import pandas as pd



x = pd.read_excel(FILE_NAME)

print(x)

#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

def hex_to_rgb(hex):
  return tuple(int(hex[i:i+2], 16) for i in (0, 2, 4))

hex_to_rgb('FFA501') # (255, 165, 1)

#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 key_of_min(d):
  return min(d, key = d.get)

key_of_min({'a':4, 'b':0, 'c':13}) # b

import anytree as at
import random as rm

# Generate a tree with node_count many nodes. Each has a number key that shows when it was made and a randomly selected color, red or white.
def random_tree(node_count):
    
    # Generates the list of nodes
    nodes = []
    for i in range(node_count):
        test = rm.randint(1,2)
        if test == 1:
            nodes.append(at.Node(str(i),color="white"))
        else:
            nodes.append(at.Node(str(i),color="red"))
    
    #Creates the various main branches
    for i in range(node_count):
        for j in range(i, len(nodes)):
            test = rm.randint(1,len(nodes))
            if test == 1 and nodes[j].parent == None and (not nodes[i] == nodes[j]):
                nodes[j].parent = nodes[i]
    
    #Collects all the main branches into a single tree with the first node being the root
    for i in range(1, node_count):
        if nodes[i].parent == None and (not nodes[i] == nodes[0]):
            nodes[i].parent = nodes[0]

    return nodes[0]