Question

问题是;我们有一个函数拿3个参数，喜欢; 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，这是显示5x5矩阵，1s表示它已满，0表示空，第二个参数（1,1）我们的起点第1行第1列，第三个参数X是;我们将统一的变量从起始点可以获得的点是（1,1）所以，如果被问到;

?- 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; 因为在（1,1）上没有向左和向上移动，如果是空的，则向右看，写下来，向下看空记录，再次进入（1,2），向右或向左或向上或向下移动，依此类推。 / p>

这里是我们没有写输出的原因，（2,4）（4,4）如果例如点（2,3）已满且（2,4）为空我们看，我们可以一个接一个地点（2,4），我的意思是，如果离开，它们的上下都已满，我们不能使用这一点来指向（2,4），因为它们已经满了。

Answer 1

我的解决方案：拿到教科书，坐在电脑前，自己搞清楚！简单地将某些东西标记为家庭作业并不能成为自己不做的理由。

Answer 2

最后我做了，这是代码;

%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).

定义5x5矩阵

2 个答案: