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 sure are currently in the same state. Is there any series of diode usage that makes this possible? Proposed by Linus Hamilton
Problem
Source: USA December TST for the 56th IMO, by Linus Hamilton
Tags: AMC, USA(J)MO, USAMO, graph theory, combinatorics, TST
17.12.2014 23:39
We reformulate the problem in terms of subsets. We begin with some subset $S$ of $A$, and at every step, a diode application from $i,j$ swaps out $a_i$ for $a_j$ in $S$ if possible. Consider the sets $\{a_1\}, \{a_1,a_2\},\cdots \{a_1,\cdots a_n\}$. These all start as possible states, and at every step applying the diode on $i,j$ either: If $i>j$ nothing happens since $a_i$ is in a set only if $a_j$ is also. If $i<j$ then in all sets $\{a_1,\cdots a_k\}$ for $i\le k<j$, $a_i$ is switched out for $a_j$. But this situation is isomorphic to the last, simply with the labels $i,j$ switched. Thus after any number of applications, we have (some permutation of) these sets which could have come from some initial state. Thus we can never be sure that two elements are both in the set or not in the set.
18.12.2014 01:12
Very nice problem in my opinion, though I think it could have easily been swapped with #2. You can come up with the solution during the test as follows: first, you try to see what the physicist is able to do. You come up pretty naturally with the idea that she can connect all the usamons to a particular usamon $U_1$, meaning either $U_1$ has an electron or none of them do. Repeat the procedure to generate a sequence $U_2, U_3, \dots, U_{2015}$, with the property that there exists an integer $k$ for which $U_1, \dots, U_k$ have electrons and the other usamons do not. Once you make this reduction it's quite natural to realize why the task is impossible, and the proof goes along these lines.
18.12.2014 05:35
pi37, what do those sets stand for?
18.12.2014 05:40
$a_i$ represent each of the usamons, $A$ is the entire set of usamons, and $S$ is some subset which corresponds to which usamons have an electron at any given time.
21.12.2014 05:34
My solution from the test is posted below. Some things I did not wrote on the actual test are italicized, and I'm afraid about some points not being that clear. Hopefully the graders will understand everything. It's not the most well written thing ever, and I did not have much time left when I finished. It's different from the solution posted by pi37, but I guess fundamentally it's similar. Without further ado: Hmm OK so First if we make the thing undirected, it's applying something in $S_n$, OK. (This is explaining some motivation, my proof is sort of like breaking an element of $S_n$ into disjoint cycles) We have our $2015$ atoms. First, we find their $0$-leaders. So say some fixed set of moves $M$ gets two vertices $A$,$B$ to have the same number of electrons. The $0$-leader of atom V is defined like this: Put an electron on $V$, put electrons nowhere else, and apply $M$. The final destination of $V$ is the $0$-leader of V. (darn, I meant the final destination of the electron on $V$, I hope this is stupid typo is clear from context but I'm extremely worried it might obfuscate my proof... fear). Since every vertex has exactly one zero leader, some cycle has $V_{01}$,..$V_{0k}$ has $V_{0{i+1}}$ as the $0$-leader of $V_{0i}$ where $V_{01}$ is the $0$-leader of $V_{0k}$. Now, call the vertices $V_{01}$,..$V_{0k}$ Class $0$. For $i \ge 1$, define $i$-leadership on vertices $V$ not already in classes $0,1,..i-1$ as such: Put an electron on all class $0,1,..i-1$ vertices as well as on $V$, and nowhere else. Now apply $M$. So clearly $M$ cycles the atoms in cycle $0$ - if an electron starts off in $V_0$ of class $0$, then an electron ends on its $0$-leader (because extra electrons cannot hurt). So Class 0 stays filled. Because Class $0$ initially was filled, similarly electrons end up on all vertices in Class $1$. So Class $1$ is filled, and then electrons end up everywhere in Class $2$. OK I should clarify this part a little. If classes $0,1,..j-1$ are filled then consider a vertex (of course I meant "atoms," I crossed out vertices and wrote atoms earlier in my solution - I guess I wanted to do graph theory ) in class j. An electron starts there and classes $0,1,..j-1$ are full and so after $M$ an electron lands on the $j$-leader of that vertex - the other electrons introduced cannot prevent an electron from going somewhere - it can only help. So then class $j$ is filled afterward. So yeah. So if we have $M$ fixing classes $0,1,..i-1$ and taking an electron of $V$ to $V'$ (on my actual solution the prime in $V'$ is not as visible as I would like it to be, scary since earlier I made it look like the $k$ leader was the same vertex - ok then there's an asteresk) - (clarify- take $V$ to $V'$ means go from electrons on class $0,1,..j-1$ and $V$ to ones on $0,1,..j-1$ and $V'$- it might not be the same electron) when classes $0,1,..i-1$ are full then $V'$ is the $i$-leader of $V$ (yeah, I took way too long to define an i-leader, oops). So we can now get an $i$-leadership cycle eventually on vertices not assigned a class. This is class $i$, where each one's $i$-leader is the next vertex in the cycle. Eventually, we will partition all $2015$ atoms into different classes $0,1,...r-1$. We have two fixed vertices $A,B$ (again I said vertices, but earlier I did cross out vertices and write atoms, so ofc that's what I meant) that we want to have the same number of electrons. Say there are in different classes, $A$ in class $a$ and $B$ in class $b$, $a<b$. So fill classes $0,1,...a$ with electrons and apply $M$. So then $0,1,..a$ remain filled with electrons and there are none for elsewhere. So $A,B$ have different number of electrons, bad. If $A,B$ are in class $k$ then fill classes $0,..k-1$ with electrons, and put a vertex (of course I meant electron) on $V$ where $V$'s $k$-leader is $A$. After $M$, $A$ has an electron but not $B$. So it's impossible. Done.
21.12.2014 06:06
Here's my official solution, a.k.a. the slickest and least motivated way to do it. Instead of 0 or 1 electrons, allow usamons to have any real number. When the diode points from a real number $x$ to a real number $y$, it swaps them iff $x>y$. Think of it as a sorting machine. This is consistent with the behavior of the diode on 0 and 1. Now, give each usamon a distinct real number. Let the physicist do her stuff. She comes back with two usamons, one with $a$ and one with $b$. Since the numbers are distinct, $a \neq b$. WLOG $a<b$. Go back and say that the usamons with $\geq b$ had an electron, and the others didn't. Boom.
23.12.2014 19:53
Post #7 was also my solution; here's a way to motivate it. First, you do v_Enhance's ordering construction, at which point it becomes clear the answer is probably that you can't do it. Next, you realize it doesn't hurt for the physicist to start out doing the ordering construction. Reason: imagine some cosmic entity provided her the 2015 usamons and gave each one a number 1 to 2015 such that for $i \leq j$, usamon $i$ being charged implied usamon $j$ is charged. The physicist can choose to ignore that information, but it doesn't hurt to have it. So now we have numbers 1 to 2015 on every single usamon, and with the ordering construction we have so far, we know that if she compares usamon $i$ with usamon $j$, then having exactly $2015 - \max(i,j) + 1$ charged usamons means they are different. When you think about how to preserve this property, the idea on post #7 about having numbers trade places comes up, and you're home. In the writeup you can also realize you don't need to assume the use of the ordering construction anywhere.
06.01.2015 20:44
Sarah the physicist has an annoying friend Bob. Before Sarah applies any diodes, Bob gives her an ordered list of all the usamons. The list is increasing in # of electrons, and Bob tells Sarah this. Sarah can't resist the temptation and looks at the list (she loses no information by doing so). It seems Bob has helped Sarah, but he has actually destroyed her dreams. Assume the list is $U_1, ... U_n$ with $n=2015$. If Sarah applies diode $i \rightarrow j$ with $i<j$, nothing will change and she knows this. If she applies the diode $j \rightarrow i$, then $U_i$ and $U_j$ have swapped electrons. This means that she can make a new list where $U_i$ and $U_j$ are switched. No extra information is gained. So the information Bob gave her at the beginning is all Sarah will ever know! If $a < b$, Sarah can never prove that $U_a$ and $U_b$ are in the same state because it's possible that $U_1, ..., U_a$ are all electronless and $U_{a+1}, ..., U_n$ are all electronful. Bob wins! lokitos, I challenge your statement that your solution is the slickest.
