We already know that 7x7 board cannot have a closed Knight's tour, and it cannot start or end at a white square if R1C1 is black. But our knowledge about 7x7 Knight's tour is still limited. So here is the final piece:

Construct exactly three Knight's tours on a 7x7 board, so that their six endpoints cover all the distinct black squares up to rotation and symmetry. This will prove the theorem "One can start a 7x7 Knight's tour from any black square" by rotation, reflection, and reversing the order from the three presented tours.

For reference, here is the image showing the six distinct endpoints for the Knight's tours:


2 Answers 2


Helper picture:

enter image description here

Tour 1 with endpoints type 1 and 6:


Tour 2 with endpoints type 2 and 5:


Tour 3 with endpoints type 3 and 4:

21-26,x,27-42,7 43-48,1-6,27-39,8-20,13:

Strategy used:

Create 3 different size 48 closed loops first

Closed loop 1: (type 6 square missing)
1 Create symmetric closed loops; ignore the middle for now (relatively simple)
2 Find connection points between the yellow-orange, red and purple loop (see bigger numbers in the drawing) to make one size 48 closed loop.

Closed loop 2 (type 5 square missing, see blue arrows):
3 Connect the missing (centre) point into the loop, 26-x-40 in the drawing
4a Note that the now loose part can be connected to 2 square adjacent to the center X
4b Note that we can connect the loose part to 6 and 8 , making 7 the lone missing square

Closed loop 3 (type 3 csquare missing, see purple arrows):
5 We can simply swap 12,13,14 to 12,7,14 to get the third closed loop

Break the 3 loops at the right spots to add the missing points secondly

13 and 21 can be endpoints if we break 20-21
7 and 43 can be endpoints if we break 42-43
X and 48 can be endpoints if we break 48-1


enter image description here This is based on the templates in A,B and C. Each of these can be rotated by 90°,180°,270° yielding disjoint paths which can be linked together. Doing this with A covers everything except the center and connecting the end point with the center gives a solution for 4-6.
Similarly, B and C plus rotations together cover everything except the center. B can also be linked into two pairs and these can be joined at the center (E). Joining this up with C and its rotations (G) gives solution F for 2-5. Finally, two simple modifications (H,J) to this solution provide solution I for 1-3


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