我有一个谓词,找到正确的解决方案,但接着找到不正确的解决方案.
?- data(D),data_threshold_nonredundantbumps(D,5,Bs),write(D).
[3,6,7,8,2,4,5,6,9,4,7,3]
D = [3, 6, 7, 8, 2, 4, 5, 6, 9|...],
Bs = [bump([11], [7]), bump([8, 9], [6, 9]), bump([2, 3, 4], [6, 7, 8])] ;
[3,6,7,8,2,4,5,6,9,4,7,3]
D = [3, 6, 7, 8, 2, 4, 5, 6, 9|...],
Bs = [bump([8, 9], [6, 9]), bump([2, 3, 4], [6, 7, 8])] ;
[3,6,7,8,2,4,5,6,9,4,7,3]
D = [3, 6, 7, 8, 2, 4, 5, 6, 9|...],
Bs = [bump([8], [6]), bump([2, 3, 4], [6, 7, 8])] ;
[3,6,7,8,2,4,5,6,9,4,7,3]
D = [3, 6, 7, 8, 2, 4, 5, 6, 9|...],
Bs = [bump([9], [9]), bump([2, 3, 4], [6, 7, 8])] ;
[3,6,7,8,2,4,5,6,9,4,7,3]
D = [3, 6, 7, 8, 2, 4, 5, 6, 9|...],
Bs = [bump([11], [7]), bump([2, 3, 4], [6, 7, 8])] ;
[3,6,7,8,2,4,5,6,9,4,7,3]
D = [3, 6, 7, 8, 2, 4, 5, 6, 9|...],
Bs = [bump([2, 3, 4], [6, 7, 8])] ;
等等
这个想法是,它会发现在数据,其中一个凸块是一个连续的子列表的所有非冗余凸点data即上面threshold,返回的有序(由尺寸)列表bump/2s,其中凸块/ 2的第一arg是从indicies的列表数据和第二个arg是值列表.因此bump([2, 3, 4], [6, 7, 8]),在数据指数2,3和4高于5时,它们是6,7,8.
如何添加条件以便找不到这些额外的解决方案？- 没有使用once/1.
如果我的代码可以通过其他方式简化,请告诉我.这似乎有点复杂,它正在尝试做什么.
所以:
这是我的代码:
:-use_module(library(clpfd)).
fd_length(L, N) :-
N #>= 0,
fd_length(L, N, 0).
fd_length([], N, N0) :-
N #= N0.
fd_length([_|L], N, N0) :-
N1 is N0+1,
N #>= N1,
fd_length(L, N, N1).
equidistant_stride([],_).
equidistant_stride([Z|Zs],D) :-
foldl(equidistant_stride_(D),Zs,Z,_).
equidistant_stride_(D,Z1,Z0,Z1) :-
Z1 #= Z0+D.
consecutive_ascending_integers(Zs) :-
equidistant_stride(Zs,1).
consecutive_ascending_integers_from(Zs,Z0) :-
Zs = [Z0|_],
consecutive_ascending_integers(Zs).
bool01_t(1,true).
bool01_t(0,false).
if_(C_1,Then_0,Else_0) -->
{ call(C_1,Truth) },
{ functor(Truth,_,0) }, % safety check
( { Truth == true } -> phrase(Then_0)
; { Truth == false }, phrase(Else_0)
).
if_(If_1, Then_0, Else_0) :-
call(If_1, T),
( T == true -> call(Then_0)
; T == false -> call(Else_0)
; nonvar(T) -> throw(error(type_error(boolean,T),_))
; /* var(T) */ throw(error(instantiation_error,_))
).
#=<(X,Y,Truth) :- X #=< Y #<==> B, bool01_t(B,Truth).
#<( X,Y,Truth) :- X #< Y #<==> B, bool01_t(B,Truth).
#>( X,Y,Truth) :- X #> Y #<==> B, bool01_t(B,Truth).
#>=(X,Y,Truth) :- X #>= Y #<==> B, bool01_t(B,Truth).
tinclude(P_2,Xs,Zs) :-
list_tinclude_list(Xs,P_2,Zs).
list_tinclude_list([], _P_2,[]).
list_tinclude_list([i_v(E0,E1)|Es],P_2,Fs0) :-
if_(call(P_2,E1), Fs0 = [i_v(E0,E1)|Fs], Fs0 = Fs),
list_tinclude_list(Es,P_2,Fs).
tfilter(P_2,As,Bs) :-
tinclude(P_2,As,Bs).
