# question3.py
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('')

# this is the logic for part A (p\/q\/r) /\ (p\/q\/~r) /\ (p\/~q\/r) /\ (p\/~q\/~r) /\ (~p\/q\/r) /\ (~p\/q\/~r) /\ (~p\/~q\/r) /\ (~p\/~q\/~r)

def parta(p,q,r) :
  
  a=(p or q or r) and (p or q or not r) and (p or not q or r)and (p or not q or  not r)
  b=(not p or q or r ) and (not p or q or not r) and (not p or not q or r) and (not p or not q or not r)
  c= a and b
  return c

def partb(p,q,r) :
  a=(p or q and r) and (p or not q or not r) and (p or not q or not  r)and (p or q or  not r)
  b=(not p or q or r ) and (not p or q or not r) and (not p or not q or r) and (not p or not q or not r)
  c= a and b
  return c

print("part A:")
tt(parta,3)

print("part B:")
tt(partb,3)

def byte_size(s):
  return len(s.encode('utf-8'))
  
byte_size('😀') # 4
byte_size('Hello World') # 11

# Python program for Bitonic Sort. Note that this program
# works only when size of input is a power of 2.

# The parameter dir indicates the sorting direction, ASCENDING
# or DESCENDING; if (a[i] > a[j]) agrees with the direction,
# then a[i] and a[j] are interchanged.*/
def compAndSwap(a, i, j, dire):
	if (dire==1 and a[i] > a[j]) or (dire==0 and a[i] > a[j]):
		a[i],a[j] = a[j],a[i]

# It recursively sorts a bitonic sequence in ascending order,
# if dir = 1, and in descending order otherwise (means dir=0).
# The sequence to be sorted starts at index position low,
# the parameter cnt is the number of elements to be sorted.
def bitonicMerge(a, low, cnt, dire):
	if cnt > 1:
		k = cnt/2
		for i in range(low , low+k):
			compAndSwap(a, i, i+k, dire)
		bitonicMerge(a, low, k, dire)
		bitonicMerge(a, low+k, k, dire)

# This funcion first produces a bitonic sequence by recursively
# sorting its two halves in opposite sorting orders, and then
# calls bitonicMerge to make them in the same order
def bitonicSort(a, low, cnt,dire):
	if cnt > 1:
		k = cnt/2
		bitonicSort(a, low, k, 1)
		bitonicSort(a, low+k, k, 0)
		bitonicMerge(a, low, cnt, dire)

# Caller of bitonicSort for sorting the entire array of length N
# in ASCENDING order
def sort(a,N, up):
	bitonicSort(a,0, N, up)

# Driver code to test above
a = [3, 7, 4, 8, 6, 2, 1, 5]
n = len(a)
up = 1

sort(a, n, up)
print ("\n\nSorted array is")
for i in range(n):
	print("%d" %a[i]),

#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}
'''

color2 = (60, 74, 172)
color1 = (19, 28, 87)

percent = 1.0
for i in range(101):
    resultRed = round(color1[0] + percent * (color2[0] - color1[0]))
    resultGreen = round(color1[1] + percent * (color2[1] - color1[1]))
    resultBlue = round(color1[2] + percent * (color2[2] - color1[2]))
    print((resultRed, resultGreen, resultBlue))
    percent -= 0.01

def format_timestamp(timestamp_epoch):
    """
    Convert epoch timestamp to formatted datetime string without using datetime package.
    
    Args:
        timestamp_epoch (int/float): Unix epoch timestamp (seconds since 1970-01-01 00:00:00 UTC)
        
    Returns:
        str: Formatted datetime string in 'YYYY-MM-DD HH:MM:SS' format
    """
    # Constants for time calculations
    SECONDS_PER_DAY = 86400
    SECONDS_PER_HOUR = 3600
    SECONDS_PER_MINUTE = 60
    
    # Handle negative timestamps and convert to integer
    timestamp = int(timestamp_epoch)
    
    # Calculate days since epoch and remaining seconds
    days_since_epoch = timestamp // SECONDS_PER_DAY
    remaining_seconds = timestamp % SECONDS_PER_DAY
    
    # Calculate hours, minutes, seconds
    hours = remaining_seconds // SECONDS_PER_HOUR
    remaining_seconds %= SECONDS_PER_HOUR
    minutes = remaining_seconds // SECONDS_PER_MINUTE
    seconds = remaining_seconds % SECONDS_PER_MINUTE
    
    # Calculate date (simplified, ignoring leap seconds)
    year = 1970
    days = days_since_epoch
    while days >= 365:
        is_leap = (year % 4 == 0 and year % 100 != 0) or (year % 400 == 0)
        days_in_year = 366 if is_leap else 365
        if days >= days_in_year:
            days -= days_in_year
            year += 1
    
    # Month lengths (non-leap year for simplicity, adjusted later for leap years)
    month_lengths = [31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31]
    if (year % 4 == 0 and year % 100 != 0) or (year % 400 == 0):
        month_lengths[1] = 29
    
    month = 0
    while days >= month_lengths[month]:
        days -= month_lengths[month]
        month += 1
    
    # Convert to 1-based indexing for month and day
    month += 1
    day = days + 1
    
    # Format the output string
    return f"{year:04d}-{month:02d}-{day:02d} {hours:02d}:{minutes:02d}:{seconds:02d}"

# Example timestamp (Unix epoch seconds)
timestamp = 1697054700
formatted_date = format_timestamp(timestamp)
print(formatted_date + " UTC")  # Output: 2023-10-11 18:45:00