Recursion 使用递归的Erlang素数分解

我试图用素数分解来计算一个数的因子。我想使用if语句来测试条件，如果factor是factor，则创建一个列表。它不起作用，我想不出任何其他可行的方法，建议

factor( N ) ->
    if
        N rem factor(N-1) == 0 ->
            [N|factor(N-1)];
        true -> false;
    end.

---

这段代码进行分解，但我不使用if语句（在Erlang IMHO中不太常见）：

我记录了一个示例的执行跟踪。请注意，decomp/3是尾部递归的，因此长返回阶段实际上是单个返回

1> help_num:decomp(70610).
++ (3) <147> decomp(70610,[],2)
++ (4) <150> decomp(35305,[2],2)
++ (5) <151> decomp(35305,[2],3)
++ (6) <152> decomp(35305,[2],5)
++ (7) <150> decomp(7061,[5,2],5)
++ (8) <152> decomp(7061,[5,2],7)
++ (9) <152> decomp(7061,[5,2],9)
++ (10) <152> decomp(7061,[5,2],11)
++ (11) <152> decomp(7061,[5,2],13)
++ (12) <152> decomp(7061,[5,2],15)
++ (13) <152> decomp(7061,[5,2],17)
++ (14) <152> decomp(7061,[5,2],19)
++ (15) <152> decomp(7061,[5,2],21)
++ (16) <152> decomp(7061,[5,2],23)
++ (17) <150> decomp(307,[23,5,2],23)
-- (17) [307,23,5,2]
-- (16) [307,23,5,2]
-- (15) [307,23,5,2]
-- (14) [307,23,5,2]
-- (13) [307,23,5,2]
-- (12) [307,23,5,2]
-- (11) [307,23,5,2]
-- (10) [307,23,5,2]
-- (9) [307,23,5,2]
-- (8) [307,23,5,2]
-- (7) [307,23,5,2]
-- (6) [307,23,5,2]
-- (5) [307,23,5,2]
-- (4) [307,23,5,2]
-- (3) [307,23,5,2]
++ (3) <147> lists:reverse([307,23,5,2])
[2,5,23,307]
2>

1>帮助编号：decomp（70610）。
++（3）反编译（70610，[]，2）
++（4）反编译（35305[2]，2）
++（5）反编译（35305[2]，3）
++（6）反编译（35305[2]，5）
++（7）反编译（7061[5,2]，5）
++（8）反编译（7061[5,2]，7）
++（9）反编译（7061[5,2]，9）
++（10）反编译（7061[5,2]，11）
++（11）反编译（7061[5,2]，13）
++（12）反编译（7061[5,2]，15）
++（13）反编译（7061[5,2]，17）
++（14）反编译（7061[5,2]，19）
++（15）反编译（7061，[5,2]，21）
++（16）反编译（7061[5,2]，23）
++（17）反编译（307[23,5,2]，23）
-- (17) [307,23,5,2]
-- (16) [307,23,5,2]
-- (15) [307,23,5,2]
-- (14) [307,23,5,2]
-- (13) [307,23,5,2]
-- (12) [307,23,5,2]
-- (11) [307,23,5,2]
-- (10) [307,23,5,2]
-- (9) [307,23,5,2]
-- (8) [307,23,5,2]
-- (7) [307,23,5,2]
-- (6) [307,23,5,2]
-- (5) [307,23,5,2]
-- (4) [307,23,5,2]
-- (3) [307,23,5,2]
++（3）列表：反向（[307,23,5,2]）
[2,5,23,307]
2>

乍看之下，我发现代码中有两件事不起作用：

函数应该返回一个列表。因此，测试“N rem factor（N-1）==0”将计算列表的剩余部分除以一个数字，这毫无意义。不能对列表执行除法

if-guard中有一个递归调用，这是不受支持的

---

除非您更详细地解释代码应该如何工作，否则我认为我无法提供更好的帮助。

这里有一个实现示例。你不必关注

下一步\u可能是\u prime/2

。您可以使用simple

Old+2

。这只是优化

-module(factor).

