我正在学习Erlang。作为练习，我学习了生成质数的Eratosthenes筛分算法。这是我的代码：

-module(seed2).
-export([get/1]).

get(N) -> WorkList = lists:duplicate(N, empty),
          get(2, N, WorkList, []).

get(thats_the_end, _N, _WorkList, ResultList) -> lists:reverse(ResultList);
get(CurrentPrime, N, WorkList, ResultList) -> ModWorkList = markAsPrime(CurrentPrime, N, WorkList),
                                              NextPrime = findNextPrime(CurrentPrime + 1, N, WorkList),
                                              get(NextPrime, N, ModWorkList, [CurrentPrime|ResultList]).

markAsPrime(CurrentPrime, N, WorkList) when CurrentPrime =< N -> WorkListMod = replace(CurrentPrime, WorkList, prime),
                                                                 markAllMultiples(CurrentPrime, N, 2*CurrentPrime, WorkListMod).

markAllMultiples(_ThePrime, N, TheCurentMark, WorkList) when TheCurentMark > N -> WorkList;
markAllMultiples(ThePrime, N, TheCurrentMark, WorkList) -> WorkListMod = replace(TheCurrentMark, WorkList, marked),
                                                           markAllMultiples(ThePrime, N, TheCurrentMark + ThePrime, WorkListMod).

findNextPrime(Iterator, N, _WorkList) when Iterator > N -> thats_the_end;
findNextPrime(Iterator, N, WorkList) -> I = lists:nth(Iterator, WorkList),
                                        if
                                          I =:= empty -> Iterator;
                                          true -> findNextPrime(Iterator + 1, N, WorkList)
                                        end.

replace(N, L, New)-> {L1, [_H|L2]} = lists:split(N - 1, L),
                     lists:append(L1, [New|L2]).

这段代码实际上是有效的：)。问题是我感觉这不是最好的实现。

我的问题是实施" Eratosthenes筛"的"古怪"方式是什么？

编辑：好的，Andreas解决方案很好，但是很慢。有什么想法可以改善这一点吗？

解决方案

我通过使用并发处理解决了这个问题。

来源

这是一个简单(但并非很快)的筛子实现：

-module(primes).
-export([sieve/1]).
-include_lib("eunit/include/eunit.hrl").

sieve([]) ->
    [];
sieve([H|T]) ->          
    List = lists:filter(fun(N) -> N rem H /= 0 end, T),
    [H|sieve(List)];
sieve(N) ->
    sieve(lists:seq(2,N)).

到目前为止，我最快的代码(比Andrea更快)是使用array：

-module(seed4).
-export([get/1]).

get(N) -> WorkList = array:new([{size, N}, {default, empty}]),
          get(2, N, WorkList, []).

get(thats_the_end, _N, _WorkList, ResultList) -> lists:reverse(ResultList);
get(CurrentPrime, N, WorkList, ResultList) -> ModWorkList = markAsPrime(CurrentPrime, N, WorkList),
                                              NextPrime = findNextPrime(CurrentPrime + 1, N, WorkList),
                                              get(NextPrime, N, ModWorkList, [CurrentPrime|ResultList]).

markAsPrime(CurrentPrime, N, WorkList) when CurrentPrime =< N -> WorkListMod = replace(CurrentPrime, WorkList, prime),
                                                                 markAllMultiples(CurrentPrime, N, 2*CurrentPrime, WorkListMod).

markAllMultiples(_ThePrime, N, TheCurentMark, WorkList) when TheCurentMark > N -> WorkList;
markAllMultiples(ThePrime, N, TheCurrentMark, WorkList) -> WorkListMod = replace(TheCurrentMark, WorkList, marked),
                                                           markAllMultiples(ThePrime, N, TheCurrentMark + ThePrime, WorkListMod).

findNextPrime(Iterator, N, _WorkList) when Iterator > N -> thats_the_end;
findNextPrime(Iterator, N, WorkList) -> I = array:get(Iterator - 1, WorkList),
                                        if
                                          I =:= empty -> Iterator;
                                          true -> findNextPrime(Iterator + 1, N, WorkList)
                                        end.

replace(N, L, New) -> array:set(N - 1, New, L).

我没有详细研究这些内容，但是我已经在下面测试了我的实现(这是我为欧拉计划挑战而写的)，并且比上述两个实现快了几个数量级。直到我取消了一些自定义功能，而是寻找列表：可以做到相同的功能，这实在是太慢了。最好学习该课程，以便始终查看是否需要执行某项操作的库实现，这通常会更快！在2.8GHz iMac上，这可以在3.6秒内计算出多达200万个质数的总和。

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%Sum of all primes below Max. Will use sieve of Eratosthenes 
sum_primes(Max) ->
    LastCheck = round(math:sqrt(Max)),
    All = lists:seq(3, Max, 2), %note are creating odd-only array
    Primes = sieve(All, Max, LastCheck),
    %io:format("Primes: ~p~n", [Primes]),
    lists:sum(Primes) + 2. %adding back the number 2 to the list

%sieve of Eratosthenes
sieve(All, Max, LastCheck) ->
    sieve([], All, Max, LastCheck).

sieve(Primes, All, Max, LastCheck) ->
    %swap the first element of All onto Primes 
    [Cur|All2] = All,
    Primes2 = [Cur|Primes],
    case Cur > LastCheck of 
        true ->
            lists:append(Primes2, All2); %all known primes and all remaining from list (not sieved) are prime
        false -> 
            All3 = lists:filter(fun(X) -> X rem Cur =/= 0 end, All2),
            sieve(Primes2, All3, Max, LastCheck)

    end.