15.01.2015 16:07
MellowMelon wrote: First, you do v_Enhance's ordering construction, at which point it becomes clear the answer is probably that you can't do it. Next, you realize it doesn't hurt for the physicist to start out doing the ordering construction. Reason: imagine some cosmic entity provided her the 2015 usamons and gave each one a number 1 to 2015 such that for $i \leq j$, usamon $i$ being charged implied usamon $j$ is charged. The physicist can choose to ignore that information, but it doesn't hurt to have it. . . . In the writeup you can also realize you don't need to assume the use of the ordering construction anywhere. Yeah, it was for these reasons I felt the problem was not very difficult, and I was surprised it was placed as #3. But I guess from how people did on this problem that perhaps this idea was not as direct as I personally found it.
15.01.2015 22:17
It's my impression that its placement at #3 was at least a nontrivial factor in low scores; in particular, if it was placed earlier, I think that people would have attempted it more, or have spent more time on it vs. spending a lot of time on 1 and 2. But it's one of those problems that seems obvious once you've found the solution, but perhaps less so before solving it.
15.01.2015 22:46
That's probably also true. I can testify from experience that if you have a problem that looks scary (which this one kind of does, since the solution is not especially standard), you can make it artificially harder just by placing it near the end of the test. USAMO 2012 #6 was another example of this.
02.03.2015 00:50
Will the January TST be posted?
16.04.2017 05:43
This feels really wrong and simple, but I can't find any mistakes... Suppose, at some moment, the physicist is sure that usamons $a$ and $b$ are currently in the same state. However, I could have made usamon $a$ and $b$ have $0$ and $1$ electrons, respectively, and the other usamons have $1$ electrons each, and "reversed" each diode usage to the start. Thus, you get a possible starting position with usamons $a$ and $b$ in different states, i.e. the physicist cannot be sure that usamons $a$ and $b$ are currently in the same state, as required. Edit: Whoops, it is wrong.
16.04.2017 05:55
@MathPanda1 I think it's impossible to have $x=1,y=0$ after connecting a diode $x\to y$; so if ,say, her last step was connecting some usamon $x\ne b$ to $a$, you can't reverse the move starting from $a=0,x=1$.
16.04.2017 05:57
Oh whoops, I see where I went wrong. Thanks!
16.04.2017 10:01
lokitos wrote: Instead of 0 or 1 electrons, allow usamons to have any real number. When the diode points from a real number $x$ to a real number $y$, it swaps them iff $x>y$. Think of it as a sorting machine. This is consistent with the behavior of the diode on 0 and 1.[\quote] How is it equivalent with the original exchange rules?
18.11.2018 06:02
This (especially Linus's solution) is related to the 0-1 principle.
03.12.2018 17:28
Suppose, for the sake of contradiction, there is an algorithm $A$ that makes it possible. One can think of $A$ as a written list of series of diode usage plus two fixed usemons $u_1,u_2$. Let $U$ be the set of all $2015$ usamons. Thus, $A$ takes as inputs a subset $E$ of $U$ (interpreted as the set of those usamons that have electrons) and gives as an output a set $A(E)$ (interpreted as the set of those usamons that have electrons after applying the list of operations). Additionally $A$ has the following property. $1)$ For any $E\subset U$, either $A(E)$ contains both $u_1,u_2$ or $A(E)$ contains neither of them. We exploit only two easy properties of the algorithm $A$: \begin{align*} &\text{2) } |A(E)|= |E| \text{ for any } E\subset U.\\ &\text{3) } E_1\subset E_2\subset U \implies A(E_1)\subset A(E_2). \end{align*}Property $2)$ just means no electron is lost. For property $3)$ it is easily checked that starting from $E_1$ and $E_2$ at every next step, applying a diode from the list of the algorithm, the inclusion still holds. Now, let $\emptyset=E_0\subset E_1\subset\dotso\subset E_{2015}=U$ be an increasing chain of subsets with $|E_i|=i,i=0,1,\dots,2015$. Denote $F_i=A(E_i)$. Then, by $2), 3)$ it follows $\emptyset=F_0\subset F_1\subset\dotso\subset F_{2015}=U$ and $|F_i|=i,i=0,1,\dots,2015$. Hence, there exists $i,1\leq i<2015$ such that one and only one of the usamons $u_1,u_2$ is in $F_i=A(E_i)$. It contradicts the basic property $1)$ of the algorithm $A$.