-export([factor/1, make_wheels/0]).

factor(X) when is_integer(X), X > 0 ->
    factor(X, 2, first_primes_steps()).

factor(X, X, _) -> [X];
factor(X, Prime, Wheel) when X rem Prime =:= 0 ->
    [Prime|factor(X div Prime, Prime, Wheel)];
factor(X, Old, Wheel) when Old*Old < X ->
    {MayBePrime, Wheel2} = next_may_be_prime(Old, Wheel),
    factor(X, MayBePrime, Wheel2);
factor(X, _, _) -> [X].

%%% Prime candidates generator

next_may_be_prime(X, [Add|T]) ->
    {X+Add, case T of [] -> wheel(); _ -> T end}.

first_primes_steps() -> [1,2,2,4,2].

wheel() ->
    [4,2,4,6,2,6,4,2,4,6,6,2,6,4,2,6,4,6,8,4,2,4,2,4,14,4,6,2,
     10,2,6,6,4,2,4,6,2,10,2,4,2,12,10,2,4,2,4,6,2,6,4,6,6,6,2,6,
     4,2,6,4,6,8,4,2,4,6,8,6,10,2,4,6,2,6,6,4,2,4,6,2,6,4,2,6,10,
     2,10,2,4,2,4,6,8,4,2,4,12,2,6,4,2,6,4,6,12,2,4,2,4,8,6,4,6,
     2,4,6,2,6,10,2,4,6,2,6,4,2,4,2,10,2,10,2,4,6,6,2,6,6,4,6,6,
     2,6,4,2,6,4,6,8,4,2,6,4,8,6,4,6,2,4,6,8,6,4,2,10,2,6,4,2,4,
     2,10,2,10,2,4,2,4,8,6,4,2,4,6,6,2,6,4,8,4,6,8,4,2,4,2,4,8,6,
     4,6,6,6,2,6,6,4,2,4,6,2,6,4,2,4,2,10,2,10,2,6,4,6,2,6,4,2,4,
     6,6,8,4,2,6,10,8,4,2,4,2,4,8,10,6,2,4,8,6,6,4,2,4,6,2,6,4,6,
     2,10,2,10,2,4,2,4,6,2,6,4,2,4,6,6,2,6,6,6,4,6,8,4,2,4,2,4,8,
     6,4,8,4,6,2,6,6,4,2,4,6,8,4,2,4,2,10,2,10,2,4,2,4,6,2,10,2,
     4,6,8,6,4,2,6,4,6,8,4,6,2,4,8,6,4,6,2,4,6,2,6,6,4,6,6,2,6,6,
     4,2,10,2,10,2,4,2,4,6,2,6,4,2,10,6,2,6,4,2,6,4,6,8,4,2,4,2,
     12,6,4,6,2,4,6,2,12,4,2,4,8,6,4,2,4,2,10,2,10,6,2,4,6,2,6,4,
     2,4,6,6,2,6,4,2,10,6,8,6,4,2,4,8,6,4,6,2,4,6,2,6,6,6,4,6,2,
     6,4,2,4,2,10,12,2,4,2,10,2,6,4,2,4,6,6,2,10,2,6,4,14,4,2,4,
     2,4,8,6,4,6,2,4,6,2,6,6,4,2,4,6,2,6,4,2,4,12,2,12].

%%% Auxiliary functions for making prime wheels

make_wheels() ->
    %%%%%%%%%%%%%%%%
    Primes = [2,3,5,7,11],
    Next = 13,
    %%%%%%%%%%%%%%%%
    Period = lists:foldl(fun(X, A) -> X*A end, 1, Primes),
    Max = Period + Next,
    F = fun(X, A) -> array:set(X, false, A) end,
    G = fun(X, A) -> lists:foldl(F, A, lists:seq(X, Max, X)) end,
    Sieve = lists:foldl(G, array:new([{default, true}]), Primes),
    {
        diffs(Primes ++ [Next]),
        diffs([ X || X <- lists:seq(Next, Max), array:get(X, Sieve)])
        }.

diffs([H|T]) ->
    diffs(H, T).

diffs(_, []) -> [];
diffs(A, [B|T]) -> [B-A|diffs(B, T)].

