Goal: write a game to play “rock, paper, scissors”
The user chooses one of these, the computer chooses the other
Program prints computer’s selection, who wins
At end, computer prints number of games human won and it won
High-level design:
initialize score
loop
ask user for choice
if quit, exit loop
computer selects one
endloop
print number of games user won, computer won, ties
Part #1: Data
Represent the rock, paper, scissors using strings: “rock”, “paper”, “scissors” (sequence things)
Represent commands as strings as above, plus “quit” (sequence cmdlist)
Store the scores in a dictionary with keys “user”, “computer”, “tie” and integer values (initially set to 0)
Part #2: Functions
Part #3: Refine algorithm
We can now put this into Python (see rps-1.py):
while True: userchoice = getuser(); if (userchoice == "quit"): break compchoice = getcomp(); winner = whowins(userchoice, compchoice) score[winner] += 1 print("You won", score["user"], "game(s), the computer won", end=" ") print(score["computer"], "game(s), and you two tied", score["tie"], "game(s)")
Represent (object1, object2) where object1 beats object2 as a list of pairs called winlist. To see if user won, check if the user-chosen object is the same as the computer-chosen object; if so, there is no winner. Otherwise, check whether (user-chosen object, computer-chosen object) pair is in winlist; if so, the user wins. Otherwise the user loses.
This leads to rps-2.py:
def whowins(user, comp): if user == comp: win = "tie" elif (user, comp) in winlist: win = "user" else: win = "computer" return win
Given the three objects in the sequence things, choose randomly.
This leads to rps-3.py:
def getcomp(): pick = random.choice(things) print("Computer picks", pick) return pick
Loop until you get a valid input. If the user types an end of file (usually control-D) or an interrupt (usually control-C), act as though the user typed “quit”; report any other exceptions and then act as though the user typed “quit”.
This leads to rps-4.py:
def getuser(): while True: try: n = input("Human: enter rock, paper, scissors, quit: ") except (EOFError, KeyboardInterrupt): n = "quit" break except Exception as msg: print("Unknown exception:", msg, "– quitting") n = "quit" break *** check input *** return n
To check input, we need to be sure it’s a valid command, so see if it’s in cmdlist:
if n not in cmdlist: print("Bad input; try again") else: break
Put these together to get the user input routine.
The program now works correctly, but it’s rather unfriendly— the “game(s)” should be “game” or “games” as appropriate, and it should tell the user who wins each round. So we need to add something to the while True loop in the main routine, and change the print statements at the end.
Telling the user who wins is straightforward. Simply put in an if statement at the end of the loop. One tricky point is that there are actually four conditions: winner can take on three known values (“user”, “computer”, and “tie”), and any other unknown value. It should never do the latter, but just in case, we program defensively and put a special case in to catch that. The resulting code is:
if winner == "user": print("You win") elif winner == "computer": print("Computer wins") elif winner == "tie": print("Tie") else: print( "*** INTERNAL ERROR *** winner is", winner) break
Next, the program should distinguish between 1 “game” and any other number of “games” (note you say “0 games” in English). Again, we use an if statement to handle it. Both the computer’s number of games, the user’s number of game, and the number of tie games have to be handled.
print("You won", end=”) if score["user"] == 1: print("1 game, the computer won",end=’ ’) else: print(score["user"], "games, the computer won", end=’ ’) if score["computer"] == 1: print("1 game, and you two tied", end=’ ’) else: print(score["computer"], "games, and you two tied", end=’ ’) if score["tie"] == 1: print("1 game.") else: print(score["tie"], "games.")
The resulting program is rps-5.py.
|
ECS 235A, Computer and Information Security Version of October 27, 2024 at 9:47PM
|
You can also obtain a PDF version of this. |