在其数字中拆分数字，以所有可能的方式分组

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Split Number in its digits, grouped in all possble ways

我有一个数字，假设是 123，我想生成一组所有可能的拆分方式：

[[1, 23], [12, 3], [1, 2, 3]].

?- findall(Powerset, powerset([1,2,3], Powerset), Z).

Z = [[1], [1, 2], [1, 2, 3], [2], [2, 3], [3]].

[[1], [2, 3]] -> [1, 23]

[[1, 2], [3]] -> [12, 3]

[[1], [2], [3]] -> [1, 2, 3]
powerset(L, [H|T]):-

 append([H|T], _, L).

powerset([_|L], P):-

 powerset(L, P).
parts(C, [C]).

parts(L, Ps) :-

  append(H, T, L), H \\= [], T \\= [],

  parts(T, Ts),

  append([H], Ts, Ps).

?- number_codes(123,Cs),parts(Cs,Ps),maplist(number_codes,Ns,Ps).

Cs = [49, 50, 51],

Ps = [[49, 50, 51]],

Ns = [123] ;

...etc...

?- findall(Ns,(number_codes(123,Cs),parts(Cs,Ps),maplist(number_codes,Ns,Ps)),Rs).

Rs = [[123], [1, 23], [1, 2, 3], [12, 3]].
int_split(X) -->

 int_split__aux(X,10,[]).



int_split__aux(X,P,Ds) -->

 ( { X =:= 0 }

 -> [Ds]

 ; { X < P }

 -> [[X|Ds]]

 ; { X0 is X mod P,

    X1 is X div P },

   int_split__aux(X1,10,[X0|Ds]),

   { P1 is P*10 },

   int_split__aux(X,P1,Ds)

 ).

?- phrase(int_split(123),Zss).

Zss = [[1,2,3], [12,3], [1,23], [123]].



?- phrase(int_split(1234),Zss).

Zss = [[1,2,3,4], [12,3,4], [1,23,4], [123,4], [1,2,34], [12,34], [1,234], [1234]].

我想过创建 [1,2,3] 的幂集：

[[1, 23], [12, 3], [1, 2, 3]].

?- findall(Powerset, powerset([1,2,3], Powerset), Z).

Z = [[1], [1, 2], [1, 2, 3], [2], [2, 3], [3]].

[[1], [2, 3]] -> [1, 23]

[[1, 2], [3]] -> [12, 3]

[[1], [2], [3]] -> [1, 2, 3]
powerset(L, [H|T]):-

 append([H|T], _, L).

powerset([_|L], P):-

 powerset(L, P).
parts(C, [C]).

parts(L, Ps) :-

  append(H, T, L), H \\= [], T \\= [],

  parts(T, Ts),

  append([H], Ts, Ps).

?- number_codes(123,Cs),parts(Cs,Ps),maplist(number_codes,Ns,Ps).

Cs = [49, 50, 51],

Ps = [[49, 50, 51]],

Ns = [123] ;

...etc...

?- findall(Ns,(number_codes(123,Cs),parts(Cs,Ps),maplist(number_codes,Ns,Ps)),Rs).

Rs = [[123], [1, 23], [1, 2, 3], [12, 3]].
int_split(X) -->

 int_split__aux(X,10,[]).



int_split__aux(X,P,Ds) -->

 ( { X =:= 0 }

 -> [Ds]

 ; { X < P }

 -> [[X|Ds]]

 ; { X0 is X mod P,

    X1 is X div P },

   int_split__aux(X1,10,[X0|Ds]),

   { P1 is P*10 },

   int_split__aux(X,P1,Ds)

 ).

?- phrase(int_split(123),Zss).

Zss = [[1,2,3], [12,3], [1,23], [123]].



?- phrase(int_split(1234),Zss).

Zss = [[1,2,3,4], [12,3,4], [1,23,4], [123,4], [1,2,34], [12,34], [1,234], [1234]].

然后将集合组合在一起并用append/3检查它们的连接是否是初始集合[1,2,3]，即

[[1, 23], [12, 3], [1, 2, 3]].

?- findall(Powerset, powerset([1,2,3], Powerset), Z).

Z = [[1], [1, 2], [1, 2, 3], [2], [2, 3], [3]].