-模块（系数）。
-出口（[系数/1，制造车轮/0]）。
因子（X）为_整数（X），X>0->
因子（X，2，first_primes_steps（））。
因子（X，X，ux）->[X]；
当X rem Prime=：=0->
[Prime |因子（X div Prime，Prime，Wheel）]；
旧*旧
{MayBePrime，Wheel2}=下一个可能是（旧的，Wheel），
因子（X，可能是素数，轮2）；
因子（X，，-->[X]。
%%%素候选者生成器
下一个可能是素数（X，[Add | T]）->
{X+Add，[]->wheel（）；u->T end}的案例T。
第一个素数步骤（）->[1,2,2,4,2]。
轮子（）->
[4,2,4,6,2,6,4,2,4,6,6,2,6,4,2,6,4,6,8,4,2,4,2,4,14,4,6,2,
10,2,6,6,4,2,4,6,2,10,2,4,2,12,10,2,4,2,4,6,2,6,4,6,6,6,2,6,
4,2,6,4,6,8,4,2,4,6,8,6,10,2,4,6,2,6,6,4,2,4,6,2,6,4,2,6,10,
2,10,2,4,2,4,6,8,4,2,4,12,2,6,4,2,6,4,6,12,2,4,2,4,8,6,4,6,
2,4,6,2,6,10,2,4,6,2,6,4,2,4,2,10,2,10,2,4,6,6,2,6,6,4,6,6,
2,6,4,2,6,4,6,8,4,2,6,4,8,6,4,6,2,4,6,8,6,4,2,10,2,6,4,2,4,
2,10,2,10,2,4,2,4,8,6,4,2,4,6,6,2,6,4,8,4,6,8,4,2,4,2,4,8,6,
4,6,6,6,2,6,6,4,2,4,6,2,6,4,2,4,2,10,2,10,2,6,4,6,2,6,4,2,4,
6,6,8,4,2,6,10,8,4,2,4,2,4,8,10,6,2,4,8,6,6,4,2,4,6,2,6,4,6,
2,10,2,10,2,4,2,4,6,2,6,4,2,4,6,6,2,6,6,6,4,6,8,4,2,4,2,4,8,
6,4,8,4,6,2,6,6,4,2,4,6,8,4,2,4,2,10,2,10,2,4,2,4,6,2,10,2,
4,6,8,6,4,2,6,4,6,8,4,6,2,4,8,6,4,6,2,4,6,2,6,6,4,6,6,2,6,6,
4,2,10,2,10,2,4,2,4,6,2,6,4,2,10,6,2,6,4,2,6,4,6,8,4,2,4,2,
12,6,4,6,2,4,6,2,12,4,2,4,8,6,4,2,4,2,10,2,10,6,2,4,6,2,6,4,
2,4,6,6,2,6,4,2,10,6,8,6,4,2,4,8,6,4,6,2,4,6,2,6,6,6,4,6,2,
6,4,2,4,2,10,12,2,4,2,10,2,6,4,2,4,6,6,2,10,2,6,4,14,4,2,4,
2,4,8,6,4,6,2,4,6,2,6,6,4,2,4,6,2,6,4,2,4,12,2,12].
%%%制造主轮的辅助功能
制造车轮（）->
%%%%%%%%%%%%%%%%
素数=[2,3,5,7,11]，
下一个=13，
%%%%%%%%%%%%%%%%
周期=列表：foldl（乐趣（X，A）->X*A结束，1，素数），
最大值=周期+下一个，
F=fun（X，A）->array:set（X，false，A）end，
G=fun（X，A）->列表：foldl（F，A，列表：seq（X，Max，X））结束，
Sieve=list:foldl（G，数组：new（[{default，true}]），Primes），
{
diff（Primes++[Next]），
差异（[X | | X
差分（H，T）。
差异（u，[]）->[]；
微分（A[B | T]）->[B-A |微分（B，T）]。

我想使用一个N数，并返回一个所有N的因子列表。我想做的是让每个数小于N，然后用N除以该数，循环生成它们。例如，如果我有3，我会得到（N-1）=2…3/2…不会除法，所以我将转到下一个数字。您可以在

if

语句中进行递归。没有这样的限制。在if语句的子句体中，是的。在保护表达式中，不是。有这样的限制。与

if

语句无关，所有保护都不能包含用户函数。相同的缺点train对于函数子句卫士和case语句子句卫士也是有效的。为什么最初要反转列表？列表反转是在分解求值结束时完成的，这只是因为当我执行此函数时，我希望得到从最小到最大排序的素因子。算法这是构建列表的常用方法，因为它避免了在每一步都创建一个新的副本；最后进行反向操作的成本比那些多个副本要低。

-module(factor).

