Right on the heals of 21-30 comes 31-40. These problems began to delve into mathematics, with a great emphasis on prime numbers and their generation. Quite interesting to work with, since prime numbers are the basis behind encryption of any non-trivial strength. Enjoy:

%% Determine whether a given integer number is prime. %% Example: (is-prime 7) -> T p31(2) -> true; p31(N) when N rem 2 : 0 -> false; p31(N) -> p31is\_prime(N, 3, N div 2 ). p31is\_prime(\_N, K, Limit) when K > Limit -> true; p31is\_prime(N, K, Limit) -> case N rem K of 0 -> false; \_Else -> p31is\_prime(N, K+2, Limit) end. %% Determine the greatest common divisor of two positive integer numbers. %% Use Euclid's algorithm. %% Example: (gcd 36 63) -> 9 p32(A, 0) -> A; p32(A, B) when B > A -> p32(B, A); p32(A, B) -> p32(B, A rem B). %% Determine whether two positive integer numbers are coprime. %% Two numbers are coprime if their greatest common divisor equals 1. %% Example: (coprime 35 64) -> T p33(A, B) -> p32(A, B) : 1. %% Calculate Euler's totient function phi(m). %% Euler's so-called totient function phi(m) is defined as the number of positive integers r (1 < = r < m) that are coprime to m. %% Example: m = 10: r = 1,3,7,9; thus phi(m) = 4. Note the special case: phi(1) = 1. %% (totient-phi 10) -> 4 %% Find out what the value of phi(m) is if m is a prime number. %% psuedo-code if is\_prime(m), phi(m) = m-1, else compute phi(m). %% Euler's totient function plays an important role in one of the most widely used public key cryptography methods (RSA). %% In this exercise you should use the most primitive method to calculate this function (there are smarter ways that we shall discuss later). p34(1) -> 1; p34(M) -> p34totient\_phi(M, 1, []). p34totient\_phi(M, M, L) -> length(L); p34totient\_phi(M, R, L) -> case p33(M, R) of true -> p34totient\_phi(M, R+1, [R | L]); false -> p34totient\_phi(M, R+1, L) end. %% Determine the prime factors of a given positive integer. %% Construct a flat list containing the prime factors in ascending order. %% Example: (prime-factors 315) -> (3 3 5 7) p35(N) -> p35prime\_factors(N, 2, []). p35prime\_factors(1, \_C, PF) -> lists:reverse(PF); p35prime\_factors(N, 2, PF) -> case (N rem 2) : 0 of true -> p35prime\_factors(N div 2, 2, [2 | PF]); false -> p35prime\_factors(N, 3, PF) end; p35prime\_factors(N, C, PF) -> case (N rem C) : 0 of true -> p35prime\_factors(N div C, C, [C | PF]); false -> p35prime\_factors(N, C+2, PF) end. %% Determine the prime factors of a given positive integer (2). %% Construct a list containing the prime factors and their multiplicity. %% Example: (prime-factors-mult 315) -> ((3 2) (5 1) (7 1)) %% Hint: The problem is similar to problem P13. p36(N) -> p10(p35(N)). %% Calculate Euler's totient function phi(m) (improved). %% See problem P34 for the definition of Euler's totient function. %% If the list of the prime factors of a number m is known in the form of problem P36 then the function phi(m) can be efficiently calculated as follows: %% Let ((p1 m1) (p2 m2) (p3 m3) …) be the list of prime factors (and their multiplicities) of a given number m. Then phi(m) can be calculated with the following formula: %% phi(m) = (p1 - 1) * [p1 * (m1 - 1)] * (p2 - 1) * [p2 * (m2 - 1)] * (p3 - 1) * [p3 * (m3 - 1)] + … %% Note that a * b stands for the b'th power of a. p37(M) -> p37phi(p36(M), 1). p37phi([], Phi) -> Phi; p37phi([[M, P] | L], Phi) -> p37phi(L, Phi * (P - 1) * round(math:pow( P, M-1 )) ). %% Compare the two methods of calculating Euler's totient function. %% Use the solutions of problems P34 and P37 to compare the algorithms. Take the number of logical inferences as a measure for efficiency. Try to calculate phi(10090) as an example. p38(M) -> {T34, V34} = timer:tc(lp, p34, [M]), {T37, V37} = timer:tc(lp, p37, [M]), io:fwrite("p34 took ~p micro seconds and returned ~p.~n", [T34, V34]), io:fwrite("p37 took ~p micro seconds and returned ~p.~n", [T37, V37]). %% A list of prime numbers. %% Given a range of integers by its lower and upper limit, construct a list of all prime numbers in that range. p39(High) -> p39(1, High). p39(Low, High) when Low > High -> p39(High, Low); p39(Low, High) -> [X || X < - p39primes(lists:seq(2, High) , [1]), X >= Low, X =< High]. p39primes([], Primes) -> lists:reverse(Primes); p39primes([1 | Sieve], Primes) ->      p39primes(Sieve, [1 | Primes]); % pops 1 off the sieve p39primes([2 | Sieve], Primes) ->      p39primes([X || X < - Sieve, (X rem 2) > 0], [2 | Primes]); % pops 2 off the sieve and removes all multiples of 2 p39primes([Curr | Sieve], Primes) -> p39primes([X || X < - Sieve, (X rem Curr) > 0], [Curr | Primes]). % pops the next value off the sieve and removes all multiples %% Goldbach's conjecture. %% Goldbach's conjecture says that every positive even number greater than 2 is the sum of two prime numbers. %% Example: 28 = 5 + 23. It is one of the most famous facts in number theory that has not been proved to be correct in the general case. %% It has been numerically confirmed up to very large numbers (much larger than we can go with our Prolog system). %% Write a predicate to find the two prime numbers that sum up to a given even integer. %% Example: (goldbach 28) -> (5 23) p40(N) -> p40goldbach(N, p39(N), []). p40goldbach(0, \_Primes, Result) when length(Result) : 2 -> lists:reverse(Result); p40goldbach(N, \_Primes, Result) when length(Result) : 2, N / 0 -> false; p40goldbach(\_N, [], \_Result) -> false; p40goldbach(N, [P | Primes], Result) -> Sol = p40goldbach(N-P, Primes, [P | Result]), case is\_list(Sol) of true -> Sol; \_else -> p40goldbach(N, Primes, Result) end.

