8-puzzle在使用曼哈顿距离的序言中提供了解决方案 [英] 8-puzzle has a solution in prolog using manhattan distance

问题描述

8个拼图将由列表位置的3x3列表表示，其中空框将由值9表示，如下所示:[[9,1,3]，[5,2,6]， [4,7,8]]

The 8-puzzle will be represented by a 3x3 list of lists positions where the empty box will be represented by the value 9, as shown below: [[9,1,3],[5,2,6],[4,7,8]]

可能的解决方案:8拼图的初始位置中只有一半是可解决的.有一个公式可以让您从一开始就知道是否可以解决难题:要确定8难题是否可解决，对于每个包含值N的正方形，将计算出当前单元格之后比N少多少个数字.例如，以初始状态:

Possibility Solution: Only half of the initial positions of the 8-puzzle are solvable. There is a formula that allows to know from the beginning if you can solve the puzzle.To determine whether an 8-puzzle is solvable, for each square containing a value N is calculated how many numbers less than N there after the current cell. For example, to the initial status:

• 1个数字不少于= 0
• 空(9)-随后必须3,5,2,6,4,7,8 = 7
• 3 have = 1 to 2
• 5随后为2,4 = 2
• 2下没有任何数字发生= 0
• 6随后是4 = 1
• 4个不小于0的数字
• 7个零之后没有次要数字
• 8个不小于0的数字

在那之后，我们计算出空的位置和 位置(3.3).对于上面的示例，空框位于位置(1.2)，因此 曼哈顿距离为: d =绝对(3-1)+绝对(3-2)= 3 最后，将所有计算出的值相加.如果结果是偶数，则表示 难题是可以解决的，但奇怪的是无法解决. 0 +7 +1 +2 +0 +1 +0 +0 +0 +3 = 14

After that, we calculate the Manhattan distance between the position of the empty and position (3.3). For the above example, the empty box is in the position (1.2), so Manhattan distance that is: d = abs (3-1) + abs (3-2) = 3 Finally, add up all the calculated values​​. If the result is even, implies that the puzzle is solvable, but it is odd not be resolved. 0 +7 +1 +2 +0 +1 +0 +0 +0 +3 = 14

该解决方案旨在创建一个具有板上所有可能状态的知识库，我们将在当前位置之后看到比N少多少个数字.

The solution is designed to create a knowledge base with all possible states of a number on the board and we'll see how many numbers less than N there after the current position.

这是我的代码:

%***********************Have Solution*********************************

posA(9,8). posA(8,7). posA(7,6). posA(6,5). posA(5,4). posA(4,3). posA(3,2). posA(2,1). posA(1,0).

posB(9,7). posB(8,7). posB(8,6). posB(7,6). posB(7,5). posB(7,4). 
posB(6,5). posB(6,4). posB(6,3). posB(6,2). posB(5,4). posB(5,3). posB(5,2). posB(5,1).  posB(5,0). 
posB(4,3). posB(4,2). posB(3,2). posB(3,1).  posB(2,1). posB(2,0). posB(1,0).

posC(9,6). posC(8,6). posC(8,5). posC(7,6). posC(7,5). posC(7,4). posC(6,5). posC(6,4). posC(6,3).
posC(5,4). posC(5,3). posC(5,2). posC(4,3). posC(4,2). posC(4,1). posC(4,0).
posC(3,2). posC(3,1). posC(3,0). posC(2,1). posC(1,0).

posD(9,5). posD(8,5). posD(8,4). posD(7,5). posD(7,4). posD(7,3). posD(6,5). posD(6,4). posD(6,3).
posD(6,2). posD(5,4). posD(5,3). posD(5,2). posD(5,1). posD(4,3). posD(4,2). posD(4,1). posD(5,0).
posD(3,2). posD(3,1). posD(3,0). posD(2,1). posD(1,0).

posE(9,4). posE(8,4). posE(8,3). posE(7,4). posE(7,3). posE(7,2). posE(6,4). posE(6,3). posE(6,2). posE(6,1).
posE(5,4). posE(5,3). posE(5,2). posE(5,1). posE(5,0). posE(4,3). posE(4,2). posE(4,1). posE(4,0).
posE(3,2). posE(3,1). posE(3,0). posE(2,1). posE(2,0). posE(1,0).

posF(9,3). posF(8,3). posF(8,2). posF(7,1). posF(7,2). posF(7,3). posF(6,0). posF(6,1). posF(6,2). 
posF(6,3). posF(5,0). posF(5,1). posF(5,2). posF(5,3). posF(4,0). posF(4,1). posF(4,2). posF(4,3).
posF(2,0). posF(2,1). posF(3,0). posF(3,1). posF(3,2). posF(1,0).

posG(9,2). posG(8,0). posG(8,1). posG(8,2).  posG(7,0). posG(7,1). posG(7,2).
posG(6,0). posG(6,1). posG(6,2). posG(5,0).  posG(5,1). posG(5,2). posG(4,0). posG(4,1). posG(4,2).
posG(3,0). posG(3,1). posG(3,2). posG(2,0).  posG(2,1). posG(1,0).

posH(9,1). posH(8,0). posH(8,1). posH(7,0). posH(7,1). posH(6,0). posH(6,1). posH(5,0). posH(5,1). 
posH(4,0). posH(4,1). posH(3,0). posH(3,1). posH(2,0). posH(1,1). posH(1,0).

posI(9,0). posI(8,0). posI(7,0). posI(6,0). posI(5,0). posI(4,0). posI(3,0). posI(2,0). posI(1,0).  

haveSolution([[A,B,C],[D,E,F],[G,H,I]]):- distManhattan([A,B,C,D,E,F,G,H,I], Z),
                                         posA(A,Pa), posB(B,Pb), posC(C,Pc),
                                         posD(D,Pd), posE(E,Pe), posF(F,Pf),
                                         posG(G,Pg), posH(H,Ph), posI(I,Pi),
                                         P is Pa+Pb+Pc+Pd+Pe+Pf+Pg+Ph+Pg+Pi+Z, 0 is P mod 2,
                                         write('The 8-puzzle have solution').

%%*************************Manhattan distance***********************
distManhattan([A,B,C,D,E,F,G,H,I], Dist):-  A=9, Dist is abs(3-1)+abs(3-1), !;
                                            B=9, Dist is abs(3-1)+abs(3-2), !;
                                            C=9, Dist is abs(3-1)+abs(3-3), !;
                                            D=9, Dist is abs(3-2)+abs(3-1), !;
                                            E=9, Dist is abs(3-2)+abs(3-2), !;
                                            F=9, Dist is abs(3-2)+abs(3-3), !;
                                            G=9, Dist is abs(3-3)+abs(3-1), !;
                                            H=9, Dist is abs(3-3)+abs(3-2), !;
                                            I=9, Dist is abs(3-3)+abs(3-3).

问题是我犯了一个错误，因为在某些情况下我可以有多个选择，例如:

The problem is that I am making a mistake because there are situations where I can have more than one alternative, eg>:

|  1 |  9 | 3  |
|  5 |  2 | 6  |
|  4 |  7 | 8  |    


posA(1,0)+posB(9,7)+posC(3,1)+posD(5,2)+posE(2,0)+posF(6,1)+posG(4,0)+posH(7,0)+posI(8,0).

posC(C，Pc)的正确解是posC(3,1)，即1；但是还有其他后果有时会导致不正确的输出...我的代码在做什么错以及如何更改它?

The right solution for posC(C,Pc) is posC(3,1), that is 1; but there are other ramifications that sometimes cause incorrect outputs ... what am I doing wrong in my code and how I can change it?

推荐答案

此答案从不同的角度看问题:

This answer looks at the problem from a different point of view:

• 使用复合结构board/9表示单板配置.
• 等于可滑动单个零件的配置通过关系m/2连接.

• Single board configurations are represented using the compound structure board/9.
• Configurations that are equal up to sliding a single piece are connected by relation m/2.

所以我们定义m/2！

m(board(' ',B,C,D,E,F,G,H,I), board(D, B ,C,' ',E,F,G,H,I)).
m(board(' ',B,C,D,E,F,G,H,I), board(B,' ',C, D ,E,F,G,H,I)).

m(board(A,' ',C,D,E,F,G,H,I), board(' ',A, C , D, E ,F,G,H,I)).
m(board(A,' ',C,D,E,F,G,H,I), board( A ,C,' ', D, E ,F,G,H,I)).
m(board(A,' ',C,D,E,F,G,H,I), board( A ,E, C , D,' ',F,G,H,I)).

m(board(A,B,' ',D,E,F,G,H,I), board(A,' ',B,D,E, F ,G,H,I)).
m(board(A,B,' ',D,E,F,G,H,I), board(A, B ,F,D,E,' ',G,H,I)).

m(board(A,B,C,' ',E,F,G,H,I), board(' ',B,C,A, E ,F, G ,H,I)).
m(board(A,B,C,' ',E,F,G,H,I), board( A ,B,C,E,' ',F, G ,H,I)).
m(board(A,B,C,' ',E,F,G,H,I), board( A ,B,C,G, E ,F,' ',H,I)).

m(board(A,B,C,D,' ',F,G,H,I), board(A, B ,C,' ',D, F ,G, H ,I)).
m(board(A,B,C,D,' ',F,G,H,I), board(A,' ',C, D ,B, F ,G, H ,I)).
m(board(A,B,C,D,' ',F,G,H,I), board(A, B ,C, D ,F,' ',G, H ,I)).
m(board(A,B,C,D,' ',F,G,H,I), board(A, B ,C, D ,H, F ,G,' ',I)).

m(board(A,B,C,D,E,' ',G,H,I), board(A,B,' ',D, E ,C,G,H, I )).
m(board(A,B,C,D,E,' ',G,H,I), board(A,B, C ,D,' ',E,G,H, I )).
m(board(A,B,C,D,E,' ',G,H,I), board(A,B, C ,D, E ,I,G,H,' ')).

m(board(A,B,C,D,E,F,' ',H,I), board(A,B,C,' ',E,F,D, H ,I)).
m(board(A,B,C,D,E,F,' ',H,I), board(A,B,C, D ,E,F,H,' ',I)).

m(board(A,B,C,D,E,F,G,' ',I), board(A,B,C,D,' ',F, G ,E, I )).
m(board(A,B,C,D,E,F,G,' ',I), board(A,B,C,D, E ,F,' ',G, I )).
m(board(A,B,C,D,E,F,G,' ',I), board(A,B,C,D, E ,F,  G,I,' ')).

m(board(A,B,C,D,E,F,G,H,' '), board(A,B,C,D,E,' ',G, H ,F)).
m(board(A,B,C,D,E,F,G,H,' '), board(A,B,C,D,E, F ,G,' ',H)).

快要完成了！ 要连接这些步骤，我们使用meta-predicate 一起 path/4 使用length/2进行迭代加深.

Almost done! To connect the steps, we use the meta-predicate path/4 together with length/2 for performing iterative deepening.

以下问题实例来自@CapelliC的答案:

The following problem instances are from @CapelliC's answer:

?- length(Path,N), path(m,Path,/* from */ board(1,' ',3,5,2,6,4,7, 8 ),
                               /*  to  */ board(1, 2 ,3,4,5,6,7,8,' ')).
N =  6, Path = [board(1,' ',3,5,2,6,4,7,8), board(1,2,3,5,' ',6,4,7,8),
                board(1,2,3,' ',5,6,4,7,8), board(1,2,3,4,5,6,' ',7,8),
                board(1,2,3,4,5,6,7,' ',8), board(1,2,3,4,5,6,7,8,' ')] ? ;
N = 12, Path = [board(1,' ',3,5,2,6,4,7,8), board(1,2,3,5,' ',6,4,7,8),
                board(1,2,3,5,7,6,4,' ',8), board(1,2,3,5,7,6,' ',4,8),
                board(1,2,3,' ',7,6,5,4,8), board(1,2,3,7,' ',6,5,4,8),
                board(1,2,3,7,4,6,5,' ',8), board(1,2,3,7,4,6,' ',5,8),
                board(1,2,3,' ',4,6,7,5,8), board(1,2,3,4,' ',6,7,5,8),
                board(1,2,3,4,5,6,7,' ',8), board(1,2,3,4,5,6,7,8,' ')] ? ;
...

?- length(Path,N), path(m,Path,/* from */ board(8,7,4,6,' ',5,3,2, 1 ),
                               /*  to  */ board(1,2,3,4, 5 ,6,7,8,' ')).
N = 27, Path = [board(8,7,4,6,' ',5,3,2,1), board(8,7,4,6,5,' ',3,2,1),
                board(8,7,4,6,5,1,3,2,' '), board(8,7,4,6,5,1,3,' ',2),
                board(8,7,4,6,5,1,' ',3,2), board(8,7,4,' ',5,1,6,3,2),
                board(' ',7,4,8,5,1,6,3,2), board(7,' ',4,8,5,1,6,3,2),
                board(7,4,' ',8,5,1,6,3,2), board(7,4,1,8,5,' ',6,3,2),
                board(7,4,1,8,5,2,6,3,' '), board(7,4,1,8,5,2,6,' ',3),
                board(7,4,1,8,5,2,' ',6,3), board(7,4,1,' ',5,2,8,6,3),
                board(' ',4,1,7,5,2,8,6,3), board(4,' ',1,7,5,2,8,6,3),
                board(4,1,' ',7,5,2,8,6,3), board(4,1,2,7,5,' ',8,6,3),
                board(4,1,2,7,5,3,8,6,' '), board(4,1,2,7,5,3,8,' ',6),
                board(4,1,2,7,5,3,' ',8,6), board(4,1,2,' ',5,3,7,8,6),
                board(' ',1,2,4,5,3,7,8,6), board(1,' ',2,4,5,3,7,8,6),
                board(1,2,' ',4,5,3,7,8,6), board(1,2,3,4,5,' ',7,8,6),
                board(1,2,3,4,5,6,7,8,' ')] ? ;
N = 29, Path = [...] ? ;
...

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