– h1:错位的瓷砖数量
– h2:总曼哈顿距离
– h3:以上的总和
移动的图块称为0.
我的目标是解决这些问题:
4 1 2 5 8 3 7 0 6
和
8 6 7 2 5 4 3 0 1
我遇到的问题是,通过我目前的A *实现,它能够解决第一个问题,但不能解决第二个问题.
所以请帮助我理解我的A *代码有什么问题:
int [,] current =从控制台输入为字符串(412583706)并转换为表示拼图的2D int.
正确相同,其中0位于右下角.
var openList = new List<int[,]> { current }; var closedList = new List<int[,]>(); while (openList.Count > 0) { steps++; current = GetBestNodeFromList(correct,dimensions,openList,useHeuristic); // "GetBestNodeFromList()" finds the cheapest node in the openList. // cheapest node: lowest value of h3. openList.Remove(current); h1 = getHeuristic1b(current,correct,dimensions); h2 = getHeuristic2b(current,dimensions); h3 = h1 + h2; if (h1 == 0 && h2 == 0) { break; } openList = Puzzle_PossibleNext(current,closedList); // the method "PossibleNext()" finds possible next moves from the current // position. if the next move exists in the closedList,it is discarded. // Drawing the puzzle and showing heuristics. DrawCurrentState(h1,h2,h3,current,steps); // adding last visited position to the closedlist. closedList.Add(current); }
第一个问题通过7个步骤解决.
根据我测试的不同程序,下一个问题可以通过32个步骤解决.
我的程序与另一个程序的不同之处在于前4个步骤是相同的,然后另一个程序选择不同的路径,而我的程序一直在继续,无法找到解决方案.
看起来我的程序确实选择了最便宜的节点,所以这就是为什么我无法理解什么是错的.
这是我第一次使用寻路算法,所以我想解决它.
我已经有这个问题3天,我觉得我已经尝试了很多解决方案,但没有一个工作T_T
最好的祝福.
– – 编辑 – – –
附加代码:
// Put heuristic value from all in list,then return list item with lowest h-value. static int[,] GetBestNodeFromList(int[,] correct,int d,List<int[,]> list,string useHeuristic) { int[,] n = new int[d,d]; if (list.Count > 0) { List<Int32> heuristicsValueList = new List<Int32>(); for (int i = 0; i < list.Count; i++) { if (useHeuristic == "h1") { heuristicsValueList.Add(getHeuristic1b(list[i],d)); } else if (useHeuristic == "h2") { heuristicsValueList.Add(getHeuristic2b(list[i],d)); } else { heuristicsValueList.Add(getHeuristic3(list[i],d)); } } n = list[heuristicsValueList.IndexOf(heuristicsValueList.Min())]; } return n; }
———编辑2 ——–
改变了我的代码,但仍然没有运气
拼图设置/节点及其启发式都在PuzzleNode对象中.
//从当前节点返回下一个可能移动的列表.
//不包括在closednodeList中找到的移动.
static List<PuzzleNode> Puzzle_GenerateNextNodes(PuzzleNode node,List<PuzzleNode> closednodeList) { List<PuzzleNode> nextList = new List<PuzzleNode>(); Point isNow = new Point(0,0); // 1) Find where [0] is. int dimensions = (int)Math.Sqrt((double)node.n.Length); for (int x = 0; x < dimensions; x++) { for (int y = 0; y < dimensions; y++) { if (node.n[x,y] == 0) { isNow.X = y; isNow.Y = x; break; } } } // 2) Check possible moves. bool moveUp = false,moveDown = false,moveLeft = false,moveRight = false; if (isNow.X == 0) { moveRight = true; if (isNow.Y == 0) { moveDown = true; } else if (isNow.Y == 1) { moveUp = true; moveDown = true; } else if (isNow.Y == 2) { moveUp = true; } } else if (isNow.X == 1) { moveRight = true; moveLeft = true; if (isNow.Y == 0) { moveDown = true; } else if (isNow.Y == 1) { moveUp = true; moveDown = true; } else if (isNow.Y == 2) { moveUp = true; } } else if (isNow.X == 2) { moveLeft = true; if (isNow.Y == 0) { moveDown = true; } else if (isNow.Y == 1) { moveUp = true; moveDown = true; } else if (isNow.Y == 2) { moveUp = true; } } // 3) Create list of possible moves. // Add moved puzzle node to list over next moves if (moveRight) { int[,] right = new int[dimensions,dimensions]; Array.copy(node.n,right,node.n.Length); PuzzleNode tmp = new PuzzleNode( PuzzleMoveRight(right,isNow.X,isNow.Y) ); if (!ListHasThisValue(tmp.n,closednodeList,dimensions)) { nextList.Add(tmp); } } // moveleft,up,down,same structure as moveRight if (moveLeft) { .. } if (moveUp) { .. } if (moveDown) { .. } return nextList; }
———–编辑3 —————-
顺便说一句,我想问一下,如果我对A *的不同步骤的实现得到了正确的理解.
目前,我的程序的A *搜索执行此操作:
>创建初始列表OPEN和CLOSED,将起始节点添加到OPEN
>启动循环,从OPEN中删除最便宜的节点,将其添加到CLOSED
*最便宜的节点由其曼哈顿距离值决定.
>使用节点查找邻居/子节点/下一步移动,添加这些移动
到SUCCESSOR列表.
>探索SUCCESSOR列表,检查其中是否包含目标状态,
否则添加到OPEN列表
>重复2-4,探索列表中的节点.
当我用Q1尝试这些步骤时,我得到7个步骤的解决方案,这是正确的.这也可以手工找到.
但是对于Q2,它一直持续到OPEN列表为空,没有其他东西可以探索.
那我错过了什么?
解决方法
var set = new int[,] { { 1,2,3 },{ 4,5,6 },{ 7,8,0 } }; var clone = (int[,])set.Clone(); var foo = clone == set; // foo is false var bar = clone.Equals(set); // bar is false var closedStates = new List<int[,]>(); closedStates.Contains(state); // wrong - contains is using Equals closedStates.Any(cs => AreEqual(cs,state)); // correct static bool AreEqual(int[,] stateA,int[,] stateB) { for (var x = 0; x < DIMENSIONS; x++) { for (var y = 0; y < DIMENSIONS; y++) { if (stateA[x,y] != stateB[x,y]) { return false; } } } return true; }
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