Re: [Tutor] Cube root
> On Sat, Sep 15, 2012 at 5:36 PM, Dave Angel wrote:
>> On 09/15/2012 05:28 PM, Amanda Colley wrote:
>>> Ok, I have to get input from a user ('enter a number') and then get
>>> the
>>> cube root of that number. I am having trouble with the code to get the
>>> cube root. If anyone can help me solve this I would greatly appreciate
>>> it.
>>> ('enter a number')
>>> n=number
>>> ??? cube root??
>>>
>>>
>>
>> The operator to raise a number to a particular power is **, and it's not
>> constrained to integer powers. So 5*2 is 25. See what's next?
>>
> This is a great place to ask questions. But don't forget google (or bing)
>
> try googling python cube root
>
> see how you do and come back with your code!
> --
> Joel Goldstick
OOPS!!
Should have been:
>>> 27.0**(1.0/3.0)
3.0
>>>
Which works better than expected!
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Re: [Tutor] Cube root
> On Sat, Sep 15, 2012 at 5:36 PM, Dave Angel wrote:
>> On 09/15/2012 05:28 PM, Amanda Colley wrote:
>>> Ok, I have to get input from a user ('enter a number') and then get
>>> the
>>> cube root of that number. I am having trouble with the code to get the
>>> cube root. If anyone can help me solve this I would greatly appreciate
>>> it.
>>> ('enter a number')
>>> n=number
>>> ??? cube root??
>>>
>>>
>>
>> The operator to raise a number to a particular power is **, and it's not
>> constrained to integer powers. So 5*2 is 25. See what's next?
>>
> This is a great place to ask questions. But don't forget google (or bing)
>
> try googling python cube root
>
> see how you do and come back with your code!
> --
> Joel Goldstick
Why does:
"""
>>> import math
>>> 9**(1/3.0)
2.080083823051904
>>> 9.0**(1.0/3.0)
2.080083823051904
>>>
"""
not seem to work?
>
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Re: [Tutor] Cube root
On 16/09/12 07:28, Amanda Colley wrote:
Ok, I have to get input from a user ('enter a number') and then get the
cube root of that number. I am having trouble with the code to get the
cube root. If anyone can help me solve this I would greatly appreciate it.
('enter a number')
n=number
??? cube root??
The definition of "cube root of x" is "x to the power of one third".
To calculate cube root, you need to get the number from the user and raise
it to the power of one third. This is actually a subtle problem:
* input you get from the user is a string, you need to turn it into a
number
* you have a choice between using the ** operator or the pow() function
for raising x to a power
* the obvious solution:
x**1/3
is wrong. Can you see why? (Hint: there are two operators there, power
and divide. What order do they get executed?)
* even more subtle: there is a difference in how Python does division
depending on the version you use. Calculating one third correctly is
slightly trickier than it seems.
I hope this is enough hints for you to solve the problem. If not, show
us the code you have and ask for more help.
--
Steven
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Re: [Tutor] Cube root
On Sat, Sep 15, 2012 at 5:36 PM, Dave Angel wrote:
> On 09/15/2012 05:28 PM, Amanda Colley wrote:
>> Ok, I have to get input from a user ('enter a number') and then get the
>> cube root of that number. I am having trouble with the code to get the
>> cube root. If anyone can help me solve this I would greatly appreciate it.
>> ('enter a number')
>> n=number
>> ??? cube root??
>>
>>
>
> The operator to raise a number to a particular power is **, and it's not
> constrained to integer powers. So 5*2 is 25. See what's next?
>
This is a great place to ask questions. But don't forget google (or bing)
try googling python cube root
see how you do and come back with your code!
--
Joel Goldstick
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Re: [Tutor] Cube root
On 09/15/2012 05:28 PM, Amanda Colley wrote:
> Ok, I have to get input from a user ('enter a number') and then get the
> cube root of that number. I am having trouble with the code to get the
> cube root. If anyone can help me solve this I would greatly appreciate it.
> ('enter a number')
> n=number
> ??? cube root??
>
>
The operator to raise a number to a particular power is **, and it's not
constrained to integer powers. So 5*2 is 25. See what's next?
--
DaveA
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[Tutor] Cube root
Ok, I have to get input from a user ('enter a number') and then get the
cube root of that number. I am having trouble with the code to get the
cube root. If anyone can help me solve this I would greatly appreciate it.
('enter a number')
n=number
??? cube root??
