Problem:
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8 Puzzle

The goal of the 8 puzzle problem is to complete pieces on $3 \times 3$ cells where one of the cells is empty space.

In this problem, the space is represented by 0 and pieces are represented by integers from 1 to 8 as shown below.

1 3 0
4 2 5
7 8 6

You can move a piece toward the empty space at one step. Your goal is to make the pieces the following configuration in the shortest move (fewest steps).

1 2 3
4 5 6
7 8 0

Write a program which reads an initial state of the puzzle and prints the fewest steps to solve the puzzle.

Input

The $3 \times 3$ integers denoting the pieces or space are given.

Output

Print the fewest steps in a line.

Constraints

There is a solution.

Sample Input

1 3 0
4 2 5
7 8 6

Sample Output

4

Solution in Python3 (judge verdict: Wrong Answer)

import sys;
import heapq

def iterative(i,j):
    q = []
    heapq.heappush(q,(sumcost,(0,i,j,0,puz)))
    global finding

    while len(q):
        c_dWithH, items = heapq.heappop(q)
        c_depth = items[0]
        _i = items[1]
        _j = items[2]
        prev_move = items[3]
        c_puz = items[4]
        _sum_cost = c_dWithH - c_depth

        if(c_dWithH - c_depth == 0):
            finding = 1
            print(c_depth)
            break
        if(c_dWithH > depth):
            continue

        c_cost = simpleHS(c_puz,_i,_j)

        print(c_depth)
        if(_i != 0 and prev_move != 1):
            swap_puz = swapPuz(c_puz[0:],_i,_j,_i-1,_j)
            heapq.heappush(q,(c_depth+1+_sum_cost+checkCost(c_cost,simpleHS(c_puz,_i-1,_j),simpleHS(swap_puz,_i,_j),simpleHS(swap_puz,_i-1,_j)),(c_depth+1,_i-1,_j,2,swap_puz)))
        if(_i != 2 and prev_move != 2):
            swap_puz = swapPuz(c_puz[0:],_i,_j,_i+1,_j)
            heapq.heappush(q,(c_depth+1+_sum_cost+checkCost(c_cost,simpleHS(c_puz,_i+1,_j),simpleHS(swap_puz,_i,_j),simpleHS(swap_puz,_i+1,_j)),(c_depth+1,_i+1,_j,1,swap_puz)))
        if(_j != 0 and prev_move != 3):
            swap_puz = swapPuz(c_puz[0:],_i,_j,_i,_j-1)
            heapq.heappush(q,(c_depth+1+_sum_cost+checkCost(c_cost,simpleHS(c_puz,_i,_j-1),simpleHS(swap_puz,_i,_j),simpleHS(swap_puz,_i,_j-1)),(c_depth+1,_i,_j-1,4,swap_puz)))
        if(_j != 2 and prev_move != 4):
            swap_puz = swapPuz(c_puz[0:],_i,_j,_i,_j+1)
            heapq.heappush(q,(c_depth+1+_sum_cost+checkCost(c_cost,simpleHS(c_puz,_i,_j+1),simpleHS(swap_puz,_i,_j),simpleHS(swap_puz,_i,_j+1)),(c_depth+1,_i,_j+1,3,swap_puz)))

def checkCost(c_cost,m_cost,c2_cost,m2_cost):
    return c2_cost - c_cost + m2_cost - m_cost
    

def sumCost(costs):
    value = 0
    for i in range(len(costs)):
        value += costs[i]
    return value

def simpleHS(c_puz,i,j):
    if(correctPuz[i*3+j] != c_puz[i*3+j]):
        return 1
    return 0

def HS(i,j,num):
    k = correctPuz.index(num)
    ki = (int)(k/3)
    kj = k - ki*3
    value = abs(_i-ki)+abs(_j-kj) 
    return value


def swapPuz(c_puz, i, j, i2,j2):
    c_puz[i2*3+j2],c_puz[i*3+j] = c_puz[i*3+j],c_puz[i2*3+j2]
    return c_puz

def puzzleCheck(c_puz):
    if(c_puz[8] == 0):
        for i in range(len(correctPuz)):
            if(correctPuz[i] != c_puz[i]):
                return False
        return True
    return False

correctPuz = [i+1 for i in range(9)]
correctPuz[8] = 0
puz = [0 for i in range(9)]

i_start = 0
j_start = 0

for i in range(3):
    puz[i*3],puz[i*3+1],puz[i*3+2] = map(int, input().split());
    if(puz[i*3] == 0):
        i_start,j_start = i,0
    elif(puz[i*3+1] == 0):
        i_start,j_start = i,1
    elif(puz[i*3+2] == 0):
        i_start,j_start = i,2

sumcost=0
for i in range(len(correctPuz)):
    if(correctPuz[i] != puz[i]):
        sumcost+=1

finding = 0
depth = 0
while True:
    if(finding == 1):
        break
    iterative(i_start,j_start)
    depth+=1