Source: USATST-2015-P3
Category: Operations and Strategies

Problem:
A physicist encounters $2015$ atoms called usamons. Each usamon either has one electron or zero electrons, and the physicist can't tell the difference. The physicist's only tool is a diode. The physicist may connect the diode from any usamon $A$ to any other usamon $B$. (This connection is directed.) When she does so, if usamon $A$ has an electron and usamon $B$ does not, then the electron jumps from $A$ to $B$. In any other case, nothing happens. In addition, the physicist cannot tell whether an electron jumps during any given step. The physicist's goal is to isolate two usamons that she is 100 % sure are currently in the same state. Is there any series of diode usage that makes this possible?

Solution:
The answer is no.

Let $m = 2015$ and label the usamons $U_1, \dots, U_m$. Consider a family of $m+1$ initial configurations, which we will call models $M_0, M_1, \dots, M_m$. In model $M_k$ (for $0 \le k \le m$), the usamons $U_1, \dots, U_k$ are charged (each has one electron) and the remaining usamons $U_{k+1}, \dots, U_m$ are uncharged. Note that in this set of models, for any pair of distinct usamons, there is a model where they are in different states by construction. Specifically, if $a < b$, then in model $M_a$, $U_a$ is charged and $U_b$ is uncharged.

We can consider the physicist as acting on these $m+1$ models simultaneously. To be 100% sure that two usamons are in the same state, the physicist must be able to reach a state where some pair of usamons has the same charge across all possible models. Thus, reaching a state where a pair is identical across our $m+1$ models is a necessary condition for a winning strategy to exist.

We claim that any diode operation $U_i \to U_j$ results in the $m+1$ models being an isomorphic copy of the previous set. Suppose the current set of models is defined by some ordering $V_1, \dots, V_m$ of the usamons, where in model $M_k$, the first $k$ usamons $V_1, \dots, V_k$ are charged and the rest are uncharged. Consider a diode operation directed from $V_i$ to $V_j$:
- If $i < j$: 
  - In models $M_0, \dots, M_{i-1}$, neither $V_i$ nor $V_j$ is charged, so nothing happens.
  - In models $M_i, \dots, M_{j-1}$, $V_i$ is charged and $V_j$ is uncharged. The electron jumps from $V_i$ to $V_j$, meaning $V_i$ becomes uncharged and $V_j$ becomes charged.
  - In models $M_j, \dots, M_m$, both $V_i$ and $V_j$ are charged, so nothing happens.
  The net effect across all models is exactly the same as swapping the positions of $V_i$ and $V_j$ in the ordering. The new set of models is simply defined by the permuted ordering, which preserves the overall structure of the set of models.
- If $i > j$:
  - In models $M_0, \dots, M_{j-1}$, neither is charged, so nothing happens.
  - In models $M_j, \dots, M_{i-1}$, $V_j$ is charged and $V_i$ is uncharged. The diode allows an electron to jump from $V_i$ to $V_j$, but since $V_i$ is empty, nothing happens.
  - In models $M_i, \dots, M_m$, both are charged, so nothing happens.
  Thus, if $i > j$, the operation never does anything in any of the $m+1$ models.

By induction, after any sequence of diode operations, the set of $m+1$ possible states of the system is isomorphic to the initial set of prefix models, simply corresponding to a permuted ordering of the usamons. Because this structure is preserved, it remains true that for any pair of usamons, there is always some model where one is charged and the other is uncharged. Therefore, no pair of usamons can ever be guaranteed to be in the same state, and the physicist cannot achieve her goal.


Grading guidelines:
(Partial) 1. Introduced (explicitly) a finite family of candidate initial configurations/models, e.g. the “prefix” models with 0,1,2,…,2015 electrons in a fixed order, and observed that for any fixed pair of usamons there is a model in which they are in different states. 2. Showed that to be 100% sure two given usamons are equal, they must be equal in every configuration still consistent with the physicist’s observations (equivalently, across all tracked models, since no feedback is received from diode uses). 3. Proved a correct invariant/transition rule for these models under one diode operation, e.g. that applying a diode between positions i<j swaps the two positions in the prefix-order family (or otherwise preserves the “all prefixes of an ordering” structure), and/or that for i>j the operation does nothing on this family. (Almost) 1. Gave the prefix-model counterexample and correctly argued that every diode operation preserves the family up to relabeling (permuting the underlying ordering), but did not fully justify one of the two direction cases (i<j vs i>j) or did not clearly state why the family remains of the same form after the operation. 2. Proved the preservation up to permutation and concluded that no pair can ever be forced equal, but left a small logical gap connecting “no pair is equal in all models” to “physicist can never be 100% sure”, or did not explicitly argue that the absence of feedback means all tracked models remain possible. 3. Solution is almost complete, but made minor mistakes which are not negligible.