Handling odd numbers

ID:978281
 
Sep 15 2012, 10:04 am
I'm stumped when it comes to handling odd numbers in tournaments and other competitive systems that require at least two. I would like to know some of your idea's for what to do with the leftovers?
Sep 15 2012, 10:05 am
Have a match with three people?
Sep 15 2012, 10:21 am
add an AI?
Sep 15 2012, 10:26 am
Give the last person a bye into the next round.
Sep 15 2012, 10:45 am
In response to Oasiscircle
Could still be odd.
Sep 15 2012, 10:53 am
Nice ideas, I'd thought up similar to Oasiscircle only removing them but it seems unfair for those to not even have a chance even though its down to luck.
Sep 15 2012, 11:12 pm
I just wanted to add another idea in. You can delay the third person from entering until ((numOfPlayers % 2) == 0)
Sep 16 2012, 2:48 am
In response to Red Hall Dev
Red Hall Dev wrote:
I just wanted to add another idea in. You can delay the third person from entering until ((numOfPlayers % 2) == 0)

I like this idea best lol, thanks for all your ideas they were great.
Sep 16 2012, 1:25 pm
In response to Tyrannicide
=)
Sep 18 2012, 11:54 am
In response to Red Hall Dev
Red Hall Dev wrote:
I just wanted to add another idea in. You can delay the third person from entering until ((numOfPlayers % 2) == 0)

Are you saying they shouldn't even be able to list themselves as in until it's an even number? If so you are only going to be able to have 1 person enter...

Sep 19 2012, 12:10 am
In response to GreatFisher
Context:
competitive systems that require at least two

The context of the question suggests they already have a way of handling what to do with the first and second player.

Question:
I would like to know some of your idea's for what to do with the leftovers?

Then the question is about what to do with leftovers. The point at which another player becomes a leftover starts at the third player. They are leftover number one.

This is why my answer simply says what to do with the third player. It naturally doesn't apply to the first and second because they're not in the range stated by the question. They're not leftovers.

Sep 19 2012, 9:21 am
In response to Red Hall Dev
I think you have those flipped. You want (numOfPlayers%2 != 0), which would mean the current number of players is odd, and adding another player would make it even.
Sep 19 2012, 12:43 pm
In response to DarkCampainger
I think you mistranslated me. I said to delay the third player from entering the tournament until there is an even number of players.

(numOfPlayers % 2 == 0) checks for even numbers.

Example:
http://img4.imageshack.us/img4/1943/evennumbers.png


In that example you can see that it would check to see if numOfPlayers is even when the fourth person enters, thus returning true and letting both players enter.
Sep 21 2012, 8:10 am
Logic with players%2 is flawed for multiple rounds

One --
        } One
Two --

            } One
Three   --
        } Four
Four    --
                }

Five    --
        } Six
Six     --

            } Six
BYE (denyed)
        } BYE
BYE     --
Sep 21 2012, 5:38 pm
In response to Pirion
This isn't a discussion about building a tournament system, in fact it's about odd numbers in "competitive systems". I don't understand how handling someone dropping out of an tournament has anything to do with (numOfPlayers % 2 == 0) not functioning.

It's not my fault if you don't know where to use it, or how many times to use it. It works, final, conclusion, done, dusted and unless you want to start another topic of how to use it in a specific system, don't bother here. That's just trying to create problems where there were none, and never should be any.