The next installment of the 99 Lisp problems. 27 and 28 are incomplete as I have not sat down to actually work through them yet. More research is needed to do multinomial coefficients. This batch of problems posed some challenge and required a bit of research and dusting off math skills, as well as getting familiar with Erlang's List Comprehension syntax. For your viewing pleasure:

%% Insert an element at a given position into a list.
%% Example:  (insert-at 'alfa '(a b c d) 2) -> (A ALFA B C D)
p21(Elem, L, Pos) ->
    p21insert(Elem, L, Pos, []).

p21insert(Elem, L, 1, NewL) ->
    lists:reverse([Elem | NewL]) ++ L;
 p21insert(Elem, [H | L], Pos, NewL) ->
    p21insert(Elem, L, Pos-1, [H | NewL]).

%% Create a list containing all integers within a given range.
%% If first argument is smaller than second, produce a list in decreasing order.
%% Example: (range 4 9) -> (4 5 6 7 8 9)
p22(Start, End) when Start < End ->
    p22range_asc(End-Start, [Start]);
p22(Start, End) ->
    p22range_desc(Start-End, [Start]).

p22range_asc(0, L) ->
    lists:reverse(L);
p22range_asc(Count, [H | L]) ->
    p22range_asc(Count-1, [H+1, H | L]).

p22range_desc(0, L) ->
    lists:reverse(L);
p22range_desc(Count, [H | L]) ->
    p22range_desc(Count-1, [H-1, H | L]).

