isEvens

"Writing bad code can only be achieve through deep knowledge of good code" ~ Socrates (Maybe)

Canonical Way

Boring, simple, standard. Weak.

def isEven(num):
  return num % 2 == 0

The Typecast Way:

Floats? What floats?

def isEven(num):
  return int(num / 2) == num / 2

The Loop way

Just leave it, and grab a cup of tea

def isEven(num):
  while num >= 1:
    print(f"isEven({num})")
    num -= 2
  return (num == 0)

The Mathematical Way

Proved by induction, hence it makes it smarter

def isEven(num):
  '''
  Suppose x is an even number. This means we can write x = 2k for some integer k.
  '''
  for k in range(1,num/2):
    if (num == 2 * k):
      return True
  return False

def isEvenInduction(num):
  '''
  Proof by induction
  '''
  nums = [i for i in range(0, num, 2)]
  result = map(isEvenBigBrain, nums)
  return all(result) and num in nums

The Last Character Way:

This one will actually work with negative numbers!

def isEven(num):
  possible_endings = ['0','2','4','6','8']
  last_char = str(num)[-1]
  if last_char in possible_endings:
    return True
  return False

Recursive way

Recursion depth is merely recommendation.

def isEven(num):
  if (num == 0):
    return True
  elif (num == 1):
    return False
  else:
    return isEven(int(num)-2)

The Dictionary Way

Its technically O(1), you may or may not smell your RAM

# Preprocessing the Dictionary
def pre_process_dict():
  even_dict = {}
  for i in range(0,sys.maxsize-1,2):
    even_dict[i] = True
    even_dict[i+1] = False
  
  return even_dict

# Is Even with the Dict, O(1) average
def isEven(num,even_dict):
  return even_dict[int(num)]

The Outsourced way:

No, how about figure it out for yourself

def isEven(num):
    while (True):
        res = input(f"Is this {num} an even number? (y/n): ")
        if (res == 'y' and (int(num) % 2 == 0)):
            return True
        elif (res == 'n' and not (int(num) % 2 == 0)):
            return False
        else:
            print("Are you sure?")

HaoCheong/isEvens