这是您所说内容的一个实现,完全按照您的规范在prolog中完成。您的“Merge”实际上是一个名为transpose的函数(在矩阵的上下文中),它恰好位于一个SWI Prolog库中!
1 ?- [user].
|: mergeandmap(Bss,Ls) :- transpose(Bss,BssT), write('Result: '), maptostrings(BssT,Ls), write('\n').
|: maptostrings([],_).
|: maptostrings([Bs|Bss],Ls) :- write(' '), zipstringbits(Bs,Ls), maptostrings(Bss,Ls).
|: zipstringbits([],_).
|: zipstringbits([0|Bs],[_|Ls]) :- zipstringbits(Bs,Ls).
|: zipstringbits([1|Bs],[L|Ls]) :- write(L), zipstringbits(Bs,Ls).
|: :- use_module(library(clpfd)).
% library(error) compiled into error 0.01 sec, 10,056 bytes
% library(apply) compiled into apply 0.02 sec, 16,256 bytes
% library(lists) compiled into lists 0.00 sec, 13,404 bytes
% library(pairs) compiled into pairs 0.01 sec, 4,772 bytes
% library(clpfd) compiled into clpfd 0.32 sec, 367,328 bytes
|:
% user://1 compiled 0.44 sec, 369,348 bytes
true.
2 ?- mergeandmap([[0,0,1],[1,0,1],[0,1,1]],[a,b,c]).
Result: b c abc
true .
3 ?- mergeandmap([[0,1,1],[1,0,1],[1,1,1]],[a,b,c]).
Result: bc ac abc
true .
4 ?- mergeandmap([[0,1],[1,0]],[a,b]).
Result: b a
true .
5 ?- mergeandmap([[0,1,1,1],[1,0,1,0],[1,1,1,0],[1,1,1,0]],[a,b,c,d]).
Result: bcd acd abcd a
true .
6 ?- mergeandmap([[0,1],[1,0],[1,0]],[a,b,c]).
Result: bc a
true .
7 ?- transpose([[A,B,C],[D,E,F],[G,H,I]],X).
X = [[A, D, G], [B, E, H], [C, F, I]].
1 ?- [user].
|: transpose(Rs,Cs) :- emptylists(Rs),emptylists(Cs).
|: transpose([[X|R]|Rs],[[X|C]|Cs]) :- headsandtails(Rs,C,NRs), headsandtails(Cs,R,NCs), transpose(NRs,NCs).
|: headsandtails(Xss,[],[]) :- emptylists(Xss).
|: headsandtails([[X|Xs]|Xss],[X|Hs],[Xs|Ts]) :- headsandtails(Xss,Hs,Ts).
|: emptylists([]) :- !.
|: emptylists([[]|L]) :- emptylists(L).
|:
% user://1 compiled 0.07 sec, 1,208 bytes
true.
2 ?- transpose([[1,2,3,4],[5,6,7],[8,9],[10]],Result).
Result = [[1, 5, 8, 10], [2, 6, 9], [3, 7], [4]] ;
false.
3 ?- transpose([[1,2,3,4],[5,6,7],[8,9,10,11],[10]],Result).
false.
4 ?- transpose([[1,2,3,4,5,7,8,3],[5,6,7],[8,9,10,11],[10]],Result).
false.
5 ?- transpose([[1,2,3,4,5,7,8,3],[5,6,7],[8,9,99],[10]],Result).
Result = [[1, 5, 8, 10], [2, 6, 9], [3, 7, 99], [4], [5], [7], [8], [3]] ;
false.
基本上这是如何工作的,它检查转置的左上角元素和原始元素是否相同,检查其中一个元素的左上角行是否与另一个元素的左上角列匹配,反之亦然,然后迭代检查所有其他元素,直到无需检查为止。
切割操作员!在EmptyList上,谓词阻止答案向右扩展,并使用更多的空列表,这意味着它可以在尝试转置没有空列表的内容时终止(它只是不断添加它们,并且总是找不到答案)。
总而言之,对于一体式解决方案,以下是您所需的全部代码:
mergeandmap(Bss,Ls) :- transpose(Bss,BssT), write('Result: '), maptostrings(BssT,Ls), write('\n').
maptostrings([],_).
maptostrings([Bs|Bss],Ls) :- write(' '), zipstringbits(Bs,Ls), maptostrings(Bss,Ls).
zipstringbits([],_).
zipstringbits([0|Bs],[_|Ls]) :- zipstringbits(Bs,Ls).
zipstringbits([1|Bs],[L|Ls]) :- write(L), zipstringbits(Bs,Ls).
transpose(Rs,Cs) :- emptylists(Rs),emptylists(Cs).
transpose([[X|R]|Rs],[[X|C]|Cs]) :- headsandtails(Rs,C,NRs), headsandtails(Cs,R,NCs), transpose(NRs,NCs).
headsandtails(Xss,[],[]) :- emptylists(Xss).
headsandtails([[X|Xs]|Xss],[X|Hs],[Xs|Ts]) :- headsandtails(Xss,Hs,Ts).
emptylists([]) :- !.
emptylists([[]|L]) :- emptylists(L).
