Euler Problem 26
Miano Njoka
mianonjoka at gmail.com
Tue Jul 1 12:34:19 BST 2008
I've written a solution using C, I got d=831, one cycle is 69 digits
long. What is the answer?
On Jul 1, 2008, at 12:02 PM, Andrew Mathenge wrote:
> This email is from a ubuntu-ke subscriber to other subscribers
> inluding you:)
> Only glad I could be of assistance. Let me know if you find a python
> solution to the problem. I wrote a solution in perl, but I'm
> interested in learning python.
>
> Good luck!
>
> Andrew.
>
> On Tue, Jul 1, 2008 at 2:50 AM, kinuthiA muchanE
> <muchanek at gmail.com> wrote:
>>
>> On Mon, 2008-06-30 at 18:33 -0400, Andrew Mathenge wrote:
>>> Actually, I think that understanding the question is the easiest
>>> part.
>> Not for me it seems, I think it is one of those days for me when
>> even
>> the obvious escapes you :-)
>>> Here's the question as copied from the website:
>>>
>>> Find the value of d < 1000 for which 1/d contains the longest
>>> recurring cycle in its decimal fraction part.
>>>
>>> Let's take the example that's given where d <= 10.
>>>
>>> When d = 7, the value of 1/d (1/7) is:
>>>
>>> 0.142857142857142857142857142857142857142857....
>>
>> Ah...okay
>>>
>>> As you can see, (142857) repeats indefinitely. It has a sequence
>>> of 6
>>> digits that repeat. This is longer than any of the others where d <=
>>> 10.
>> That was the crux of the matter!
>>>
>>> Therefore, the question asks, for d = 1 to 1000, which value of d
>>> has
>>> the longest sequence of repeating digits in the fractional part?
>>>
>>> Quite rightfully so, you can't limit the number of digits to 9 as in
>>> your example since there will be a number with a series of repeating
>>> digits greater than 9.
>>>
>>> For example, 1/17 is:
>>>
>>> 0.0588235294117647058823529411764705882352941176470588235294117647
>>> .....
>>>
>>> or
>>>
>>> 0.(0588235294117647)
>>>
>>> The number of repeating digits in this sequence is 16.
>>>
>>> Hope that helps. I'm not a python programmer so I really can't help
>>> debug your code.
>> It is now crystal clear, Anrdew, and thank you very much for
>> bailing me
>> out, do not worry about the code, I just wanted a nudge in the right
>> direction ;-)
>>>
>>> Andrew.
>>>
>>> On Mon, Jun 30, 2008 at 4:03 PM, kinuthiA muchanE <muchanek at gmail.com
>>> > wrote:
>>>> This email is from a ubuntu-ke subscriber to other subscribers
>>>> inluding you:)
>>>>
>>>> On Mon, 2008-06-30 at 22:04 +0300, Tony White wrote:
>>>>> Hi
>>>>>
>>>>> My guess, your pattern is limiting you to uniquely the digits
>>>>> 0-9, so
>>>>
>>>> ...and that is the whole idea IMHO, actually "input-ting" nothing
>>>> would
>>>> match my regular expression. There are only ten non-recurring
>>>> digits in
>>>> the world ie zero to nine, :-).
>>>>> that's where it stops. Consider - (1011011101234101111109) could
>>>>> also
>>>>> be a valid sequence. ie, digits may be used more than once.
>>>>
>>>> My dear Tony, so what is the question? I thought they wanted the
>>>> reciprocal of a number, that will give you the digits "123456789"
>>>> in a
>>>> non-repeating order. And 1011011101234101111109 would not match the
>>>> regular expression I had! 1's are and 0's occur more than once!
>>>>
>>>>
>>>>
>>>> Mugumo, I think there is always a mathematical angle to
>>>> everything! And
>>>> I am okay...
>>>>
>>>> Asante kwa kuweza kupata wasaa wa kunijibu, natumai tutaendelea
>>>> kuwasiliana.
>>>>
>>>> Kinuthia...
