my_list = [1, 2, 3, 4, 5]
removed_element = my_list.pop(2)  # Remove and return element at index 2
print(removed_element)  # 3
print(my_list)  # [1, 2, 4, 5]

last_element = my_list.pop()  # Remove and return the last element
print(last_element)  # 5
print(my_list)  # [1, 2, 4]

def to_roman_numeral(num):
  lookup = [
    (1000, 'M'),
    (900, 'CM'),
    (500, 'D'),
    (400, 'CD'),
    (100, 'C'),
    (90, 'XC'),
    (50, 'L'),
    (40, 'XL'),
    (10, 'X'),
    (9, 'IX'),
    (5, 'V'),
    (4, 'IV'),
    (1, 'I'),
  ]
  res = ''
  for (n, roman) in lookup:
    (d, num) = divmod(num, n)
    res += roman * d
  return res

to_roman_numeral(3) # 'III'
to_roman_numeral(11) # 'XI'
to_roman_numeral(1998) # 'MCMXCVIII'

# 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)

from functools import partial

def curry(fn, *args):
  return partial(fn, *args)
  
add = lambda x, y: x + y
add10 = curry(add, 10)
add10(20) # 30

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 for Bitonic Sort. Note that this program
# works only when size of input is a power of 2.

# The parameter dir indicates the sorting direction, ASCENDING
# or DESCENDING; if (a[i] > a[j]) agrees with the direction,
# then a[i] and a[j] are interchanged.*/
def compAndSwap(a, i, j, dire):
	if (dire==1 and a[i] > a[j]) or (dire==0 and a[i] > a[j]):
		a[i],a[j] = a[j],a[i]

# It recursively sorts a bitonic sequence in ascending order,
# if dir = 1, and in descending order otherwise (means dir=0).
# The sequence to be sorted starts at index position low,
# the parameter cnt is the number of elements to be sorted.
def bitonicMerge(a, low, cnt, dire):
	if cnt > 1:
		k = cnt/2
		for i in range(low , low+k):
			compAndSwap(a, i, i+k, dire)
		bitonicMerge(a, low, k, dire)
		bitonicMerge(a, low+k, k, dire)

# This funcion first produces a bitonic sequence by recursively
# sorting its two halves in opposite sorting orders, and then
# calls bitonicMerge to make them in the same order
def bitonicSort(a, low, cnt,dire):
	if cnt > 1:
		k = cnt/2
		bitonicSort(a, low, k, 1)
		bitonicSort(a, low+k, k, 0)
		bitonicMerge(a, low, cnt, dire)

# Caller of bitonicSort for sorting the entire array of length N
# in ASCENDING order
def sort(a,N, up):
	bitonicSort(a,0, N, up)

# Driver code to test above
a = [3, 7, 4, 8, 6, 2, 1, 5]
n = len(a)
up = 1

sort(a, n, up)
print ("\n\nSorted array is")
for i in range(n):
	print("%d" %a[i]),