Re: [Haskell-cafe] Fwd: Re: Period of a sequence

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 To: Steffen Schuldenzucker

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 //*wrote:

From: Steffen Schuldenzucker Subject: Re: [Haskell-cafe] Period of a sequence To: "michael rice" Cc: haskell-cafe@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

_______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe