I came across an interesting question in a competition (it is long over by now, but I haven't found a solution yet):

In a chess tournament $2^n$ ($n \in \mathbb{N}$) players play each other round-robin (everybody against everybody). If a game draws it is repeated until one of the player wins. Show that it is possible to pick a sequence of $n+1$ players in which each player won over his predecessors.

Obviously this is true for $n = 1$. But I'm stuck even to show this for any higher number.

  • 5
    $\begingroup$ You should get an award for the most non-descriptive question title ever. And it's not like others hadn't tried... $\endgroup$ – joriki Nov 12 '11 at 20:52
  • $\begingroup$ Took it upon myself to provide a more informative title. $\endgroup$ – Gerry Myerson Nov 12 '11 at 22:32

There must be some player who lost against half the players. Take that player as the first in the sequence, keep $2^{n-1}$ players that she lost against, and discard the remaining players. Now repeat the same procedure with the $2^{n-1}$ until you're left with only one player. At each stage, you're left with only players who won against all the players you've selected for the sequence so far.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.