gvaughn / elixir_kata Goto Github PK

Hi!

I've found this Erlang solution of Euler's problem 54 "Poker hands", that seems to me very inspiring:

-module(problem54_poker).

-compile([export_all]).

-define(HC,0). % High Card: Highest value card.
-define(OP,1). % One Pair: Two cards of the same value.
-define(TP,2). % Two Pairs: Two different pairs.
-define(TK,3). % Three of a Kind: Three cards of the same value.
-define(ST,4). % Straight: All cards are consecutive values.
-define(FL,5). % Flush: All cards of the same suit.
-define(FH,6). % Full House: Three of a kind and a pair.
-define(FK,7). % Four of a Kind: Four cards of the same value.
-define(SF,8). % Straight Flush: All cards are consecutive values of same suit.

problem54_poker() ->
{ok,F} = file:open("C:\erl5.10.1\bin\poker.txt",[read]),
R = problem54_poker(F,file:read_line(F),0),
file:close(F),
R.

problem54_poker(_,eof,I) -> I;
problem54_poker(F,{ok,D},I) -> problem54_poker(F,file:read_line(F),I+readline(D)).

readline(S) ->
{S1,S2} = lists:split(15, S),
case (readline(S1,[],undefined) > readline(S2,[],undefined)) of
false -> 0;
true -> 1
end.

readline([],R,Colour) -> hand({lists:sort(R),Colour =/= false});
readline([C,K,|Q],R,K) -> readline(Q,[convert(C)|R],K);
readline([C,K,|Q],R,undefined) -> readline(Q,[convert(C)|R],K);
readline([C,,|Q],R,_) -> readline(Q,[convert(C)|R],false).

convert($2) -> 2;
convert($3) -> 3;
convert($4) -> 4;
convert($5) -> 5;
convert($6) -> 6;
convert($7) -> 7;
convert($8) -> 8;
convert($9) -> 9;
convert($T) -> 10;
convert($J) -> 11;
convert($Q) -> 12;
convert($K) -> 13;
convert($A) -> 14.

hand({[A,A,A,A,],}) -> {?FK,A};
hand({[,A,A,A,A],}) -> {?FK,A};
hand({[B,B,A,A,A],}) -> {?FH,A};
hand({[A,A,A,B,B],}) -> {?FH,A};
hand({[A,A,A,,],}) -> {?TK,A};
hand({[,A,A,A,],}) -> {?TK,A};
hand({[,,A,A,A],}) -> {?TK,A};
hand({[,B,B,A,A],}) -> {?TP,A};
hand({[B,B,,A,A],}) -> {?TP,A};
hand({[B,B,A,A,],}) -> {?TP,A};
hand({[A,A,,,],}) -> {?OP,A};
hand({[,A,A,,],}) -> {?OP,A};
hand({[,,A,A,],}) -> {?OP,A};
hand({[,,,A,A],}) -> {?OP,A};
hand({[2,3,4,5,14],true}) -> {?SF,5};
hand({[2,3,4,5,14],false}) -> {?ST,5};
hand({[A,,,,B],true}) when (B-A) == 4 -> {?SF,B};
hand({[A,,,,B],false}) when (B-A) == 4 -> {?ST,B};
hand({[,,,,A],true}) -> {?FL,A};
hand({[,,,,A],}) -> {?HC,A}.