如何使用 diff/2 防止生成序列中的重复? [英] How to prevent duplicates in generated sequences by using dif/2?
问题描述
在回答关于 StackOverflow 的另一个问题关于(概括一点)生成由有限元素集形成的所有序列,没有重复出现.
正如鲍里斯在评论中正确指出的那样,这个问题有许多现有的解决方案.但是,我对不使用累加器(即,要与新选择的元素进行比较的已选择元素的列表)但使用 dif/2
语句的解决方案感兴趣.
为了说明,在我的以下程序中,我有 4 个元素,并且在 4 次递归调用之后,有几个 div/2
语句表明到目前为止已选择的 4 个元素是成对不同的.由此可以推断,继续递归并查找第五个元素是没有意义的,因为给定 div/2
语句,已经没有元素了.有没有办法将这种知识"编码到程序中,使其不再循环?
:- use_module(library(apply)).:- use_module(library(dif)).序列([]).序列([H|T]):-地图列表(差异(H),T),之间(1, 4, H),序列(T).
当前的循环行为:
?- 序列(X).X = [] ;X = [1] ;...X = [4, 3, 1, 2] ;X = [4, 3, 2, 1] ;<循环>
从 — 开始的小问题名称:sequences/1
建议一个序列列表(无论序列是什么),它应该是 sequence/1
.
您需要很多糟糕的 Prolog 系统:您需要更强的一致性.我想不惜任何代价.
我的即时反应(使用library(clpfd)
!)不起作用,让我们看看为什么
?- length(Xs,N),Xs ins 1..4, all_distinct(Xs).
它的循环次数和你的版本一样多,最好用这个 失败切片:
<预>?- length(Xs,N), false,所以只有 length/2
是错误的.也许我重申一下您的程序,并尝试找出您的程序没有终止的原因:
我们最亲爱的声明式海报孩子maplist/2
被抓到了!好吧,也许我们不应该那么苛刻.毕竟,诚实地不终止谓词总是比肆无忌惮地不健全或不完整的黑客更可取.
我们需要了解的是,all_distinct/1
需要知道列表的长度,并且所有域也必须存在.
sequence(Xs) :-序列辅助(Xs,[]).序列辅助([],_).sequence_aux([X|Xs], Ys) :-1..4 中的 X,all_distinct([X|Ys]),序列辅助(Xs,[X|Ys]).?-序列(X).
现在终止.
@mat 可能会注意到 all_distinct([_])
可能会被删除.可能还不止这些.
如果你不喜欢这个解决方案,因为它使用了一个额外的参数,你需要实现一个更安全的maplist/2
.
fmaplist(C_1, Xs) :-冻结(Xs,fmaplist_aux(C_1,Xs)).fmaplist_aux(_C_1, []).fmaplist_aux(C_1, [X|Xs]) :-呼叫(C_1,X),冻结(Xs,fmaplist_aux(C_1,Xs)).
现在您可以逐字使用您的原始程序.但我感觉不太好.了解冻结程序中非终止的精确边界要困难得多.
<小时>顺便说一句:您可能会尝试在 SWI 中获取正确的变量名称以进行答案替换,因为类似于 _G772
的编号不允许将答案重新粘贴回顶级 shell 并获得正确答案结果.
This question came up while answering another question on StackOverflow on (generalizing a bit) generating all sequences formed out of a finite set of elements with no duplicate occurrences.
As Boris rightly indicated in the comments, there are many existing solutions to this problem. However, I am interested in a solution that does not use an accumulator (i.e., a list of already picked elements against which a newly selected element is to be compared) but that uses dif/2
statements instead.
To illustrate, in my following program I have 4 elements and, after 4 recursive calls, a couple of div/2
statements which state that the 4 elements that have been chosen until now are pairswise dissimilar. From this one can deduce that it makes no sense to continue the recursion and look for a fifth element, since there are no elements left given the div/2
statements. Is there a way to encode this 'knowledge' into the program so that it no longer loops?
:- use_module(library(apply)).
:- use_module(library(dif)).
sequences([]).
sequences([H|T]):-
maplist(dif(H), T),
between(1, 4, H),
sequences(T).
Current, looping behavior:
?- sequences(X).
X = [] ;
X = [1] ;
...
X = [4, 3, 1, 2] ;
X = [4, 3, 2, 1] ;
<LOOP>
Tiny issue to start with — the name: sequences/1
suggests a list of sequences (whatever a sequence is), it should be rather sequence/1
.
You are demanding quite a lot of a poor Prolog system: You are demanding stronger consistency. At any price, I presume.
My immediate reactio (use library(clpfd)
!) does not work, let's see why
?- length(Xs,N),Xs ins 1..4, all_distinct(Xs).
It loops just as much as your version, which can be best seen with this failure-slice:
?- length(Xs,N), false,Xs ins 1..4, all_distinct(Xs).
So already length/2
alone is wrong. Maybe I reiterate to your program, and try to identify why your program does not terminate:
sequences([]) :- false. sequences([H|T]):- maplist(dif(H), T), falsebetween(1, 4, H),sequences(T). ?- sequences(X), false.
Our dearest declarative poster child maplist/2
caught in flagranti! OK, maybe we should not be that harsh. After all, honest non-termination of a predicate is always preferable to an unscrupulously unsound or incomplete hack.
What we need to understand is that all_distinct/1
requires the length of the list to be known, and all domains must be present, too.
sequence(Xs) :-
sequence_aux(Xs, []).
sequence_aux([], _).
sequence_aux([X|Xs], Ys) :-
X in 1..4,
all_distinct([X|Ys]),
sequence_aux(Xs, [X|Ys]).
?- sequence(X).
Now terminates.
@mat may note that all_distinct([_])
might be removed. Maybe even more than that.
If you do not like this solution because it uses an extra argument, you will need to implement a safer maplist/2
.
fmaplist(C_1, Xs) :-
freeze(Xs, fmaplist_aux(C_1, Xs)).
fmaplist_aux(_C_1, []).
fmaplist_aux(C_1, [X|Xs]) :-
call(C_1, X),
freeze(Xs, fmaplist_aux(C_1, Xs)).
Now you can use your original program verbatim. But I do not feel very good at it. Understanding the precise borders of non-termination in a program with freeze is much more difficult.
As an aside: you might try to get correct variable names in SWI for answer substitutions because the _G772
-like numbering does not permit to re-paste an answer back into the toplevel shell and get correct results.
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