Erlang solution for Euler problem 18

In project Euler problem number 18 we are asked to determine the maximum sum traveling from the top of a triangle to the base.

This time, I did not solve the problem using the brute-force approach of trying all possible roots from the top to the bottom.

Instead, I rely on the observation that the cost from a to the bottom = cost from a to the bottom + max(cost from b to the bottom, cost from c to the bottom):

   x
  x x
 x x x
a x x x
b c x x x
x x x x x x

This is used to "collapse" the bottom two rows of the triangle into one as follows:

   x                         x
  x x                       x x
 x x x         ==>         x x x
x x x x                   x x x x
a x x x x                 d x x x x
b c x x x x

where d = a + max(b,c).

This process is repeated until there is one remaining row which contains the answer.

In Erlang this is implemented as follows:

%% Project Euler, problem 18
%%
%% Find the maximum sum travelling from the top of the triangle to the base.

-module(euler18).
-author('Cayle Spandon').

-export([solve/0]).

-define(TRIANGLE, [[04, 62, 98, 27, 23, 09, 70, 98, 73, 93, 38, 53, 60, 04, 23],
                   [63, 66, 04, 68, 89, 53, 67, 30, 73, 16, 69, 87, 40, 31],
                     [91, 71, 52, 38, 17, 14, 91, 43, 58, 50, 27, 29, 48],
                       [70, 11, 33, 28, 77, 73, 17, 78, 39, 68, 17, 57],
                         [53, 71, 44, 65, 25, 43, 91, 52, 97, 51, 14],
                           [41, 48, 72, 33, 47, 32, 37, 16, 94, 29],
                             [41, 41, 26, 56, 83, 40, 80, 70, 33],
                               [99, 65, 04, 28, 06, 16, 70, 92],
                                 [88, 02, 77, 73, 07, 63, 67],
                                   [19, 01, 23, 75, 03, 34],
                                     [20, 04, 82, 47, 65],
                                       [18, 35, 87, 10],
                                         [17, 47, 82],
                                           [95, 64],
                                             [75]].

% Reduce the triangle to a smaller triangle by collapsing the 1st and 2nd row into one.
% If we have collapsed the triangle to a single row with one number, we are done.
%
collapse_triangle([[N]]) ->
  N;
collapse_triangle([FirstRow, SecondRow | RemainingRows]) ->
  CollapsedRow = collapse_rows(FirstRow, SecondRow),
  collapse_triangle([CollapsedRow | RemainingRows]).

% Collapse two rows into one as follows:
%    A B
%     C  --> C' = C + max(A,B)
% When the first row is reduced to a single number, we are done.
%
collapse_rows([_A], []) ->
  [];
collapse_rows([A, B | Tail1], [C | Tail2]) ->
  [C + max(A, B)] ++ collapse_rows([B | Tail1], Tail2).

max(A, B) ->
  if
      A > B ->
          A;
      true ->
          B
  end.
    
solve() ->
  collapse_triangle(?TRIANGLE).