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  1. Practice
  2. Functional Programming
  3. Recursion
  4. Pascal's Triangle
  5. Discussions

Pascal's Triangle

  • csixteen + 1 comment

    Erlang:

    -module(solution).
-export([main/0]).

%% Factorial with tail recursion
fac(N) -> fac(N, 1).
fac(0, Acc) -> Acc;
fac(N, Acc) when N > 0 -> fac(N-1, N*Acc).

%% Calculates the value of a column in a row.
column(N, M) -> fac(N) div (fac(M) * fac(N-M)).

%% Builds a row
build_row(N) -> build_row(N,1).
build_row(N,M) when N rem 2 == 1, M == N div 2 + 1 -> 
    [column(N-1,M-1)];
build_row(N,M) when N rem 2 == 0, M-1 == N div 2 -> [];
build_row(N,M) when N > M ->
	C = column(N-1,M-1),
	[C] ++ build_row(N,M+1) ++ [C].

%% Prints a row
print_row([]) -> io:format("~n");
print_row([X|Xs]) ->
	io:format("~p ", [X]),
	print_row(Xs).

%% Builds the triangle
pascal(N,M) when N < M -> ok;
pascal(N,I) ->
	R = build_row(I),
	print_row(R),
	pascal(N,I+1).

main() ->
	{ok, [N]} = io:fread("", "~d"),
	pascal(N,1)
	.
    • Andrei_Leontev + 0 comments

      Another Erlang solution:

      main() ->
	{ ok, [N]} = io:fread("", "~d"),
    pascal(0, N, []).

pascal(C, N, A) ->
    case N-C of
        0 -> ok;
        _ -> A1 = generate(C, A), 
            print_line(A1), 
            pascal(C+1, N, A1)
    end.

generate(0, _A) ->
    [1];
generate(1, _A) ->
    [1, 1];
generate(_N, A) -> 
    {_, A1} = lists:foldl(fun(E, {P, AN}) -> {E, AN ++ [P + E]} end, {0, []}, A), 
    A1 ++ [1].

print_line(A) -> 
    lists:foreach(fun(E) -> io:format("~p ", [E]) end, A), 
    io:format("~n", []).