--
Amanda Colley
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Re: [Tutor] cube root
On Monday 19 January 2009 18:56, col speed wrote: > Wow! I seem to have caused a great deal of comments! > I actually am looking to see if a number is a "perfect cube". I will try > out all the suggestions. > Thanks a lot. > Colin The reliable way to do this is to store a list of cubes. If the number you want to check is not in your list, then compute more cubes until you exceed that number. Unless you are dealing with absurdly huge numbers, this won't be very onerous on computing time. Cheers ___ Tutor maillist - Tutor@python.org http://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] cube root
Wow! I seem to have caused a great deal of comments! I actually am looking to see if a number is a "perfect cube". I will try out all the suggestions. Thanks a lot. Colin P.S. I have a small programme that changes decimal to binary (I don't know if it can be changed to work with fractions or not), maybe it will help. def dec2bin(dec): bin = '' while dec > 0: bin = str(dec & 1) + bin dec >>= 1 return bin 2009/1/19 Vicent > > > > > On Mon, Jan 19, 2009 at 07:41, Brett Wilkins wrote: > >> >> What you're running into here is the limited accuracy of floating point >> values... >> You'll likely find this happens a lot in python.. CPython, at least. (I >> know, as I do) >> I'm not sure, as I've never used it... but perhaps Numeric/Numpy handle >> this kinda stuff better (for accuracy's sake) >> >> > Maybe this issue can be overcome by using symbolic notation when possible > (maybe by using SymPy)? Has anyone any experience with this? > > -- > Vicent > ___ Tutor maillist - Tutor@python.org http://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] cube root
-Original Message- >From: Andre Engels >Sent: Jan 19, 2009 7:22 AM >To: spir >Cc: tutor@python.org >Subject: Re: [Tutor] cube root > >On Mon, Jan 19, 2009 at 1:13 PM, spir wrote: >> Do you know any common algorithm to convert decimal (in the sense of >> fractional) decimals (in the sense of base 10 numbers) into binaries? >> >> 123.456 --> 011.bbb... >> and/or >> 123456 * 10**(-3) --> bbb... * 2**(-bbb...) >> >> How do python/C achieve that? > >There's probably more efficient methods, but a simple method to >convert a decimal fraction to a binary would be the following >(untested): > >def tobinary(n,precision=12) ># n is a number between 0 and 1 that should be converted, >precision is the number of binary digits to use. >assert 0.0 <= n < 1.0 >outcome = "0." >compare = 0.5 >for i in xrange(precision): >if n > compare: >outcome += "1" >n -= compare >if n == 0.0: break >else: >outcome += "0" >compare /= 2 >return outcome > >-- >André Engels, andreeng...@gmail.com I hope my memory serves. To convert decimal fraction into binary fraction, do the following repeatedly to the decimal until desired accuracy is achieved. Multiply by 2, if result is less than 1, next digit is 0 else next digit is 1 and you drop the whole number part of the result. Continue... .456 * 2 = .912(first digit is 0) .912 * 2 = 1.824 (next digit is 1) .824 * 2 = 1.648 (next digit is 1) .648 * 2 = 1.296 (next digit is 1) .296 * 2 = .592(next digit is 0) 0.456 ~ 0.01110 ___ Tutor maillist - Tutor@python.org http://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] cube root
On Mon, Jan 19, 2009 at 1:13 PM, spir wrote: > Do you know any common algorithm to convert decimal (in the sense of > fractional) decimals (in the sense of base 10 numbers) into binaries? > > 123.456 --> 011.bbb... > and/or > 123456 * 10**(-3) --> bbb... * 2**(-bbb...) > > How do python/C achieve that? There's probably more efficient methods, but a simple method to convert a decimal fraction to a binary would be the following (untested): def tobinary(n,precision=12) # n is a number between 0 and 1 that should be converted, precision is the number of binary digits to use. assert 0.0 <= n < 1.0 outcome = "0." compare = 0.5 for i in xrange(precision): if n > compare: outcome += "1" n -= compare if n == 0.0: break else: outcome += "0" compare /= 2 return outcome -- André Engels, andreeng...@gmail.com ___ Tutor maillist - Tutor@python.org http://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] cube root
Hmm, Well I thought it was both, but the latter seems untrue (now that I test a bit more) (expt 64 (/ 1 3)) gives the value 4, but turning any of those into floating point numbers seems to give me the infamous 3.996 thing all over again. I was originally thinking that scheme would handle exponents as floating point numbers, the way that C does (ie 2e20 is a fp number...), and I tested 2e20+1-2e20, which stumps most languages (python just gives 0.0, gcc 4.0.1 gives the same). Oh well, sorry to get your hopes up! Btw, I'm using MIT/GNU scheme, in case that makes a difference. Cheers --Brett Kent Johnson wrote: > On Mon, Jan 19, 2009 at 6:11 AM, Brett Wilkins wrote: >> The only language I've run into so far (I haven't used many, mind) that >> doesn't have this issue is Scheme :) > > It doesn't have an issue with cube roots or with floating point > inaccuracies in general? If the latter, I would like to know how they > do that... > > Kent ___ Tutor maillist - Tutor@python.org http://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] cube root
