Henry Ernest Dudeney/Modern Puzzles/199 - Golf Competition Puzzle/Solution

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Modern Puzzles by Henry Ernest Dudeney: $199$

Golf Competition Puzzle
I was asked to construct some schedules for players in American golf competitions.
The conditions are:
$(1)$ Every player plays every player once, and once only.
$(2)$ There are half as many links as players, and every player plays twice on every links except one, on which he plays but once.
$(3)$ All the players play simultaneously in every round, and the last round is the one in which every player is playing on a links for the first time.
I have written out schedules for a long series of even numbers of players up to $26$,
but the problem is too difficult for this page except for in its most simple form -- for six players.
Can the reader, calling the players $A$, $B$, $C$, $D$, $E$, and $F$,
and pairing these in all possible ways, such as $AB$, $CD$, $EF$, $AF$, $BD$, $CE$, etc.,
complete this table for six players?


$\qquad \begin {array} {r | c |} \text {Rounds} & 1 & 2 & 3 & 4 & 5 \\ \hline \text{$1$st Links} & & & & & \\ \hline \text{$2$nd Links} & & & & & \\ \hline \text{$3$rd Links} & & & & & \\ \hline \end{array}$


Solution

Here is one arrangement:

$\qquad \begin {array} {r | c |} \text {Rounds} & 1 & 2 & 3 & 4 & 5 \\ \hline \text{$1$st Links} & CE & DE & DF & CF & AB \\ \hline \text{$2$nd Links} & AF & BF & AE & BE & CD \\ \hline \text{$3$rd Links} & BD & AC & BC & AD & EF \\ \hline \end{array}$


Derivation

Fill up the final column with any pairings.

Then each of the remaining four columns of each row consists of both possible pairings of the other four players.

The first row can again be assigned arbitrarily.

The second and third row follow after some straightforward trial and error.


Sources

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