# 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")

#84 48 13 20 61 20 33 97 34 45 6 63 71 66 24 57 92 74 6 25 51 86 48 15 64 55 77 30 56 53 37 99 9 59 57 61 30 97 50 63 59 62 39 32 34 4 96 51 8 86 10 62 16 55 81 88 71 25 27 78 79 88 92 50 16 8 67 82 67 37 84 3 33 4 78 98 39 64 98 94 24 82 45 3 53 74 96 9 10 94 13 79 15 27 56 66 32 81 77
# xor a list of integers to find the lonely integer
res = a[0]
    for i in range(1,len(a)):
        res = res ^ a[i]

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'

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('')

# p \/ (q /\ r) = (p \/ q) /\ (p \/ r)

def prob1(p,q,r) :
  x=p or (q and r)
  y= (p or q) and (p or r)
  return x==y

tt(prob1,3)

# p/\(q\/r)=(p/\q)\/(p/\r)

def prob2(p,q,r) :
  x=p and ( q or r )
  y=(p and q) or (p and r)
  return x==y

tt(prob2,3)

#~(p/\q)=(~p\/~q)

def prob3(p,q) :
  x=not (p and q)
  y=(not p) or (not q)
  return x==y
tt(prob3,2)

#(~(p\/q))=((~p)/\~q)

def prob4(p, q):
   x = not(p or q)
   y = not p and not q
   return x == y

tt(prob4, 2)

#(p/\(p=>q)=>q)

def prob5(p,q):
  x= p and ( not p or q)
  return not x or q
    
tt(prob5,2)

# (p=>q)=((p\/q)=q)

def prob6(p,q) :
  x = (not p or q)
  y=((p or q) == q)
  return x==y
 
tt(prob6,2)


#((p=>q)=(p\/q))=q
def prob7(p,q):
  if ((not p or q)==(p or q))==q:
    return 1
 
tt(prob7,2)



#(p=>q)=((p/\q)=p)
def prob8(p,q):
  if (not p or q)==((p and q)==p):
    return 1
 
tt(prob8,2)


#((p=>q)=(p/\q))=p

def prob9(p,q):
  if ((not p or q)==(p and q))==p:
    return '1'

tt(prob9,2)



#(p=>q)/\(q=>r)=>(p=>r)
def prob10(p,q,r) :
  x = not ((not p or q) and (not q or r)) or (not p or r)
  return x
 
tt(prob10, 3)


# (p = q) /\ (q => r)  => (p => r)
#answer 1
def prob11(p,q,r) :
  x = not((p is q) and (not q or r)) or (not p or r)
  return x
 
tt(prob11, 3)

#(p=q)/\(q=>r)=>(p=>r)
#answer 2
def prob11(p,q,r):
  x=(p==q) and (not q or r)
  y=not p or r
  return not x or y

tt(prob11,3)

#((p=>q)/\(q=r))=>(p=>r)

def prob12(p,q,r):
  x=(not p or q) and ( q==r )
  y=not p or r
  return not x or y

tt(prob12,3)

#(p=>q)=>((p/\r)=>(q/\r))

def prob13(p,q,r):
  x=not p or q
  y=(not(p and r) or ( q and r))
  return not x or y

tt(prob13,3)

#Question#2----------------------------------------

#(p=>q)=>r=p=>(q=>r)

def prob14(p,q,r):
  x=(not(not p or q) or r)
  y=(not p or (not q or r))
  return x==y

tt(prob14,3) 


def prob15(p, q):
    x = not(p and q)
    y = not p and not q
    return x == y

tt(prob15, 2)


def prob16(p, q):
    x = not(p or q)
    y = not p or not q
    return x == y

tt(prob16, 2)



def prob17(p):
    x = p
    y = not p
    return x == y

tt(prob17, 1)

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 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