[Haskell-cafe] Fwd: Re: Period of a sequence
michael rice
nowgate at yahoo.com
Mon Jun 27 16:54:55 CEST 2011
Thanks, all.
I have an evaluation copy of Mathematica and have been looking for problems to feed it.
Michael
--- On Mon, 6/27/11, Brent Yorgey <byorgey at seas.upenn.edu> wrote:
From: Brent Yorgey <byorgey at seas.upenn.edu>
Subject: Re: [Haskell-cafe] Fwd: Re: Period of a sequence
To: haskell-cafe at haskell.org
Date: Monday, June 27, 2011, 9:56 AM
I've attached some code I wrote a while ago for playing with repeating
decimal expansions, perhaps you'll find some of it useful.
-Brent
On Mon, Jun 27, 2011 at 02:21:55PM +0200, Steffen Schuldenzucker wrote:
>
> Michael,
>
> On 06/27/2011 01:51 PM, Steffen Schuldenzucker wrote:
> >
> > Forwarding to -cafe
> >
> > -------- Original Message --------
> > Subject: Re: [Haskell-cafe] Period of a sequence
> > Date: Mon, 27 Jun 2011 04:46:10 -0700 (PDT)
> > From: michael rice <nowgate at yahoo.com>
> > To: Steffen Schuldenzucker <sschuldenzucker at uni-bonn.de>
> >
> > Hi Steffen,
> >
> > Repeating decimals.
> >
> > 5/7 == 0.714285 714285 7142857 ... Period = 6
> >
> > It does seem like a difficult problem.
> >
> > This one is eventually repeating, with Period = 3
> >
> > 3227/555 = 5.8144 144 144…
>
> why not use the well-known division algorithm: (I hope this is readable)
>
> 3227 / 555
> = 3227 `div` 555 + 3227 `mod` 555 / 555
> = 5 + 452 / 555
> = 5 + 0.1 * 4520 / 555
> = 5 + 0.1 * (4520 `div` 555 + (4520 `mod` 555) / 555)
> = 5 + 0.1 * (8 + 80 / 555)
> = 5 + 0.1 * (8 + 0.1 * (800 / 555))
> = 5 + 0.1 * (8 + 0.1 * (800 `div` 555 + (800 `mod` 555) / 555))
> = 5 + 0.1 * (8 + 0.1 * (1 + 245 / 555))
> = 5 + 0.1 * (8 + 0.1 * (1 + 0.1 * 2450 / 555))
> = 5 + 0.1 * (8 + 0.1 * (1 + 0.1 * (4 + 230 / 555)))
> = 5 + 0.1 * (8 + 0.1 * (1 + 0.1 * (4 + 0.1 * 2300 / 555)))
> = 5 + 0.1 * (8 + 0.1 * (1 + 0.1 * (4 + 0.1 * (4 + 80 / 555))))
> *whoops*, saw 80 already, namely in line 6. Would go on like that
> forever if I continued like this, so the final result has to be:
>
> vvv Part before the place where I saw the '80' first
> 5.8 144 144 144 ...
> ^^^ Part after I saw the '80'
>
> So you could write a recursive function that takes as an accumulating
> parameter containing the list of numbers already seen:
>
> -- periodOf n m gives the periodic part of n/m as a decimal fraction.
> -- (or an empty list if that number has finitely many decimal places)
> > periodOf :: (Integral a) => a -> a -> [a]
> > periodOf = periodOfWorker []
> > where
> > periodOfWorker seen n m
> > | n `mod` m == 0 = ...
> > | (n `mod` m) `elem` seen = ...
> > | otherwise = ...
>
> >--- On *Mon, 6/27/11, Steffen Schuldenzucker
> >/<sschuldenzucker at uni-bonn.de>/*wrote:
> >
> >
> >From: Steffen Schuldenzucker <sschuldenzucker at uni-bonn.de>
> >Subject: Re: [Haskell-cafe] Period of a sequence
> >To: "michael rice" <nowgate at yahoo.com>
> >Cc: haskell-cafe at haskell.org
> >Date: Monday, June 27, 2011, 4:32 AM
> >
> >
> >
> >On 06/26/2011 04:16 PM, michael rice wrote:
> > > MathWorks has the function seqperiod(x) to return the period of
> >sequence
> > > x. Is there an equivalent function in Haskell?
> >
> >Could you specify what exactly the function is supposed to do? I am
> >pretty sure that a function like
> >
> >seqPeriod :: (Eq a) => [a] -> Maybe Integer -- Nothing iff non-periodic
> >
> >cannot be written. If "sequences" are represented by the terms that
> >define them (or this information is at least accessible), chances
> >might be better, but I would still be interested how such a function
> >works. The problem seems undecidable to me in general.
> >
> >On finite lists (which may be produced from infinite ones via
> >'take'), a naive implementation could be this:
> >
> > >
> > > import Data.List (inits, cycle, isPrefixOf)
> > > import Debug.Trace
> > >
> > > -- Given a finite list, calculate its period.
> > > -- The first parameter controls what is accepted as a generator.
> >See below.
> > > -- Set it to False when looking at chunks from an infinite sequence.
> > > listPeriod :: (Eq a) => Bool -> [a] -> Int
> > > listPeriod precisely xs = case filter (generates precisely xs)
> >(inits xs) of
> > > -- as (last $ init xs) == xs, this will always suffice.
> > > (g:_) -> length g -- length of the *shortest* generator
> > >
> > > -- @generates prec xs g@ iff @g@ generates @xs@ by repitition. If
> >@prec@, the
> > > -- lengths have to match, too. Consider
> > > --
> > > -- >>> generates True [1,2,3,1,2,1,2] [1,2,3,1,2]
> > > -- False
> > > --
> > > -- >>> generates False [1,2,3,1,2,1,2] [1,2,3,1,2]
> > > -- True
> > > generates :: (Eq a) => Bool -> [a] -> [a] -> Bool
> > > generates precisely xs g = if null g
> > > then null xs
> > > else (not precisely || length xs `mod` length g == 0)
> > > && xs `isPrefixOf` cycle g
> > >
> >
> >-- Steffen
> >
> >
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