Solution Outline: (1). Prove that in any sequence in $\mathcal{A}$, we must not have a player with $2$ draws. (Otherwise, the matches outcome is not unique.) (2). Consider the player with the maximum number of points. Prove that this player must not be defeated by any other player. (Hence, this player must win all his matches or draw against one other player which has exactly the same point as him ($2N - 3$ points)). (3). From (2), we deduce that the number of sequences $\mathcal{A}$ such that the outcome of all matches is uniquely determined is $F_{n+1}$ where $F_{i}$ is the $i$-th fibonacci number with $F_1 = 1, F_2 = 2,$ and $F_n = F_{n-1} + F_{n-2}$ $\forall n \geq 3$.

Question: what if 3 points is awarded for each win, i.e., Quote: $N$ teams take part in a league. Every team plays every other team exactly once during the league, and receives 3 points for each win, 1 point for each draw, and 0 points for each loss. At the end of the league, the sequence of total points in descending order $\mathcal{A} = (a_1 \ge a_2 \ge \cdots \ge a_N )$ is known, as well as which team obtained which score. Find the number of sequences $\mathcal{A}$ such that the outcome of all matches is uniquely determined by this information.

Call the full information of all the matches a tournament. For any tournament $T$, let $d(T)$ denote the final score distribution, along with which teams received which score. We say a tournament $T$ is good if there is no other tournament $T'$ such that $d(T)=d(T')$. Suppose $T$ is a good tournament. We then have the following lemmas. Lemma 1: There is no directed cycle of wins in $T$, and there is no cycle of draws. Proof: We may interchange cycles of wins with cycles of draws while leaving $d(T)$ unchanged, which violates the goodness of $T$. $\blacksquare$ Thus, there is a ranking $1,\ldots,N$ (not necessarily unique) of the players such that $i<j$ means $i$ beats $j$ or $i$ draws $j$. From now on, we refer to players by their ranking. Lemma 2: If $1$ beats $2$, then $1$ beats everyone. Proof: Say $1$ instead draws $x$ for some $x\ge 3$. If $2$ beats $x$, then we may replace \[\{1\to 2,2\to x,1\sim x\}\iff\{1\sim 2,2\sim x,1\to x\}\]and leave the score distribution unchanged. Here $i\to j$ means $i$ beats $j$ and $i\sim j$ means $i$ draws $j$. This violates the goodness of $T$, so we have that $2$ also draws $x$. Note that we may also replace \[\{1\to 2,2\sim x,1\sim x\}\iff\{1\sim 2,x\to 2,1\to x\}\]and leave the score unchanged, so again this violates the goodness of $T$. Therefore, $1$ beats all $x$ for $x\ge 3$, as desired. $\blacksquare$ Lemma 3: If $1$ draws $2$, then $1$ and $2$ both beat all $x\ge 3$. Proof: First, suppose that $2$ draws $x$ for some $x\ge 3$. Then, $1$ beats $x$ so as to not create a cycle of draws. We may now replace \[\{1\sim 2, 2\sim x,1\to x\}\iff\{1\to 2,2\to x,1\sim x\}\]and leave the score unchanged. This violates goodness of $T$, so we must have that $2$ beats $x$ for all $x\ge 3$. Now suppose $1$ draws $x$ for $x\ge 3$. From the above, we see that $2$ beats $x$. But now we may replace \[\{1\sim 2, 2\to x, 1\sim x\}\iff\{2\to 1,2\sim x,1\to x\}\]and leave the score unchanged. This violates goodness of $T$, so we have that $1$ beats $x$ for all $x\ge 3$, as desired. $\blacksquare$ Now suppose that a good tournament $T$ generates the score sequence $\mathcal{A} = (a_1 \ge a_2 \ge \cdots \ge a_N )$. Lemma 4: Either $a_1=2N-2$ or $a_1=a_2=2N-3$. Proof: We have two cases. Case 1: Suppose $1$ beats $2$. By lemma 2, we see that $1$ beats everyone, so they have a score of $2N-2$. Everyone else has at least one loss, so their score is at most $2N-4$. Thus, $a_1=2N-2$. Case 2: Suppose $1$ draws $2$. By lemma $3$, we see that $1$ and $2$ beat everyone else, so they both have score $2N-3$. Everyone else has at least two losses, so their score is at most $2(N-3)=2N-6$, so $a_1=a_2=2N-3$. $\blacksquare$ We are now ready to count the number of sequences that generate a unique tournament. Indeed, let $f(N)$ be the number of such sequences. If $a_1=2N-2$, then we know that the person who got that score beat everyone. Now, $(a_2\ge \cdots\ge a_N)$ is the exact score report for the rest of the players, so we can determine all the information if and only if $(a_2\ge\cdots\ge a_N)$ generates a unique tournament. So in this case, there are exactly $f(N-1)$ such sequences. If $a_1=a_2=2N-3$, then we know there are two players who both have $N-2$ wins and one draw. As they have no losses, we see that they must have in fact drawn each other, and beat everyone else. Then, $(a_3\ge \cdots\ge a_N)$ is the exact score report for the remaining $N-2$ players, so as before, there are $f(N-2)$ sequences from this case. Therefore, we have $f(N)=f(N-1)+f(N-2)$, as well as $f(1)=1$ and $f(2)=2$. Thus, $f(N)=\boxed{F_{n+1}}$.

I claim that the answer is $F_{n + 1}$ where $F_n$ denote the $n$th Fibonacci number: $F_1 = F_2 = 1, F_{i + 2} = F_{i + 1} + F_i, \forall i \ge 1$. I'll change the rule of the game a bit: Tsukuyomi wrote: $N$ teams take part in a league. Every team plays every other team exactly once during the league, and receives 1 point for each win, 0 point for each draw, and -1 points for each loss. At the end of the league, the sequence of total points in descending order $\mathcal{A} = (a_1 \ge a_2 \ge \cdots \ge a_N )$ is known, as well as which team obtained which score. Find the number of sequences $\mathcal{A}$ such that the outcome of all matches is uniquely determined by this information. Proposed by Dominic Yeo, United Kingdom. We'll start off with the following lemma. First, let the number of sets satisfying the criteria of the problem be $A_n$. We'll prove $A_n = F_{n + 1}$ Interpret this in the obvious directed graph representation: undirected when there's a draw and pointing to the winner otherwise. $\textbf{Lemma 01.}$ There aren't any cycle of draws. $\textit{Proof.}$ Suppose that there exists 3 players such that $A-B, B-C, C-A$ are all draw. Then it is not unique because we can have another configuration: $\{ (A,B), (B,C), (C,A) \} = \{ (1,-1), (1,-1),(1,-1) \} = \{ (0,0), (0,0), (0,0) \}$. $\textbf{Lemma 02.}$ There doesn't exist any player that has two draws. $\textit{Proof.}$ This is a stronger claim of lemma 01. Now suppose $A$ draws with $B$ and $C$, and WLOG $B$ wins $C$. Then \[ \{ (A,B), (B,C), (C,A) \} = \{ (0,0), (0,0), (1,-1) \} = \{ (1,-1), (1,-1), (0,0) \} \] Now, consider the player with the maximal number of points. Let it be $A$. Suppose that $A$ loses to $B$ in a match. By Lemma 01, we may not have any directed cycle of triangles. Now, since $A$ loses to $B$, any players that $A$ beats must be beaten by $B$ as well, or there exists a directed cycle. Now, if any players $D$ that has a draw with $A$, then $B$ has to either draw or wins from $D$. Otherwise, this is not unique. \[ \{ (A,B), (B,D), (D,A) \} = \{ (-1,1), (-1,1), (0,0) \} = \{ (0,0), (0,0), (-1,1) \} \]So $B$ must have more points than $A$, which contradicts the maximality of $A$. Therefore, $A$ either beats all $n - 1$ other participants or $n - 2$ participants and leaving 1 draw. If $A$ beats all participants, any match paired with $A$ has an obvious outcome. Therefore, erasing $A$ from the list won't affect anything. (Basically just erase the score $n - 1$ immediately, since it must be certain that it beats any other player). Therefore, we just have to deal with the remaining $n - 1$ participants, giving us $A_{n - 1}$ ways. If $A$ beats $n - 2$ participants, leaving one draw, let say with $B$, We'll prove that $A$ and $B$ must have the same score. Suppose otherwise, if $B$ lose to $C$, then \[ \{ (A,B),(B,C), (C,A) \} = \{ (0,0), (-1,1), (-1,1) \} = \{ (1,-1), (0,0), (0,0) \} \]If $B$ draws to $C$, then \[ \{ (A,B),(B,C),(C,A) \} = \{ (0,0), (0,0), (-1,1) \} = \{ (1,-1), (1,-1), (0,0) \} \]Then $A$ and $B$ must have the same score of both $n - 2$. We could then rule out both of them that have $n - 2$ as a score and deal with the remaining $n - 2$ participants. This gives us $A_{n - 2}$ ways. Therefore, we establish \[ A_n = A_{n - 1} + A_{n - 2} \]A quick induction and bash of small case yields the result.

This is a beautiful problem We claim that the answer is $F_{n+1}$ , where $F_n$ denotes the $n$-th Fibonacci Number. Let the desired number be $f(n)$. For two teams $x,y$ we will write $x\rightarrow y$ if $x$ defeats $y$ and $x\leftrightarrow y$ if $x$ draws with $y$. Call the sequence of scores on a tournament its leaderboard. Call a tournament to be proper if it can be distinguished from another tournament by its leaderboard. We aim to find the number of possible leaderboards of a proper tournament. Lemma : A proper tournament can never have three teams $x,y,z$ satisfying the following sub tournament between them : $x\rightarrow y\rightarrow z \rightarrow x$ $x\leftrightarrow y\leftrightarrow z\leftrightarrow x$ $x\rightarrow y \rightarrow z \leftrightarrow x$ $x\leftrightarrow y \leftrightarrow z$ ; $x \rightarrow z$ Proof : It is easy to see that the first and second tournaments offer the same score distribution and similarly for the third and fourth conditions.$\blacksquare$. We now present some immediate corollaries from this lemma : Corollary 1 : There exists no directed cycle of wins or no cycle of draws Proof : If there was such a directed cycle of wins , we could replace all wins by draws without affecting score distribution (and vice-versa). Corollary 2 : Every team has atmost $1$ draws. Proof : This is obvious due to the third and fourth criterions. Corollary 3 : If $x\leftrightarrow x'$ then $x'$ mimics $x$. In other words for some $y \notin \{x,x'$\} , we have : $$x\leftrightarrow y \iff x'\leftrightarrow y\quad \text{and} \quad x\rightarrow y \iff x'\rightarrow y$$Proof : Proof is obvious by casework on a contrary vertex, keeping in mind the second,third and fourth conditions. Call $x,y$ to be clones iff $x\leftrightarrow y$. We will now provide an explicit construction of a proper tournament having $k\leq \left \lfloor \frac N2 \right \rfloor$ pairs of clowns : Consider a set $S$ having $k$ disjoint pairs of adjacent elements from $[n]^2$. Then we have $i\leftrightarrow i+1$ if $(i,i+1)\in S$ and otherwise $i\mapsto j \iff i<j$. Note that the score sequence of the $i$-th team is $a_i=a_{i+1}2n-{2i+1}$ if $(i,i+1)\in S$ and $a_i=2n-2i$ otherwisw. It is obvious to see (say by strong induction on number of teams) that this tournament is proper. We now claim that every proper tournament is of the above form. Note that if for some proper tournament we have that $t_j\rightarrow t_i$, yet $a_i\geq a_j$ , then $\exists k$ such that the result obtained by $t_k$ in the game with $t_j$ is strictly better than the one with $t_i$. Again $(t_i,t_j,t_k)$ poses contradiction. This means that $a_i<a_j \iff t_j \rightarrow t_i$. This combined with Corollary 3, implies the claim. Finally we are ready to compute $f(n)$. First we compute the number of possible score sequences of a proper tournament with $n$ teams with $k$ clowns. This is equivalent to choosing $k$ non adjacent members of $\{1,2,\dots,n-1\}$. By Stars and Bars, its easy to get that this number is $\tbinom {n-k}{k}$. Hence we have: $$f(n)= \sum_{k=0}^{\left \lfloor \frac N2 \right \rfloor} \binom {n-k}{k}=F_{n+1}$$ where we have used te well known identity : $\sum_{k\geq 0}\binom{n-k}{k}= F_{n+1}$. We are done. $\blacksquare$

I finally figured out what RSL is The answer is the $2019^\text{th}$ Fibbonacci number $F_{2019}$. Interpret the tournament as a graph such that a directed edge goes from each winning team to each losing team and an undirected edge goes from each pair of teams that tied. Call a graph good if it satisfies the condition. Main claim: There are no cycles. Proof: Construct a graph such that each undirected edge becomes a directed edge in the opposite direction of the cycle, and each directed edge becomes undirected. It's not too hard to see that this preserves the score of each team. Now, turn each undirected edge into a directed edge (direction is chosen arbitrarily). Notice that this new graph has no cycles, so it is transitive. Thus, every good graph can be created from a transitive graph, making some of the edges undirected. Suppose we label each vertex with a number from $1$ to $2018$ and draw an arrow from each higher number to each lower number. Two more facts to finish the problem: 1. The undirected edges must connect consecutive numbers. Proof: Anything else would result in a cycle. 2. No two undirected edges can share a vertex. Proof: If they do, then there is a cycle. Also, notice that if a graph satisfies these two conditions, then it must be good. Therefore, the number of good graphs is the same as the number of ways to cover a $1 \cdot 2018$ strip with dominos and squares, which is $F_{2019}$.