%% Extract a given number of randomly selected elements from a list.
%% The selected items shall bereturned in a list.
%% Example: (rnd-select '(a b c d e f g h) 3) -> (E D A)
%% Hint: Use the built-in random number generator and the result of
%% problem P20.
p23(L, Count) when length(L) < Count ->
    p23rnd_select(L, length(L), []);
p23(L, Count) ->
    p23rnd_select(L, Count, []).

p23rnd_select(_L, 0, RandL) -> RandL;
p23rnd_select(L, Count, RandL) ->
    Rnd = random:uniform(length(L)),
    V = lists:nth(Rnd, L),
    p23rnd_select(p20(L, Rnd), Count-1, [V | RandL]).

%% Lotto: Draw N different random numbers from the set 1..M.
%% The selected numbers shall be returned in a list.
%% Example: (lotto-select 6 49) -> (23 1 17 33 21 37)
%% Hint: Combine the solutions of problems P22 and P23.
p24(Num, High) -> p23(p22(1, High), Num).

%% Generate a random permutation of the elements of a list.
%% Example: (rnd-permu '(a b c d e f)) -> (B A D C E F)
%% Hint: Use the solution of problem P23.
p25(L) -> p23(L, length(L)).

%% Generate the combinations of K distinct objects chosen from the N elements of a list
%% In how many ways can a committee of 3 be chosen from a group of 12 people? We all know that there are C(12,3) = 220 possibilities (C(N,K) denotes the well-known binomial coefficients).
%% For pure mathematicians, this result may be great. But we want to really generate all the possibilities in a list.
%% Example: (combination 3 '(a b c d e f)) -> ((A B C) (A B D) (A B E) ... ) % Not the most efficient way, but generate the power set for the list and remove all but the sub-lists of length K
powerset([]) -> [[]];
powerset([H | T]) ->
    SubPS = powerset(T),
    lists:append(SubPS, lists:map(fun(E) -> [H | E] end, SubPS)).

p26(K, L) -> [X || X < - powerset(L), length(X) == K].

%% Group the elements of a set into disjoint subsets.
%% a) In how many ways can a group of 9 people work in 3 disjoint
%% subgroups of 2, 3 and 4 persons? Write a function that generates
%% all the possibilities and returns them in a list.
%% Example: (group3 '(aldo beat carla david evi flip gary hugo ida))
%% -> ( ( (ALDO BEAT) (CARLA DAVID EVI) (FLIP GARY HUGO IDA) ) ... )
p27([]) -> [].

%% b) Generalize the above predicate in a way that we can specify a
%% list of group sizes and the predicate will return a list of groups.
%% Example: (group '(aldo beat carla david evi flip gary hugo ida) '(2
%% 2 5)) -> ( ( (ALDO BEAT) (CARLA DAVID) (EVI FLIP GARY HUGO IDA) )
%% ... )
%% Note that we do not want permutations of the group members;
%% i.e. ((ALDO BEAT) ...) is the same solution as ((BEAT ALDO) ...).
%% However, we make a difference between ((ALDO BEAT) (CARLA DAVID)
%% ...) and ((CARLA DAVID) (ALDO BEAT) ...).
%% You may find more about this combinatorial problem in a good book
%% on discrete mathematics under the term "multinomial coefficients".
p28() -> [].

%% Sorting a list of lists according to length of sublists

%% a) We suppose that a list contains elements that are lists
%% themselves. The objective is to sort the elements of this list
%% according to their length. E.g. short lists first, longer lists
%% later, or vice versa.

%% Example: (lsort '((a b c) (d e) (f g h) (d e) (i j k l) (m n) (o)))
%% -> ((O) (D E) (D E) (M N) (A B C) (F G H) (I J K L))
p29(L) -> p29lsort(L).

