I’m sure this information is available elsewhere, but I couldn’t find it with a quick google search — the two hardest 8-puzzle instances are:
8 6 7 2 5 4 3 . 1 6 4 7 8 5 . 3 2 1 |
Both require at least 31 moves to solve. All other instances can be solved in 30 moves or less. (Found using breadth-first graph search, starting at the goal state). Somewhat interestingly, they are not obviously symmetric, but differ only in the locations of the even pieces.
Mati
But which is the final state for those puzzles?
Cause there are two disconnected parts in the search space for this problem.
Thx!
Jason Wolfe
Good point. The post was referring to the standard version where the final state is:
1 2 3
4 5 6
7 8 .
mimighost
I got your blog from google.
This is indeed a very difficult puzzle, i wrote a python program and using the A* to solve it, and it takes a really long time.
So, my question is how long does it take for your program to find the answer? Is there any special optimization? I rsearch the web but can’t find many useful information.
Jason Wolfe
There are only (9 !) / 2 = 181,440 reachable states in the 8-puzzle, so you should be able to solve any instance pretty quickly (on the order of seconds or less) even using brute force, with a decently fast implementation. Repeated-state checking (i.e., a closed list) and proper data structures are essential, of course.
solnit
These two states are the same state really. Let’s take first position.
You can reflect this position about main diagonal.
Then you can rename pieces. More precisely, you can swap pieces 2-4, 3-7 and 6-8. It is because in goal state these pieces are symmetric.
Note that this is second hardest position.
Now try to do the same two steps for, say, the following position:
After reflection:
After renaming pieces 2-4, 3-7 and 6-8:
Position A can be solved in three moves: U L L (U means “move piece up”).
Position B also can be solved in three moves: L U U.
Moreover, solution to the position B can be obtained by replacing moves U,L,D,R with moves L,U,R,D respectively.
The following position is self-symmetric:
If you reflect this position and then rename tiles you will return to the same position.
Era Ahmadian
Hi Solnit
we are working on an Optimezd Program which is solved 8 puzzle problem with A* algorithm and with OOP methods in C# .
Our Program Can solve the 31-depth instance easily!
but i have a question of you?!
why u said The hardest eight-puzzle instances take 31 moves to solve?how can u prove it?!
we have of instances that they needs more moves for solving!!!
and the second question is the reflection instances are the same state with every goals position or just the standard one?
truly yours,Era
Nihal
For some reason my BFS algorithm got the first one in 27 states. Ive penned them down as a list.
[3, 4, 0, 6, 1, 8, 2, 5, 7]
[3, 4, 8, 6, 1, 0, 2, 5, 7]
[3, 4, 8, 6, 0, 1, 2, 5, 7]
[3, 0, 8, 6, 4, 1, 2, 5, 7]
[0, 3, 8, 6, 4, 1, 2, 5, 7]
[6, 3, 8, 0, 4, 1, 2, 5, 7]
[6, 3, 8, 2, 4, 1, 0, 5, 7]
[6, 3, 8, 2, 4, 1, 5, 0, 7]
[6, 3, 8, 2, 0, 1, 5, 4, 7]
[6, 3, 8, 2, 1, 0, 5, 4, 7]
[6, 3, 0, 2, 1, 8, 5, 4, 7]
[6, 0, 3, 2, 1, 8, 5, 4, 7]
[6, 1, 3, 2, 0, 8, 5, 4, 7]
[6, 1, 3, 0, 2, 8, 5, 4, 7]
[0, 1, 3, 6, 2, 8, 5, 4, 7]
[1, 0, 3, 6, 2, 8, 5, 4, 7]
[1, 2, 3, 6, 0, 8, 5, 4, 7]
