Re: Dice probability problem
On 2006-04-05, Tomi Lindberg <[EMAIL PROTECTED]> wrote:
> Antoon Pardon wrote:
>
>> def __rmul__(self, num):
>> tp = num * [self]
>> return reduce(operator.add, tp)
>>
>> sum3d6 = 3 * D(6)
>
> One basic question: is there any particular reason not to
> use __mul__ instead (that would allow me to use both 3 *
> D(6) and D(6) * 3, while __rmul__ raises an AttributeError
> with the latter)?
Well 3 * D(6) is similar to the notation used in roleplaying,
while D(6) * 3 would make me think of the distribution
{3:1, 6:1, 9:1, 12:1, 15:1, 18:}
> Difference between the two methods is
> slightly unclear to me.
I have to look it up myself regularly. But in this case
it was more a matter of some intuition that 3 * D(6)
was not the same as D(6) * 3. You may have a different
intuition.
--
Antoon Pardon
--
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Re: Dice probability problem
Antoon Pardon wrote: > def __rmul__(self, num): > tp = num * [self] > return reduce(operator.add, tp) > > sum3d6 = 3 * D(6) One basic question: is there any particular reason not to use __mul__ instead (that would allow me to use both 3 * D(6) and D(6) * 3, while __rmul__ raises an AttributeError with the latter)? Difference between the two methods is slightly unclear to me. Thanks, Tomi Lindberg -- http://mail.python.org/mailman/listinfo/python-list
Re: Dice probability problem
Tomi Lindberg <[EMAIL PROTECTED]> writes:
> # Adds another die to results.
> def add_dice(sums, die):
> # If first die, all values appear once
I'd add something like
sums = sums or {}
because otherwise your function will sometimes mutate sums and sometimes
return a fresh object, which usually is a very bad thing as it can easily lead
to quite nasty bugs.
> if not sums:
> for face in die:
> sums[face] = 1
> # Calculating the number of appearances for additional
> # dice
> else:
> new_sums = {}
> for k in sums.keys():
> for f in die:
> if new_sums.has_key(k+f):
> new_sums[k+f] += sums[k]
> else:
> new_sums[k+f] = sums[k]
> sums = new_sums
> return sums
'as
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Re: Dice probability problem
Antoon Pardon wrote: > IMO you are making things too complicated and not general > enough. I believe that the above is very likely more than just your opinion :) Programming is just an occasional hobby to me, and I lack both experience and deeper (possibly a good chunk of shallow as well) knowledge on the subject. I'll study the code you posted, and make further questions if something remains unclear afterwards. Thanks, Tomi Lindberg -- http://mail.python.org/mailman/listinfo/python-list
Re: Dice probability problem
Op 2006-04-04, Tomi Lindberg schreef <[EMAIL PROTECTED]>:
> First, thanks to Antoon and Alexander for replying.
>
> Antoon Pardon wrote:
>
>> It would be better to construct distributions for one
>> die and make a function that can 'add' two distributions
>> together.
>
> As both replies pointed to this direction, I tried to take
> that route. Here's the unpolished code I came up with. Does
> it look even remotely sane way to accomplish my goal?
> -- code begins --
>
> # A die with n faces
> D = lambda n: [x+1 for x in range(n)]
>
> # A new die with 6 faces
> d6 = D(6)
>
> # Adds another die to results.
> def add_dice(sums, die):
> # If first die, all values appear once
> if not sums:
> for face in die:
> sums[face] = 1
> # Calculating the number of appearances for additional
> # dice
> else:
> new_sums = {}
> for k in sums.keys():
> for f in die:
> if new_sums.has_key(k+f):
> new_sums[k+f] += sums[k]
> else:
> new_sums[k+f] = sums[k]
> sums = new_sums
> return sums
>
> sums = add_dice({}, d6)
> sums = add_dice(sums, d6)
> sums = add_dice(sums, d6)
>
> -- code ends --
IMO you are making things too complicated and not general
enough. Here is my proposal.
