Knight's tour backtrack implementation choosing the step array

Question

So I came up with this implementation for solving knights tour for a 8*8 chess board. But seems like it is taking a long time running (so long that I have to stop it). But i

Answer 1

There are eight valid knight's moves in chess:

The two arrays list those eight moves.

The two versions of the code try the moves in different order. It so happens that one version comes across a valid solution much quicker than the other.

That's all there is to it.

Answer 2

Summary

The reason the commented-out dx/dy array works better than your initial array is because it performs the depth-first search for a solution in a different order -- an order that was chosen with a specific solution in mind and which therefore is able to find that solution relatively quickly.

Details

Depth-first search starts at the root of a tree and examines every path to a leaf. For example, a depth-first search of this tree would first examine the path that visits only a nodes (a -> a -> a) then it would backtrack slightly and examine a -> a -> b, then a -> a -> c, etc.

This can take a lot of time if the tree is large and there are no solutions that start by visiting a, because you have to waste a lot of time examining all of the paths that start with a before you can move on to better paths.

If you happen to know that there is a good solution that starts with d, you can speed things up a little by re-ordering the nodes of your tree such that you start by examining paths that begin with d:

Right off the bat you've removed 7/8ths of the work your program has to do, because you never have to bother searching for paths that start with something other than d! By choosing a good ordering for the rest of the nodes, you can get similar speedups.

You can see this happening if you look at the output of your program:

(0,7)  (1,5)  (3,4)  (1,3)  (0,1)  (2,0)  (4,1)  (6,0)  (7,2)  (5,3)  (7,4)
(6,2)  (7,0)  (5,1)  (4,3)  (3,1)  (5,0)  (7,1)  (5,2)  (7,3)  (6,1)  (4,0)
(3,2)  (4,4)  (2,3)  (0,2)  (1,0)  (2,2)  (3,0)  (1,1)  (0,3)  (2,4)  (1,2)
(0,4)  (1,6)  (3,7)  (2,5)  (3,3)  (5,4)  (6,6)  (4,5)  (6,4)  (7,6)  (5,5)
(4,7)  (2,6)  (0,5)  (1,7)  (3,6)  (5,7)  (6,5)  (7,7)  (5,6)  (3,5)  (1,4)
(0,6)  (2,7)  (4,6)  (6,7)  (7,5)  (6,3)  (4,2)  (2,1)  (0,0)

The first move (reading from the bottom) is from (0,0) to (2,1), which corresponds to dx=2 and dy=1 -- and sure enough, in the commented-out dx/dy list, that's the first possibility that is examined. In fact, the first three moves of this solution use dx=2 and dy=1, which effectively means you only have to search a tiny little sub-tree instead of the whole thing.