What the chess knight is doing...

The knight moves over the chess board in the way that is displayed on the left. If the knight happens to be on the field marked with an S, he can access any of the X-fields in the next step.

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Trouble for the knight

The idea of looking at a single the chess knight on the board goes all the way back to the 18th century, when Leonhard Euler, a swiss matematician, introduced a problem in Berlin, 1758. The problem, that is usually referred to as Springerproblem or The Knight's Tour, is given by

Starting with an empty chess board, is there a path that has a knight visit all the fields (black and white) of the board exactly one time?

The knight is paticularly interesting because of his strange way of moving. It would be much less interesting to try that with the queen, for example, since most attempts should lead to a solution. The knight in turn has a very limited way of moving that keeps the number of accessible fields per move below eight.

When Euler first thought of that problem, he imagined an 8x8 board, the regular chess board, while today's mathematical and computerized methods make us want to take a look at bigger boards, too.
But in order to find out about the chances of solving the problem, we have to take a closer look at it first.

Different flavours...

During the last almost 250 years, different ways of looking at the problem have evolved. It is remarkable, that even the smallest one among them took about 230 years to be sufficiently solved in a way that is scalable even for huge boards.
The problem of the Knight's Tours is often used as an example in graph theory, since it is easy to imagine and to try out but nevertheless a very big problem with many different flavours to emphasize on. It can be used to show the idea of symmetry, of directed and indirected tours/graphs and of paths and closed tours. Though it is a Hamiltonian problem, solutions can easily be found.
Since it was proven to have solutions for all boards >= 5x5 in the early 1990s, it has now become even more popular and commenly known and there were several attempts and applications in this field that reach from simple Java-Applets for demonstration and visualisation purposes up to parallel computation and neural networks (why ever...)

All under a general headline

The Knight's Tour represents a special case of one of the biggest problems in computer science: The Hamilton Cycle.
R.W. Hamilton was an irish (cheerz!) mathematician of the 19th century. The problem of the Hamilton Cycle describes the search for a path in a graph to connect all knodes so that they build a cycle, similar to "connect the dots" (Malen-nach-Zahlen) adding the slight difficulty that the numbers are missing and you have to find the right order yourself - and usually there are quite a few dots to connect... The Hamilton Cycles belong to the class of NP-complete problems and are therefore believed to be only solvable with an afford that grows exponentially with the size of the problem.

To find a Knight's Tour is by far more simple since the limitation to a chess board provides some benefits like indirectness, symmetry and (here we go:) recursion. In fact, finding a Knight's Tour is so "easy" that todays home computers do this for giant boards with some billions of fields in an almost unmeasurably short period of time. You can just wait for it...

Finding a Knight's Tour

The problem of finding a single solution for the Knight's Tour was solved in the early 1990s by a group of students as a project for the german scientific contest "Jugend forscht". Their algorithm finds a single solution on a chess board of any size (>=5x5) within an almost unmeasurably short period of time.
The basic idea is quite simply generic recursion. They cut the big boards into very small ones (5x5, 6x6, 6x7, 6x8, 7x7, 7x8, 8x8), for which a solution can easily be found, and then reconnected the small boards to fill the bigger (and really big) boards. That way they only had to find solutions for small boards that match certain criteria, like a given starting and ending position (easy enough) that allowed the knight to move from the end position on one board to the starting point of the next one.
Using this algorithm, they could prove that any board bigger than or equal to 5x5 has at least one solution, as all boards with longer sides can be split into boards with the sizes mentioned above (length N = a*6+b*7+c*8 for all N>17, N<=17 works anyway).
Now some of those solutions can be found in no-time. Normally there are quite a few to be found, so it's not that hard to find at least one.

Since this algorithm can be used for parallel computation, it is impossible (and unneccessary, BTW) to find a faster one.

Coming back to where we started

A variation of this problem makes the knight return to the starting point on the board. It's not really more afford to find a special solution like that since it only constraints the algorithm to end up on one of a certain number of fields, which is always possible if the knight actually can end up on those fields. The exceptions are given by the following idea:
A 5x5 board has 25 fields. On each move, the color of the field the knight is standing on changes because of his way of moving. So if the first field was white, the second one will be black and so will be every field the knight accesses with an even number. Therefore the 25th field, the last one he moves on, will be white again. On this board, the first position and the last one always have the same color, so there is no way for the knight to move from the last one directly back to the starting position. Therefore re-entrant paths (real cycles) are only possible on boards with an even number of fields.

All boards (>5x5) with an even number of fields provide at least one solution for a cycle. Boards of sizes 3xN or 4xN may vary.

The Knight's Paths

Well, I didn't like all those at leasts when I wrote this text, since it means that we do not actually know the exact number of solutions that can be found for a given board.
Now we are coming to the biggest problem of this little family: We are trying to count the number of paths the knight can move along in order to access every field of the board exactly one time. This really involves finding all solutions and stupidly counting them. To get an idea of the size of the problem, you may take a look at the following table, that gives the number of directed paths on different squared chess boards.