On Mon, Jan 19, 2009 at 12:11 PM, Brett Wilkins wrote: > The only language I've run into so far (I haven't used many, mind) that > doesn't have this issue is Scheme :) > (Just learning it at the moment.) It doesn't? That would surprise me. The only one that I know to do this kind of thing correctly is Matlab. -- André Engels, andreeng...@gmail.com ___ Tutor maillist - Tutor@python.org http://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] cube root
Do you know any common algorithm to convert decimal (in the sense of fractional) decimals (in the sense of base 10 numbers) into binaries? 123.456 --> 011.bbb... and/or 123456 * 10**(-3) --> bbb... * 2**(-bbb...) How do python/C achieve that? denis -- la vida e estranya ___ Tutor maillist - Tutor@python.org http://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] cube root
On Mon, Jan 19, 2009 at 6:11 AM, Brett Wilkins wrote: > The only language I've run into so far (I haven't used many, mind) that > doesn't have this issue is Scheme :) It doesn't have an issue with cube roots or with floating point inaccuracies in general? If the latter, I would like to know how they do that... Kent ___ Tutor maillist - Tutor@python.org http://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] cube root
The only language I've run into so far (I haven't used many, mind) that doesn't have this issue is Scheme :) (Just learning it at the moment.) Cheers, --Brett P.S. Forgive me if this email doesn't sort properly, sending through webmail, as I don't have a relaying SMTP available to me currently. Alan Gauld wrote: > > "Brett Wilkins" wrote > >> What you're running into here is the limited accuracy of floating point >> values... You'll likely find this happens a lot in python.. CPython, >> at least. > > In fact it happens with virtually every programming language available. > The problem traces back to the way that computers represent floating > point numbers in hardware. The integrated circuits cannot hold infinitely > long numbers so they need to use a condensed representation and > that inherently introduces small inaccuracies. (Think of 1/3 in > decimal - an infinite series of 1...., the same is true in binary, > some numbers just cannot be represented accurately) > > In the specific case being discussed you can disguise the problem > quite easily by displaying the result with a smaller number of decimal > places using round() or string formatting: > 64**(1.0/3) > 3.9996 print 64**(1.0/3) > 4.0 round(64**(1.0/3)) > 4.0 "%5.3f" % 64**(1.0/3) > '4.000' > > When you need to compare floating point numbers you also > need to be careful since: > 4.0 == 64**(1.0/3) ## is false! > > You need to introduce an error range, typically: > e = 0.0001 4.0-e < 64**(1.0/3) < 4.0+e # is true > > You can also use the decimal module for exact decimal > arithmetic but it introduces some other complications. > > HTH, > > ___ Tutor maillist - Tutor@python.org http://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] cube root
Wow! Everybody jumped on the "floating point inaccuracy" topic, I'm
surprised no one tried to find out what the OP was trying to do with his
cube root solver in the first place. Of course, the first-cut approach to
solving the cube root is to raise to the 1/3 power, but this is not the only
possible approach. Assuming that the OP wanted to try to find cube roots
for integers, or raise an exception if the given number is not a perfect
cube, I came up with this:
def cube_root(n):
"A modified Newton's Method solver for integral cube roots."
if int(n) != n:
raise ValueError("must provide an integer")
if n in (-1,0,1): return n
offset = (1,-1)[n > 0]
x = n/2
seen = set()
steps = 0
while 1:
y = x**3
if y == n:
#~ print "Found %d ^ 1/3 = %d in %d steps" % (n,x,steps)
return x
dydx = 3.0*x*x
x += int((n-y)/dydx)+offset
x = x or 1
if x in seen:
raise ValueError("%d is not a perfect cube" % n)
seen.add(x)
steps += 1
This will correctly give 4 for 64, 5 for 125, etc., not 3.999.
Then I went googling for "python cube root", and found this curious
solution:
def root3rd(x):
y, y1 = None, 2
while y!=y1:
y = y1
y3 = y**3
d = (2*y3+x)
y1 = (y*(y3+2*x)+d//2)//d
return y
This only works for perfect cubes, but could be made more robust with this
test just before returning y:
if y*y*y != x:
raise ValueError("%d is not a perfect cube" % x)
I was intrigued at this solution - timing it versus mine showed it to be 5x
faster. I tried to see some sort of Newton's method implementation, but
instead its more of a binary search. This was confirmed by adding this line
inside the while loop:
print y,y1
And to solve for the cube root of 1860867 (123**3), I get this output:
None 2
2 4
4 8
8 16
16 32
32 62
62 105
105 123
with the value 123 returned as the answer. Note that the guess (variable
y1) is initially doubled and redoubled, until the solution is neared. I
tried initializing y1 to a different guess, the original number, and got
this:
None 1860867
1860867 930434
930434 465217
465217 232609
232609 116305
116305 58153
58153 29077
29077 14539
14539 7270
7270 3635
3635 1818
1818 909
909 456
456 235
235 141
141 123
Now the guess is halved each time until nearing the solution, then again
converging to the solution 123.