我有点喜欢这个主题，因此我开始修改BarryE的代码，并通过创建自己的list_filter函数使它快了70％，并且可以利用两个CPU。我也很容易在两个版本之间切换。测试运行显示：

61> timer:tc(test,sum_primes,[2000000]).
{2458537,142913970581}

代码：

-module(test).

%%-export([sum_primes/1]).
-compile(export_all).

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%Sum of all primes below Max. Will use sieve of Eratosthenes 
sum_primes(Max) ->
    LastCheck = round(math:sqrt(Max)),
    All = lists:seq(3, Max, 2), %note are creating odd-only array
    %%Primes = sieve(noref,All, LastCheck),
    Primes = spawn_sieve(All, LastCheck),
    lists:sum(Primes) + 2. %adding back the number 2 to the list

%%sieve of Eratosthenes
sieve(Ref,All, LastCheck) ->
    sieve(Ref,[], All, LastCheck).

sieve(noref,Primes, All = [Cur|_], LastCheck) when Cur > LastCheck ->
    lists:reverse(Primes, All); %all known primes and all remaining from list (not sieved) are prime    
sieve({Pid,Ref},Primes, All=[Cur|_], LastCheck) when Cur > LastCheck ->
    Pid ! {Ref,lists:reverse(Primes, All)}; 
sieve(Ref,Primes, [Cur|All2], LastCheck) ->
    %%All3 = lists:filter(fun(X) -> X rem Cur =/= 0 end, All2),
    All3 = lists_filter(Cur,All2),
    sieve(Ref,[Cur|Primes], All3,  LastCheck).

lists_filter(Cur,All2) ->
    lists_filter(Cur,All2,[]).

lists_filter(V,[H|T],L) ->
    case H rem V of
    0 ->
        lists_filter(V,T,L);
    _ ->
        lists_filter(V,T,[H|L])
    end;
lists_filter(_,[],L) ->
    lists:reverse(L).

%% This is a sloppy implementation ;)
spawn_sieve(All,Last) ->
    %% split the job
    {L1,L2} = lists:split(round(length(All)/2),All),
    Filters = filters(All,Last),
    %%io:format("F:~p~n",[Filters]),
    L3 = lists:append(Filters,L2),
    %%io:format("L1:~w~n",[L1]),
    %%    io:format("L2:~w~n",[L3]),
    %%lists_filter(Cur,All2,[]).
    Pid = self(),
    Ref1=make_ref(),
    Ref2=make_ref(),
    erlang:spawn(?MODULE,sieve,[{Pid,Ref1},L1,Last]),
    erlang:spawn(?MODULE,sieve,[{Pid,Ref2},L3,Last]),
    Res1=receive
         {Ref1,R1} ->
         {1,R1};
         {Ref2,R1} ->
         {2,R1}
     end,
    Res2= receive
          {Ref1,R2} ->
          {1,R2};
          {Ref2,R2} ->
          {2,R2}
      end,
    apnd(Filters,Res1,Res2).

filters([H|T],Last) when H 
    [H|filters(T,Last)];
filters([H|_],_) ->
    [H];
filters(_,_) ->
    [].

apnd(Filters,{1,N1},{2,N2}) ->
    lists:append(N1,subtract(N2,Filters));
apnd(Filters,{2,N2},{1,N1}) ->
    lists:append(N1,subtract(N2,Filters)).

subtract([H|L],[H|T]) ->
    subtract(L,T);
subtract(L=[A|_],[B|_]) when A > B ->
    L;
subtract(L,[_|T]) ->
    subtract(L,T);
subtract(L,[]) ->
    L.

我以前的帖子未正确格式化。这是代码的转贴。很抱歉发送垃圾邮件...

-module(test).

%%-export([sum_primes/1]).
-compile(export_all).

