E-99: 1-10

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.