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Re: Josephus Circle problem

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  Re: Josephus Circle problem "WJ" <w_a_x_man@yahoo.com> - 2014-12-11 17:57 +0000

#34873 — Re: Josephus Circle problem

Paul Rubin wrote:

> Tristan Plumb <firth@trstn.net> writes:
> > However! Combinators can do amazing things with quotations. Take nrec:...
> > : nrec ( n F F0 -- Fn ) rot 1+ 1 do over i swap execute loop nip ;
> 
> Neat ;-)
> 
> > : (josephus) swap 2 + swap mod 1+ ;
> > : josephus ' (josephus) 1 nrec ;
> 
> ' should say ['] .
> 
> The Haskell equivalent of Marcel's algorithm is:
> 
>   josephus k n = 1 + foldl (\a b->(a+k)`mod`b) 0 [1..n]
> 
> I guess it's possible to concoct a reduce combinator in Forth, but it
> just doesn't seem in the Forth style.  It would also want to use a
> coroutine to generate the input stream.

In OCaml, using recursion:

let rec jo a = function
    41 -> 1+a
  |  b -> jo ((a+3) mod b) (b+1) ;;

jo 0 1 ;;

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