Output #1
6
Let us represent a ball as $($Color$,$ Integer$)$. The initial situation is $(1,3)$, $(5,2)$, $(2,1)$, $(2,2)$, $(1,1)$. Here is a possible sequence of operations for Takahashi:
* Swap the $1$\-st ball (Color $1$) and $2$\-nd ball (Color $5$). Now the balls are arranged in the order $(5,2)$, $(1,3)$, $(2,1)$, $(2,2)$, $(1,1)$.
* Swap the $2$\-nd ball (Color $1$) and $3$\-rd ball (Color $2$). Now the balls are arranged in the order $(5,2)$, $(2,1)$, $(1,3)$, $(2,2)$, $(1,1)$.
* Swap the $3$\-rd ball (Color $1$) and $4$\-th ball (Color $2$). Now the balls are in the order $(5,2)$, $(2,1)$, $(2,2)$, $(1,3)$, $(1,1)$.
* Swap the $4$\-th ball (Color $1$) and $5$\-th ball (Color $1$). Now the balls are in the order $(5,2)$, $(2,1)$, $(2,2)$, $(1,1)$, $(1,3)$.
* Swap the $3$\-rd ball (Color $2$) and $4$\-th ball (Color $1$). Now the balls are in the order$(5,2)$, $(2,1)$, $(1,1)$, $(2,2)$, $(1,3)$.
* Swap the $1$\-st ball (Color $5$) and $2$\-nd ball (Color $2$). Now the balls are in the order $(2,1)$, $(5,2)$, $(1,1)$, $(2,2)$, $(1,3)$.
* Swap the $2$\-nd ball (Color $5$) and $3$\-rd ball (Color $1$). Now the balls are in the order $(2,1)$, $(1,1)$, $(5,2)$, $(2,2)$, $(1,3)$.
After the last operation, the numbers written on the balls are $1,1,2,2,3$ from left to right, which achieves Takahashi's objective.
The $1$\-st, $2$\-nd, $3$\-rd, $5$\-th, $6$\-th, and $7$\-th operations incur a cost of $1$ each, for a total of $6$, which is the minimum. Note that the $4$\-th operation does not incur a cost since the balls are both in Color $1$.