Functional programming and lists go together like Fred and Ginger. This little exercise is one of Werner Hett's "Ninety-Nine Prolog Problems". The idea is to implement the *run length encoding data compression method*.

Here's how we start. First we write a function that packs consecutive duplicates of a list into sublists e.g.

# B.pack ['a'; 'a'; 'a'; 'b'; 'c'; 'c'; 'd'] ;; - : char list list = [['a'; 'a'; 'a']; ['b']; ['c'; 'c']; ['d'];]Then, consecutive duplicates of elements are encoded as terms $(N, E)$ where $N$ is the number of duplicates of the element $E$ e.g.

# B.rle (B.pack ['a'; 'a'; 'a'; 'b'; 'c'; 'c'; 'd'; 'e'; 'e']) ;; - : (int * char) list = [(3, 'a'); (1, 'b'); (2, 'c'); (1, 'd'); (2, 'e')]We will of course require a function to decode compressed data e.g.

# B.unrle(B.rle( B.pack ['a'; 'a'; 'a'; 'b'; 'c'; 'c'; 'd'; 'e'; 'e']) ) ;; - : char list = ['a'; 'a'; 'a'; 'b'; 'c'; 'c'; 'd'; 'e'; 'e'](Credit goes to Harvey Stein for the names

`rle`

and `unrle`

by the way).
So that's it for the first iteration - here's some code that aims to implement these specifications.
module B = struct let pack (x : α list) : α list list = let f (acc : α list list) (c : α) : α list list = match acc with | (((b :: _) as hd) :: tl) when c = b -> (c :: hd) :: tl | _ -> [c] :: acc in List.fold_left f [] x let rle (x : α list list) : (int * α) list = let f (acc : (int * α) list) (l : α list) : (int * α) list = (List.length l, List.hd l) :: acc in List.fold_left f [] x let unrle (data : (int * α) list) = let repeat ((n : int), (c : α)) : α list = let rec aux acc i = if i = 0 then acc else aux (c :: acc) (i - 1) in aux [] n in let f (acc : α list) (elem : (int * α)) : α list = acc @ (repeat elem) in List.fold_left f [] data end

Now, `pack`

is just a device of course. We don't really need it so here's the next iteration that does away with it.

module E = struct let rle (x : α list) : (int * α) list = let f (acc : (int * α) list) (c : α) : (int * α) list = match acc with | ((n, e) :: tl) when e = c -> (n + 1, c):: tl | _-> (1, c) :: acc in List.rev (List.fold_left f [] x) let unrle (data : (int * α) list) = let repeat ((n : int), (c : α)) : α list = let rec aux acc i = if i = 0 then acc else aux (c :: acc) (i - 1) in aux [] n in let f (acc : α list) (elem : (int * α)) : α list = acc @ (repeat elem) in List.fold_left f [] data endNifty!

Ok, the next idea is that when a singleton byte is encountered, we don't write the term $(1, E)$ instead, we just write $E$. Now OCaml doesn't admit heterogenous lists like Prolog appears to do so we need a sum type for the two possibilities. This then is the final version.

module F = struct type α t = | S of α | C of (int * α) let rle (bytes : α list) : α t list = let f (acc : α t list) (b : α) : α t list = match acc with | ((S e) :: tl) when e = b -> (C (2, e)) :: tl | ((C (n, e)) :: tl) when e = b -> (C (n + 1, b)) :: tl | _-> S b :: acc in List.rev (List.fold_left f [] bytes) let unrle (data : (α t) list) = let rec aux (acc : α list) (b : α) : (int -> α list) = function | 0 -> acc | i -> aux (b :: acc) b (i - 1) in let f (acc : α list) (e : α t) : α list = acc @ (match e with | S b -> [b]| C (n, b) -> aux [] b n) in List.fold_left f [] data end

Having worked out the details in OCaml, translation into C++ is reasonably straight-forward. One economy granted by this language is that we can do away with the data constructor `S`

in this version.

#include <boost/variant.hpp> #include <boost/variant/apply_visitor.hpp> #include <boost/range.hpp> #include <boost/range/numeric.hpp> #include <list> //Representation of the encoding template <class A> struct C { std::pair <int, A> item; }; template <class A> using datum = boost::variant <A, C<A>>; template <class A> using encoding = std::list<datum<A>>; //Procedural function object that updates an encoding given a //datum template <class A> struct update : boost::static_visitor<> { A c; encoding<A>& l; update (A c, encoding<A>& l) : c (c), l (l) {} void operator ()(A e) const { if (e == c) { l.back () = C<A>{ std::make_pair(2, c) }; return; } l.push_back (c); } void operator ()(C<A> const& elem) const { if (elem.item.second == c) { l.back () = C<A>{ std::make_pair (elem.item.first + 1, c) }; return; } l.push_back (c); } }; template <class R> encoding<typename boost::range_value<R>::type> rle (R bytes) { typedef boost::range_value<R>::type A; auto f = [](encoding<A> acc, A b) -> encoding<A> { if (acc.size () == 0) acc.push_back (b); else { boost::apply_visitor (update<A>(b, acc), acc.back ()); } return acc; }; return boost::accumulate (bytes, encoding<A> (), f); }

I've left implementing `unrle ()`

as an exercise. Here's a little test though that confirms that we are getting savings in from the compression scheme as we hope for.

int main () { std::string buf= "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa" "bbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbb" "c" "ddddddddddddddddddddddddddddddddddddddddddddddddddddddddddd" "ddddddddddddddddddddddddddddddddddddddddddddddddddddddddddd" "ddddddddddddddddddddddddddddddddddddddddddddddddddddddddddd" "ddddddddddddddddddddddddddddddddddddddddddddddddddddddddddd" "ddddddddddddddddddddddddddddddddddddddddddddddddddddddddddd" "ddddddddddddddddddddddddddddddddddddddddddddddddddddddddddd" "eeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeee" "z"; std::list<char> data(buf.begin (), buf.end()); encoding<char> compressed = rle (data); std::cout << sizeof (char) * (data.size ()) << std::endl; std::cout << sizeof (datum <char>) * (compressed.size ()) << std::endl; return 0; }On my machine, this program prints the values $484$ and $72$.