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
end
Nifty!
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 $72$.