-
import operator
class Distribution(dict):
'''A distribution is a dictionary where the keys are dice
totals and the values are the number of possible ways
this total can come up '''
def __add__(self, term):
'''Take two distributions and combine them into one.'''
result = Distribution()
for k1, v1 in self.iteritems():
for k2, v2 in term.iteritems():
k3 = k1 + k2
v3 = v1 * v2
try:
result[k3] += v3
except KeyError:
result[k3] = v3
return result
def __rmul__(self, num):
tp = num * [self]
return reduce(operator.add, tp)
def D(n):
''' One die has a distribution where each result has
one possible way of coming up '''
return Distribution((i,1) for i in xrange(1,n+1))
sum3d6 = 3 * D(6)
sum2d6p2d4 = 2 * D(6) + 2 * D(4)
-
--
Antoon Pardon
--
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Re: Dice probability problem
That looks reasonable. The operation you are implementing is known as 'convolution' and is equivalent to multiplying polynomials. It would be a little more general if you had the input 'die' be a sequence of the count for each outcome, so d6 would be [1]*6 (or [0]+[1]*6 if you prefer). That would allow allow you to represent unfair dice and also to add not just a die to a distribution, but to add any two distributions, so you can play tricks like computing 16d6 as (d6)*2*2*2*2. (The code above is a convolution that restricts the second distribution 'die' to have only 0 and 1 coefficients.) The general convolution can be implemented much like what you have, except that you need another multiplication (to account for the fact that the coefficient is not always 0 or 1). My not particularly efficient implementation: def vsum(seq1, seq2): return [ a + b for a, b in zip(seq1, seq2) ] def vmul(s, seq): return [ s * a for a in seq ] # Convolve 2 sequences # equivalent to adding 2 probabililty distributions def conv(seq1, seq2): n = (len(seq1) + len(seq2) -1) ans = [ 0 ] * n for i, v in enumerate(seq2): vec = [ 0 ] * i + vmul(v, seq1) + [ 0 ] * (n - i - len(seq1)) ans = vsum(ans, vec) return ans # Convolve a sequence n times with itself # equivalent to multiplying distribution by n def nconv(n, seq): ans = seq for i in range(n-1): ans = conv(ans, seq) return ans print nconv(3, [ 1 ] * 6) print nconv(3, [ 1.0/6 ] * 6) print nconv(2, [ .5, .3, .2 ]) [1, 3, 6, 10, 15, 21, 25, 27, 27, 25, 21, 15, 10, 6, 3, 1] [0.0046296296296296294, 0.013888, 0.027776, 0.046296296296296294, 0.069448, 0.097238, 0.11574074074074074, 0.125, 0.125, 0.11574074074074073, 0.097224, 0.069448, 0.046296296296296294, 0.027776, 0.013888, 0.0046296296296296294] [0.25, 0.2, 0.29004, 0.12, 0.040008] -- http://mail.python.org/mailman/listinfo/python-list
Re: Dice probability problem
Tomi Lindberg wrote: > > # A die with n faces > D = lambda n: [x+1 for x in range(n)] > That can be written: D = lambda n : range(1,n+1) Gerard -- http://mail.python.org/mailman/listinfo/python-list
Re: Dice probability problem
First, thanks to Antoon and Alexander for replying.
Antoon Pardon wrote:
> It would be better to construct distributions for one
> die and make a function that can 'add' two distributions
> together.
As both replies pointed to this direction, I tried to take
that route. Here's the unpolished code I came up with. Does
it look even remotely sane way to accomplish my goal?
-- code begins --
# A die with n faces
D = lambda n: [x+1 for x in range(n)]
# A new die with 6 faces
d6 = D(6)
