$\textbf{Solution}$

Three weighs are sufficient.

First, equally divide these balls into three groups of $4$ balls each. Then, weigh the first and second group. There are two possible outcomes:

$\underline{These\ two\ groups\ weigh\ the\ same}$

Then we know the outlier must be in the third group and all the balls in the first two groups are normal. Label the four balls in the third group as $A$, $B$, $C$, and $D$, respectively. Get a normal ball from the first two groups, call it $N$. Then weigh $A+B$ and $C+N$.

• If they are balanced, then we immediately know $D$ is the outlier.
• If they are not balanced, say, $A+B$ is heavier than $C+N$, then either $C$ is lighter or one of $A$ and $B$ is heavier. The next step is to weigh $A$ against $B$. If they are balanced, then $C$ is the outlier. Otherwise, the heavier between $A$ and $B$ is the outlier.
• If $A+B$ is lighter than $C+N$, then either $C$ is heavier or one of $A$ and $B$ is lighter. By the similar logic, it takes just another weigh, $A$ v.s. $B$, to find out the outlier.

$\underline{These\ two\ groups\ weigh\ differently}$

• If they weigh the same, then the outlier must be among $A_4$, $B_3$ and $B_4$. Furthermore, we know either $A_4$ is heavier or one of $B_3$ and $B_4$ is lighter. Therefore, another weigh between $B_3$ and $B_4$ is sufficient to identify the outlier.
• If $A_1+A_2+B_1$ is heavier than $A_3+B_2+N$, then we know either $A_1$ or $A_2$ is heavier than normal or $B_2$ is lighter than normal. Either case can be solved by measuring $A_1$ against $A_2$.
• Similarly, if  $A_1+A_2+B_1$ is lighter than $A_3+B_2+N$, then we know either $A_3$ is heavier, or $B_1$ is lighter. Either way, this can be determined by weighing $A_3$ against $N$.

Therefore, in total, three weighs are sufficient.

$\textbf{Note}$

Similar problems appear often in various tests and interviews. They are more challenging than # 4643. It can be shown that $n$ weighs are sufficient to find one outlier among up to $(3^n-3)/2$ objects.