Euler Problem 26
kinuthiA muchanE
muchanek at gmail.com
Tue Jul 1 07:54:12 BST 2008
On Tue, 2008-07-01 at 03:06 +0300, Tony White wrote:
> Excellently put Andrew - a much better example than mine ;)
And you were telling me! Thanks Tony (c:
>
> 2008/7/1 Andrew Mathenge <mathenge at gmail.com>:
> > This email is from a ubuntu-ke subscriber to other subscribers inluding you:)
> > Actually, I think that understanding the question is the easiest part.
> > 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....
> >
> > 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.
> >
> > 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.
> >
> > 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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