# Adds another die to results.
def add_dice(sums, die):
# If first die, all values appear once
if not sums:
for face in die:
sums[face] = 1
# Calculating the number of appearances for additional
# dice
else:
new_sums = {}
for k in sums.keys():
for f in die:
if new_sums.has_key(k+f):
new_sums[k+f] += sums[k]
else:
new_sums[k+f] = sums[k]
sums = new_sums
return sums
sums = add_dice({}, d6)
sums = add_dice(sums, d6)
sums = add_dice(sums, d6)
-- code ends --
Thanks,
Tomi Lindberg
--
http://mail.python.org/mailman/listinfo/python-list
Re: Dice probability problem
Op 2006-04-04, Tomi Lindberg schreef <[EMAIL PROTECTED]>: > Hi, > > I'm trying to find a way to calculate a distribution of > outcomes with any combination of dice. I have the basics > done, but I'm a bit unsure how to continue. My main concern > is how to make this accept any number of dice, without > having to write a new list comprehension for each case? IMO you are looking at it from the wrong side. It would be better to construct distributions for one die and make a function that can 'add' two distributions together. So for 3D6 you first add the distribution of a D6 to the distribution of a D6 and to this result you add the distribution of a D6 again. If you need more to start, just ask. > Here's a piece of code that shows the way I'm doing things > at the moment. > > -- code begins -- > > # A die with n faces > D = lambda n: [x+1 for x in range(n)] > > # A pool of 3 dice with 6 faces each > pool = [D(6)] * 3 > > # A List of all outcomes with the current 3d6 pool. > results = [x+y+z for x in pool[0] for y in pool[1] > for z in pool[2]] This is very inefficient. I wouldn't want to calculate the distribution of 10D10 this way. Try to think how you would do this with only D2's. (Triangle of Pascal) and generalize it. -- Antoon Pardon -- http://mail.python.org/mailman/listinfo/python-list
Re: Dice probability problem
Alexander Schmolck <[EMAIL PROTECTED]> writes: > addDice(resultFor1, pool[1]) > addDice(pool[0], pool[1]) sorry should have spelled out that successive lines are meant to be equivalent, i.e. addDice(resultFor1, pool[1]) == addDice(pool[0], pool[1]) 'as -- http://mail.python.org/mailman/listinfo/python-list
Re: Dice probability problem
Tomi Lindberg <[EMAIL PROTECTED]> writes: > I'm trying to find a way to calculate a distribution of outcomes with any > combination of dice. I have the basics done, but I'm a bit unsure how to > continue. My main concern is how to make this accept any number of dice, > without having to write a new list comprehension for each case? You need to think about the right way to break the problem down into some operation that can be repeated one fewer times than there are dice (if you just have a single dice, nothing needs to be done) and then repeat it. An obvious candidate is adding a single dice to the sums computed so far: def addDice(sums, dice): return [x+y for x in dice for y in sums] If you have less than 1 dice the answer is # len(pool) == 1 pool[0] After that, each time you add a dice you need to call addDice on the sum computed for all the previous dice and the new dice: # len(pool) == 2 addDice(resultFor1, pool[1]) addDice(pool[0], pool[1]) then # len(pool) == 3 addDice(resultFor2, pool[2]) addDice(addDice(resultFor1, pool[1]), pool[2]) addDice(addDice(pool[0], pool[1]), pool[2]) finally you get # len(pool) == n addDice(addDice(addDice(..., pool[n-3]), pool[n-2]) pool[n-1]) OK, so how do we get the repetition? Conveniently the pattern f(...f(f(x[0],x[1]),x[2])...,x[n-1]) or equivalently, if we write the infix operator * for f: x[0]*x[1]*...*x[n-1], can just be written as reduce(f, x) in python. So we get: reduce(addDice, pool) == reduce(lambda sums, dice: [x+y for x in dice for y in sums], pool) You should presumably also try writing this out as a single function, without using reduce (but recognizing the (left) reduction pattern is useful, even if you don't use python's reduce). 'as -- http://mail.python.org/mailman/listinfo/python-list
Dice probability problem
Hi,
I'm trying to find a way to calculate a distribution of
outcomes with any combination of dice. I have the basics
done, but I'm a bit unsure how to continue. My main concern
is how to make this accept any number of dice, without
having to write a new list comprehension for each case?
Here's a piece of code that shows the way I'm doing things
at the moment.
-- code begins --
# A die with n faces
D = lambda n: [x+1 for x in range(n)]
# A pool of 3 dice with 6 faces each
pool = [D(6)] * 3
# A List of all outcomes with the current 3d6 pool.
results = [x+y+z for x in pool[0] for y in pool[1]
for z in pool[2]]
# A dictionary to hold the distribution
distribution = {}
# If outcome is already a key, adds 1 to its value.
# Otherwise adds outcome to keys and sets its value
# to 1.
def count(x):
if distribution.has_key(x): distribution[x] += 1
else: distribution[x] = 1
# Maps the results with above count function.
map(count, results)
-- code ends --
Thanks,
Tomi Lindberg
--
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