Problem:
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15 Puzzle

The goal of the 15 puzzle problem is to complete pieces on $4 \times 4$ 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 15 as shown below.

1 2 3 4
6 7 8 0
5 10 11 12
9 13 14 15

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
9 10 11 12
13 14 15 0

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

Input

The $4 \times 4$ integers denoting the pieces or space are given.

Output

Print the fewest steps in a line.

Constraints

The given puzzle is solvable in at most 45 steps.

Sample Input

1 2 3 4
6 7 8 0
5 10 11 12
9 13 14 15

Sample Output

8

Solution in Python3 (judge verdict: Wrong Answer)

import heapq

def move(P,p):
    if p > 3:
        tmp = P[:]
        tmp[p],tmp[p-4] = tmp[p-4],tmp[p]
        tmpp = p - 4
        yield tmp,tmpp
    if p < 12:
        tmp = P[:]
        tmp[p],tmp[p+4] = tmp[p+4],tmp[p]
        tmpp = p + 4
        yield tmp,tmpp
    if p%4 > 0:
        tmp = P[:]
        tmp[p],tmp[p-1] = tmp[p-1],tmp[p]
        tmpp = p - 1
        yield tmp,tmpp
    if p%4 < 3:
        tmp = P[:]
        tmp[p],tmp[p+1] = tmp[p+1],tmp[p]
        tmpp = p + 1
        yield tmp,tmpp

def evaluate(P):
    mht = 0
    for i,j in enumerate(P):
        if i+1 == 0:
            a,b = 3,3
        else:
            a,b = i//4,i%4
        if j == 0:
            c,d = 3,3
        else:
            c,d = (j-1)//4,(j-1)%4
        mht += abs(a-c) + abs(b-d)
    return mht
        
A = []
for i in range(4):
        A+=[int(i) for i in input().split()]
dp = {str(A) : 1}
h = [(evaluate(A),A,0,A.index(0))]
heapq.heapify(h)
flag = True
while(flag):
    mht,tmp,count,p = heapq.heappop(h)
    for i,j in move(tmp,p):
        key = str(i)
        if key in dp:
            continue
        elif key == "[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 0]":
            ans = count + 1
            flag = False
        else:
            dp[key] = 1
            mht = evaluate(i)
            if count+mht//2 < 46:
                heapq.heappush(h,(mht*3//2+count,i,count+1,j))
if str(A) == "[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 0]":
    ans = 0
print(ans)