Chewing The Cud

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 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).

I don’t know why, but Cingular apparently does not want me paying my bill. When I log into my account to pay online, I was taken to an intermediate page where I was asked, in a mandatory fashion, for my email address. I’m not a big fan of giving that sucker out for no reason, so I simply clicked the ‘Continue’ button. Unfortunately, the button was a link whose href was the lovely octothorpe.

This of course meant I was going nowhere by means of that button. I was quite peeved at this point because, not only did I not get an explanation from the page telling my why my email was required, but I also had no explanation for why the continue button failed to do so. The age-old adage, “Fail to communicate with me properly once, you suck; fail twice, now I’m pissed” came to mind.

So what’s a web geek to do? Obviously search the page’s javascript for the form submission line and execute it directly through Firebug commandline! Should any Cingular users get frustrated as well, get Firebug (and Firefox, which I have to mention since Google Analytics tells me the internets are alive with IE users still – shame on you all), and execute this line of code:

document.updateRequiredRegisteredInfoActionForm.submit();

Not only did this forward me on to my account page, but it also bypassed the need to fill in my email for no good reason. I get enough spam as it is. So there it is my friends; simple, one-liner to defeat Cingular’s attempts to be poor communicators. A silver lining here, besides getting in, is the lesson learned from Cingular’s mistake: Don’t suck at communicating with your users!

So long, and thanks for all the fish.