# 8 chess players in a round robin tournament

Eight people played chess in a round robin tournament.

No match ended in a tie and both A and B won the first place.

Since C and D were rivals, C beat all players who beat D and D beat all players who beat C.

What is the final score of A and B (the number of wins and defeats)?

• Did the winner of C vs. D then play against and beat himself? – KSmarts Feb 2 '15 at 15:28
• @KSmarts Probably. But, it was not recorded on the score board. :-) – P.-S. Park Feb 2 '15 at 15:35

With $8$ players, there have been $(8 \times 7)/2 = 28$ matches and therefore $28$ points.

If A and B scored 4 points or less, the other players would have scored at most 3 points which brings a total of at $2\times 4+6\times 3 = 26$ points or less. But there are 28 points altogether. Therefore A and B scored at least 5 points.

One of A or B must have lost against the other, and both have lost against C or D. Therefore one of them has lost 2 matches and scores at most 5 points.

As a result, A and B can have scored only 5 points. They won 5 and lost 2 matches each.

(I assume that $A, B, C, D$ represent names, and denote four pairwise distinct players; I call the other four players $E,F,G,H$. Furthermore, I assume that the statement "Since $C$ and $D$ were rivals ..." does not cover the particular game played between $C$ and $D$.)

By the problem statement, no player in $\{A,B,E,F,G,H\}$ has won both games against $C$ and $D$. Hence $\{A,B,E,F,G,H\}$ can be partitioned into three disjoint sets:

• the set $L(C)$ of players who have lost to $C$ and won against $D$
• the set $L(D)$ of players who have lost to $D$ and won against $C$
• the set $L(CD)$ of players who have lost against both $C$ and $D$

By symmetry we now assume that $C$ ended up with at least as many points as $D$.

(1) $C$ scored at least $4$ points.

Proof: Suppose for the sake of contradiction that $C$ scored at most $3$ points. If $C$ won against $D$, this implies $|L(C)|\le2$ and hence $|L(D)|+|L(CD)|\ge4$; then $D$ has scored at least $4$ points; contradiction. IF $C$ lost to $D$, this implies $|L(C)|+|L(CD)|\le3$ and hence $|L(D)|\ge3$; then $D$ has scored at least $4$ points; another contradiction.

(2) $A$ and $B$ each scored at least $5$ points.

Proof: $A$ and $B$ won the first place, and hence scored more points than $C$.

(3) $A$ and $B$ each scored at most $5$ points.

Proof: Suppose for the sake of contradiction that $A$ and $B$ both scored at least $6$ points. They played one game against each other, and altogether $12$ games against the other six players. Then one of them must have won all six matches against ${C,D,E,F,G,H}$, and in particular his matches against $C$ and $D$; contradiction.

So the answer is: $A$ and $B$ each lost two matches and each won five matches.

For something like this, simply constructing a specific outcome that meets the requirements isn't actually that hard, and I think is an often overlooked way of solving some puzzles. Here's an example:

  a b c d e f g h total
a - 1 0 1 0 1 1 1   5
b 0 - 1 0 1 1 1 1   5
c 1 0 - 1 0 1 1 0   4
d 0 1 0 - 1 0 0 1   3
e 1 0 1 0 - 1 0 0   3
f 0 0 0 1 0 - 1 1   3
g 0 0 0 1 1 0 - 1   3
h 0 0 1 0 1 0 0 -   2


I started out with A beating everyone except D. It then immediately became obvious that B could not tie A. So I added another loss to A (against E). From there, I made sure C and D had mirror outcomes, and filled out the rest of the table making sure to keep everyone below 5 wins. The whole thing only required backtracking once.