#Sets
U = {0,1,2,3,4,5,6,7,8,9}
P = {1,2,3,4}
Q = {4,5,6}
R = {3,4,6,8,9}

def set2bits(xs,us) :
    bs=[]
    for x in us :
        if x in xs :
            bs.append(1)
        else:
            bs.append(0)
    assert len(us) == len(bs)
    return bs

def union(set1,set2) :
    finalSet = set()
    bitList1 = set2bits(set1, U)
    bitList2 = set2bits(set2, U)

    for i in range(len(U)) :
        if(bitList1[i] or bitList2[i]) :
            finalSet.add(i)

    return finalSet

def intersection(set1,set2) :
    finalSet = set()
    bitList1 = set2bits(set1, U)
    bitList2 = set2bits(set2, U)

    for i in range(len(U)) :
        if(bitList1[i] and bitList2[i]) :
            finalSet.add(i)

    return finalSet

def compliment(set1) :
    finalSet = set()
    bitList = set2bits(set1, U)

    for i in range(len(U)) :
        if(not bitList[i]) :
            finalSet.add(i)

    return finalSet

def implication(a,b):
    return union(compliment(a), b)

###########################################################################################
######################         Problems 1-6         #######################################
###########################################################################################

#p \/ (q /\ r) = (p \/ q) /\ (p \/ r)
def prob1():
    return union(P, intersection(Q,R)) == intersection(union(P,Q), union(P,R))

#p /\ (q \/ r) = (p /\ q) \/ (p /\ r)
def prob2():
    return intersection(P, union(Q,R)) == union(intersection(P,Q), intersection(P,R))

#~(p /\ q) = ~p \/ ~q
def prob3():
    return compliment(intersection(P,R)) == union(compliment(P), compliment(R))

#~(p \/ q) = ~p /\ ~q
def prob4():
    return compliment(union(P,Q)) == intersection(compliment(P), compliment(Q))

#(p=>q) = (~q => ~p)
def prob5():
    return implication(P,Q) == implication(compliment(Q), compliment(P))

#(p => q) /\ (q => r)  =>  (p => r)
def prob6():
    return implication(intersection(implication(P,Q), implication(Q,R)), implication(P,R))


print("Problem 1: ", prob1())
print("Problem 2: ", prob2())
print("Problem 3: ", prob3())
print("Problem 4: ", prob4())
print("Problem 5: ", prob5())
print("Problem 6: ", prob6())

'''
Problem 1:  True
Problem 2:  True
Problem 3:  True
Problem 4:  True
Problem 5:  True
Problem 6:  {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}
'''

def get_ldap_user(member_cn, user, passwrd):
    '''
    Get an LDAP user and return the SAMAccountName
    '''
    #---- Setting up the Connection
    #account used for binding - Avoid putting these in version control
    bindDN = str(user) + "@unt.ad.unt.edu"
    bindPass = passwrd

    #set some tuneables for the LDAP library.
    ldap.set_option(ldap.OPT_X_TLS_REQUIRE_CERT, ldap.OPT_X_TLS_ALLOW)
    #ldap.set_option(ldap.OPT_X_TLS_CACERTFILE, CACERTFILE)
    
    conn = ldap.initialize('ldaps://unt.ad.unt.edu')
    conn.protocol_version = 3
    conn.set_option(ldap.OPT_REFERRALS, 0)

    #authenticate the connection so that you can make additional queries
    try:
        result = conn.simple_bind_s(bindDN, bindPass) 
    except ldap.INVALID_CREDENTIALS:
        result = "Invalid credentials for %s" % user
        sys.exit()