11.01.2019 10:04
Label the usamons $1,\ldots,n=2015$. Let $x_i=0$ if usamon $i$ has no electron, and $x_i=1$ otherwise. Lemma: If there exists $\sigma\in S_n$ such that the physicist's knowledge is exactly \[x_{\sigma(1)}\le x_{\sigma(2)}\le\cdots\le x_{\sigma(n)},\]then a firing of a diode does not change this fact (note that $\sigma$ may be different). Proof of Lemma: If $i\to j$ is fired where $\sigma(i)<\sigma(j)$, then the physicist knows the charge distribution won't change. However, if $i\to j$ is fired with $\sigma(i)>\sigma(j)$, then the physicist knows that the charges on $i$ and $j$ will swap. Thus, if $\sigma'\in S_n$ such that $\sigma'(j)=\sigma(i)$ and $\sigma'(i)=\sigma(j)$, and otherwise $\sigma'(x)=\sigma(x)$, then the physicist's information is of the form \[x_{\sigma'(1)}\le x_{\sigma'(2)}\le\cdots\le x_{\sigma'(n)}.\]Thus, the lemma is proven. $\blacksquare$ This actually implies that if the physicist has information \[x_{\sigma(1)}\le x_{\sigma(2)}\le\cdots\le x_{\sigma(n)},\]then she can never win, because whatever she does, she'll end up at the information \[x_{\sigma'(1)}\le x_{\sigma'(2)}\le\cdots\le x_{\sigma'(n)}.\]At this point if she presents usamons $i$ and $j$ with $\sigma'(i)<\sigma'(j)$, simply set $x_i=0$ and $x_j=1$, and the physicist loses. But the physicist starts with no information, but we showed that even if she knew the $\sigma\in S_n$ such that \[x_{\sigma(1)}\le x_{\sigma(2)}\le\cdots\le x_{\sigma(n)},\]then she still couldn't win. Therefore, with no information to start of with, she certainly cannot win. $\blacksquare$ Remarks Originally, I was trying to show that the physicist's knowledge at any time was a poset on the $x_i$s. This proved to be very tricky, but in the process of trying to do so, I was trying to split the poset into chains and doing a similar argument to the above. Eventually, I realized that the solution was starting me in the face if I simply took the knowledge poset to be a full chain. This is not hard to guess since it is very easy to see (and was in fact the first thing I noticed while solving this problem) that the physicist can attain the knowledge that is in the form of a chain (fire diodes $i\to j$ for $i<j$ in lexicographic order on $(i,j)$).
07.04.2021 19:40
dgrozev wrote: Suppose the last move of the list is $A\to C,$ where $C\ne A.$ Ohh the moves are directed. I should really start reading more carefully...
15.05.2021 20:04
dame dame
02.04.2022 03:56
Suppose that Alice labels the usamons $U_1,\ldots,U_{2015}$. Further, she is told that there exists some $0 \leq k \leq n$ such that for all $i\leq k$, $U_i$ has one electron, while for all $i>k$, $U_i$ has zero electrons, which only makes her goal easier. However, I claim that even with this, she still cannot isolate two usamons that are in the same state. Indeed, by case checking we can find that if she links $U_i \to U_j$ with $i>j$, then the state of both usamons doesn't change, while if $i<j$, then the state of the two usamons is swapped. Thus this operation gives no new information, as it's just equivalent to swapping the places of $U_i$ and $U_j$ in the list. Then if Sarah can isolate two usamons in the same state at some point, she can also do so before she makes any moves. But if these usamons are $U_a$ and $U_b$ with $a<b$, it is possible that $U_1,\ldots,U_a$ all have one electron and the rest have none: contradiction. Thus Sarah cannot complete her goal. $\blacksquare$ Remarks: kind of like USAMO 2016/6?
