def evalpostfix(postfixexpr):
nums = Stack()
tokenlist = list(postfixexpr.replace(' ', ''))
for token in tokenlist:
if token not in prec.keys():
nums.push(token)
else:
right = nums.pop()
left = nums.pop()
expression = left + token + right
result = eval(expression)
nums.push(str(result))
if nums.size() == 1:
return nums.pop()
class DiceResolver:
def __init__(self):
# BEDMAS ==> d()!^/%*+-
# PEMDAS ==> d()!^*/%+-
# Precedence weighting reflects rule; higher means priority
# Close bracket ')' not included; it is a special case (collapses stack to '(')
self.precedence = {
"(": 0,
"+": 3,
"-": 3,
"/": 5,
"*": 5,
"%": 5,
"C": 5,
"c": 5,
"^": 7,
"!": 8,
"d": 9
}
# 's' is a stack used for rearranging infix arguments
self.s = Stack()
# 'q' is a queue used to store the postfix arguments
self.q = Queue()
self.error = False
# Converts a valid infix expression (mathematical expression)
# to postfix using Reverse Polish Notation (RPN). Infix exp-
# ression must be valid; this function can not check validi-
# ty. Note that by design, this only supports integer expr-
# ession (no floating point support). FP support can be add-
# ed if while building numbers, the '.' character is accepted.
# Example: Expression="1 + 2 * 3" --> 7, NOT 9
# RPN="1 2 3 * +" --> 7
# Note that the order of operations is preserved in the RPN.
def infixToRPN(self, expression):
# Since a number may be multiple characters, we start with an empty string,
# and while each character is numeric, we append the number until a non-
# numeric value is encountered.
num = ""
# Tokenize expression character by character
for c in expression:
token = str(c)
# Case: we had accumulated a number but this character is not a
# numeric value; so save accumulated number, and reset accumulator.
if (num != "" and not token.isnumeric()):
self.q.enqueue(num)
num = ""
# We aren't a number; so handle the token
# '(' start brackets are simply markers of what point to return to when
# a ')' close bracket is encountered.
if (token == "("):
self.s.push(token)
# Special case; we look for this first -> it means we have to pop all
# previous values off stack into the RPN queue until we find the '('
elif (token == ")"):
# pop up until the bracket
while (self.s.peek() != "("):
self.q.enqueue(self.s.pop())
# pop the bracket / throw it away (it was just a marker, we're done with it)
self.s.pop()
# Casee: operator handling
# we are done handling brackets, check for a valid operator
elif (token in self.precedence):
while self.s.size() != 0 and (self.precedence[token] <=
self.precedence[self.s.peek()]):
self.q.enqueue(self.s.pop())
self.s.push(token)
# Case: character is numeric.
# Append to accumulator and continue parsing
elif (token.isnumeric()):
num += token
# Did token end on a number? If so store accumulated number in RPN queue
if (num != ""):
self.q.enqueue(num)
# Now pop items from stack to the queue to cleanup
while (self.s.size() != 0):
self.q.enqueue(self.s.pop())
# At this point, we have a valid RPN in the 'q' queue
# (if the infix expression was valid)
# Let's return a string version:
q_cp = self.q.copy()
rpn = ""
for c in q_cp:
rpn += c + " "
return (rpn)
# Routine to calculate a factorial
def factorial(self, value):
if (value < 0):
return (0)
elif (value == 0 or value == 1):
return (1)
elif (value == 2):
return (2)
product = value
for x in range(2, value):
product = product * x
return (product)
# Routine to calculate "choose" (combinatorics)
# Formula:
# nCr (n Choose r) = n! / r!(n-r)!
def choose(self, n, r):
numerator = self.factorial(n)
denominator = self.factorial(r) * self.factorial(n - r)
# Sanity
if (denominator == 0):
return (0)
# Compute
# NOTE: Should always be an integer result, but cast
# it anyways to be safe
return (int(numerator / denominator))
