我目前正在研究一个使用适应度函数解决 15 个难题的项目。有3种健身功能可以使用,
- 适应度函数1= 1 - (错位棋子数/棋子总数);
- 适应度函数2 = 1-(所有错位瓷砖距离目标位置的距离之和/64)
- 适应度函数3=(适应度2+适应度1)/2
求解器旨在搜索提高适应度函数 VALUE 的可能移动,直到解决整个拼图板(其中适应度函数 = 1)。我在尝试生成动作时备货,因为求解器会打印不成功的搜索路线以及实际路线,例如 W,S,W,N,S,N,S,N,S,N,S,N,S,N,S ,N,S,N (顺序相反),表示 NWSW。然而,求解器来回搜索多次,并打印出不需要的路线。我想在递归中排除以前访问过的位置,并且只打印有效的移动而不是不成功的移动。代码如下:
<pre><code>
enter code here
package Definitions;
public class Puzzle {
public TreeNode boardsTree;
public Boolean searching;
public int lastPos;
Puzzle(Board b) {
boardsTree = new TreeNode(b);
lastPos = b.Blank_pos;
}
public String solver(Board b, String route, int level, double fitVal){
//System.out.println("route in the level " + level +": " +route + " Fitness: ");
//System.out.print(b.f1.getFitnessVal());
//If the depth of the Tree is level 15 or Board is solved return route(moves e.g. WNS)
if(level == 15||b.solved){
//if(route != ""){
//if(route.length()>1){
// return route.substring(0, route.length()-1);
//}else{
return route;
//}
}else{
//Create a temporary store for the last position
//Create four auxilliary puzzle boards has child puzzles
//Perform the different moves on the blank space on each board in different directions
//(N=0,S=1,E=2,W=3)
lastPos = b.Blank_pos;
Board auxN = new Board(4,b.tilesList);
Board auxS = new Board(4,b.tilesList);
Board auxE = new Board(4,b.tilesList);
Board auxW = new Board(4,b.tilesList);
//Builds the tree
//MOVES TO NORTH
//b.isValidMove(North=0,South=1,East=2,West=3)
if( b.isValidMove(0)){
//If the lastPosition (blankPosition in the parent board) is not the position of the
//cell in the north MoveBlank to North
if(lastPos != (lastPos - 4)){
auxN.MoveBlank(0);
boardsTree.addSon(0,auxN);
if(fitVal > boardsTree.getSon(0).getState().f1.getFitnessVal()){
auxN.print();
searching = false;
return route + "N";
}
}
}
//MOVES TO SOUTH
//b.isValidMove(North=0,South=1,East=2,West=3)
if( b.isValidMove(1) ){
//If the lastPosition (blankPosition in the parent board) is not the position of the
//cell in the north MoveBlank to North
if(lastPos != (lastPos + 4)){
auxS.MoveBlank(1);
boardsTree.addSon(1,auxS);
if(fitVal > boardsTree.getSon(1).getState().f1.getFitnessVal()){
auxS.print();
searching = false;
return route + "S";
}
}
}
//MOVES TO EAST
if( b.isValidMove(2) ){
if(lastPos != (lastPos + 1)){
auxE.MoveBlank(2);
boardsTree.addSon(2,auxE);
if(fitVal > boardsTree.getSon(2).getState().f1.getFitnessVal()){
auxE.print();
searching = false;
return route + "E";
}
}
}
//MOVES TO WEST
if( b.isValidMove(3) ){
if(lastPos != (lastPos -1)){
auxW.MoveBlank(3);
boardsTree.addSon(3,auxW);
if(fitVal > boardsTree.getSon(3).getState().f1.getFitnessVal()){
auxW.print();
searching = false;
return route + "W";
}
}
}
//EVALUATE NEW STATES
if(searching && b.isValidMove(0)){
System.out.println("Actual: " +auxN.Blank_pos+ " Previo: "+lastPos );
if(lastPos != auxN.Blank_pos){
lastPos = auxN.Blank_pos;
route = solver(auxN,route + "N", level+1, fitVal); //NORTH
}else{
//If a solution is not generated enter a recursion to find further solutions at a
//further depth of the tree
solver(auxN,route, level+1, fitVal); //NORTH
}
}
if(searching && b.isValidMove(1)){
//System.out.println("Actual: " +auxS.Blank_pos+ " Previo: "+lastPos );
if(lastPos != auxS.Blank_pos){
lastPos = auxS.Blank_pos;
route =solver(auxS,route + "S", level+1, fitVal); //NORTH
}else{
solver(auxS,route, level+1, fitVal); //NORTH
}
}
if(searching && b.isValidMove(2)){
//System.out.println("Actual: " +auxE.Blank_pos+ " Previo: "+lastPos );
if(lastPos != auxE.Blank_pos){
lastPos = auxE.Blank_pos;
route = solver(auxE,route + "E", level+1, fitVal); //NORTH
}else{
solver(auxE,route, level+1, fitVal); //NORTH
}
}
if(searching && b.isValidMove(3)){
//System.out.println("Actual: " +auxW.Blank_pos+ " Previo: "+lastPos );
if(lastPos != auxW.Blank_pos){
lastPos = auxW.Blank_pos;
route =solver(auxW,route + "W", level+1, fitVal); //NORTH
}else{
solver(auxW,route, level+1, fitVal); //NORTH
}
}
}
return route;
}
}