Chewing The Cud

The second set of 10 problems are now in the bag. Feeling the Erlang-fu! The first set can be found here.

%% Modified run-length encoding.
%% Modify the result of problem P10 in such a way that if an element has no duplicates it is simply copied into the result list.
%% Only elements with duplicates are transferred as (N E) lists.
%% Example: (encode-modified ‘(a a a a b c c a a d e e e e)) -> ((4 A) B (2 C) (2 A) D (4 E))
p11([]) -> [];
p11(L) -> p11encode(p09(L), []).

p11encode([], Encoded) -> lists:reverse(Encoded);
p11encode([[Elem | SubL] | L], Encoded) ->
case length(SubL) of
0 -> p11encode(L, [Elem | Encoded]);
_Else -> p11encode(L, [[length(SubL)+1, Elem] | Encoded])
end.

%% Decode a run-length encoded list.
%% Given a run-length code list generated as specified in problem P11. Construct its uncompressed version.
p12([]) -> [];
p12(L) -> p12decode(L, []).

p12decode([], Decoded) -> lists:reverse(Decoded);
p12decode([[N, Elem] | L], Decoded) -> p12decode(L, lists:duplicate(N, Elem) ++ Decoded);
p12decode([Elem | L], Decoded) -> p12decode(L, [Elem | Decoded]).

%% Run-length encoding of a list (direct solution).
%% Implement the so-called run-length encoding data compression method directly.
%% I.e. don’t explicitly create the sublists containing the duplicates, as in problem P09, but only count them.
%% As in problem P11, simplify the result list by replacing the singleton lists (1 X) by X.
%% Example: (encode-direct ‘(a a a a b c c a a d e e e e)) -> ((4 A) B (2 C) (2 A) D (4 E))
p13([]) -> [];
p13([H | L]) -> p13encode(L, H, 1, []).

%% p13encode(The Remaining List,
%% The Current Element,
%% The Number Of Repetitions Thus Far,
%% The Encoded List Thus Far)
p13encode([], Curr, Rep, Encoded) ->
case Rep of
1 -> lists:reverse([Curr | Encoded]);
_Else -> lists:reverse([[Rep, Curr] | Encoded])
end;
p13encode([H | L], Curr, Rep, Encoded) ->
case H of
Curr -> p13encode(L, Curr, Rep+1, Encoded);
_Else ->
case Rep of
1 -> p13encode(L, H, 1, [Curr | Encoded]);
_Other -> p13encode(L, H, 1, [[Rep, Curr] | Encoded])
end
end.

%% Duplicate the elements of a list.
%% Example: (dupli ‘(a b c c d)) -> (A A B B C C C C D D)
p14([]) -> [];
p14([H | L]) -> p14duplicate(L, [H, H]).

p14duplicate([], Cloned) -> lists:reverse(Cloned);
p14duplicate([H | L], Cloned) -> p14duplicate(L, [H, H | Cloned]).

%% Replicate the elements of a list a given number of times.
%% Example: (repli ‘(a b c) 3) -> (A A A B B B C C C)
p15([], _Num) -> [];
p15([H | L], Reps) -> p15replicate(L, Reps, [lists:duplicate(Reps, H)]).

p15replicate([], _Reps, Replicated) -> lists:append(lists:reverse(Replicated));
p15replicate([H | L], Reps, Replicated) -> p15replicate(L, Reps, [lists:duplicate(Reps, H) | Replicated]).

%% Drop every N’th element from a list.
%% Example: (drop ‘(a b c d e f g h i k) 3) -> (A B D E G H K)
p16([], _Nth) -> [];
p16([H | L], Nth) -> p16drop(L, Nth, Nth-1, [H]).

p16drop([], _Nth, _Curr, Dropped) -> lists:reverse(Dropped);
p16drop([_H | L], Nth, Curr, Dropped) when Curr == 1 -> p16drop(L, Nth, Nth, Dropped);
p16drop([H | L], Nth, Curr, Dropped) -> p16drop(L, Nth, Curr-1, [H | Dropped]).

%% Split a list into two parts; the length of the first part is given.
%% Do not use any predefined predicates.
%% Example: (split ‘(a b c d e f g h i k) 3) -> ( (A B C) (D E F G H I K))
p17([], _SplitPoint) -> [];
p17(L, SplitPoint) -> p17split(L, SplitPoint, []).

p17split([], _SplitPoint, List) -> lists:reverse(List);
p17split(L, SplitPoint, Head) when SplitPoint == 0 -> [lists:reverse(Head) | [L]];
p17split([H | L], SplitPoint, Head) -> p17split(L, SplitPoint-1, [H | Head]).

%% Extract a slice from a list.
%% Given two indices, I and K, the slice is the list containing the elements
%% between the I’th and K’th element of the original list (both limits included).
%%Start counting the elements with 1.
%% Example: (slice ‘(a b c d e f g h i k) 3 7) -> (C D E F G)
p18([], _Start, _End) -> [];
p18(L, Start, End) -> p18slice(L, Start, End-Start, []).

p18slice([], _Start, _Count, Sliced) -> lists:reverse(Sliced);
p18slice([H | _L], 1, 0, Sliced) -> lists:reverse([H | Sliced]);
p18slice([H | L], 1, Count, Sliced) -> p18slice(L, 1, Count-1, [H | Sliced]);
p18slice([_H | L], Start, Count, Sliced) -> p18slice(L, Start-1, Count, Sliced).

