Euler Problem 26

[Tony White]

Hi

My guess, your pattern is limiting you to uniquely the digits 0-9, so
that's where it stops. Consider - (1011011101234101111109) could also
be a valid sequence. ie, digits may be used more than once.

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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-- 
Tony White
SpeechNet Technologies Ltd