我正在为一个机器人项目做一个前端(一个’自主’汽车,它使用一些传感器和一个地图 – 从SVG文件生成本地化).

为了使机器人可控,我们必须在其当前位置和目标之间生成路径.我使用了最简单的算法：A *.

我得到了一些奇怪的结果：汽车倾向于达到45°的倍数,还有一个特别恼人的问题：一些生成的路径非常嘈杂！

在这种情况下,请查看橙色矩形附近的嘈杂路径：

反正有没有避免那些奇怪/嘈杂的结果？最终我们想要建立一个具有最小数量的航向角变化的路径. (汽车可以在不移动的情况下转弯,因此我们不需要任何路径’平滑’).

这是我的A *实现：

def search(self, begin, goal):
    if goal.x not in range(self.width) or goal.y not in range(self.height):
        print "Goal is out of bound"
        return []
    elif not self.grid[begin.y][begin.x].reachable:
        print "Beginning is unreachable"
        return []
    elif not self.grid[goal.y][goal.x].reachable:
        print "Goal is unreachable"
        return []
    else:

        self.cl = set()
        self.ol = set()

        curCell = begin
        self.ol.add(curCell)

        while len(self.ol) > 0:

            # We choose the cell in the open list having the minimum score as our current cell
            curCell = min(self.ol, key = lambda x : x.f)

            # We add the current cell to the closed list
            self.ol.remove(curCell)
            self.cl.add(curCell)

            # We check the cell's (reachable) neighbours :
            neighbours = self.neighbours(curCell)

            for cell in neighbours:
                # If the goal is a neighbour cell :
                if cell == goal:
                    cell.parent = curCell
                    self.path = cell.path()
                    self.display()
                    self.clear()
                    return self.path
                elif cell not in self.cl:
                    # We process the cells that are not in the closed list
                    # (processing <-> calculating the "F" score)
                    cell.process(curCell, goal)

                    self.ol.add(cell)

编辑1：通过popuplar需求,这里是分数计算功能(过程)：

def process(self, parent, goal):
    self.parent = parent
    self.g = parent.distance(self)
    self.h = self.manhattanDistance(goal)
    self.f = self.g + self.h

编辑这是邻居方法(根据user1884905的回答更新)：

def neighbours(self, cell, radius = 1, unreachables = False, diagonal = True):
    neighbours = set()
    for i in xrange(-radius, radius + 1):
        for j in xrange(-radius, radius + 1):
            x = cell.x + j
            y = cell.y + i
            if 0 <= y < self.height and 0 <= x < self.width and ( self.grid[y][x].reachable or unreachables ) and (diagonal or (x == cell.x or y == cell.y)) :
                neighbours.add(self.grid[y][x])

    return neighbours

(这看起来很复杂,但它只给出了一个单元的8个邻居 – 包括对角线邻居;它也可以采用与1不同的半径,因为它用于其他功能)

和距离计算(取决于对角线的使用与否：)

def manhattanDistance(self, cell):
    return abs(self.x - cell.x) + abs(self.y - cell.y)

def diagonalDistance(self, cell):

    xDist = abs(self.x - cell.x)
    yDist = abs(self.y - cell.y)

    if xDist > yDist:
        return 1.4 * yDist + (xDist - yDist)
    else:
        return 1.4 * xDist + (yDist - xDist)

最佳答案 如果无法看到你如何实现邻居和距离函数,我仍然可以很好地猜测出现了什么问题：

如果允许对角线遍历,则不应使用曼哈顿距离.

目标函数中的曼哈顿距离应该是距离目标的最短距离的度量. (如果你可以沿着对角线穿过积木,那就不是了.)

解决此问题的最简单方法是将曼哈顿距离保持为目标函数,并将邻居的定义更改为仅包括四个左右上下相邻单元格.

编辑

您的代码仍然存在问题.以下伪代码取自wikipedia.我已标记了您必须检查的重要行.如果找到更好的解决方案,则必须确保i)更新开放集中的节点; ii)您始终考虑先前行驶的距离.

function A*(start,goal)
     closedset := the empty set    // The set of nodes already evaluated.
     openset := {start}    // The set of tentative nodes to be evaluated, initially containing the start node
     came_from := the empty map    // The map of navigated nodes.

     g_score[start] := 0    // Cost from start along best known path.
     // Estimated total cost from start to goal through y.
     f_score[start] := g_score[start] + heuristic_cost_estimate(start, goal)

     while openset is not empty
         current := the node in openset having the lowest f_score[] value
         if current = goal
             return reconstruct_path(came_from, goal)

         remove current from openset
         add current to closedset
         for each neighbor in neighbor_nodes(current)
             // -------------------------------------------------------------------
             // This is the way the tentative_g_score should be calculated.
             // Do you include the current g_score in your calculation parent.distance(self) ?
             tentative_g_score := g_score[current] + dist_between(current,neighbor)
             // ------------------------------------------------------------------- 
             if neighbor in closedset
                 if tentative_g_score >= g_score[neighbor]
                     continue

             // -------------------------------------------------------------------
             // You never make this comparrison
             if neighbor not in openset or tentative_g_score < g_score[neighbor]
             // -------------------------------------------------------------------
                 came_from[neighbor] := current
                 g_score[neighbor] := tentative_g_score
                 f_score[neighbor] := g_score[neighbor] + heuristic_cost_estimate(neighbor, goal)
                 if neighbor not in openset
                     add neighbor to openset

     return failure

 function reconstruct_path(came_from, current_node)
     if current_node in came_from
         p := reconstruct_path(came_from, came_from[current_node])
         return (p + current_node)
     else
         return current_node