In a Swiss tournament for adults with cash prizes - the prizes are split and are not determined by tie breaks.
Tie breaks are only used for non-divisible prizes.
You are proposing a solution for something that most do not seem to view as a problem - especially for most of the chess played in this country.
The first part of that statement is true (provided log2(N) is a whole number, i.e. N is a power of 2). The second and third parts are definitely not true.
The clear “winner” (actually clear leader) after log2(N) rounds will, in round log2(N)+1, be paired against one of the players with N-1 points. If the latter wins, there is no longer a clear anything. There is at least a tie for 1st and 2nd.
Example with 16 players, after 4 rounds:
at 4-0: 1 player
at 3-1: 4 players
at 2-2: 6 players
at 1-3: 4 players
at 0-4: 1 player
In round 5, the 4-0 will have to play a 3-1. Two others at 3-1 will play each other. The remaining 3-1 will play a 2-2. One possible result (not the only one) is:
at 5-0: nobody
at 4-1: 4 players
at 3-2: 4 players
at 2-3: 5 players
at 1-4: 2 players
at 0-5: 1 player
So much for clear 2nd. Even without that “upset” on board 1, there would still be a two-way tie for 2nd-3rd.
Your math needs a little work, Brian.
Bill Smythe
10 players
Results A / Results B
5 @ 1-0 / 5 @ 1-0
5 @ 0-1 / 5 @ 0-1
…
3 @ 2-0 / 2 @ 2-0
4 @ 1-1 / 6 @ 1-1
3 @ 0-2 / 2 @ 0-2
…
2 @ 3-0 / 1 @ 3-0
3 @ 2-1 / 4 @ 2-1
3 @ 1-2 / 4 @ 1-2
2 @ 0-3 / 1 @ 0-3
…
1 @ 4-0 / 4 @ 3-1
3 @ 3-1 / 3 @ 2-2
2 @ 2-2 / 2 @ 1-3
3 @ 1-3 / 1 @ 0-4
1 @ 0-4
Unless the number is an exact power of 2, even with no draws you cannot predict when you will have a sole leader.
That was implied by saying the number of rounds is the log2 of the number of players but it is not blatantly clear to people who are not used to thinking in terms of a tournament with a non-integer number of rounds (the number would be, poetically, irrational).
Also, the way many large prize fund events are run, you could have multiple perfect scores in a section even though the number of rounds is well above log2 of the number of players.
One example is two players having a decisive game in the first round, the loser re-entering in a quicker schedule, and both players winning out after that. They wouldn’t be paired again because that would result in the re-entering player playing somebody twice.