I'm stuck trying to write a function that generates a factorial of a
number using iteration and not recursion. Any simple ideas would be
appreciated.
Oct 22 '07
59 5656
Anurag <an**********@gmail.comwrote:
What about this no map, reduce, mutiplication or even addition
Its truly interative and You will need to interate till infinity if
you want correct answer ;)
def factorial(N):
"""
Increase I ...and go on increasing...
"""
import random
myNumer = range(N)
count = 0
I = 10000 #10**(N+1)
for i in range(I):
bucket = range(N)
number = []
for i in range(N):
k = bucket[ random.randrange(0,len(bucket))]
bucket.remove(k)
number.append(k)
if number == myNumer:
count+=1
return int(I*1.0/count+.5)
;-)
Note you can write your middle loop as
for i in range(I):
number = myNumer[:]
random.shuffle(number)
if number == myNumer:
count+=1
--
Nick Craig-Wood <ni**@craig-wood.com-- http://www.craig-wood.com/nick
Nick Craig-Wood wrote:
Note you can write your middle loop as
for i in range(I):
number = myNumer[:]
random.shuffle(number)
if number == myNumer:
count+=1
Nice. Try 'em all, then count 'em.
Another wtfery would be a SQLAlchemy solution, generating dynamic
queries, using only OUTER JOINs and COUNT(). Could be a way to justify
hardware upgrades.
Py-Fun wrote:
I'm stuck trying to write a function that generates a factorial of a
number using iteration and not recursion. Any simple ideas would be
appreciated.
fact = lambda n : len(map([1].__imul__,range(1,n+1))[0])
hth :)
BB
On Oct 30, 3:30 pm, Nick Craig-Wood <n...@craig-wood.comwrote:
Anurag <anuraguni...@gmail.comwrote:
What about this no map, reduce, mutiplication or even addition
Its truly interative and You will need to interate till infinity if
you want correct answer ;)
deffactorial(N):
"""
Increase I ...and go on increasing...
"""
import random
myNumer = range(N)
count = 0
I = 10000 #10**(N+1)
for i in range(I):
bucket = range(N)
number = []
for i in range(N):
k = bucket[ random.randrange(0,len(bucket))]
bucket.remove(k)
number.append(k)
if number == myNumer:
count+=1
return int(I*1.0/count+.5)
;-)
Note you can write your middle loop as
for i in range(I):
number = myNumer[:]
random.shuffle(number)
if number == myNumer:
count+=1
--
Nick Craig-Wood <n...@craig-wood.com--http://www.craig-wood.com/nick- Hide quoted text -
- Show quoted text -
good :) i thinkif go on improving this we will have worlds' best
useless factorial function.
actually number = myNumer[:] can be moved out of the loop.
On Tue, Oct 30, 2007 at 01:09:38PM +0100, Boris Borcic wrote regarding Re: Iteration for Factorials:
>
Py-Fun wrote:
I'm stuck trying to write a function that generates a factorial of a
number using iteration and not recursion. Any simple ideas would be
appreciated.
fact = lambda n : len(map([1].__imul__,range(1,n+1))[0])
hth :)
Nice one. I was trying to grok it, and started out by breaking it down:
>>[1].__imul__(2)
[1, 1]
>>map([1].__imul__,range(1,3))
[[1, 1], [1, 1]]
So far so good, but I tried to break it down a little more:
>>[1].__imul__(1), [1].__imul__(2), [1].__imul__(3)
([1], [1, 1], [1, 1, 1])
Hmm. Wasn't what I was expecting.
Then it hit me:
>>L = [1] L.__imul__(1), L.__imul__(2), L.__imul__(3)
([1, 1, 1, 1, 1, 1], [1, 1, 1, 1, 1, 1], [1, 1, 1, 1, 1, 1])
Pretty sneaky, sis.
Cheers,
Cliff
P.S. Regards to those who lack a grounding in American pop-culture, or who are to young to remember the origins of "Pretty sneaky, sis."
On Tue, Oct 30, 2007 at 01:09:38PM +0100, Boris Borcic wrote regarding Re: Iteration for Factorials:
>
Py-Fun wrote:
I'm stuck trying to write a function that generates a factorial of a
number using iteration and not recursion. Any simple ideas would be
appreciated.
fact = lambda n : len(map([1].__imul__,range(1,n+1))[0])
OK. Now I've been sucked in. How about this:
def fact(x):
def f(x):
if int(x) != x:
raise ValueError
elif x 1:
return f(x-1) ** x
elif x == 1:
return 10
else:
raise ValueError
return len(str(f(x))) -1
The great part about this recursive solution is that you don't have to worry about the stack limit because performance degrades so quickly on the conversion to string! fact(8) takes a little less than a second, fact(9) takes about a minute, and fact(10) takes more time than I had patience to wait for!
Cheers,
Cliff
On Oct 30, 10:25 am, "J. Clifford Dyer" <j...@sdf.lonestar.orgwrote:
On Tue, Oct 30, 2007 at 01:09:38PM +0100, Boris Borcic wrote regarding Re: Iteration for Factorials:
Py-Fun wrote:
I'm stuck trying to write a function that generates a factorial of a
number using iteration and not recursion. Any simple ideas would be
appreciated.
fact = lambda n : len(map([1].__imul__,range(1,n+1))[0])
OK. Now I've been sucked in. How about this:
def fact(x):
def f(x):
if int(x) != x:
raise ValueError
elif x 1:
return f(x-1) ** x
elif x == 1:
return 10
else:
raise ValueError
return len(str(f(x))) -1
The great part about this recursive solution is that you don't have to worry about the stack limit because performance degrades so quickly on the conversion to string! fact(8) takes a little less than a second, fact(9) takes about a minute, and fact(10) takes more time than I had patience to wait for!
