Euler Problem 26
kinuthiA muchanE
muchanek at gmail.com
Thu Jul 3 08:53:59 BST 2008
Ronald,
Bed at 2am, that is the spirit! :-) And yes, that is longest recurring
sequence for d < 1000. And I am curious, why do you want to count the
decimals manually?
Kinuthia...
On Thu, 2008-07-03 at 09:18 +0300, Ronald Odero wrote:
> Hi all,
> I went to bed at 2am because of this. Even a scientific calculator
> cannot give the complete permutation and so I programmed in C++ to get
> it. I can say that I have completed the calculation and now the
> remaining part is counting the decimal places - manually.
> Kinuthia,are you sure that this is the highest permutation? If so,
> then you will have saved me.
> Ronald.
>
> On Wed, Jul 2, 2008 at 1:46 PM, kinuthiA muchanE <muchanek at gmail.com>
> wrote:
> This email is from a ubuntu-ke subscriber to other subscribers
> inluding you:)
>
> Habari wajamaa,
>
> I got it but I am not enthusiastic about it because it was
> largely trial
> and error. I have never done so much long division with a pen
> and paper
> trying to look for a pattern in my life! What I was sure about
> was that
> it had to be a prime.
>
> Using the python's decimal module I set the decimal precision
> to 1000
> places! Then using my eyes and maybe a bit of luck, I started
> with 997,
> 991, and then 983... bingo!
>
> 00101729399796541200406917599186164801627670396744659206510681586978636826042726347914547304170905391658189216683621566632756866734486266531027466937945066124109867751780264496439471007121057985757884028484231943031536113936927772126144455747711088504577822990844354018311291963377416073245167853509664292980671414038657171922685656154628687690742624618514750762970498474059003051881993896236012207527975584944048830
> 11190233977619532044760935910478128179043743641912512716174974567650050864699898270600203458799593082400813835198372329603255340793489318413021363173957273652085452695829094608341810783316378433367243133265513733468972533062054933875890132248219735503560528992878942014242115971515768056968463886063072227873855544252288911495422177009155645981688708036622583926754832146490335707019328585961342828077314343845371312309257375381485249237029501525940996948118006103763987792472024415055951169888097660223804679552390640895218718209562563580874872838250254323499491353
>
> 982 digits!
>
> Andrew, I do not know if you would like to see the code, if
> you can call
> it that! How did you guys do it?
>
> Kinuthia...
>
> On Tue, 2008-07-01 at 18:06 +0300, Miano Njoka wrote:
> > 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...
> > >>>>>>>>
> > >>>>>>>>
> > >>>>>>>> --
> > >>>>>>>> Ubuntu-ke mailing list
> > >>>>>>>> Ubuntu-ke at lists.ubuntu.com
> > >>>>>>>> https://lists.ubuntu.com/mailman/listinfo/ubuntu-ke
> > >>>>>>>>
> > >>>>>>>
> > >>>>>>>
> > >>>>>>>
> > >>>>>>
> > >>>>>>
> > >>>>>> --
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> > >>>>>> 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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