    #build query in the form of (uid=user)
    ldap_query = '(|(displayName=' + member_cn + ')(cn='+ member_cn + ')(name=' + member_cn + '))'
    ldap_info = conn.search_s('DC=unt,DC=ad,DC=unt,DC=edu', ldap.SCOPE_SUBTREE, filterstr=ldap_query)
    sAMAccountName = str(ldap_info[0][1]['sAMAccountName']).replace("[b'", "").replace("']","")
    return sAMAccountName

def when(predicate, when_true):
  return lambda x: when_true(x) if predicate(x) else x

double_even_numbers = when(lambda x: x % 2 == 0, lambda x : x * 2)
print(double_even_numbers(2)) # 4
print(double_even_numbers(1)) # 1

# Python program for implementation of Radix Sort
	
# A function to do counting sort of arr[] according to
# the digit represented by exp.
def countingSort(arr, exp1):
	
	n = len(arr)
	
	# The output array elements that will have sorted arr
	output = [0] * (n)
	
	# initialize count array as 0
	count = [0] * (10)
	
	# Store count of occurrences in count[]
	for i in range(0, n):
		index = (arr[i]/exp1)
		count[int((index)%10)] += 1
	
	# Change count[i] so that count[i] now contains actual
	# position of this digit in output array
	for i in range(1,10):
		count[i] += count[i-1]
	
	# Build the output array
	i = n-1
	while i>=0:
		index = (arr[i]/exp1)
		output[ count[ int((index)%10) ] - 1] = arr[i]
		count[int((index)%10)] -= 1
		i -= 1
	
	# Copying the output array to arr[],
	# so that arr now contains sorted numbers
	i = 0
	for i in range(0,len(arr)):
		arr[i] = output[i]

# Method to do Radix Sort
def radixSort(arr):

	# Find the maximum number to know number of digits
	max1 = max(arr)

	# Do counting sort for every digit. Note that instead
	# of passing digit number, exp is passed. exp is 10^i
	# where i is current digit number
	exp = 1
	while max1/exp > 0:
		countingSort(arr,exp)
		exp *= 10

# Driver code to test above
arr = [ 170, 45, 75, 90, 802, 24, 2, 66]
radixSort(arr)

for i in range(len(arr)):
	print(arr[i]),

def generate_pascals_triangle(num_rows):
    triangle = []
    for row in range(num_rows):
        # Initialize the row with 1
        current_row = [1]

        # Calculate the values for the current row
        if row > 0:
            previous_row = triangle[row - 1]
            for i in range(len(previous_row) - 1):
                current_row.append(previous_row[i] + previous_row[i + 1])

        # Append 1 at the end of the row
        current_row.append(1)

        # Add the current row to the triangle
        triangle.append(current_row)

    return triangle


def display_pascals_triangle(triangle):
    for row in triangle:
        for number in row:
            print(number, end=" ")
        print()


# Prompt the user for the number of rows
num_rows = int(input("Enter the number of rows for Pascal's Triangle: "))

# Generate Pascal's Triangle
pascals_triangle = generate_pascals_triangle(num_rows)

# Display Pascal's Triangle
display_pascals_triangle(pascals_triangle)

# Python program to reverse a linked list
# Time Complexity : O(n)
# Space Complexity : O(n) as 'next'
#variable is getting created in each loop.

# Node class


class Node:

	# Constructor to initialize the node object
	def __init__(self, data):
		self.data = data
		self.next = None


class LinkedList:

	# Function to initialize head
	def __init__(self):
		self.head = None

	# Function to reverse the linked list
	def reverse(self):
		prev = None
		current = self.head
		while(current is not None):
			next = current.next
			current.next = prev
			prev = current
			current = next
		self.head = prev

	# Function to insert a new node at the beginning
	def push(self, new_data):
		new_node = Node(new_data)
		new_node.next = self.head
		self.head = new_node

	# Utility function to print the linked LinkedList
	def printList(self):
		temp = self.head
		while(temp):
			print temp.data,
			temp = temp.next


# Driver program to test above functions
llist = LinkedList()
llist.push(20)
llist.push(4)
llist.push(15)
llist.push(85)

print "Given Linked List"
llist.printList()
llist.reverse()
print "\nReversed Linked List"
llist.printList()