05.06.2022 22:38
No. Let the physicist's series of diodes usage be represented by the mapping $E\colon \{0,1\}^{2015} \to \{0,1\}^{2015}$. Then, consider \[E(0,0,\ldots,0), E(1,0,\ldots,0), E(1,1,0\ldots,0),\cdots, E(1,1,\ldots,1,0), E(1,1,\ldots,1,1)\]Let $\mathbb{1}_k$ be the vector consisting of $k$ ones and then $2015-k$ zeroes. Then, note that $E(\mathbb{1}_{k+1})$ is equal to $E(\mathbb{1}_k)$ with one 0 turned into a 1 due to the logic of the diode. Thus, if the physicist chooses usamons $A$ and $B$, at some point $E(\mathbb{1}_m)$ has a 1 for $A$ and 0 for $B$ since each swap happens at most once. (discrete IVT, same idea as in USAMO 2022/1) Thus, the physicist cannot guarantee that two usamons are in the same state because she cannot tell the difference between different $\mathbb{1}_k$ initializations. Remark: The idea here is to slowly add in more electrons, and the new electron gets mapped somewhere, which means that there's an ordering of usamons who have different levels of priority. Then, we fill to some point between the priority of A and B to make sure that they have different number of electrons.
05.01.2023 02:01
This is so janky and I absolutely love it $\boxed{\text{She cannot.}}$ If the physicist performs $2014\to 2015$, $2013\to 2014\to 2015$, $2012\to 2013\to 2014\to 2015$, \dots, $1\to 2\to 3\to \dots 2014\to 2015$. The the physicist know that the usamons are in one of the following forms (where the first number is the number of electrons on usamon $1$, and the last is the number of electrons on usamon $2015$): \[ 11111\dots 11, 01111\dots 11, 00111\dots 11, 00011\dots 11, \dots, 00000\dots 00 \]Notice that for any two usamons $i$ and $i+j$, there are $j$ of the above arrangements where they are in different states, and so she cannot conclude anything from this. Now, if she performs some series of diode usages on $i$ or $i+j$, it results in a permutation of one of the above arrangements which we can relabel to become one of the above, and she gains no more information.
08.09.2023 02:29
Consider each of $11...1, 011...1,\dots, 00\dots 0$ distributions of electrons; note that any swapping is just a permutation of this, since swapping usamon A to B with A>B does nothing, while A<B just swaps their spots. The configuration is invariant, so the physicist can't get anything from it.
12.11.2023 01:04
The answer is no. Label the usamons $U_1$ through $U_{2015}$, and say $U_i = 1$ if and only if usamon $i$ has an electron, and $U_i = 0$ otherwise. Consider the following $2015$ states: take states $1$, $2$, $\dots$, $2015$, where state $n$ consists of $U_1 = \cdots = U_n = 1$. $\textbf{Claim.}$ After any diode usage, the usamons can be relabeled in such a way that the original $2015$ states are returned. This is really easy to see, actually; consider laying out a $2015 \times 2015$ grid, with rows labeled $1$ through $2015$, and columns labeled $1$ through $2015$, where each row corresponds to state $i$, each column corresponding to usamon $i$. Each diode usage either does nothing, or swaps two columns. Thus it is clear. $\square$ Since these states have no pair of usamons which are the same state in all configurations, she cannot ever certainly isolate two usamons with the same state.
14.01.2024 23:53
Here is a solution that I haven't seen in this thread (maybe I've just missed it). Assume for contradiction that the physicist has a strategy. For each possible starting position assign a value of 0 if after the physicist did her strategy the two isolated usamons have zero electrons each and a value of 1 if the isolated usamons each have an electron. It's easy to prove that we have a discrete continuity - if we change the value of one usamon at the start, at most one of the two isolated usamons at the end of the process will change state. Therefore all possible starting states must have the same value, which is a contradiction as the state with all particles having an electron has a different value than the state with no particles having an electron.