Boardsize Solutions Number of valid board configurations Max. number of board configurations
4x4 0 29976 2.69*10^7
5x5 1728 38010697 7.13*10^13
6x6 6637920 ? 1.3*10^17
7x7 ? ? 1.1*10^26
8x8 8,121,130,233,753,702,400 by (Löbbing,Wegener 1996) ? 4.5*10^36

Heuristics and estimates

The maximal number of field configurations given above allows to estimate the size of the problem since those have to be tested in order to find all solutions. This number is nevertheless much bigger than the number of allowed paths since the knight normally runs into dead ends long before he can actually finish his journey across the board. The number is easily calculated since we can count the number of accessible fields for each position on the board and multiply them. This results in the maximal number of paths that may allow the knight to access all fields, but still includes everything that whould lead to a dead end situation.
An example: The knight starts in the top-left corner of the chess board. From that position he can access two other fields, making it two possible paths. Selecting one of them the knight now can access five new fields and so on. To find the maximal number of paths we can now multply the number of accessible fields on all his ways: 2 * 5 * ...
Since the knight always has to access each field once, this is just the same as if we multiplied the number of successors for each field on the board. We may argue that we have to decrease all factors (but the first one) by 1 since the knight always comes from one of those fields, so there's always at least one less accessible, but since we search the solutions beginning on any of the fields on the board, any of them can be the first one. So we really end up multiplying all counters.
Examples for 5x5 and 6x6 are given below.


Some test calculations show an average depth for backtracking at about two third of the way (60-70%) if no optimizations are used. This leads to the following calculation:
The maximal branching factor of the problem (i.e. the maximal number of successors that have to be tested in the next iteration) is seven, since one field has at most eight successors one of which the Knight previously came from.
As it can be seen from the boards above, if the board grows, the middle part will fill with a successor count of eight while the two rows and columns on both sides stay basically the same. Since the middle part is a square, it will grow much faster than the borders that grow linearly, so for big boards, a branching factor of 7 will be a good assumption. That's why counting the Knight's Tours becomes such a big problem even if the board is only slightly growing.
Now we see that there are (NxN) valid configurations in the first iteration, not more than (NxN)*7 in the second, then (NxN)*7*7 and so on. That makes it (NxN)*7^(N*N-1) in the last iteration. The resulting exponential function is c(i) = (N*N) * 7^(i-1). We see that most of the possible configurations reside at the end of the paths, because of the high branching factor, and so we can integrate the function to see how many of them do. The integral of the function c is given by I(i) = (N*N)/ln(7) * 7^(i-1). We already said that after about two third of the path the knight has to turn back and do backtracking. For generalisation purposes I will name this ratio r in the following text. So now we take the integral of that part to have an approximation for the valid board configurations: Ptried = I(r*N*N) - I(1) = (N*N)/ln(7) * (7^(r*N*N-1) - 1) and the integral of the part that represents the invalid configurations since the knight had to return earlier on: Pbacktracked = I(N*N) - I(r*N*N) = (N*N)/ln(7) * (7^(N*N-1) - 7^(r*N*N-1)) We now can calculate the relation between the two: Ptried / Pbacktracked = (7^(r*N*N-1) - 1) / (7^(N*N-1) - 7^(r*N*N-1)) ~ 7^(r*N*N-1) / (7^(N*N-1) - 7^(r*N*N-1)) This is a constant for a given board. It gives us an approximate relation between the maximum number of board configurations and the number of valid ones and allows us to estimate the size of the problem more exactly.
So now here is a new table which gives the new approximation of the above mentioned boards.

Boardsize Solutions Number of valid board configurations New estimate between Max. number of board configurations
4x4 0 29976 (=3.00*10^4) 2.34*10^3 2.69*10^7
5x5 1728 38010697 (=3.80*10^7) 1.49*10^7 7.13*10^13
6x6 6637920 ? 1.3*10^17
7x7 ? ? 1.1*10^26
8x8 8,121,130,233,753,702,400 by (Löbbing,Wegener 1996) ? 4.5*10^36

I also wrote an essai about this estimate. The PDF file can be found here.


Since symmetry tends to decrease the number of field configurations that have to be tested by factors, it is usually one of the first ideas to implement. The following symmetries exist on any board:


Different areas with the same numbers are symmetric and therefore provide the same number of solutions. The areas marked with 2 only exist on boards with an odd height, the areas 3 only on boards with odd width. If both width and height are odd, area 4 has to be taken into account, too. On evenly sided boards there are only the areas marked with 1 one of which has to be tested.
On square boards, the areas 2 and 3 provide the same number of solutions.


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