Knights and liars around a table

Constraints.
- A knight tells the truth, so both his neighbours are liars: no two knights are adjacent.
- A liar's statement is false, so at least one of his neighbours is a knight: no three consecutive liars.

Conversely, these two conditions are sufficient for consistency.

Counting. There is at least one knight. Let there be $k\ge1$ knights. Going around the circle, the knights split the remaining $10-k$ people into $k$ groups of consecutive liars. Each group has $1$ liar or $2$ liars (not $0$: knights are not adjacent; not $3$ or more: no three consecutive liars). Hence
$$k\le10-k\le2k\ \Longrightarrow\ \frac{10}3\le k\le5\ \Longrightarrow\ k\in\{4,5\}.$$

Both values occur.
- $k=5$: alternate knight, liar, knight, liar, ...
- $k=4$: groups of liars of sizes $2,2,1,1$ between the four knights ($4+2+2+1+1=10$).

Answer. $4$ or $5$ knights.

Check. A computer search over all $2^{10}$ assignments finds exactly the solutions with $4$ or $5$ knights.

Lesson. Turn each statement into a local constraint, then count blocks between the special elements.