15.01.2024 03:30
Solved with dolphinday. $\boxed{\text{Such a series of moves does not exist}}$ To see this begin by randomly labeling the usamons $\{u_1, u_2, \dots, u_{2015}\}$. Now consider the following algorithm. Iterate over all pairs $(u_i, u_j)$ with $i < j$, and apply the diode directing any electrons from $u_i \mapsto u_j$. Begin with $j = 2015$, as $i$ ranges from $2014$ to $1$ in that order. Then continue with $j = 2014$ as $i$ ranges over $2013$ to $1$ in that order and so on. Claim: There do not exist $x$ and $y$ with $x < y$ such that $u_x$ has an electron and $u_y$ does not. Proof. Note that in the phrasing of our algorithm, when we had $j = y$ and were iterating over $i$ in the range of $j - 1$ to $1$, if there were any electrons in the usamons $\{u_1, u_2, \dots, u_{j-1} \}$ they must have been sent to $u_j$. Hence if $u_i$ had an electron, at this step in our algorithm $u_j$ would take that electron. $\square$ Now note that the physicist's current information consists of the fact that the last $k$ usamons have an electron. Namely they are in one of the states, \begin{align*} 111 &\dots 111\\ 011 &\dots 111\\ 001 &\dots 111\\ &\vdots\\ 000 &\dots 011\\ 000 &\dots 001\\ 000 &\dots 000 \end{align*}where a $0$ and $1$ depict not having, or having an electron respectively. Now in this state if the physicist applies the diode to any two electrons $u_a \mapsto u_b$, she either, Gains no new information, if $u_a$ and $u_b$ are in the same state. Breaks the ordering if $a < b$ and she sends an electron from $u_b \mapsto u_a$. This effectively ``loses" information in the sense that the physicist can no longer state that all usamons $u_k$ with $k > \ell$ for some $\ell$ possess an electron. Alternatively, she may just renumber the usamons, sending her back to the same state, hence giving no new information. Thus the physicist will never be able to guarantee that she can isolate two usamons of the same state. $\blacksquare$ Remark: While group-solving this, I felt that the problem was extremely intuitive in the sense that you only really have one thing to do with the usamons, which is apply some sort of algorithm to sort them out. In fact, I felt that the entire difficulty was arriving at whether the answer was that the physicist could not make a valid series of moves. However even for this as dolphinday stated ``If the electrons jump the scientist has no way of telling", and hence cannot gain any actual, helpful information from any set of moves. On another note, $15$ minute solve, which is something I'm extremely happy with
04.04.2024 01:22
Non-constructive solution or something. FTSOC suppose not. Then the physicist has a set of fixed moves $F$ before saying what the two usamons are, say WLOG let them be $u_1$ and $u_2$. We construct a counter example to the physicist as follows. Take a starting configuration $C_0$ with no usamons which maps to a final configuration $F_0$. Then one at a time, place a new usamon in $C_i$ to get $C_{i+1}$. We can check that after each move, the configuration is the same but with a new usamon somewhere. In other words, $F_0 \subsetneq F_1 \dots$ when considered as sets. Then some $F_k$ has a usamon in exactly one of $\{u_1, u_2\}$. This is our counterexample.
19.05.2024 19:15
While solving this problem i felt like it could be represented as some logical structure, anyone have a solution using logic?
06.06.2024 11:23
The answer is no. Number the usamons $a_1, a_2, \dots, a_{2015}$. WLOG, assume that usamons are sorted so that all usamons with $0$ electrons are before usamons with $1$ electron. Let the number of usamons with $0$ electrons be $k$. Since the physicist has no way of figuring out how many electrons there are, we can freely pick $k$. If the diode is used from $a_i$ to $a_j$, then $a_i$ and $a_j$ are swapped if $i > j$, otherwise there is no change. Suppose that after some number of diode operations the physicist claims to have to usamons $a_m$ and $a_n$ that are in the same state, where $m > n$. Then we can set $k = n$, so $a_m$ has one electron and $a_n$ has zero electrons. Therefore, the physicist cannot be $100\%$ sure that two usamons are in the same state.
22.07.2024 19:45
Let the usamons be $u_1,u_2,\dots,u_{2015}$. Assume a diode is placed between $A$ and $B$. Let $a$ and $b$ denote whether or not $A$ and $B$ respectively have an electron before the connection. No matter which direction the diode goes in, we get usamons equivalent to $a\land b$ and $a\lor b$. So, assume that all connections go from a higher numbered diode to a lower numbered diode. Let $i<j$. Then, if we start with $u_1,\dots,u_i$ having electrons and all other usamons not having electrons, it will stay that way after every connection. Then, $u_i$ and $u_j$ will be in different states after any sequence of connections. So, the physicist can't guarantee $u_i$ and $u_j$ being the same.
25.08.2024 05:42
Note that we can order the atoms from least electrons to most electrons. However, after doing this we cannot attain more information, because we can simply reorder the list if we sent a usamon with greater than or equal to electrons to another usamon, we can simply reorder the list. Therefore the task is impossible$.\blacksquare$