How to solve this puzzle by invariance principle?

Question

I have the following puzzle-

You initially have $15$ $\color{red}{\text{red}}$ and $12$ $\color{brown}{\text{yellow}}$ balls.You can do one of the two things with the balls-

1. $\text{Swap}-$You can swap the balls.This means if you start with $x$ red and $y$ yellow balls,after the transformation you have $y$ red and $x$ yellow balls.

2. $\text{Exchange}-$You can exchange $3$ red balls for $2$ yellow balls provided that you have atleast $3$ red balls.

   • Now,if possible after how many transformations (you can use either of the rules mentioned above as $1$ or $2$ any number of times and in any order) will you have $5$ red balls and $5$ yellow balls?

   • Also,what is the total number of combination of the number of balls you can have?

How to solve this with invariance preferably.If not possible other methods are welcome too!

Thanks!!

Answer 1

After each exchange the parity between red and yellow changes (from same parity to different parity, and vice versa). Moreover, each exchange reduces the total number of balls by $1$.

At the start, we have $27$ balls and different parities. If the final state has a total of $k$ balls, we will need to remove a total of $27-k$ balls; in other words, we will need to perform $27-k$ exchanges. If a state has $k$ balls, with $k_r$ red balls and $k_y$ yellow, then at the end one of the following must hold

• $k$ is even $\iff$ $27-k$ is odd $\implies$ $k_r$ and $k_y$ are the same parity
• $k$ is odd $\iff$ $27-k$ is even $\implies$ $k_r$ and $k_y$ are different parities

Now, if $k_r$ and $k_y$ satisfy the above relations, is the final state achievable?

Observe that a state $(k_r,k_y)$ is reachable if and only if the state $(k_y,k_r)$ is reachable, so it's fair to assume without loss of generality that $k_r\geq k_y$. For any given configuration, let $d=k_r-k_y$, so that in the beginning we have $d=3$.

An exchange makes $d\mapsto d-5$ and a swap makes $d\mapsto -d$. In particular, $d\pmod5$ is either $3$ or $-3\equiv 2$, so it can never be $0$. The state with $k_r=k_y=5$ is hence not reachable.