Question

所以我想对一系列数字进行排序，以便每个连续数字的总和是一个平方数（4,9,16,25，...）我被告知我应该使用Depth first search但是我无法正确地实现它，特别是回溯部分。这是正确的方法，我没有看到，做错了什么？ DFS甚至是正确的方法吗？

class Node:                         #Define node Object
    def __init__(self, value):
        self.value = value         #Node "Name"
        self.adjacentNodes=list()  #Node Neighbours

#Set limit
limit=16
squares=list() #list for squared numbers in rang smaller than 2Xlimit
tree=list() #Tree containing nodes
path=list() #current path taken

def calculate(limit):   #Population control legal squares
    for i in range(1,(limit+1)):
        i=i*i
        if i<(limit+limit):
            squares.append(i)

calculate(limit)    #Populate squares list

for a in range(1,(limit+1)): #creating the nodes
    newnode=Node(a)
    for s in squares:
        legalsum=s-a
        if legalsum>0 and legalsum<limit and legalsum!=a: #zero and lower can't play,keeping non-exiting nodes out of node.adjacentNodes, don't connect to selve.
            newnode.adjacentNodes.append(legalsum)  #Add adjacentnode
    tree.append(newnode)    #add newborn to tree

for b in range(0,limit):
    print tree[b].value,tree[b].adjacentNodes #Quick checkup
#Now the tree is build and can be used to search for paths.
'''
Take a node and check adjnodes
is it already visited? check path
yes-> are all nodes visited (loop over tree(currentnode).adjnodes) Maybe use len()
        yes->pop last node from path and take other node -> fear for infinte loops, do I need a current node discard pile?
        no -> visit next available adje node next in line from loop got from len(tree.adjn)
no-> take this as node and check adj nodes.

'''
def dfs (node):
    path.append(node)
    #load/check adjacent nodes
    tree[node].adjacentNodes
    for adjnode in tree[node-1].adjacentNodes:
        if adjnode in path:
            #skip to next adj node
            continue


        else:
            print path #Quick checkup
            dfs(adjnode)
dfs(8)    # start with 8 since 1-16 should be {8,1,15,10,6,3,13,12,4,5,11,14,2,7,9,16}

Answer 1

我不知道你在说什么广场，但这里有一些问题

def dfs (node):
    #always assume the path starts with this node
    path=[node]
    if node == GoalState: return path #we win!!
    #load/check adjacent nodes
    #this line did nothing `tree[node].adjacentNodes`
    for adjnode in tree[node-1].adjacentNodes:
    # if adjnode in path: ****this will never be true, lets eliminate it
    #         #skip to next adj node
    #        continue

    #   else:
            subpath = dfs(adjnode)
            if subpath and subpath.endswith(GoalState): 
               return path + subpath #return solution

    #if we get here there is no solution from this node to the goal state

何时使用深度优先搜索

1 个答案: