print("test")

# 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)

def bitonic_sort(arr, low, cnt, direction):
    ...

from time import sleep

def delay(fn, ms, *args):
  sleep(ms / 1000)
  return fn(*args)

delay(lambda x: print(x), 1000, 'later') # prints 'later' after one second

prime_lists=[] # a list to store the prime numbers

def prime(n): # define prime numbers
    if n <= 1:
        return False
    # divide n by 2... up to n-1
    for i in range(2, n):
        if n % i == 0:  # the remainder should'nt be a 0
            return False
    else:
        prime_lists.append(n)
        return True



for n in range(30,1000):  # calling function and passing starting point =30 coz we need primes >30
    prime(n)

check=0 # a var to limit the output  to 10 only
for n in prime_lists:
    for x in prime_lists:
        val=  n *x
        if (val > 1000 ):
            check=check +1
        if (check <10) :
            print("the num is:", val , "=",n , "* ", x )
        break

# Python Program to calculate the square root

num = float(input('Enter a number: '))

num_sqrt = num ** 0.5
print('The square root of %0.3f is %0.3f'%(num ,num_sqrt))