### Question

In this exercise, it explains that ties can occur in the classifications. Is it possible to avoid ties entirely?

### Answer

No, unfortunately, ties cannot be avoided entirely. To account for this fact, we implement ways to “break” a tie, like choosing one at random or several other methods.

In very few specific cases, they might be avoidable. For instance, for the movie classification, there are only two classifications: “good” or “bad”. If we choose an odd value for `k`

, then ties will never occur, since it cannot have exactly half “good” and half “bad” neighbors. One will have an odd number and the other, an even number, never causing a tie.

In general, for more than two classifications of a dataset, ties are almost impossible to avoid entirely. For example, say that there are three classifications: “A”, “B”, or “C”. No matter what value of `k`

, greater than 1, that we choose, ties can happen. If we choose an even `k`

, like 8, then it might end up as

`["A", "A", "A", "A", "B", "B", "B", "B"]`

with a tie between “A” and “B”.

Or, if we choose an odd `k`

, say 7, then it might end up as

`["A", "B", "B", "B", "C", "C", "C"]`

As you can see, we still get a tie, but this time with “B” and “C”.