Rock Paper Scissor GAME

[perfringo]

I like ThomasL solution for using dictionary but at the same time I am vary of using long handwritten datastructures (matrix dictionary). There is good probability that some typo or simple error will slip in while entering this data. Therefore alternative approaches for consideration focusing on how to determine winner. Following does not provide fully functional code but just ideas how problem can be approached.

If taking away all validation and output formatting it boils down to how to determine winner. Basically there can be two outcomes: (1) draw or (2) either side wins.

Draw is simple - if choices are equal its draw. For winners there is actually only three winning combinations:

>>> first_wins = [('Rock', 'Scissors'), ('Paper', 'Rock'), ('Scissors', 'Paper')]

So it can be expressed:

if user_choice == computer_choice:
    # draw
elif (user_choice, computer_choice) in first_wins:
    # user wins
else:
    # computer wins

Another approach can be considered 'too clever'. One can have possible choices ordered so that next beats previous:

>>> choices = ['Rock', 'Paper', 'Scissors']

The problem is, that 'previous' to 'Rock' should be 'Scissors' which is actually last in list.

Winning combinations using indices are:

0,2 (Rock beat Scissors)
1,0 (Paper beat Rock)
2,1 (Scissors beat Paper)

Loosing combinations using indices are:

2,0 (Scissors defeated by Rock)
0,1 (Rock defeated by Paper)
1,2 (Paper defeated by Scissors)

Doing %3 on these combinations:

>>> (0 - 2) % 3
1
>>> (1 - 0) % 3
1
>>> (2 - 1) % 3
1
>>> (2 - 0) % 3
2
>>> (0 - 1) % 3
2
>>> (1 - 2) % 3
2

As one can see, winning combination is always 1 and loosing one is 2. This can be used something like this:

choices = ['Rock', 'Paper', 'Scissors']
players = ['user', 'computer'] 
user_choice = # get user choice and return corresponding index in choices
computer_choice = random.choice(range(3))  # or random.choice(len(choices))

if user_choice == computer_choice:
    # draw
else:
    # winner = players[(user_choice - computer_choice) % 3 - 1]

Of course, draw is always 3 as same indices return 0 (0-0, 1-1, 2-2) and 0 % 3 is 3. It can be used if deemed useful (a la result = ['user wins', 'computer wins', 'draw']).

I would not argue that these are better options. They are alternative options