the number is an Armstrong number
if is_armstrong_number(number):
    print(number, "is an Armstrong number.")
else:
    print(number, "is not an Armstrong number.")

import os

# Get the current directory
current_dir = os.getcwd()

# Loop through each file in the current directory
for filename in os.listdir(current_dir):

    # Check if the file name starts with a number followed by a period and a space
    if filename[0].isdigit() and filename[1] == '.' and filename[2] == ' ':
        
        # Remove the number, period, and space from the file name
        new_filename = filename[3:]
        
        # Rename the file
        os.rename(os.path.join(current_dir, filename), os.path.join(current_dir, new_filename))

def clamp_number(num, a, b):
  return max(min(num, max(a, b)), min(a, b))
  
clamp_number(2, 3, 5) # 3
clamp_number(1, -1, -5) # -1

# Given a number n, print all primes smaller than or equal to n. It is also given that n is a small number.
# For example, if n is 10, the output should be “2, 3, 5, 7”. If n is 20, the output should be “2, 3, 5, 7, 11, 13, 17, 19”.


# Python program to print all primes smaller than or equal to
# n using Sieve of Eratosthenes

def SieveOfEratosthenes(n):
	
	# Create a boolean array "prime[0..n]" and initialize
	# all entries it as true. A value in prime[i] will
	# finally be false if i is Not a prime, else true.
	prime = [True for i in range(n + 1)]
	p = 2
	while (p * p <= n):
		
		# If prime[p] is not changed, then it is a prime
		if (prime[p] == True):
			
			# Update all multiples of p
			for i in range(p * 2, n + 1, p):
				prime[i] = False
		p += 1
	prime[0]= False
	prime[1]= False
	# Print all prime numbers
	for p in range(n + 1):
		if prime[p]:
			print (p)

# driver program
if __name__=='__main__':
    n = 30
    print("Following are the prime numbers smaller")
    print("than or equal to ", n)
    print("than or equal to ", n)
    SieveOfEratosthenes(n)

# Python binary search function
def binary_search(arr, target):
    left = 0
    right = len(arr) - 1
    while left <= right:
        mid = (left + right) // 2
        if arr[mid] == target:
            return mid
        elif arr[mid] < target:
            left = mid + 1
        else:
            right = mid - 1
    return -1

# Usage
arr = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
target = 7
result = binary_search(arr, target)
if result != -1:
    print(f"Element is present at index {result}")
else:
    print("Element is not present in array")