%% code taken from Erlang Programming Examples -> List Comprehensions
%% -> QuickSort and adapted
p29lsort([]) -> [];
p29lsort([Pivot | L]) ->
    p29lsort([ X || X < - L,
		    length(X) < length(Pivot)
	     ])
	++ [Pivot | p29lsort([ X || X = length(Pivot)])].

%% b) Again, we suppose that a list contains elements that are lists
%% themselves. But this time the objective is to sort the elements of
%% this list according to their length frequency;
%% i.e., in the default, where sorting is done ascendingly, lists with
%% rare lengths are placed first, others with a more frequent length
%% come later.
%% Example: (lfsort '((a b c) (d e) (f g h) (d e) (i j k l) (m n)
%% (o))) -> ((i j k l) (o) (a b c) (f g h) (d e) (d e) (m n))
%% Note that in the above example, the first two lists in the result
%% have length 4 and 1, both lengths appear just once. The third and
%% forth list have length 3 which appears twice (there are two list of
%% this length).

%% And finally, the last three lists have length 2. This is the most
%% frequent length.
p30(L) -> p30lfsort(L).

p30lfsort([]) -> [];
p30lfsort([H | L]) ->
    Grouped = p30group(L, [[H]]),
    Sorted = p29(Grouped),
    p30ungroup(Sorted, []).

p30group([], Grouped) -> Grouped;
p30group([H | L], Grouped) -> p30group(L, p30insert(H, Grouped, [])).

%% insert Elem in to a group with the same length, or create a new
%% group for the length of Elem
p30insert(Elem, [], PrevGroups) ->
    lists:reverse([[Elem] | PrevGroups]);
p30insert(Elem, [ [GroupH | GroupT] | RestGrouped], PrevGroups)
  when length(GroupH) == length(Elem) ->
    PrevGroups ++ [ [GroupH, Elem | GroupT] | RestGrouped];
p30insert(Elem, [ [GroupH | GroupT] | RestGrouped], PrevGroups) ->
    p30insert(Elem, RestGrouped, [[GroupH | GroupT] | PrevGroups]).

%% remove length-specific groupings
p30ungroup([], Ungrouped) -> lists:reverse(Ungrouped);
p30ungroup([H | L], Ungrouped) when not is_list(H) ->
    [[H | L] | Ungrouped];
p30ungroup([Group | L], Ungrouped) ->
    p30ungroup(Group, []) ++ p30ungroup(L, Ungrouped).

Working through the 99 Lisp Problems in Erlang and just did a comparison of two different methods for computing Euler's totient function. The results:

> lp:p39(10090). p34 took 10864 micro seconds and returned 4032. p37 took 72 micro seconds and returned 4032.

I'll leave it to the reader to work the problems themselves, but it does illustrate the importance of good algorithm design in mission-critical parts of your code. Were my application's lifeblood to be in calculating phi, and I was naive and took the p34 algorithm to solve for phi, I would be 150x slower at doing the job. I guess the moral is, always research mission-critical portions of code for optimized algorithms. A sub-moral would be to keep abreast of the latest and greatest in algorithms and math in general. I like to have a friend in grad school studying quantum mechanics to keep me informed!

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.

Holy ballz this guy needs a job or something. Still cool what he does with dice.

At least, that's what the hope is. This article from The Christian Science Monitor is rasing awareness of new micro-combined heat-and-power untis, or micro-CHP, that have recently entered the market in the New England area. These little babies supposedly generate heat and electricity as a by-product, helping reduce your electric bills while still heating your home. If anything, it is important for technologies like this to be exposed to a wide audience to let people know there are alternatives to relying on the utility companies for all your heating and power needs. Let's hope something big comes of this!

Pictures of cats. What else needs be said?

Simple, yet addictive games are my productivity kryptonite. I play Mahjongg, Minesweeper, and a host of other flash-based games on the web. The latest installment came to me via Reddit. Play Orbit and have fun doing it. I got up to twelve planets before finally pulling myself away.