#### Answer: no

#### First Solution

Each step is reversible. In fact, each step is its own reverse, so we can reach all-white from a configuration if and only if we can reach that configuration from an all-white board.

There are $2^9$ possible colorings of the square. We shall prove that there are at most $2^6$ configurations from an all-white board.

There are 6 elementary steps, say $A_1, \cdots, A_6$. Each of these steps are commutative. So we can write a sequence of steps as $A_1^{a_1} A_2^{a_2} \cdots A_6^{a_6}$. Furthermore, $A_i^2$ is the identity step, so each $a_i$ can be replaced by either 0 or 1, depending on whether it was odd or even. Now, the number of reachable configurations is the number of different arrangements we can give to $a_i$s, and that is $2^6$, which completes our proof.

*Remark: the number of reachable configurations is in fact less than $2^6$ because some distinct arrangements might still arrive at the same configuration.*

#### Second Solution

Let $S$ be the number of black squares in the four corner cells. Then $S$ always changes by an even number. If we start with a configuration where $S$ is odd, we will never reach an all-white configuration where $S$ is even.

*Remark: this beautiful solution is due to Johan Gunardi, the winner of KBB3.*

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