3. A physicist discovered 2015 atoms called "American Fu," each of which either has an electron or does not have an electron. The physicist cannot distinguish between the two. The only tool the physicist has is a diode, which can connect any American Fu \( A \) with any other American Fu \( B \) (this connection is directed). When he does this, if American Fu \( A \) has an electron and American Fu \( B \) does not have an electron, the electron will quickly transfer from American Fu \( A \) to American Fu \( B \). For other cases, there is no change. Additionally, the physicist cannot determine at any step whether an electron has moved. The physicist's goal is to isolate two American Fu such that he can be 100% certain that these two American Fu are in the same state. Question: Can the physicist achieve this goal using several diodes?

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3. Cannot.

Let the American soap be $U_{1}, U_{2}, \cdots, U_{m}(m=2015)$.
Consider the following pattern $M_{k}$: for some integer $k(0 \leqslant k \leqslant m)$, $U_{1}, U_{2}, \cdots, U_{k}$ all have electrons, and other American soaps do not have electrons. Then, for any two different patterns, they have different states.

Consider the physicist simultaneously operating on these $m+1$ patterns and attempting to achieve a state: in all patterns, there exist two American soaps with the same state (this is a necessary condition for winning the existence strategy).

Thus, in these $m+1$ patterns, any operation of the diode from $U_{i} \rightarrow U_{j}$ is an isomorphic copy of the set before the operation. If $i=j$, the operation does not change anything. Therefore, the physicist cannot achieve the goal.