-export([factor/1, make_wheels/0]).

factor(X) when is_integer(X), X > 0 ->
    factor(X, 2, first_primes_steps()).

factor(X, X, _) -> [X];
factor(X, Prime, Wheel) when X rem Prime =:= 0 ->
    [Prime|factor(X div Prime, Prime, Wheel)];
factor(X, Old, Wheel) when Old*Old < X ->
    {MayBePrime, Wheel2} = next_may_be_prime(Old, Wheel),
    factor(X, MayBePrime, Wheel2);
factor(X, _, _) -> [X].

%%% Prime candidates generator

next_may_be_prime(X, [Add|T]) ->
    {X+Add, case T of [] -> wheel(); _ -> T end}.

first_primes_steps() -> [1,2,2,4,2].

wheel() ->
    [4,2,4,6,2,6,4,2,4,6,6,2,6,4,2,6,4,6,8,4,2,4,2,4,14,4,6,2,
     10,2,6,6,4,2,4,6,2,10,2,4,2,12,10,2,4,2,4,6,2,6,4,6,6,6,2,6,
     4,2,6,4,6,8,4,2,4,6,8,6,10,2,4,6,2,6,6,4,2,4,6,2,6,4,2,6,10,
     2,10,2,4,2,4,6,8,4,2,4,12,2,6,4,2,6,4,6,12,2,4,2,4,8,6,4,6,
     2,4,6,2,6,10,2,4,6,2,6,4,2,4,2,10,2,10,2,4,6,6,2,6,6,4,6,6,
     2,6,4,2,6,4,6,8,4,2,6,4,8,6,4,6,2,4,6,8,6,4,2,10,2,6,4,2,4,
     2,10,2,10,2,4,2,4,8,6,4,2,4,6,6,2,6,4,8,4,6,8,4,2,4,2,4,8,6,
     4,6,6,6,2,6,6,4,2,4,6,2,6,4,2,4,2,10,2,10,2,6,4,6,2,6,4,2,4,
     6,6,8,4,2,6,10,8,4,2,4,2,4,8,10,6,2,4,8,6,6,4,2,4,6,2,6,4,6,
     2,10,2,10,2,4,2,4,6,2,6,4,2,4,6,6,2,6,6,6,4,6,8,4,2,4,2,4,8,
     6,4,8,4,6,2,6,6,4,2,4,6,8,4,2,4,2,10,2,10,2,4,2,4,6,2,10,2,
     4,6,8,6,4,2,6,4,6,8,4,6,2,4,8,6,4,6,2,4,6,2,6,6,4,6,6,2,6,6,
     4,2,10,2,10,2,4,2,4,6,2,6,4,2,10,6,2,6,4,2,6,4,6,8,4,2,4,2,
     12,6,4,6,2,4,6,2,12,4,2,4,8,6,4,2,4,2,10,2,10,6,2,4,6,2,6,4,
     2,4,6,6,2,6,4,2,10,6,8,6,4,2,4,8,6,4,6,2,4,6,2,6,6,6,4,6,2,
     6,4,2,4,2,10,12,2,4,2,10,2,6,4,2,4,6,6,2,10,2,6,4,14,4,2,4,
     2,4,8,6,4,6,2,4,6,2,6,6,4,2,4,6,2,6,4,2,4,12,2,12].

%%% Auxiliary functions for making prime wheels

make_wheels() ->
    %%%%%%%%%%%%%%%%
    Primes = [2,3,5,7,11],
    Next = 13,
    %%%%%%%%%%%%%%%%
    Period = lists:foldl(fun(X, A) -> X*A end, 1, Primes),
    Max = Period + Next,
    F = fun(X, A) -> array:set(X, false, A) end,
    G = fun(X, A) -> lists:foldl(F, A, lists:seq(X, Max, X)) end,
    Sieve = lists:foldl(G, array:new([{default, true}]), Primes),
    {
        diffs(Primes ++ [Next]),
        diffs([ X || X <- lists:seq(Next, Max), array:get(X, Sieve)])
        }.

diffs([H|T]) ->
    diffs(H, T).

diffs(_, []) -> [];
diffs(A, [B|T]) -> [B-A|diffs(B, T)].