A) Instead of requiring each Knight to be unobserved, suppose we relax the condition by allowing each Knight to be either unobserved or observed by just one other Knight. How many Knights can the 8-by-8 board accommodate now?

B) What if we require that each Knight be observed by

**A**

A) Still 32 (though with more possible configurations).

B) Again 32 — this time again with a configuration
unique up to board symmetries!

For (A), divide the board into four 4-by-4 squares. In each of these, divide the 16 squares into four knight circuits of length 4. At most two Knights can be put into each circuit if no Knight is to be observed by two or more others. Hence the 8-by-8 board can accommodate no more than 4·4·2=32 Knights.

Given part (A), clearly the answer to part (B) is at most 32. So, it is exactly 32, as long as there's at least one configuration of 32 Knights that works. Can you find this configuration? Can you prove its uniqueness?

(See my joint article “The Mathematical Knight” with Richard Stanley,
in the *Math. Intelligencer*, Vol. 25 (2003), #1, pages 22-34.)