Over on NewsTrust, a report on a 17 year old’s documentary film has caused quite a discussion on race. Mimicking the experiment performed back in the 1950′s to prove “separate but equal” was not equal, the young filmmaker found how little things have changed in 50 years. The newsstory is here.

## Archive for January, 2007

### Sobering Video

Friday, January 26th, 2007### Futuristic interaction with computers…now!

Sunday, January 21st, 2007### E-99: 31-40

Sunday, January 7th, 2007Right 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.

### E-99: 21-30

Sunday, January 7th, 2007The 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 be returned 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 <- L, length(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).

### The Importance of good algorithm design

Saturday, January 6th, 2007Working 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!