Dear PHPers,

I'm currently developing a cupsystem for a new gaming ladder called "ClanFight". Now, I hit a snag.

I am trying to determine how many brackets should be played before the end is 'near', so to speak. For example:

Now, of course I know that we have to play:
64 players, remainder 32, remainder 16, remainder 8, remainder 4, remainder 2, winner 1. A total of 127 brackets (64 + 32 + 16 + 8 + 4 + 2 + 1 = 127).
Now, this means, that there will be played 7 ROUNDS.
Round 1 :: 64 (32 matches, 32 dropouts)
Round 2 :: 32 (16 matches, 16 dropouts)
Round 3 :: 16 (8 matches, 8 dropouts)
Round 4 :: 8 (4 matches, 2 dropouts)
Round 5 :: 4 (2 matches, 2 dropouts
Round 6 :: 2 (1 match, 1 winner)
Round 7 :: 1 (cookie?)

Anyhow, this all looks rather simple because we can calculate the yet unknown variable. We know by seeing the nummers 64 to 1, that 7 rounds are played, with the 7th round being the winner. So most fairly, you could say 6 rounds with the result 1 winner. Either way. I rather have it displayed in 6 rounds though, because the 6th round, 1 on1 will spit out the winner anyway.

Now, how am I supposed to calculate it? 8 players requires 3 rounds. (8 - 4 - 2 (- 1)) So how am I supposed to draw from 8, 3 and from 64, 6.

64 / 8 = 8 - 2 = 6... But that doesn't work on the 8 players.
8 / 8 = 0 - 2 = -2...

I am still ignorant to this variable or calculationmethod. Perhaps you guys could take a crack at it.

The total players will increase to 512 players total, 1023 brackets with A LOT of ROUNDS. :)

Help? :)

__________________

"Life is a bitch, take that bitch on a ride"

To find the number of rounds from players, it looks like you could use the log2 value. In PHP we don't have this function available (though it is in C) but we do have a more general log function which can take a variable base value and for our purposes that base should be 2.

To quote the PHP Manual page:

float log ( float $arg [, float $base ] )

If the optional base parameter is specified, log() returns logbase arg , otherwise log() returns the natural logarithm of arg.

In practice, we can use it like so:

I must say, I didn't know that there was a function like that. However, the code you supplied displays 5 instead of 6. :)

Perhaps any other methods you care to try out?

__________________

"Life is a bitch, take that bitch on a ride"

I think you must have copy/pasted incorrectly and/or altered something because log(64, 2) is 6, fact. :)

I'm not kidding. :) It seriously returns 5. :) Running version 5.2.4.

http://markernst.eu/clanfight/test.php

__________________

"Life is a bitch, take that bitch on a ride"

OK, looks like there may be a rounding problem on your machine or version of PHP (mine's 5.2.5). Simply round the value rather than casting it to integer.

$rounds = round(log($players, 2)); // 6

As a fun check try: echo (int) 5.999999999999999; that's what PHP might be seeing rather than 6 as we expect.

Worked like a charm Salathe. Thanks a loOot.

__________________

"Life is a bitch, take that bitch on a ride"