And the not-so-great part is that it raises an exception
on fact(0) which makes it utterly useless for calculating
combinations of m things taken n at a time: m!/n!*(m-n)!
Why is it that no one seems able to get this right?
>
Cheers,
Cliff
On Tue, Oct 30, 2007 at 11:37:57AM -0700, me********@aol.com wrote regarding Re: Iteration for Factorials:
>
On Oct 30, 10:25 am, "J. Clifford Dyer" <j...@sdf.lonestar.orgwrote:
On Tue, Oct 30, 2007 at 01:09:38PM +0100, Boris Borcic wrote regarding Re: Iteration for Factorials:
Py-Fun wrote:
I'm stuck trying to write a function that generates a factorial of a
number using iteration and not recursion. Any simple ideas would be
appreciated.
fact = lambda n : len(map([1].__imul__,range(1,n+1))[0])
OK. Now I've been sucked in. How about this:
def fact(x):
def f(x):
if int(x) != x:
raise ValueError
elif x 1:
return f(x-1) ** x
elif x == 1:
return 10
else:
raise ValueError
return len(str(f(x))) -1
The great part about this recursive solution is that you don't have to worry about the stack limit because performance degrades so quickly on the conversion to string! fact(8) takes a little less than a second, fact(9) takes about a minute, and fact(10) takes more time than I had patience to wait for!
And the not-so-great part is that it raises an exception
on fact(0) which makes it utterly useless for calculating
combinations of m things taken n at a time: m!/n!*(m-n)!
Why is it that no one seems able to get this right?
I can't speak for everyone, but my excuses are as follows:
* I haven't studied math or done math-heavy work in 8 years.
* I'm not at my home computer, and most of the thread (wherein, I now recall, that particular rule was specified) is downloaded to my home computer, so I was working from my feeble memory.
* I didn't care enough to google for it.
That said, s/elif x == 1:/elif x in (0,1):/ should solve the problem
On Oct 30, 1:52 pm, "J. Clifford Dyer" <j...@sdf.lonestar.orgwrote:
On Tue, Oct 30, 2007 at 11:37:57AM -0700, mensana...@aol.com wrote regarding Re: Iteration for Factorials:
On Oct 30, 10:25 am, "J. Clifford Dyer" <j...@sdf.lonestar.orgwrote:
On Tue, Oct 30, 2007 at 01:09:38PM +0100, Boris Borcic wrote regarding Re: Iteration for Factorials:
Py-Fun wrote:
I'm stuck trying to write a function that generates a factorial of a
number using iteration and not recursion. Any simple ideas would be
appreciated.
fact = lambda n : len(map([1].__imul__,range(1,n+1))[0])
OK. Now I've been sucked in. How about this:
def fact(x):
def f(x):
if int(x) != x:
raise ValueError
elif x 1:
return f(x-1) ** x
elif x == 1:
return 10
else:
raise ValueError
return len(str(f(x))) -1
The great part about this recursive solution is that you don't have to worry about the stack limit because performance degrades so quickly on the conversion to string! fact(8) takes a little less than a second, fact(9) takes about a minute, and fact(10) takes more time than I had patience to wait for!
And the not-so-great part is that it raises an exception
on fact(0) which makes it utterly useless for calculating
combinations of m things taken n at a time: m!/n!*(m-n)!
Why is it that no one seems able to get this right?
I can't speak for everyone, but my excuses are as follows:
* I haven't studied math or done math-heavy work in 8 years.
Fair enough. I primarily do math-heavy work, and am familiar
with such matters. But that's just me.
* I'm not at my home computer, and most of the thread (wherein,
I now recall, that particular rule was specified) is downloaded
to my home computer, so I was working from my feeble memory.
Well, I've only had to point it out a dozen times already in
this thread. Nice to see that all that effort has been for nought.
* I didn't care enough to google for it.
Quoting from Monty Python:
"It's just as easy to get these things right, you know."
>
That said, s/elif x == 1:/elif x in (0,1):/ should solve the problem
Sure, it's always easily solvable. I just hope someone learns the
lesson on how to test properly, to make sure things work the way
they should and to fail the way they should so that one can actually
trust the algorithms they write.
En Tue, 30 Oct 2007 15:37:57 -0300, me********@aol.com
<me********@aol.comescribió:
On Oct 30, 10:25 am, "J. Clifford Dyer" <j...@sdf.lonestar.orgwrote:
>The great part about this recursive solution is that you don't have to worry about the stack limit because performance degrades so quickly on the conversion to string! fact(8) takes a little less than a second, fact(9) takes about a minute, and fact(10) takes more time than I had patience to wait for!
And the not-so-great part is that it raises an exception
on fact(0) which makes it utterly useless for calculating
combinations of m things taken n at a time: m!/n!*(m-n)!
Why is it that no one seems able to get this right?
Uhmmm... don't be so serious, most of those recipes are a joke :)
Who cares about fact(0) when fact(10) can't be computed...? At least,
that's how I see it, Nimo dixit.
--
Gabriel Genellina ### This discussion thread is closed Replies have been disabled for this discussion. |