[1, 2, 3, 0, 6, 8, 5, 4, 7]
[1, 2, 3, 5, 6, 8, 0, 4, 7]
[1, 2, 3, 5, 6, 8, 4, 0, 7]
[1, 2, 3, 5, 6, 8, 4, 7, 0]
[1, 2, 3, 5, 6, 0, 4, 7, 8]
[1, 2, 3, 5, 0, 6, 4, 7, 8]
[1, 2, 3, 0, 5, 6, 4, 7, 8]
[1, 2, 3, 4, 5, 6, 0, 7, 8]
[1, 2, 3, 4, 5, 6, 7, 0, 8]
[1, 2, 3, 4, 5, 6, 7, 8, 0]
27 states in solution
Nihal
sorry my bad. was checking something else -_-
agv
[[[8, 6, 7] [2, 5, 4] [3, 0, 1] -1]
[[8, 6, 7] [2, 0, 4] [3, 5, 1] 0]
[[8, 0, 7] [2, 6, 4] [3, 5, 1] 0]
[[8, 7, 0] [2, 6, 4] [3, 5, 1] 3]
[[8, 7, 4] [2, 6, 0] [3, 5, 1] 2]
[[8, 7, 4] [2, 0, 6] [3, 5, 1] 1]
[[8, 7, 4] [0, 2, 6] [3, 5, 1] 1]
[[8, 7, 4] [3, 2, 6] [0, 5, 1] 2]
[[8, 7, 4] [3, 2, 6] [5, 0, 1] 3]
[[8, 7, 4] [3, 2, 6] [5, 1, 0] 3]
[[8, 7, 4] [3, 2, 0] [5, 1, 6] 0]
[[8, 7, 4] [3, 0, 2] [5, 1, 6] 1]
[[8, 7, 4] [0, 3, 2] [5, 1, 6] 1]
[[0, 7, 4] [8, 3, 2] [5, 1, 6] 0]
[[7, 0, 4] [8, 3, 2] [5, 1, 6] 3]
[[7, 4, 0] [8, 3, 2] [5, 1, 6] 3]
[[7, 4, 2] [8, 3, 0] [5, 1, 6] 2]
[[7, 4, 2] [8, 0, 3] [5, 1, 6] 1]
[[7, 4, 2] [8, 1, 3] [5, 0, 6] 2]
[[7, 4, 2] [8, 1, 3] [0, 5, 6] 1]
[[7, 4, 2] [0, 1, 3] [8, 5, 6] 0]
[[0, 4, 2] [7, 1, 3] [8, 5, 6] 0]
[[4, 0, 2] [7, 1, 3] [8, 5, 6] 3]
[[4, 1, 2] [7, 0, 3] [8, 5, 6] 2]
[[4, 1, 2] [7, 5, 3] [8, 0, 6] 2]
[[4, 1, 2] [7, 5, 3] [0, 8, 6] 1]
[[4, 1, 2] [0, 5, 3] [7, 8, 6] 0]
[[0, 1, 2] [4, 5, 3] [7, 8, 6] 0]
[[1, 0, 2] [4, 5, 3] [7, 8, 6] 3]
[[1, 2, 0] [4, 5, 3] [7, 8, 6] 3]
[[1, 2, 3] [4, 5, 0] [7, 8, 6] 2]
[[1, 2, 3] [4, 5, 6] [7, 8, 0] 2]]
agv
[[[6, 4, 7] [8, 5, 0] [3, 2, 1] -1]
[[6, 4, 0] [8, 5, 7] [3, 2, 1] 0]
[[6, 0, 4] [8, 5, 7] [3, 2, 1] 1]
[[6, 5, 4] [8, 0, 7] [3, 2, 1] 2]
[[6, 5, 4] [0, 8, 7] [3, 2, 1] 1]
[[6, 5, 4] [3, 8, 7] [0, 2, 1] 2]
[[6, 5, 4] [3, 8, 7] [2, 0, 1] 3]
[[6, 5, 4] [3, 0, 7] [2, 8, 1] 0]
[[6, 5, 4] [3, 7, 0] [2, 8, 1] 3]
[[6, 5, 4] [3, 7, 1] [2, 8, 0] 2]
[[6, 5, 4] [3, 7, 1] [2, 0, 8] 1]
[[6, 5, 4] [3, 0, 1] [2, 7, 8] 0]
[[6, 0, 4] [3, 5, 1] [2, 7, 8] 0]
[[0, 6, 4] [3, 5, 1] [2, 7, 8] 1]
[[3, 6, 4] [0, 5, 1] [2, 7, 8] 2]
[[3, 6, 4] [2, 5, 1] [0, 7, 8] 2]
[[3, 6, 4] [2, 5, 1] [7, 0, 8] 3]
[[3, 6, 4] [2, 0, 1] [7, 5, 8] 0]
[[3, 6, 4] [2, 1, 0] [7, 5, 8] 3]
[[3, 6, 0] [2, 1, 4] [7, 5, 8] 0]
[[3, 0, 6] [2, 1, 4] [7, 5, 8] 1]
[[0, 3, 6] [2, 1, 4] [7, 5, 8] 1]
[[2, 3, 6] [0, 1, 4] [7, 5, 8] 2]
[[2, 3, 6] [1, 0, 4] [7, 5, 8] 3]
[[2, 3, 6] [1, 4, 0] [7, 5, 8] 3]
[[2, 3, 0] [1, 4, 6] [7, 5, 8] 0]
[[2, 0, 3] [1, 4, 6] [7, 5, 8] 1]
[[0, 2, 3] [1, 4, 6] [7, 5, 8] 1]
[[1, 2, 3] [0, 4, 6] [7, 5, 8] 2]
[[1, 2, 3] [4, 0, 6] [7, 5, 8] 3]
[[1, 2, 3] [4, 5, 6] [7, 0, 8] 2]
[[1, 2, 3] [4, 5, 6] [7, 8, 0] 3]]
با الگوریتم پس گرد 🙂
Carlos
How many nodes expand your program?,
since I am trying to run it. however it couldn’t complete 🙁
Regards!
THiago
I solved it, but my algorithm took 32 moves to do it and I don’t know why yet.
mimighost you have to remove from your frontier when the state is already visited to get better performance.
Nicholas Estrada
I’m using an iterative Priority Queue with the priority being based on the number of tiles misplaced and mine gets it in 31 moves in significantly less than a second.
swap
[] + []=8
+ +
[] – []=6
= =
13 8
Nathan Burgess
The first puzzle can be solved in 27 moves and the second one in 25 moves. Perhaps your BFS algorithm was off a bit. I’ve tested this using BFS, DFS, and A*.