# Given left value, right value, and an operator, calculate.
def calculate(self, left, right, op):
if (op == "+"):
return (left + right)
elif (op == "-"):
return (left - right)
elif (op == "*"):
return (left * right)
elif (op == "/"):
return (int(left / right))
elif (op == "^"):
return (left**right)
elif (op == "%"):
return (left % right)
elif (op == "!"):
return (self.factorial(left))
elif (op == "c" or op == "C"):
return (self.choose(left, right))
# dice roll; handled with 'random'
# NOTE: expressions without 'd' are deterministic;
# expressions with 'd' are non-deterministic (variable
# outcomes).
elif (op == "d"):
sum = 0
# Left value is number of rolls; right value is die
# IE 3d6 = 3 rolls of a 6 sided die, summed.
for i in range(left):
sum += random.randint(1, right)
return (sum)
# whoops shouldn't have happened try to be graceful
return (0)
# Nifty little stack and queue algorithm for evaluating
# the RPN. Expects a valid RPN expression.
def evaluateRPN(self):
workstack = Stack()
# As we pull tokens from the queue, we validate them and if neither a number
# nor an operator, we abort with an error.
for t in self.q.copy():
if (t in self.precedence):
# As we work backwards, right value is first; validate
right = workstack.pop()
if (not str(right).isnumeric()
and not right in self.precedence):
self.error = True
break
# Now get left value, validate
# Special case: ! only takes one argument. Make them identical
if (t == "!"):
left = right
else:
left = workstack.pop()
if (not str(left).isnumeric()
and not left in self.precedence):
self.error = True
break
# Both valid, so calculate
workstack.push(self.calculate(left, right, t))
else:
workstack.push(int(t))
# answer is now on the stack
if (not self.error):
return (workstack.pop())
else:
return (0)
# One function to handle it all. How Pythonic.
def resolve(self, expression, repeat=False):
if not repeat:
self.error = False
self.q.clear()
self.s.clear()
self.infixToRPN(expression)
return (self.evaluateRPN())
else:
# Repeat=True
# This allows repeat dice rolls / calculations, without rebuilding
# the RPN queue each time.
return (self.evaluateRPN())
# Heuristic to calculate expression distribution, ment to be
# used with dice rolls (ie, 2d6). This is done by repeating rolls
# to a cap of n trials, then assessing the results.
# Returns a histogram report with trial results, mean and mode.
def getHistogram(self, expression, trials):
# Validate min/max boundaries
if (trials < 0):
trials = 1
elif (trials > 1000000):
trials = 1000000
# Initialize
sb = ""
rolls = dict()
sum = 0
pct = 1.0
max = dict()
result = dict()
# Build
for i in range(trials):
if i == 0:
roll = self.resolve(expression)
else:
# We already built the RPN, don't waste cycles
# reuilding it on every iteration, set repeat=True.
roll = self.resolve(expression, repeat=True)
# Track the recurrences of roll values
if (roll in rolls.keys()):
rolls[roll] += 1
else:
rolls[roll] = 1
# Nifty way to build a key sorted report
keys = list(rolls.keys())
keys.sort()
# Report
sb = f"DISTRIBUTION HISTORGRAM ({trials:,} trials):\n"
max[0] = max[1] = 0
for key in keys:
result[key] = rolls[key]
sum += key * rolls[key]
pct = float(rolls[key]) / float(trials, ) * 100.0
if (pct > max[0]):
max[0] = pct
max[1] = key
sb += f"[{key:3}] ==> {rolls[key]:,} ({pct:.2f}%)\n"
# Stash pct for later
rolls[key] = pct
mean = float(sum) / float(trials) + 0.5
sb += f"Mean: {mean:.2f}\n"
mode = max[1]
sb += f"Mode: {mode}\n\n"
# Scaling calculation uses a lambda function
scale = lambda x, y: int(float(x / 100) * float(y) + 0.5)
# Build histogram pictogragh
pic = "PICTORIAL HISTOGRAM:\n"
for key in keys:
pic += f"[{key:3}] "
for i in range(scale(rolls[key], 160)):
pic += "*"
pic += "\n"
# Send back the report
return (sb + pic)