Euler Problem 26
Miano Njoka
mianonjoka at gmail.com
Tue Jul 1 16:06:48 BST 2008
I was wrong the first time, got it after a couple of tries. good
luck!!! :)
On Jul 1, 2008, at 4:37 PM, kinuthiA muchanE wrote:
> Miano,
>
> Just got back to the computer, right now I am struggling with "for"
> loops :-/
> I do not know what the answer is but check here
> http://projecteuler.net/index.php?section=problems&id=26
>
> Kinuthia...
> On Tue, 2008-07-01 at 14:34 +0300, Miano Njoka wrote:
>> 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
>>>>>>>> https://lists.ubuntu.com/mailman/listinfo/ubuntu-ke
>>>>>>>>
>>>>>>>
>>>>>>>
>>>>>>>
>>>>>>
>>>>>>
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>>>>
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