Adam, Seoul, Korea
In 8 Puzzle, the hardest 2 cases take 31 moves to solve.
In 8 Puzzle, the second hardest 232 cases take 30 moves to solve.
In 8 Puzzle, the third hardest 800 cases take 29 moves to solve.
visit here : http://blog.naver.com/simadam222
Lucky
Which heuristics did you use?
I’m using Manhattan distance and my code solves the puzzle in 45 moves. Is it possible to use both hamming and Manhattan distance? If yes: how? Do I add them together or subtract?
Your help will be greatly appreciated.
Thierry Parmentelat
as for me I find these 2 boards to be at distances 23 and 25 resp.
00-> [8 6 7 – 2 5 4 – 3 0 1]
01 -> [8 6 7 – 2 0 4 – 3 5 1] (Up)
02 -> [8 0 7 – 2 6 4 – 3 5 1] (Up)
03 -> [0 8 7 – 2 6 4 – 3 5 1] (Left)
04 -> [2 8 7 – 0 6 4 – 3 5 1] (Down)
05 -> [2 8 0 – 7 6 4 – 3 5 1] (Up)
06 -> [2 8 4 – 7 6 0 – 3 5 1] (Down)
07 -> [2 8 4 – 7 6 1 – 3 5 0] (Down)
08 -> [2 8 4 – 7 6 1 – 3 0 5] (Left)
09 -> [2 8 4 – 7 0 1 – 3 6 5] (Up)
10 -> [2 8 4 – 7 1 0 – 3 6 5] (Right)
11 -> [2 8 4 – 7 1 3 – 0 6 5] (Down)
12 -> [2 8 4 – 0 1 3 – 7 6 5] (Up)
13 -> [2 8 4 – 1 0 3 – 7 6 5] (Right)
14 -> [2 0 4 – 1 8 3 – 7 6 5] (Up)
15 -> [0 2 4 – 1 8 3 – 7 6 5] (Left)
16 -> [1 2 4 – 0 8 3 – 7 6 5] (Down)
17 -> [1 2 0 – 4 8 3 – 7 6 5] (Up)
18 -> [1 2 3 – 4 8 0 – 7 6 5] (Down)
19 -> [1 2 3 – 4 8 5 – 7 6 0] (Down)
20 -> [1 2 3 – 4 8 5 – 7 0 6] (Left)
21 -> [1 2 3 – 4 0 5 – 7 8 6] (Up)
22 -> [1 2 3 – 4 5 0 – 7 8 6] (Right)
23 -> [1 2 3 – 4 5 6 – 7 8 0] (Down)
00-> [6 4 7 – 8 5 0 – 3 2 1]
01 -> [6 4 7 – 8 5 1 – 3 2 0] (Down)
02 -> [6 4 7 – 8 5 1 – 3 0 2] (Left)
03 -> [6 4 7 – 8 0 1 – 3 5 2] (Up)
04 -> [6 4 7 – 8 1 0 – 3 5 2] (Right)
05 -> [6 4 7 – 8 1 3 – 0 5 2] (Down)
06 -> [6 4 7 – 0 1 3 – 8 5 2] (Up)
07 -> [0 4 7 – 6 1 3 – 8 5 2] (Up)
08 -> [4 0 7 – 6 1 3 – 8 5 2] (Right)
09 -> [4 1 7 – 6 0 3 – 8 5 2] (Down)
10 -> [4 1 7 – 0 6 3 – 8 5 2] (Left)
11 -> [4 1 0 – 7 6 3 – 8 5 2] (Up)
12 -> [4 1 3 – 7 6 0 – 8 5 2] (Down)
13 -> [4 1 3 – 7 6 8 – 0 5 2] (Down)
14 -> [4 1 3 – 7 6 8 – 5 0 2] (Right)
15 -> [4 1 3 – 7 6 8 – 5 2 0] (Right)
16 -> [4 1 3 – 7 6 0 – 5 2 8] (Up)
17 -> [4 1 3 – 7 0 6 – 5 2 8] (Left)
18 -> [4 1 3 – 7 2 6 – 5 0 8] (Down)
19 -> [4 1 3 – 7 2 6 – 0 5 8] (Left)
20 -> [4 1 3 – 0 2 6 – 7 5 8] (Up)
21 -> [0 1 3 – 4 2 6 – 7 5 8] (Up)
22 -> [1 0 3 – 4 2 6 – 7 5 8] (Right)
23 -> [1 2 3 – 4 0 6 – 7 5 8] (Down)
24 -> [1 2 3 – 4 5 6 – 7 0 8] (Down)
25 -> [1 2 3 – 4 5 6 – 7 8 0] (Right)
now this is my first ever program written in Rust so pardon me if there are remaining bugs in there
for what it’s worth – and provided that it is bug-free of course – my program computes all the shortest distances to the 9!/2 reachable boards in 2.2s on a macbook pro