How do I logically check for a royal flush in poker? - python

Code I'm using for card checking
def pair(player_hand):
for x in range(1,len(player_hand)):
if player_hand[x-1][0] == player_hand[x][0]:
return True
return False
def triple(player_hand):
for x in range(2, len(player_hand)):
if player_hand[x][0] == player_hand[x-2][0]:
return True
return False
def flush(player_hand):
if player_hand[0][1] == player_hand[1][1]==player_hand[2][1]==player_hand[3][1]==player_hand[4][1]:
return True
else:
return False
def straight(player_hand):
for x in range(0,len(player_hand)-1):
if(rank(player_hand[x]) + 1 != rank(player_hand[x+1])):
return False
return True
I know how to do a normal flush
def flush(player_hand):
if player_hand[0][1] == player_hand[1][1]==player_hand[2][1]==player_hand[3][1]==player_hand[4][1]:
return True
else:
return False
But how do I check for a specific order of cards for royal flush?

Related

CS50P PSETS 2, Vanity plates

I'm having some issues with the psets 2 of cs50p, precisely I'm talking about the "Vanity Plates" problem, where I fulfilled all requests except one, which said:
“Numbers cannot be used in the middle of a plate; they must come at the end. For example, AAA222 would be an acceptable … vanity plate; AAA22A would not be acceptable. The first number used cannot be a ‘0’.” Can you help me? Thank's
this is the code I wrote so far:
def main():
plate = input("Plate: ")
if is_valid(plate):
print("Valid")
else:
print("Invalid")
def is_valid(s):
if s.isalnum() | s[:2].isalpha() | 2 < len(s) < 6 | :
else:
return False
main()

you have to consider all the cases one by one, this is how I solved it:
def main():
plate = input("Plate: ")
if is_valid(plate):
print("Valid")
else:
print("Invalid")
def is_valid(s):
if len(s) < 2 or len(s) > 6:
return False
elif not s[0].isalpha() or not s[1].isalpha():
return False
elif checkFirstZero(s):
return False
elif checkMiddleZero(s):
return False
elif last(s):
return False
elif worng(s):
return False
return True
def last(s):
isAlp = False
isNum = False
for w in s:
if not w.isalpha():
isNum = True
else:
if isNum:
return True
return False
def checkCuntuNNumber(s):
isFirstTry = True
isNum = False
for w in s:
if not w.isalpha():
if isFirstTry:
isNum = True
isFirstTry = False
if isNum and s[-1].isalpha():
return True
def checkMiddleZero(s):
isFirstTry = True
isNum = False
for w in s:
if not w.isalpha():
if isFirstTry:
isNum = True
isFirstTry = False
if isNum and s[-1].isalpha():
return True
else:
return False
def checkFirstZero(s):
for w in s:
if not w.isalpha():
if int(w) == 0:
return True
else:
return False
def worng(s):
for w in s:
if w in [" ", ".", ","]:
return True
return False
main()

This is how I did it. I am sure there is an easier way to do it out there but hopefully this helps :)
characters = ['A','B','C','D','E','F','G','H','I','J','K','L','M','N','O','P','Q','R','S','T','U','V','W','X','Y','Z']
numbers = ['1','2','3','4','5','6','7','8','9','0']
def main ():
plate = (input ("Plate: ")).upper()
if is_valid(plate):
print ('Valid')
else:
print ('Invalid')
def is_valid (s):
#Check whether length is between 2 and 6 included
if len(s) < 2 or len(s) > 6:
return False
elif char_check(s):
return False
elif char_start(s):
return False
elif zero_check(s):
return False
elif alpha_follow_check (s):
return False
else:
return True
#Check for valid characters
def char_check(s):
for i in s:
if not (i in characters or i in numbers):
return True
#Check whether first two are letters
def char_start (s):
for i in s[:2]:
if not i in characters:
return True
#Check if zero is first number listed
def zero_check (plate_response):
length_string = len (plate_response)
letter_position = 0
number_present = 0
zero_position = None
if any (i in numbers for i in plate_response):
for i in plate_response [0:length_string]:
if i == '0':
zero_position = letter_position
break
letter_position = letter_position + 1
for i in plate_response [0:zero_position]:
if i in numbers:
number_present = 1
if number_present == 0:
return True
else:
return False
#Check alphabet follows numbers
def alpha_follow_check (plate_response):
length_string = len (plate_response)
letter_position = 0
number_position = None
if any (i in numbers for i in plate_response):
for i in plate_response [0:length_string]:
if i in numbers:
number_position = letter_position
break
letter_position = letter_position + 1
for i in plate_response [number_position:length_string]:
if i in characters:
return True
else:
return False
main ()