[[1], [2, 3]] -> [1, 23]

[[1, 2], [3]] -> [12, 3]

[[1], [2], [3]] -> [1, 2, 3]
powerset(L, [H|T]):-

 append([H|T], _, L).

powerset([_|L], P):-

 powerset(L, P).
parts(C, [C]).

parts(L, Ps) :-

  append(H, T, L), H \\= [], T \\= [],

  parts(T, Ts),

  append([H], Ts, Ps).

?- number_codes(123,Cs),parts(Cs,Ps),maplist(number_codes,Ns,Ps).

Cs = [49, 50, 51],

Ps = [[49, 50, 51]],

Ns = [123] ;

...etc...

?- findall(Ns,(number_codes(123,Cs),parts(Cs,Ps),maplist(number_codes,Ns,Ps)),Rs).

Rs = [[123], [1, 23], [1, 2, 3], [12, 3]].
int_split(X) -->

 int_split__aux(X,10,[]).



int_split__aux(X,P,Ds) -->

 ( { X =:= 0 }

 -> [Ds]

 ; { X < P }

 -> [[X|Ds]]

 ; { X0 is X mod P,

    X1 is X div P },

   int_split__aux(X1,10,[X0|Ds]),

   { P1 is P*10 },

   int_split__aux(X,P1,Ds)

 ).

?- phrase(int_split(123),Zss).

Zss = [[1,2,3], [12,3], [1,23], [123]].



?- phrase(int_split(1234),Zss).

Zss = [[1,2,3,4], [12,3,4], [1,23,4], [123,4], [1,2,34], [12,34], [1,234], [1234]].

您是否想到了另一种更简单(更优雅)的解决方案？

我使用来自 gnu Prolog powerset 修改的这个谓词

[[1, 23], [12, 3], [1, 2, 3]].

?- findall(Powerset, powerset([1,2,3], Powerset), Z).

Z = [[1], [1, 2], [1, 2, 3], [2], [2, 3], [3]].

[[1], [2, 3]] -> [1, 23]

[[1, 2], [3]] -> [12, 3]

[[1], [2], [3]] -> [1, 2, 3]
powerset(L, [H|T]):-

 append([H|T], _, L).

powerset([_|L], P):-

 powerset(L, P).
parts(C, [C]).

parts(L, Ps) :-

  append(H, T, L), H \\= [], T \\= [],

  parts(T, Ts),

  append([H], Ts, Ps).

?- number_codes(123,Cs),parts(Cs,Ps),maplist(number_codes,Ns,Ps).

Cs = [49, 50, 51],

Ps = [[49, 50, 51]],

Ns = [123] ;

...etc...

?- findall(Ns,(number_codes(123,Cs),parts(Cs,Ps),maplist(number_codes,Ns,Ps)),Rs).

Rs = [[123], [1, 23], [1, 2, 3], [12, 3]].
int_split(X) -->

 int_split__aux(X,10,[]).



int_split__aux(X,P,Ds) -->

 ( { X =:= 0 }

 -> [Ds]

 ; { X < P }

 -> [[X|Ds]]

 ; { X0 is X mod P,

    X1 is X div P },

   int_split__aux(X1,10,[X0|Ds]),

   { P1 is P*10 },

   int_split__aux(X,P1,Ds)

 ).

?- phrase(int_split(123),Zss).

Zss = [[1,2,3], [12,3], [1,23], [123]].



?- phrase(int_split(1234),Zss).

Zss = [[1,2,3,4], [12,3,4], [1,23,4], [123,4], [1,2,34], [12,34], [1,234], [1234]].

我们可以有一个更集中的谓词来代替 powerset

[[1, 23], [12, 3], [1, 2, 3]].

?- findall(Powerset, powerset([1,2,3], Powerset), Z).

Z = [[1], [1, 2], [1, 2, 3], [2], [2, 3], [3]].

[[1], [2, 3]] -> [1, 23]

[[1, 2], [3]] -> [12, 3]

[[1], [2], [3]] -> [1, 2, 3]
powerset(L, [H|T]):-

 append([H|T], _, L).

powerset([_|L], P):-

 powerset(L, P).
parts(C, [C]).

parts(L, Ps) :-

  append(H, T, L), H \\= [], T \\= [],

  parts(T, Ts),

  append([H], Ts, Ps).