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%Sum of all primes below Max. Will use sieve of Eratosthenes 
sum_primes(Max) ->
    LastCheck = round(math:sqrt(Max)),
    All = lists:seq(3, Max, 2), %note are creating odd-only array
    %%Primes = sieve(noref,All, LastCheck),
    Primes = spawn_sieve(All, LastCheck),
    lists:sum(Primes) + 2. %adding back the number 2 to the list

%%sieve of Eratosthenes
sieve(Ref,All, LastCheck) ->
    sieve(Ref,[], All, LastCheck).

sieve(noref,Primes, All = [Cur|_], LastCheck) when Cur > LastCheck ->
    lists:reverse(Primes, All); %all known primes and all remaining from list (not sieved) are prime    
sieve({Pid,Ref},Primes, All=[Cur|_], LastCheck) when Cur > LastCheck ->
    Pid ! {Ref,lists:reverse(Primes, All)}; 
sieve(Ref,Primes, [Cur|All2], LastCheck) ->
    %%All3 = lists:filter(fun(X) -> X rem Cur =/= 0 end, All2),
    All3 = lists_filter(Cur,All2),
    sieve(Ref,[Cur|Primes], All3,  LastCheck).

lists_filter(Cur,All2) ->
    lists_filter(Cur,All2,[]).

lists_filter(V,[H|T],L) ->
    case H rem V of
    0 ->
        lists_filter(V,T,L);
    _ ->
        lists_filter(V,T,[H|L])
    end;
lists_filter(_,[],L) ->
    lists:reverse(L).

%% This is a sloppy implementation ;)
spawn_sieve(All,Last) ->
    %% split the job
    {L1,L2} = lists:split(round(length(All)/2),All),
    Filters = filters(All,Last),
    L3 = lists:append(Filters,L2),
    Pid = self(),
    Ref1=make_ref(),
    Ref2=make_ref(),
    erlang:spawn(?MODULE,sieve,[{Pid,Ref1},L1,Last]),
    erlang:spawn(?MODULE,sieve,[{Pid,Ref2},L3,Last]),
    Res1=receive
         {Ref1,R1} ->
         {1,R1};
         {Ref2,R1} ->
         {2,R1}
     end,
    Res2= receive
          {Ref1,R2} ->
          {1,R2};
          {Ref2,R2} ->
          {2,R2}
      end,
    apnd(Filters,Res1,Res2).

filters([H|T],Last) when H 
    [H|filters(T,Last)];
filters([H|_],_) ->
    [H];
filters(_,_) ->
    [].

apnd(Filters,{1,N1},{2,N2}) ->
    lists:append(N1,subtract(N2,Filters));
apnd(Filters,{2,N2},{1,N1}) ->
    lists:append(N1,subtract(N2,Filters)).

subtract([H|L],[H|T]) ->
    subtract(L,T);
subtract(L=[A|_],[B|_]) when A > B ->
    L;
subtract(L,[_|T]) ->
    subtract(L,T);
subtract(L,[]) ->
    L.

我们可以向老板展示：http://www.sics.se/~joe/apachevsyaws.html。其他一些(经典的？)erlang参数是：

-不间断操作，可以即时加载新代码。

-易于调试，不再需要分析核心转储。

-易于利用多核/ CPU

-容易利用集群吗？

谁想处理指针和东西？这不是21世纪吗？ ;)

一些陷阱：
编写某些东西可能看起来容易又快捷，但是性能却很差。如果我
想快点做，我通常最终会写2-4个相同的版本
功能。通常，我们需要采取鹰眼的态度来解决可能是一个问题的问题。
与所使用的也没有什么不同。

• 在列表中查找内容>大约1000个元素的速度很慢，请尝试使用ets表。
• 字符串" abc"占用的空间比3个字节大得多。因此，请尝试使用二进制文件(这很痛苦)。

总而言之，我认为在使用erlang编写某些内容时，始终要记住性能问题。 Erlang帅哥需要解决这个问题，我认为他们会的。

在这里可以找到4种不同的实现，以在Erlang中查找质数(其中两个是"实"筛)并获得性能测量结果：

http://caylespandon.blogspot.com/2009/01/in-euler-problem-10-we-are-asked-to.html

这是我的筛子实现，它使用列表推导并尝试进行尾递归。我在末尾反转了列表，以便对素数进行排序：

primes(Prime, Max, Primes,Integers) when Prime > Max ->
    lists:reverse([Prime|Primes]) ++ Integers;
primes(Prime, Max, Primes, Integers) ->
    [NewPrime|NewIntegers] = [ X || X <- Integers, X rem Prime =/= 0 ],
    primes(NewPrime, Max, [Prime|Primes], NewIntegers).

primes(N) ->
    primes(2, round(math:sqrt(N)), [], lists:seq(3,N,2)). % skip odds

在我的2ghz mac机上，大约需要2.8毫秒来计算最多200万的素数。

足够简单，完全实现算法，并且不使用任何库函数(仅使用模式匹配和列表理解)。
确实不是很强大。我只是试图使其尽可能简单。

-module(primes).
-export([primes/1, primes/2]).

primes(X) -> sieve(range(2, X)).
primes(X, Y) -> remove(primes(X), primes(Y)).

range(X, X) -> [X];
range(X, Y) -> [X | range(X + 1, Y)].

sieve([X]) -> [X];
sieve([H | T]) -> [H | sieve(remove([H * X || X <-[H | T]], T))].

remove(_, []) -> [];
remove([H | X], [H | Y]) -> remove(X, Y);
remove(X, [H | Y]) -> [H | remove(X, Y)].