Euler Problem 26
Andrew Mathenge
mathenge at gmail.com
Tue Jul 1 10:02:21 BST 2008
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...
>> >> >
>> >> >
>> >> > --
>> >> > Ubuntu-ke mailing list
>> >> > Ubuntu-ke at lists.ubuntu.com
>> >> > https://lists.ubuntu.com/mailman/listinfo/ubuntu-ke
>> >> >
>> >>
>> >>
>> >>
>> >
>> >
>> > --
>> > Ubuntu-ke mailing list
>> > Ubuntu-ke at lists.ubuntu.com
>> > https://lists.ubuntu.com/mailman/listinfo/ubuntu-ke
>> >
>
>
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