%% =====================================================================
%% =====================================================================
data([5,6,7,8,3,2,6,7]).
list_index_element(L,I,E):-
nth1(I,L,E).
filter(Threshold,DataPairs,FilterdPairs):-
tfilter(#<(Threshold),DataPairs,FilterdPairs).
i_v_pair(I,V,i_v(I,V)).
data_indices_indicespairs(D,Is,Pairs):-
same_length(D,Is),
consecutive_ascending_integers_from(Is,1),
maplist(i_v_pair,Is,D,Pairs).
list_ascending(List,MinLength,MaxLength):-
Max in MinLength..MaxLength,
labeling([max(Max)],[Max]),
fd_length(List,Max),
consecutive_ascending_integers(List).
region_minlength_maxlength(Region,MinLength,MaxLength,All):-
list_ascending(Region,MinLength,MaxLength),
append(_Before,End,All),
append(Region,_End2,End).
data_threshold_bumpvalues_bumplocation(Data,Threshold,Bumpvalues,Bumplocation):-
length(Data,MaxBump),
data_indices_indicespairs(Data,_Is,Pairs),
filter(Threshold,Pairs,FilteredPairs),
maplist(i_v_pair,FilteredIndices,_FilteredValues,FilteredPairs),
%Test =test(FilteredIndexes,FilteredValues),
dif(Bumplocation,[]),
region_minlength_maxlength(Bumplocation,0,MaxBump,FilteredIndices),
maplist(list_index_element(Data), Bumplocation,Bumpvalues).
list_first_last([H|T],H,L):-
last(T,L).
listoflists_firsts_lasts(Listoflists,Firsts,Lasts):-
maplist(list_first_last,Listoflists,Firsts,Lasts).
%start is not between location1 and location2
start_location1_location2(Start,Location1,Location2) :-
#\( Location1 #=< Start,
Start #=< Location2).
bumplocation_notsublist_of_any_acs(Bumplocation,Acs):-
listoflists_firsts_lasts(Acs,Firsts,Lasts),
%the start of bumplocation can not be between the start of any Acs
Bumplocation =[Bumpstart|_],
maplist(start_location1_location2(Bumpstart),Firsts,Lasts).
loc_val_bump(Location,Value,bump(Location,Value)).
data_bumplocations_bumpvalues(Data,Bumplocations,Bumpvalues):-
maplist(list_index_element(Data),Bumplocations,Bumpvalues).
%this works but finds extra solutins so needs to be refined.
data_threshold_nonredundantbumps(Data,Threshold,Bumps):-
data_threshold_nonredundantbumps_ac(Data,Threshold,Nonredundantbumpslocations,[]),
maplist(data_bumplocations_bumpvalues(Data),Nonredundantbumpslocations,Nonredundantbumps),
maplist(loc_val_bump,Nonredundantbumpslocations,Nonredundantbumps,Bumps).
data_threshold_nonredundantbumps_ac(Data,Threshold,Nonredundantbumps,Ac0):-
bumplocation_notsublist_of_any_acs(Bumplocation,Ac0),
data_threshold_bumpvalues_bumplocation(Data,Threshold,_Bumpvalues,Bumplocation),
append([Bumplocation],Ac0,Ac1),
data_threshold_nonredundantbumps_ac(Data,Threshold,Nonredundantbumps,Ac1).
data_threshold_nonredundantbumps_ac(_Data,_Threshold,Ac0,Ac0).
我的印象是你略微过度思考.对于超过阈值的数字运行,存在直接的公式,可以通过在列表的单个遍历中考虑从头到尾的元素来定义.特别是,我们没有需要append/3做到这一点.
在Prolog中描述列表时,请始终考虑使用DCG表示法(dcg).在这种情况下,需要花一点时间来决定如何最好地应用DCG,因为我们描述了两个列表:
但是,除了一些技巧和扩展之外,DCG基本上只允许我们描述单个列表,而不是同时描述单独的列表.因此,我们有我们所掌握这个强大的和可能非常适合机制,必须选择哪种类型的列表,我们要应用它主要.
在下文中,我表明,使用DCG来描述的列表的溶液凹凸/ 1条件,即,I"奉献"的机制来描述第一种上面提到的列表,并使用另一个DCG来描述第二类列表,我phrase/2从第一个DCG内部调用.
data_threshold_bumps(Ds, T, Bs) :-
phrase(bumps(Ds, 1, T), Bs).
bumps([], _, _) --> [].
bumps([D|Ds0], I0, T) -->
{ D #> T,
phrase(bump(D, T, Ds0, Ds, I0, I), Bs) },
[bump(Bs)],
bumps(Ds, I, T).
bumps([D|Ds0], I0, T) -->
{ D #=< T,
I #= I0 + 1 },
bumps(Ds0, I, T).
bump(D, T, Ds0, Ds, I0, I) --> [I0-D],
{ I1 #= I0 + 1 },
run(Ds0, Ds, T, I1, I).
run([], [], _, I, I) --> [].
run([D|Ds0], Ds, T, I0, I) --> [I0-D],
{ D #> T,
I1 #= I0 + 1 },
run(Ds0, Ds, T, I1, I).
run([D|Ds0], [D|Ds0], T, I, I) -->
{ D #=< T }.
示例查询和答案:
?- data_threshold_bumps([3,6,7,8,2,4,5,6,9,4,7,3], 5, Bs). Bs = [bump([2-6, 3-7, 4-8]), bump([8-6, 9-9]), bump([11-7])] ; false.
请注意,这不是很,你需要完全一样的数据表示,但它是微不足道的将其转换为一个.
以下是一些改进此解决方案的想法,从更容易到更难:
if_/3.bumps//3和run//5在上面的代码？在这里使用DCG比常规谓词有什么好处和缺点？顺便说一下,为了否定一个(可再生的)CLP(FD)约束,你需要(#/\)/2用来表示一个连词.它不工作,(,)/2.