This algorithm is also robust in that it can do more than just cube roots.
By changing this line:
y3 = y**3
to:
y3 = y**4
We get a 4th root solver! Unfortunately, this breaks down at the 5th root,
as I found solutions would not converge. At this point, I lost interest in
the general applicability of this algorithm, but it is still an interesting
approach - does anyone know the theoretical basis for it?
I guess I was just disappointed that taking cube roots of so small a number
as 64 quickly got us into floating-point roundoff errors, and if the OP is
doing something like looking for perfect cubes, then there are alternatives
to the one-size-fits-almost-all logarithmic implementation of the '**'
operator.
-- Paul
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Re: [Tutor] cube root
On Mon, Jan 19, 2009 at 07:41, Brett Wilkins wrote: > > What you're running into here is the limited accuracy of floating point > values... > You'll likely find this happens a lot in python.. CPython, at least. (I > know, as I do) > I'm not sure, as I've never used it... but perhaps Numeric/Numpy handle > this kinda stuff better (for accuracy's sake) > > Maybe this issue can be overcome by using symbolic notation when possible (maybe by using SymPy)? Has anyone any experience with this? -- Vicent ___ Tutor maillist - Tutor@python.org http://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] cube root
"Brett Wilkins" wrote What you're running into here is the limited accuracy of floating point values... You'll likely find this happens a lot in python.. CPython, at least. In fact it happens with virtually every programming language available. The problem traces back to the way that computers represent floating point numbers in hardware. The integrated circuits cannot hold infinitely long numbers so they need to use a condensed representation and that inherently introduces small inaccuracies. (Think of 1/3 in decimal - an infinite series of 1...., the same is true in binary, some numbers just cannot be represented accurately) In the specific case being discussed you can disguise the problem quite easily by displaying the result with a smaller number of decimal places using round() or string formatting: 64**(1.0/3) 3.9996 print 64**(1.0/3) 4.0 round(64**(1.0/3)) 4.0 "%5.3f" % 64**(1.0/3) '4.000' When you need to compare floating point numbers you also need to be careful since: 4.0 == 64**(1.0/3) ## is false! You need to introduce an error range, typically: e = 0.0001 4.0-e < 64**(1.0/3) < 4.0+e # is true You can also use the decimal module for exact decimal arithmetic but it introduces some other complications. HTH, -- Alan Gauld Author of the Learn to Program web site http://www.freenetpages.co.uk/hp/alan.gauld ___ Tutor maillist - Tutor@python.org http://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] cube root
On Mon, Jan 19, 2009 at 12:11 PM, Brett Wilkins wrote: > Hey Colin, > What you're running into here is the limited accuracy of floating point > values... Here the Python Documentation mentioning about the inherent limitations in dealing with floating point numbers: http://docs.python.org/tutorial/floatingpoint.html -- Senthil ___ Tutor maillist - Tutor@python.org http://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] cube root
Hey Colin, What you're running into here is the limited accuracy of floating point values... You'll likely find this happens a lot in python.. CPython, at least. (I know, as I do) I'm not sure, as I've never used it... but perhaps Numeric/Numpy handle this kinda stuff better (for accuracy's sake) Cheers, --Brett col speed wrote: Hi there, just a quick one. Is there a way to obtain cube roots in python? I have been trying: math.pow(n,1.0/3) This works with some, but with 64, for example I get: >>> pow(64,1.0/3) 3.9996 However: >>> 4**3 64 Any ideas? Thanks Colin ___ Tutor maillist - Tutor@python.org http://mail.python.org/mailman/listinfo/tutor ___ Tutor maillist - Tutor@python.org http://mail.python.org/mailman/listinfo/tutor
[Tutor] cube root
Hi there, just a quick one. Is there a way to obtain cube roots in python? I have been trying: math.pow(n,1.0/3) This works with some, but with 64, for example I get: >>> pow(64,1.0/3) 3.9996 However: >>> 4**3 64 Any ideas? Thanks Colin ___ Tutor maillist - Tutor@python.org http://mail.python.org/mailman/listinfo/tutor
