CODEWITHSUNDEEP
python
mathshungry
mathshungry

Reputation: 153

A very simple program:Rock Paper Scissors (python)

Usual Rock Paper Scissors, but my if-else statement doesn't work. I would to appreciate if anyone could spot my errors.

def main():

    #ask for input and create a list to check valid input
    list = ['rock', 'paper', 'scissors']
    player_A = input('Rock, paper or scissors? ').lower()
    player_B = input('Rock, paper or scissors? ').lower()

    if player_A not in list or player_B not in list:
        print('Invalid value, please enter rock, paper or scissors')
        return
    else:
        rule(player_A,player_B)

def rule(x,y):

    if (x,y == ['rock', 'paper']) or (x,y == ['paper', 'scissors']) or (x,y == ['scissors', 'rock']):
        print('{} wins'.format(y))
        print('{} loses'.format(x))
    elif (x,y == ['paper', 'rock']) or (x,y == 'scissors', 'paper') or (x,y == ['rock', 'scissors']):
        print('{} wins'.format(x))
        print('{} wins'.format(y))
    else:
        print('fair')

main()

Also, I would also like to know if there is a faster way to determine the win condition, seems quite clumsy to check case by case.

Upvotes: 0

Views: 396

Answers (3)

Ilija
Ilija

Reputation: 1604

You can define a list of all conditions when first player wins. Then you try other way around and if both are the same then its fair.

P, S, R = 'paper', 'scissors', 'rock'

WINS = ((P, R), (R, S), (S, P))


def main():
    # ask for input and create a list to check valid input
    lst = [P, R, S]

    player_A = input('Rock, paper or scissors? ').lower()
    player_B = input('Rock, paper or scissors? ').lower()

    if player_A not in lst or player_B not in lst:
        print('Invalid value, please enter rock, paper or scissors')
        return
    else:
        rule(player_A, player_B)


def rule(x, y):
    if (x, y) in WINS:
        print('{} wins'.format(x))
        print('{} loses'.format(y))
    elif (y, x) in WINS:
        print('{} wins'.format(y))
        print('{} loses'.format(x))
    else:
        print('fair')


main()

You can even make conditions checking simpler with this:

def rule(x, y):
    if x == y:
        print('fair')
    elif (x, y) in WINS:
        print('{} wins'.format(x))
        print('{} loses'.format(y))
    else:
        print('{} wins'.format(y))
        print('{} loses'.format(x))

Upvotes: 1

jpp
jpp

Reputation: 164673

I've corrected several issues / errors:

  • Never name a variable after a class, so use lst instead of list.
  • Compare tuples instead of lists, as so: (x, y) = ('scissors', 'paper')
  • In 3 of your 6 scenarios, both players win.

The below code should work as expected:

def main():

    #ask for input and create a list to check valid input
    lst = ['rock', 'paper', 'scissors']
    player_A = input('Rock, paper or scissors? ').lower()
    player_B = input('Rock, paper or scissors? ').lower()

    if player_A not in lst or player_B not in lst:
        print('Invalid value, please enter rock, paper or scissors')
        return
    else:
        rule(player_A,player_B)

def rule(x,y):
    if ((x,y) == ('rock', 'paper')) or ((x,y) == ('paper', 'scissors')) or ((x,y) == ('scissors', 'rock')):
        print('{} wins'.format(y))
        print('{} loses'.format(x))
    elif ((x,y) == ('paper', 'rock')) or ((x,y) == ('scissors', 'paper')) or ((x,y) == ('rock', 'scissors')):
        print('{} wins'.format(x))
        print('{} loses'.format(y))
    else:
        print('fair')

main()

Upvotes: 2

Matias Cicero
Matias Cicero

Reputation: 26281

Those conditions are wrong:

x, y == ['rock', 'paper'] will not result in True or False, but in a tuple, with the first value being x and the second value being False (y == ['rock', 'paper']).

This is because the comma has more precedence here.

To fix this, you need to lose the ambiguity. You can either:

  • Create an array with the values of x and y

  • Compare both of them individually.

The first option is the simplest:

[x, y] == ['rock', 'paper']

The second option is not as pretty, but it's more explicit:

x == 'rock' and y == 'paper'

Upvotes: 3

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