生成单个淘汰赛

first , sorry for my english I'll do my best to explain my problem !

So , i'm trying to generate a single elimination tournament with an unlimited number of players.

for now i'm just thinking about it , i have nothing on paper , i think i will not have problem for tournament with power of two ( 2 4 8 16 32 players..) , my brain hangs on players going directly to round 2 , i don't know how to determine this number and where to place them.

eg (with 59 players) eg (with 59 players)

I think there is a formula but I can't find it, I have some ideas but i think too specifically on a case, without knowing if it would work for another .

Thank you if you can help me !

For a given number N, find the difference between it and the smallest power of 2 at least as large as N. For 59, that'll be 5 (64 - 59). Those 5 players will be added to the tournament schedule at the second round.

This algorithm allows for all the players to be part of the game when the second round begins - i.e., as early as possible. Its explanation is very simple: imagine that originally there were 2**N players - but some just didn't come to their games, so their opponents went further without a fight. )

As a sidenote, your formula should take into account that it's strongest players that should enter the game from the second round, not the weakest ones. )

The first step apparently is calculating the number of players that will participate in the first round. Now, let's continue that 'missing players' metaphor - let's say there were 64 players originally, so the first round should have 32 games played. But 5 players (64 - 59) didn't come for those games - so the number of real games is 27 ( 64/2 - 5 ), and the number of real participants of the first round is 54 (27 * 2).

After the first round, there'll be 27 people left in the tournament - those people will be joined by those other 5 guys, so the total number of the 2nd round players is 32. The rest is trivial, I suppose. )

Actually, this is easy to commonize. Let's say we have N players, and the smallest power of 2 at least as large as N is P. Now...

  • The first round should have (N - (P - N)) (or just (2*N - P)) players.
  • The total number of the games in the first round is (N - P/2).
  • Apparently, the same is the number of players going into the 2nd round.
  • These will be joined by (P - N) players left without a play in the 1st round, so the total number of players in the 2nd round will be...

N - P/2 + P - N => P - P/2 => P/2

  • ... and from now you just go with the direct schedule of 2^N players (as P/2, as well as P, is the power of 2).

ooh thank you @raina77ow , you blew my mind , So here are my calculations:

64/2 = 32
59/2 = 29 ( rounded to the lower ) => nb of total player at left ( round 1 & 2)
32-29 = 3 => nb players at left going to round 2
29-3 = 26 => nb players at left going to round 1

59-29 = 30 => nb total players at right ( round 1 & 2 )
5-3 = 2 => nb players going to round 2 at right
30-2 = 28 = nb players round 1

` I think I can make an algorithm now if that's right for each case.