定义5x5矩阵

Question

问题是;我们有一个函数取3个参数， 就好; func（[[0,0,0,1,0]，[0,1,1,1,0]，[0,0,1,0,0]， [0,0,1,0,0 ]，[0,0,0,1,0]]，（1,1），X）第一个是嵌套列表，即 显示5x5矩阵，1s表示满，0表示空，并且 第二个参数（1,1）我们的起点第一行第一列， 第三个参数X是;变量，我们将与 统一点可以从起点访问是（1,1） 所以如果问;

?- func ([ [0,0,0,1] [0,0,1,0] [0,0,1,1] [0,0,1,0] ], (1,1), X). 
X = (1, 1); 

X = (1, 2); 

X = (1, 3); 

X = (2, 2); 

X = (3, 2); 

X = (4, 1); 

X = (4, 2); 

false. 

当我们从（1,1）开始时，我们可以向上，向下，向左和向右移动; 因为如果空（1,1）上没有向左和向上运动，空白时请向右看，写下来，向下看空写下来，再次转到（1,2），向右或向左或向上或向下移动，依此类推。 （2,4）（4,4） 如果例如点（2,3）已满并且（2,4）为空 我们看起来如果我们没有写输出，（2,4）（4,4） 我们可以一个接一个点（2,4），我的意思是， 如果离开，他们的上下是满的，我们不能用点来点（2,4），因为他们已满。

Answer 1

我的解决方案：拿到教科书，坐在电脑前，为自己找出答案！简单地将某些东西标为家庭作业并不是你自己做的原因。

Answer 2

最后我做了它，这里是代码;

%returns the nth element from the list 
nth([F|_],1,F). 
nth([_|R],N,M) :- N > 1, N1 is N-1, nth(R,N1,M). 

%returns true if cell is empty: gets the cell value at (StartRow,StartColumn) and returns whether the value is 0 
isempty(Maze,StartRow,StartColumn) :- nth(Maze,StartRow,Line),nth(Line,StartColumn,Y), Y == 0. 

%returns the head of the list 
head([Elem|_],Elem). 

%find accessible returns empty list if not in maze size (1 to N for row and column) 
findaccessible(Maze, (StartRow,StartColumn), [], _) :- head(Maze,L),length(L,N), (StartColumn > N ; StartRow > N ; StartColumn < 1 ; StartRow < 1). 

%find all empty cells and retain them in X. L retains the current found cells in order to avoid returning to visited positions. 
findaccessible(Maze, (StartRow,StartColumn), X, L) :- 
    %if cell is empty, retain position and add it to the list 
    isempty(Maze,StartRow,StartColumn) -> (union(L,[(StartRow,StartColumn)],L1),X1 = [(StartRow,StartColumn)], 

    %check right column and if element not visited, find all accessible cells from that point and unify the lists 
    SR is StartRow, SC is StartColumn+1,(member((SR,SC),L) -> union(X1,[],X2) ; (findaccessible(Maze, (SR,SC), Tmp1, L1), union(X1,Tmp1,X2))), 
    %check down row and if element not visited, find all accessible cells from that point and unify the lists 
    SR2 is StartRow+1,SC2 is StartColumn, (member((SR2,SC2),L) -> union(X2,[],X3) ; (findaccessible(Maze, (SR2,SC2), Tmp2, L1), union(X2,Tmp2,X3))), 
    %check left column and if element not visited, find all accessible cells from that point and unify the lists 

    SR3 is StartRow, SC3 is StartColumn-1, (member((SR3,SC3),L) -> union(X3,[],X4) ; (findaccessible(Maze, (SR3,SC3), Tmp3, L1), union(X3,Tmp3,X4))), 
    %check up row and if element not visited, find all accessible cells from that point and unify the lists 
    SR4 is StartRow-1, SC4 is StartColumn, (member((SR4,SC4),L) -> union(X4,[],X) ; (findaccessible(Maze, (SR4,SC4), Tmp4, L1), union(X4,Tmp4,X)))) ; X = []. 

%lists each result 
%if no more results return false 
results(_,[]) :- fail. 
%return the result or return the rest of the results 
results(X,[Head|Rest]) :- X = Head ; results(X,Rest). 

%accessible predicate that finds all empty accessible cells and then list each of them 
accessible(Maze, (StartRow,StartColumn), X) :- findaccessible(Maze, (StartRow,StartColumn), Lst, []), !, results(X,Lst). 

%sample test run 
%accessible([[0, 0, 0, 1, 0], [0, 1, 1, 1, 0], [0, 0, 1, 0, 0], [0, 0, 1, 0, 0], [0, 0, 0, 1, 0]], (1, 1), X).