idk if will help, but the part that i've had the most difficulty in this problem was: "Numbers cannot be used in the middle of a plate; they must come at the end, AAA22A would not be acceptable", then i learned that you can create a full list from the plate that the user inputed, and how to actually use it, with the:
ls = list(s)
for i in range(len(ls)):
After that, we check when the first number appears. "if == '0'" ,then returns False to the function.
After that, if the first number isn't a 0, the program checks if the next item in that list is letter, and, if it is, also return False.
i < len(ls) -1 => this part guarantee that the program will not run in the last item of the list
ls[i+1].isalpha() => and this part check that, if the item on the list was a number, and then the next item is a letter, it returns False
I hope it helps someone, i've spend a lot of time trying to figure it out what to do, and then reached this solution: "for i in range(len(ls))".
Now my code is complete and working.
My code:
def main():
plate = input("Plate: ")
if is_valid(plate):
print("Valid")
else:
print("Invalid")
def is_valid(s):
if not s.isalnum():
return False
elif len(s) < 4 or len(s) > 7:
return False
elif s[0].isdigit()or s[1].isdigit():
return False
elif s[-1].isalpha() or s[-2].isalpha():
return False
else:
ls = list(s)
for i in range(len(ls)):
if ls[i].isdigit():
if ls[i] == '0':
return False
elif i < len(ls) -1 and ls[i+1].isalpha():
return False
else:
return True
main()

Python: Why donnot I get a result in the terminal

Python: Why don't I get a result in the terminal
def is_even(number):
if number % 2 == 0:
return True
return False
is_even(10)

You need to print the output of the function. You also need to create the function.
def is_even(num):
if num % 2 == 0:
return True
else:
return False
print(is_even(10)) #True
print(is_even(7)) #False

Finding Cyclops Numbers in Python

I am working on a problem in which I need to return True or False after checking to see whether a number is a cyclops number or not. A cyclops number is made up of odd number of digits, consists of only one zero and that zero is located in the middle. Here's what I have so far:
def is_cyclops(n):
strNum = str(n)
for i, el in enumerate(strNum):
if(len(strNum) % 2 == 0):
return False
else:
# find middle number is zero
# no other zeros exist
# return True
is_cyclops(0) # True
is_cyclops(101) # True
is_cyclops(1056) # False
is_cyclops(675409820) # False
How can I find the median number (without using numpy) & ensure it is a zero, and it is the only zero that exists in that number?

This worked for me:
def is_cyclops(num: int) -> bool:
str_ = str(num)
if not len(str_) % 2:
return False
if not str_.count('0') == 1:
return False
mid_index = len(str_) // 2
if str_[mid_index] == '0':
return True
return False
print(
is_cyclops(0),
is_cyclops(101),
is_cyclops(1056),
is_cyclops(675409820)
)
Output:
True True False False

As it looks like you've had a good attempt here, I'll help out.
def is_cyclops(n):
strNum = str(n)
if(len(strNum) % 2 == 0):
return False
else:
middle_index = len(strNum)//2
if strNum[middle_index] != "0": return False # find middle number is zero
if strNum.count("0") > 1: return False # no other zeros exist
return True
is_cyclops(0) # True
is_cyclops(101) # True
is_cyclops(1056) # False
is_cyclops(675409820) # False

In a nested if loop, how to return true for multiple matching conditions?

In the below program, even though all the if conditions are matching, it returns true just once. How do i make it return true and print as many times as the conditions match?
lotto_numbers = [1,1,1]
fireball_number = 1
user_input1 = user_input2 = user_input3 = 1
def fbcheck():
if lotto_numbers == [user_input1,user_input2,fireball_number]:
return True
elif lotto_numbers == [fireball_number, user_input2, user_input3]:
return True
elif lotto_numbers == [user_input1, fireball_number, user_input3]:
return True
else:
return False
if (fbcheck() == True):
print ('you won')

You can use all:
def fbcheck():
user_data = [user_input1,user_input2,fireball_number]
lotto_numbers = [1,1,1]
print([a==b for a, b in zip(lotto_numbers, user_data)])
return all(a==b for a, b in zip(lotto_numbers, user_data))
print(fbcheck())
Output:
[True, True, True]
True

Are booleans overwritten in python?

def space_check(board, position):
return board[position] == ' '
def full_board_check(board):
for i in range(1,10):
if space_check(board, i):
return False
return True
the last line is return True
why not else: return True
if the if statement returned false, won't the last return True overwrite it??

If it was
for i in range(1,10):
if space_check(board, i):
return False
else:
return True
then after the first iteration in the for loop the function would return. This would not lead the the expected behaviour. Currently, you check every space, and not just the first one

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