%% Rotate a list N places to the left.
%% Examples: (rotate ‘(a b c d e f g h) 3) -> (D E F G H A B C)
%% (rotate ‘(a b c d e f g h) -2) -> (G H A B C D E F)
%% Hint: Use the predefined functions length and append, as well as the result of problem P17.
p19([], _Rotate) -> [];
p19(L, Rotate) ->
case Rotate of
Point when Rotate > 0 -> Point;
_LT0 -> Point = length(L) + Rotate
end,
[F, B] = lp:p17(L, Point),
B ++ F.

%% Remove the K’th element from a list.
%% Example: (remove-at ‘(a b c d) 2) -> (A C D)
p20([], _Point) -> [];
p20(L, Point) -> p20remove(L, Point, []).

p20remove([], _Point, BeforePoint) -> lists:reverse(BeforePoint);
p20remove([_H | L], 1, BeforePoint) -> lists:reverse(BeforePoint) ++ L;
p20remove([H | L], Point, BeforePoint) -> p20remove(L, Point-1, [H | BeforePoint]).

Mitch Hedberg has a bit about escalators that is hilarious. Today, while exiting the Metro station in DC, I lived that joke. First, the bit:

An escalator can never break. It can only become stairs. You would never see an “Escalator Temporarily Out Of Order” sign, just “Escalator Temporarily Stairs. Sorry for the convenience.”

Sure enough, on exiting the metro, the escalators were off. Sadly, no one in DC has either heard of Mitch Hedberg or had the stones to put up a sign like that. I think a little humor would go a long way in that city…

Using the 99 Lisp Problems as a guide, I have been writing the problems in Erlang to grow more comfortable with the non-concurrent aspects of the language. Once completed, the goal is then to add in some concurrency exercises to round out the Erlang experience. I have to say, to this point, I feel I’m beginning to grok the ole Erlang list-handling / pattern-matching capabilities.

<update date=”December 22nd, 2006″> Reading the Erlang efficiency guide I found out that appending lists is not the greatest way to do list creation, so I rewrote the applicable exercises with the suggestion of simply reversing the list at the end.</update>
Here, then, are my solutions to the first ten problems:

%% Find the last box of a list.

%% Example: (my-last ‘(a b c d)) -> (D)
p01(L) when length(L) =< 1 -> L;
p01([_ | L]) -> p01(L).

%% Find the last but one box of a list.
%% Example: (my-but-last ‘(a b c d)) -> (C D)
p02(L) when length(L) =< 2 -> L;
p02([_ | L]) -> p02(L).

%% Find the K’th element of a list. The first element in the list is number 1.
%% Example: (element-at ‘(a b c d e) 3) -> C
p03([], _) -> [];
p03(_, Pos) when Pos < 1 -> bad_pos;
p03([H | _], 1) -> H;
p03([_ | L], Pos) -> p03(L, Pos-1).

%% Find the number of elements of a list.
p04([]) -> 0;
p04(L) -> p04count(L, 0).

p04count([], C) -> C;
p04count([_ | L], C) -> p04count(L, C+1).

%% Reverse a list.
p05([]) -> [];
p05(L) -> p05reverse(L, []).

p05reverse([], RevL) -> RevL;
p05reverse([H | L], RevL) -> p05reverse(L, [H | RevL]).

%% Find out whether a list is a palindrome. A palindrome can be read forward or backward; e.g. (x a m a x).
p06([]) -> true;
p06(L) when length(L) == 1 -> true;
p06(L) -> L == p05(L).

%% Flatten a nested list structure. Transform a list, possibly holding lists as elements into a `flat’ list by replacing each list with its elements (recursively).
%% Example: (my-flatten ‘(a (b (c d) e))) -> (A B C D E)
%% Hint: Use the predefined functions list and append.
p07([]) -> [];
p07(L) when not is_list(L) -> [L];
p07([H | L]) ->
Head = p07(H),
Tail = p07(L),
Head ++ Tail.

%% Eliminate consecutive duplicates of list elements.
%% If a list contains repeated elements they should be replaced with a single copy of the element. The order of the elements should not be changed.
%% Example: (compress ‘(a a a a b c c a a d e e e e)) -> (A B C A D E)
p08([]) -> [];
p08([H | L]) -> p08compress(L, H, [H]).

p08compress([], _Curr, Compressed) -> lists:reverse(Compressed);
p08compress([H | L], Curr, Compressed) ->
case Curr of
H -> p08compress(L, Curr, Compressed);
_Else -> p08compress(L, H, [H | Compressed])
end.

%% Pack consecutive duplicates of list elements into sublists. If a list contains repeated elements they should be placed in separate sublists.
%% Example: (pack ‘(a a a a b c c a a d e e e e)) -> ((A A A A) (B) (C C) (A A) (D) (E E E E))
p09([]) -> [];
p09([H | L]) -> p09pack(L, H, [H], []).

p09pack([], _Curr, SubL, Packed) -> lists:reverse([SubL | Packed]);
p09pack([H | L], Curr, SubL, Packed) ->
case Curr of
H -> p09pack(L, Curr, [H | SubL], Packed);
_Else -> p09pack(L, H, [H], [SubL | Packed])
end.

%% Run-length encoding of a list.
%% Use the result of problem P09 to implement the so-called run-length encoding data compression method.
%% Consecutive duplicates of elements are encoded as lists (N E) where N is the number of duplicates of the element E.
%% Example: (encode ‘(a a a a b c c a a d e e e e)) -> ((4 A) (1 B) (2 C) (2 A) (1 D)(4 E))
p10([]) -> [];
p10(L) -> p10encode(p09(L), []).

p10encode([], Encoded) -> lists:reverse(Encoded);
p10encode([[Elem | SubL] | L], Encoded) -> p10encode(L, [[length(SubL)+1, Elem] | Encoded]).

Those who grok Erlang and happen to find this blog, please feel free to offer criticism of my implementations.