>>>>
>>>>>
>>>>> Tony
>>>>>
>>>>>
>>>>> 2008/6/30 kinuthiA muchanE <muchanek at gmail.com>:
>>>>>> This email is from a ubuntu-ke subscriber to other subscribers
>>>>>> inluding you:)
>>>>>> Habari wote,
>>>>>>
>>>>>>
>>>>>> I am trying to solve Problem Number 26
>>>>>> (http://projecteuler.net/index.php?section=problems&id=26) on
>>>>>> Project
>>>>>> Euler but apparently the answer I am submitting is wrong.
>>>>>>
>>>>>> Here is the problem:
>>>>>>
>>>>>>
>>>>>> A unit fraction contains 1 in the numerator. The decimal
>>>>>> representation
>>>>>> of the unit fractions with denominators 2 to 10 are given:
>>>>>>
>>>>>> 1/2
>>>>>> =
>>>>>> 0.5
>>>>>> 1/3
>>>>>> =
>>>>>> 0.(3)
>>>>>> 1/4
>>>>>> =
>>>>>> 0.25
>>>>>> 1/5
>>>>>> =
>>>>>> 0.2
>>>>>> 1/6
>>>>>> =
>>>>>> 0.1(6)
>>>>>> 1/7
>>>>>> =
>>>>>> 0.(142857)
>>>>>> 1/8
>>>>>> =
>>>>>> 0.125
>>>>>> 1/9
>>>>>> =
>>>>>> 0.(1)
>>>>>> 1/10
>>>>>> =
>>>>>> 0.1
>>>>>>
>>>>>> Where 0.1(6) means 0.166666..., and has a 1-digit recurring
>>>>>> cycle. It
>>>>>> can be seen that 1/7 has a 6-digit recurring cycle.
>>>>>>
>>>>>> Find the value of d < 1000 for which 1/d contains the longest
>>>>>> recurring
>>>>>> cycle in its decimal fraction part.
>>>>>>
>>>>>> I am giving the answer 38, because 1/38 = 0.0263157894. It
>>>>>> seems I have
>>>>>> misunderstood the question or I cant solve it! Here is the
>>>>>> code(in
>>>>>> Python) that I
>>>>>> came up with:
>>>>>>
>>>>>> def aux(num):
>>>>>> import re
>>>>>> pattern = re.compile(r"^0?1?2?3?4?5?6?7?8?9?$")
>>>>>>
>>>>>> frac ="%.9f"%(1.0/num)
>>>>>> fracSlice = frac[2:] # get the decimal
>>>>>> fractional part, ie remove
>>>>>> '0.'
>>>>>>
>>>>>> fracList = list(fracSlice) #convert string
>>>>>> to a
>>>>>> list
>>>>>> fracList.sort() # I need to
>>>>>> sort , because I will be searching by
>>>>>> increasing order
>>>>>>
>>>>>> testFrac = "".join(fracList) # convert list back to a
>>>>>> string,
>>>>>> phew!
>>>>>> if re.match(pattern,testFrac): # if the pattern matches,
>>>>>> the
>>>>>> number is
>>>>>> our candidate
>>>>>> print (num,fracSlice)
>>>>>>
>>>>>>
>>>>>> for b in xrange(1,1000):
>>>>>> aux(b)
>>>>>>
>>>>>> Er... er, that does not exactly work as expected but it narrows
>>>>>> the
>>>>>> search to only 3 candidates because of the inclusion of the zero:
>>>>>>
>>>>>> (28, '035714286')
>>>>>> (38, '026315789')
>>>>>> (81, '012345679')
>>>>>>
>>>>>> For 28, the digit, in the fractional part, after 8 is 5, so 5 is
>>>>>> repeated and as for, 81 the next digit after 7 is 0, so again 0
>>>>>> occurs
>>>>>> twice. But for 38, the next digit after 9 is 4, and because it
>>>>>> has NOT
>>>>>> occurred before, I assume 38 is the correct answer... and I am
>>>>>> wrong!
>>>>>>
>>>>>> I suspect I have completely misunderstood the question.
>>>>>>
>>>>>> Any ideas?
>>>>>> Thanks!
>>>>>>
>>>>>> Kinuthia...
>>>>>>
>>>>>>
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>>>>>> Ubuntu-ke at lists.ubuntu.com
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>>>>>>
>>>>>
>>>>>
>>>>>
>>>>
>>>>
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>>
>
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