#question1.py
def rose(n) :
  if n==0 : 
    yield []
  else :
    for k in range(0,n) :    
      for l in rose(k) :
        for r in rose(n-1-k) :      
            yield [l]+[r]+[r]
            
def start(n) :
  for x in rose(n) : 
      print(x) #basically I am printing x for each rose(n) file


print("starting program: \n")
start(2) # here is where I call the start function

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))

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

def binary_search(arr, target):
    low = 0
    high = len(arr) - 1
    while low <= high:
        mid = (low + high) // 2
        if arr[mid] == target:
            return mid
        elif arr[mid] < target:
            low = mid + 1
        else:
            high = mid - 1
    return -1

https://codecatch.net/post/06c9f5b7-1e00-40dc-b436-b8cccc4b69be

def max_n(lst, n = 1):
  return sorted(lst, reverse = True)[:n]

max_n([1, 2, 3]) # [3]
max_n([1, 2, 3], 2) # [3, 2]