将A *改为dijkstra's

Question

function A*(start, goal)

closedSet := {}

openSet := {start}

cameFrom := an empty map

gScore := map with default value of Infinity

gScore[start] := 0

fScore := map with default value of Infinity

fScore[start] := heuristic_cost_estimate(start, goal)

while openSet is not empty
    current := the node in openSet having the lowest fScore[] value
    if current = goal
        return reconstruct_path(cameFrom, current)

    openSet.Remove(current)
    closedSet.Add(current)

    for each neighbor of current
        if neighbor in closedSet
            continue    

        if neighbor not in openSet  
            openSet.Add(neighbor)

        tentative_gScore := gScore[current] + dist_between(current, neighbor)
        if tentative_gScore >= gScore[neighbor]
            continue        

        cameFrom[neighbor] := current
        gScore[neighbor] := tentative_gScore
        fScore[neighbor] := gScore[neighbor] + heuristic_cost_estimate(neighbor, goal) 

return failure

function reconstruct_path(cameFrom, current)
total_path := [current]
while current in cameFrom.Keys:
    current := cameFrom[current]
    total_path.append(current)
return total_path

如果我采用这个A * alorgithim，而不是从openset中取出最低的fScore，那么拿最低的gscore甚至不考虑fScore会有效地制作这个dijksta的吗？

Answer 1

Dijkstra的最短路径算法是A *的一个特例，其中启发式算法为零，尽管Dijkstra算法（1958）是在10年之前构思出来的。 A *根据f(v) f(v) = h(v) + g(v)贪婪地选择在搜索期间查找下一个顶点。如果将启发式算法（h(v)）设置为零，则A *将变为选择仅考虑当前成本（g(v)）的顶点，并且您可以实现将知情A *转换为未知的Dijkstra。

简而言之，我对你问题的回答是“是”。