[Haskell-cafe] Josephus problem and style
paul.hudak at yale.edu
Sun Apr 1 16:46:53 EDT 2007
Here's a solution that I think is a bit more elegant.
josephus n k =
let loop xs = let d:r = drop (k-1) xs
in d : loop (filter (/= d) r)
in take n (loop (cycle [1..n]))
Anthony Chaumas-Pellet wrote:
> I've written a function to compute the general Josephus problem,
> giving both the number of the survivor and the order in which the
> others are killed. However, I am not overly fond of my actual Haskell
> implementation, so I would like some comments on elegance. My current
> function is as follow::
> josephus k n = josephus' k 1 [1..n]  where
> josephus' k i (x:xs) sorted
> | xs ==  = (x, reverse sorted)
> | i `mod` k == 0 = josephus' k (i+1) xs ([x] ++ sorted)
> | otherwise = josephus' k (i+1) (xs ++ [x]) sorted
> The biggest difficulty I had is representing the circle (a continuous
> unit, unlike the list). The only solution I've found is to explicitly
> concatenate lists during each iteration, using an index to check
> whether the current element should be kept.
> It seems to me that manupilating lists so often makes the resulting
> function unclear, and hides the basic principle behind the Josephus
> sequence. So, I'm looking for a better way to write this function, but
> enlightenment eludes me.
> I've taken up Haskell in earnest a couple weeks ago, after a fairly
> long stay in Lisp land (perhaps it shows). My previous "Eureka!"
> moment in Haskell was matrix multiplication, along the lines of:
> matProd a b = [[sum (zipWith (*) x y) | y <- transpose b]| x <- a]
> Anthony Chaumas-Pellet
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