First, let's say we can "kill" a person. After killing someone he is out of the results and all his match would not count. denote the answer as $T_N$, easy to see $T_1=1$, $T_2=2$, $T_3=3$. First, note that there can't be three people X,Y,Z with X beating Y, Y beating Z, and Z beating X since you can replace these wins by draws. Second, note that a person can't have two draws. If X draws with Y and Z, and Y beats Z, you can replace this by X beats Z, Y beats X and Y,Z have draw. Now note that the person, say, A, has the most point. If A were to lose a game, say he lost to B, then all people losing to A must lose to B, however this would give B more points than A, contradiction. So A cannot lost a game. So there are only two cases for A: all wins, or one draw. First let's see the case where A has all wins. We can kill A and nobody lose points. This way we have $T_{n-1}$ such seqences. Second is where A has a draw with, say, B. If B were to lose to C, then the results can be changes to A draws with C, B draws with C and A beats B. So B would not lost to another person as well. Now we can kill both A and B, and none of the others will lose a point. This way we have $T_{n-2}$ sequences. So $T_n=T_{n-1}+T_{n-2}$, the result follows.

Define the fibonacci numbers with $F_1=1$, $F_2=2$, and $F_n=F_{n-1}+F_{n-2}$ for all $n \geq 3$. The answer is then $F_N$. Obviously the answer is $F_N$ when $N=1$ and $N=2$. Now suppose $N \geq 3$. View the problem in the obvious digraph interpretation, with the $N$ teams as vertices. We draw an edge $v \to w$ if $v$ beats $w$, and essentially wish to characterize digraphs which are uniquely determined by the "degree" (outdegree minus indegree) of each vertex. Call such graphs distinguishable. The key idea is to look at induced subgraphs of $3$ vertices. If the subgraph is a cycle, this is indistinguishable from the graph obtained by deleting that cycle. If this subgraph is $\{a,b,c\}$ and $a \to b \to c$ but $a$ and $c$ aren't adjacent, then this is indistinguishable from the subgraph with $a \to c$ but no edges from $b$, and vice versa. This is enough to limit the possible induced subgraphs to be those where one vertex has outdegree $2$ or indegree $2$. Now consider the vertex $v$ with maximal outdegree. If there is any vertex $w$ with $w \to v$, then any $u$ such that $v \to a$ must also have $w \to a$ based on our characterization of subgraphs (looking at $\{v,w,a\}$), contradicting maximality (since $w \to v$ is an outedge of $w$ that $v$ obviously lacks). Thus every edge of $v$ is an outedge. On the other hand, at most $1$ other vertex isn't adjacent to $v$, since if there are two vertices $a,b$ then this contradicts our characterization of subgraphs for $\{v,a,b\}$. Suppose $v$ has no vertices not adjacent to itself, so $v \to a$ for all $a \neq v$. Then the graph obtained by deleting $v$ and its edges must also be distinguishable: necessity is obvious, and conversely any distinguishable graph with $N-1$ vertices corresponds to one of these distinguishable graphs by adding the vertex $v$ and drawing an edge to every other vertex, since by looking at degrees we can clearly identify $v$ and then ignore it, and then determine the subgraph on the remaining $N-1$ vertices. Otherwise, let $w$ be the unique vertex not adjacent to $v$. By our subgraph characterization on $\{v,w,a\}$ for all $a \neq v,w$, it follows that $w$ also has an outedge to every vertex other than $v$. Then the graph obtained by deleting $v$ and $w$ must also be distinguishable for the same reason, and if this graph is distinguishable then we can identify $v$ and $w$ simply from the fact that they have degree $N-1$ (this implies $v$ and $w$ have outdegree $N-1$ and indegree $0$, hence $v$ and $w$ aren't adjacent but point to every other vertex), and then ignore them, determining the subgraph on the remaining $N-2$ vertices. Hence if the answer is $f(n)$, we have $f(1)=1$, $f(2)=2$, and from the above, the recursion $f(N)=f(N-1)+f(N-2)$ for $N \geq 3$, since there's a bijection between distinguishable graphs on $N-2$ or $N-1$ vertices and distinguishable graphs on $N$ vertices. This implies the answer. $\blacksquare$

We uploaded our solution https://calimath.org/pdf/RMM-SL2018-C3.pdf on youtube https://youtu.be/zaS6PBJzMjk.