n拼图python中的曼哈顿距离启发式

如何解决n拼图python中的曼哈顿距离启发式

我的函数在 python 中计算 8 拼图的曼哈顿距离时遇到问题。我的代码有一个问题，尤其是在特定状态下： [4,3,7,5,8,6,1,2] ，（在其他一些状态我尝试过它并且它工作......）我的代码运行但没有'不停止运行并且不打印结果那么我不知道我的函数中的问题出在哪里。

我运行的代码：

def search(n):
    s=[[4,2],""]
    print(s)
    f=frontier.create(s)
    while not frontier.is_empty(f):
        s=frontier.remove(f)
        if state.is_target(s):
            return [s,f[1],f[3]]
        ns=state.get_next(s)
        for i in ns:
            frontier.insert(f,i)
    return 0


print(search(3))

我计算曼哈顿距离的函数：

def hdistance(s):
    # the heuristic value of s
    i = 0 #represent the row number
    j = 0 #represent the column number
    sum = 0 #the manhattan distance
    n = math.sqrt(len(s[0])) #calculate the size of the n puzzle
    for k in s[0]: #run through the list
        if k != 0:
            row = int(k / n)
            column = int(k % n)
            sum = sum + abs(i - row) + abs(j - column)
            j += 1 #increment the column
            if j == n: #if I get to the end of the line
                i += 1 #increment the row
                j = 0
    return sum #return the manhattan distance

边境档案：

def create(s):
    return [[s],0]


def is_empty(f):
    return f[0]==[]

def val(s):
    return state.hdistance(s)+state.path_len(s)

# Uniform Cost: return state.path_len(s)
# Greedy Best First: return state.hdistance(s)

def insert(h,s):
    #inserts state s to the frontier
    f=h[0]
    h[1]+=1
    h[2]+=1
    if h[2]>h[3]:
        h[3]=h[2]
    f.append(s)
    i=len(f)-1
    while i>0 and val(f[i])<val(f[(i-1)//2]):
        t=f[i]
        f[i]=f[(i-1)//2]
        f[(i-1)//2]=t
        i=(i-1)//2

def remove(h):
    if is_empty(h):
        return 0
    h[2]-=1
    f=h[0]
    s=f[0]
    f[0]=f[len(f)-1]
    del f[-1]
    heapify(f,0)
    return s

def heapify(f,i):
    minSon=i
    if 2*i+1<len(f) and val(f[2*i+1])<val(f[minSon]):
        minSon=2*i+1
    if 2*i+2<len(f) and val(f[2*i+2])<val(f[minSon]):
        minSon=2*i+2
    if minSon!=i:
        t=f[minSon]
        f[minSon]=f[i]
        f[i]=t
        heapify(f,minSon)

状态文件：

def get_next(x):            # returns a list of the children states of x
    ns=[]                   # the next state list
    for i in "<>v^":
        s=x[0][:]           # [:] - copies the board in x[0]
        if_legal(s,i)       # try to move in direction i
        # checks if the move was legal and...
        if s.index(0)!=x[0].index(0) and \
           (x[1]=="" or x[1][-1]!="><^v"["<>v^".index(i)]): # check if it's the first move or it's a reverse move
            ns.append([s,x[1]+i])   # appends the new state to ns
    return ns


def path_len(x):
    return len(x[1])

def is_target(x):
    n=len(x[0])                     # the size of the board
    return x[0]==list(range(n))     # list(range(n)) is the target state

#############################
def if_legal(x,m):                  # gets a board and a move and makes the move if it's legal
    n=int(math.sqrt(len(x)))        # the size of the board is nXn
    z=x.index(0)                    # z is the place of the empty tile (0)
    if z%n>0 and m=="<":            # checks if the empty tile is not in the first col and the move is to the left
        x[z]=x[z-1]                 # swap x[z] and x[z-1]...
        x[z-1]=0                    # ...and move the empty tile to the left
    elif z%n<n-1 and m==">":        # check if the empty tile is not in the n's col and the move is to the right
        x[z]=x[z+1]
        x[z+1]=0
    elif z>=n and m=="^":           # check if the empty tile is not in the first row and the move is up
        x[z]=x[z-n]
        x[z-n]=0
    elif z<n*n-n and m=="v":        # check if the empty tile is not in the n's row and the move is down
        x[z]=x[z+n]
        x[z+n]=0

其余代码（所有其他函数）在我的作业中给出，那么问题肯定出在我的 hdistance 函数中。 如果您能在我的函数中找到问题所在，那么可能存在无限循环... 谢谢你！！