diff --git a/76 Russian Roulette/python/russianroulette.py b/76 Russian Roulette/python/russianroulette.py index a1391608..508fd15a 100644 --- a/76 Russian Roulette/python/russianroulette.py +++ b/76 Russian Roulette/python/russianroulette.py @@ -1,3 +1,21 @@ +######################################################## +# +# Russian Roulette +# +# From Basic Computer Games (1978) +# +# In this game, you are given by the computer a +# revolver loaded with one bullet and five empty +# chambers. You spin the chamber and pull the trigger +# by inputting a "1", or, if you want to quit, input +# a "2". You win if you play ten times and are still +# alive. +# Tom Adametx wrote this program while a student at +# Curtis Jr. High School in Sudbury, Massachusetts. +# +######################################################## + + from random import random NUMBER_OF_ROUNDS = 9 @@ -14,7 +32,7 @@ def parse_input(): correct_input = False while not correct_input: try: - i = int(input('?')) + i = int(input('? ')) correct_input = True except ValueError: print('Number expected...') @@ -53,3 +71,22 @@ while True: else: print(" Chicken!!!!!\n\n\n") print("...Next victim....") + + +######################################################## +# Porting Notes +# +# Altough the description says that accepts "1" or "2", +# the original game accepts any number as input, and +# if it's different of "2" the program considers +# as if the user had passed "1". That feature was +# kept in this port. +# Also, in the original game you must "pull the trigger" +# 11 times instead of 10 in orden to win, +# given that N=0 at the beginning and the condition to +# win is "IF N > 10 THEN 80". That was fixed in this +# port, asking the user to pull the trigger only ten +# times, tough the number of round can be set changing +# the constant NUMBER_OF_ROUNDS. +# +########################################################