?- number_codes(123,Cs),parts(Cs,Ps),maplist(number_codes,Ns,Ps).

Cs = [49, 50, 51],

Ps = [[49, 50, 51]],

Ns = [123] ;

...etc...

?- findall(Ns,(number_codes(123,Cs),parts(Cs,Ps),maplist(number_codes,Ns,Ps)),Rs).

Rs = [[123], [1, 23], [1, 2, 3], [12, 3]].
int_split(X) -->

 int_split__aux(X,10,[]).



int_split__aux(X,P,Ds) -->

 ( { X =:= 0 }

 -> [Ds]

 ; { X < P }

 -> [[X|Ds]]

 ; { X0 is X mod P,

    X1 is X div P },

   int_split__aux(X1,10,[X0|Ds]),

   { P1 is P*10 },

   int_split__aux(X,P1,Ds)

 ).

?- phrase(int_split(123),Zss).

Zss = [[1,2,3], [12,3], [1,23], [123]].



?- phrase(int_split(1234),Zss).

Zss = [[1,2,3,4], [12,3,4], [1,23,4], [123,4], [1,2,34], [12,34], [1,234], [1234]].

现在

[[1, 23], [12, 3], [1, 2, 3]].

?- findall(Powerset, powerset([1,2,3], Powerset), Z).

Z = [[1], [1, 2], [1, 2, 3], [2], [2, 3], [3]].

[[1], [2, 3]] -> [1, 23]

[[1, 2], [3]] -> [12, 3]

[[1], [2], [3]] -> [1, 2, 3]
powerset(L, [H|T]):-

 append([H|T], _, L).

powerset([_|L], P):-

 powerset(L, P).
parts(C, [C]).

parts(L, Ps) :-

  append(H, T, L), H \\= [], T \\= [],

  parts(T, Ts),

  append([H], Ts, Ps).

?- number_codes(123,Cs),parts(Cs,Ps),maplist(number_codes,Ns,Ps).

Cs = [49, 50, 51],

Ps = [[49, 50, 51]],

Ns = [123] ;

...etc...

?- findall(Ns,(number_codes(123,Cs),parts(Cs,Ps),maplist(number_codes,Ns,Ps)),Rs).

Rs = [[123], [1, 23], [1, 2, 3], [12, 3]].
int_split(X) -->

 int_split__aux(X,10,[]).



int_split__aux(X,P,Ds) -->

 ( { X =:= 0 }

 -> [Ds]

 ; { X < P }

 -> [[X|Ds]]

 ; { X0 is X mod P,

    X1 is X div P },

   int_split__aux(X1,10,[X0|Ds]),

   { P1 is P*10 },

   int_split__aux(X,P1,Ds)

 ).

?- phrase(int_split(123),Zss).

Zss = [[1,2,3], [12,3], [1,23], [123]].



?- phrase(int_split(1234),Zss).

Zss = [[1,2,3,4], [12,3,4], [1,23,4], [123,4], [1,2,34], [12,34], [1,234], [1234]].

所以我们可以用 findall

收集有趣的 Ns

[[1, 23], [12, 3], [1, 2, 3]].

?- findall(Powerset, powerset([1,2,3], Powerset), Z).

Z = [[1], [1, 2], [1, 2, 3], [2], [2, 3], [3]].

[[1], [2, 3]] -> [1, 23]

[[1, 2], [3]] -> [12, 3]

[[1], [2], [3]] -> [1, 2, 3]
powerset(L, [H|T]):-

 append([H|T], _, L).

powerset([_|L], P):-

 powerset(L, P).
parts(C, [C]).

parts(L, Ps) :-

  append(H, T, L), H \\= [], T \\= [],

  parts(T, Ts),

  append([H], Ts, Ps).

?- number_codes(123,Cs),parts(Cs,Ps),maplist(number_codes,Ns,Ps).

Cs = [49, 50, 51],

Ps = [[49, 50, 51]],

Ns = [123] ;

...etc...

?- findall(Ns,(number_codes(123,Cs),parts(Cs,Ps),maplist(number_codes,Ns,Ps)),Rs).

