在Prolog中计算连续出现的数字 | 在Prolog中计算连续出现的数字

本文介绍了在Prolog中计算连续出现的数字的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！

问题描述

您好，我正在尝试在 Prolog 中编写一个程序，给定一个列表，它计算列表中每个连续元素的出现次数，如下所示:

count(1,[1,1,1,2,2,2,3,1,1],0,X)

结果将是 X=[ [1,3],[2,3],[3,1][1,2] ]也就是每个子列表是 [element,occurrences]

就我而言，我认为基本情况有问题，但我无法解决.你能帮帮我吗?

% 将一个元素添加到列表中追加([]，Y，Y).附加([X|Xs]，Ys，[X|Zs]):-附加(Xs，Ys，Zs).%c 是以 0 开头的计数器数数(_，[]，_，[]).计数(X，[X]，C，[L]):-计数(X，[]，C，[L|[X，C]]).%增加计数器count(X,[X|Tail],C,L):-Z 是 C+1,count(X,Tail,Z,L).count(X,[Head|Tail],C,[L]):-append(L,[X,C],NL),count(Head,Tail,1,NL).

解决方案

这是另一个尝试运行长度编码的方法，基于 clpfd！

:- 使用模块(库(clpfd)).

基于 if_/3 和 (=)/3，我们定义list_rle/2:

list_rle([],[]).list_rle([X|Xs],[N*X|Ps]) :-list_count_prev_runs(Xs,N,X,Ps).list_count_prev_runs(Es,N,X,Ps) :-N #> 0,N #= N0+1,list_count_prev_runs_(Es,N0,X,Ps).list_count_prev_runs_([],0,_,[]).list_count_prev_runs_([E|Es],N,X,Ps0) :-if_(X=E,list_count_prev_runs(Es,N,X,Ps0),(N = 0, Ps0 = [M*E|Ps], list_count_prev_runs(Es,M,E,Ps))).

示例查询:

编码/解码 #1
?- list_rle([a,a,b,c,c,c,d,e,e],Ys).Ys = [2*a,1*b,3*c,1*d,2*e].?- list_rle(Xs,[2*a,1*b,3*c,1*d,2*e]).Xs = [a,a,b,c,c,c,d,e,e];错误的.
编码/解码#2
?- dif(A,B),dif(B,C),dif(C,D),dif(D,E), list_rle([A,A,B,C,C,C,D,E,E]，是).Ys = [2*A,1*B,3*C,1*D,2*E]，dif(A,B)，dif(B,C)，dif(C,D)，dif(D,E).?- list_rle(Xs,[2*A,1*B,3*C,1*D,2*E]).Xs = [A,A,B,C,C,C,D,E,E], dif(A,B), dif(B,C), dif(C,D), dif(D,E);错误的.
更一般的东西怎么样?
?- list_rle([A,B,C,D],Xs).Xs = [4*A]，A=B，B=C，C=D;Xs = [3*A, 1*D], A=B, B=C, dif(C,D);Xs = [2*A, 2*C], A=B, dif(B,C), C=D;Xs = [2*A, 1*C,1*D], A=B , dif(B,C), dif(C,D);Xs = [1*A,3*B], dif(A,B), B=C, C=D;Xs = [1*A,2*B, 1*D], dif(A,B), B=C , dif(C,D);Xs = [1*A,1*B,2*C]，dif(A,B)，dif(B,C)，C=D;Xs = [1*A,1*B,1*C,1*D]，dif(A,B)，dif(B,C)，dif(C,D).

Hello I am trying to make a program in Prolog that given a list it counts the occurrences of each successive element in the list as follows:

count(1,[1,1,1,2,2,2,3,1,1],0,X)

the result would be X=[ [1,3],[2,3],[3,1][1,2] ]aka each sublist is [element,occurrences]

In my case i believe there is something wrong with the base case but I cannot solve it. Can you help me?

%append an element to a list
append([ ],Y,Y).
append([X|Xs],Ys,[X|Zs]):-append(Xs,Ys,Zs).

%c is the counter beginning with 0
count(_,[],_,[]).
count(X,[X],C,[L]):-count(X,[],C,[L|[X,C]]).

%increase counter
count(X,[X|Tail],C,L):-Z is C+1,count(X,Tail,Z,L).
count(X,[Head|Tail],C,[L]):-append(L,[X,C],NL),count(Head,Tail,1,NL).

解决方案

Here's another try of doing run-length encoding, based on clpfd!

:- use_module(library(clpfd)).

Based on if_/3 and (=)/3, we define list_rle/2:

list_rle([],[]).
list_rle([X|Xs],[N*X|Ps]) :-
   list_count_prev_runs(Xs,N,X,Ps).

list_count_prev_runs(Es,N,X,Ps) :-
   N #> 0,
   N #= N0+1,
   list_count_prev_runs_(Es,N0,X,Ps).

list_count_prev_runs_([],0,_,[]).
list_count_prev_runs_([E|Es],N,X,Ps0) :-
   if_(X=E,
       list_count_prev_runs(Es,N,X,Ps0),
       (N = 0, Ps0 = [M*E|Ps], list_count_prev_runs(Es,M,E,Ps))).

Sample queries:

encode/decode #1

?- list_rle([a,a,b,c,c,c,d,e,e],Ys).
Ys = [2*a,1*b,3*c,1*d,2*e].

?- list_rle(Xs,[2*a,1*b,3*c,1*d,2*e]).
  Xs = [a,a,b,c,c,c,d,e,e]
; false.

encode/decode #2

?- dif(A,B),dif(B,C),dif(C,D),dif(D,E), list_rle([A,A,B,C,C,C,D,E,E],Ys).
Ys = [2*A,1*B,3*C,1*D,2*E], dif(A,B), dif(B,C), dif(C,D), dif(D,E).

?- list_rle(Xs,[2*A,1*B,3*C,1*D,2*E]).
  Xs = [A,A,B,C,C,C,D,E,E], dif(A,B), dif(B,C), dif(C,D), dif(D,E)
; false.

How about something a little more general?

?- list_rle([A,B,C,D],Xs).
  Xs = [4*A            ],     A=B ,     B=C ,     C=D
; Xs = [3*A,        1*D],     A=B ,     B=C , dif(C,D)
; Xs = [2*A,    2*C    ],     A=B , dif(B,C),     C=D
; Xs = [2*A,    1*C,1*D],     A=B , dif(B,C), dif(C,D)
; Xs = [1*A,3*B        ], dif(A,B),     B=C ,     C=D
; Xs = [1*A,2*B,    1*D], dif(A,B),     B=C , dif(C,D)
; Xs = [1*A,1*B,2*C    ], dif(A,B), dif(B,C),     C=D
; Xs = [1*A,1*B,1*C,1*D], dif(A,B), dif(B,C), dif(C,D).

这篇关于在Prolog中计算连续出现的数字的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！