Rs = [[123], [1, 23], [1, 2, 3], [12, 3]].
int_split(X) -->

 int_split__aux(X,10,[]).



int_split__aux(X,P,Ds) -->

 ( { X =:= 0 }

 -> [Ds]

 ; { X < P }

 -> [[X|Ds]]

 ; { X0 is X mod P,

    X1 is X div P },

   int_split__aux(X1,10,[X0|Ds]),

   { P1 is P*10 },

   int_split__aux(X,P1,Ds)

 ).

?- phrase(int_split(123),Zss).

Zss = [[1,2,3], [12,3], [1,23], [123]].



?- phrase(int_split(1234),Zss).

Zss = [[1,2,3,4], [12,3,4], [1,23,4], [123,4], [1,2,34], [12,34], [1,234], [1234]].

这里不需要使用findall/3---使用dcg！

[[1, 23], [12, 3], [1, 2, 3]].

?- findall(Powerset, powerset([1,2,3], Powerset), Z).

Z = [[1], [1, 2], [1, 2, 3], [2], [2, 3], [3]].

[[1], [2, 3]] -> [1, 23]

[[1, 2], [3]] -> [12, 3]

[[1], [2], [3]] -> [1, 2, 3]
powerset(L, [H|T]):-

 append([H|T], _, L).

powerset([_|L], P):-

 powerset(L, P).
parts(C, [C]).

parts(L, Ps) :-

  append(H, T, L), H \\= [], T \\= [],

  parts(T, Ts),

  append([H], Ts, Ps).

?- number_codes(123,Cs),parts(Cs,Ps),maplist(number_codes,Ns,Ps).

Cs = [49, 50, 51],

Ps = [[49, 50, 51]],

Ns = [123] ;

...etc...

?- findall(Ns,(number_codes(123,Cs),parts(Cs,Ps),maplist(number_codes,Ns,Ps)),Rs).

Rs = [[123], [1, 23], [1, 2, 3], [12, 3]].
int_split(X) -->

 int_split__aux(X,10,[]).



int_split__aux(X,P,Ds) -->

 ( { X =:= 0 }

 -> [Ds]

 ; { X < P }

 -> [[X|Ds]]

 ; { X0 is X mod P,

    X1 is X div P },

   int_split__aux(X1,10,[X0|Ds]),

   { P1 is P*10 },

   int_split__aux(X,P1,Ds)

 ).

?- phrase(int_split(123),Zss).

Zss = [[1,2,3], [12,3], [1,23], [123]].



?- phrase(int_split(1234),Zss).

Zss = [[1,2,3,4], [12,3,4], [1,23,4], [123,4], [1,2,34], [12,34], [1,234], [1234]].

示例使用：

[[1, 23], [12, 3], [1, 2, 3]].

?- findall(Powerset, powerset([1,2,3], Powerset), Z).

Z = [[1], [1, 2], [1, 2, 3], [2], [2, 3], [3]].

[[1], [2, 3]] -> [1, 23]

[[1, 2], [3]] -> [12, 3]

[[1], [2], [3]] -> [1, 2, 3]
powerset(L, [H|T]):-

 append([H|T], _, L).

powerset([_|L], P):-

 powerset(L, P).
parts(C, [C]).

parts(L, Ps) :-

  append(H, T, L), H \\= [], T \\= [],

  parts(T, Ts),

  append([H], Ts, Ps).

?- number_codes(123,Cs),parts(Cs,Ps),maplist(number_codes,Ns,Ps).

Cs = [49, 50, 51],

Ps = [[49, 50, 51]],

Ns = [123] ;

...etc...

?- findall(Ns,(number_codes(123,Cs),parts(Cs,Ps),maplist(number_codes,Ns,Ps)),Rs).

Rs = [[123], [1, 23], [1, 2, 3], [12, 3]].
int_split(X) -->

 int_split__aux(X,10,[]).



int_split__aux(X,P,Ds) -->

 ( { X =:= 0 }

 -> [Ds]

 ; { X < P }

 -> [[X|Ds]]

 ; { X0 is X mod P,

    X1 is X div P },

   int_split__aux(X1,10,[X0|Ds]),

   { P1 is P*10 },

   int_split__aux(X,P1,Ds)

 ).

?- phrase(int_split(123),Zss).

Zss = [[1,2,3], [12,3], [1,23], [123]].



?- phrase(int_split(1234),Zss).

Zss = [[1,2,3,4], [12,3,4], [1,23,4], [123,4], [1,2,34], [12,34], [1,234], [1234]].
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