Major Help Needed!

[PeterPigmy]

Hi guys,

I hope you don't mind me dropping into the maths forum for the first time when I just need help!

I am really stuck here.

I am trying to work out some "grids" for car racing.

Let me explain by way of an example.

70 People enter a race meeting(# of COMPETITORS). The circuit is only allowed to have 30 competitors in each race (The GRID size).

Each Competitor must have 3 races each (# of RACES).

Each competitor must be allocated "fair" starting positions.

The definition of fair is that if allocate "points" to a position, ie. 1 for 1st, 2 for 2nd etc then if you add up the points for each competitor everyone has the same (or maximum of one point difference).

So everyone would have either 1 front/middle/back start or 3 "midfield" starts.

I can work out grids easily enough when the # of competitors is smaller than the grid size. But when it's greater then you have to have (say) 6 races, with each competitor only taking part in 3 of them.

This has been driving me nuts.

The porgramming side of it i'm happy with, but coming up with the algorithm to make these grids is really annoying me now.

It will ALWAYS be three races each, but the number of competitors could be anything from (say) 30 to 200 and the grid size could be anything from 20 to 50!

I can't do it by hand, BUT I know other people have, so it's possible.

HELP!!!!!!!! (and thank you!)


Peter

[dglienna]

they usually have prior points that determine their starting position. without this, I'd think that nobody would have preferential placement. I'd guess that a random order would have to suffice. maybe for the second and third race, you could factor in the prior points for the first race.

I would use a weighting system for the second and third race. Mabe subtract original position from 70 to give the weight?

[PeterPigmy]

Actually this system is used to Define the starting order for the main race. Each driver runs three "heats" and the finishing positions of these heats are used to calculate a grid order for the "feature" race.

So the starting positions for the heats have to be absolutely fair and even.

This is a mind-melter!

Thanks for the response


Peter

[dglienna]

I don't think that there could be a truely fair and even method.
If they don't like alphabetically, then random would be method of choice for me. I'd settle for whatever position was chosen for me.

There are too many unknowns in a race to use criteria from the race. Maybe winning one would figure figuratively into where you start in the main event, but other than that, back to random?

Lets see what others think...

[dglienna]

maybe you would want to post this somewhere else, like Games?

[alkatran]

So there are 30 positions in each race, 10 front, 10 mid, 10 back, and 70 racers?

Just divide the 70 racers into 3 (almost) even groups and the problem is simplified.

[PeterPigmy]

OK. That may be a better way to look at it.

I suppose I should have though about breaking the problem down into smaller "chunks".

Thanks for the efforts guys.

I know this problem HAS been solved before because it is done quite regularly by other clubs.

But everyone is so secretive about the way they do things.

To Give you a better idea of a small "Fair" Grid.


Race 1.
---------
Driver 1
Driver 2
D3
D4
D5
D6

Race 2
--------
D6
D3
D5
D2
D4
D1

Race 3
--------
D4
D5
D6
D1
D2
D3

So if you add up their "starting positions" where 1st=1point and 2nd=2 points then

Driver 1 = 1+6+4 = 11
Driver 2 = 2+4+5 = 11
Driver 3 = 3+2+6 = 11
Driver 4 = 4+5+1 = 10
Driver 5 = 5+3+2 = 10
Driver 6 = 6+1+3 = 10

This is the "fairest" grid you can give an even numbered group of drivers (with odd numbers of drivers the "points" will be exctly the same, not one off).

But this is easy.

This is where the total number of drivers is <= the maximum permitted "grid" size.

The problem is where there are MORE drivers than are allowed on the grid at any one time. So drivers will take part in 3 races of the 4/5/6 ... necessary to give everyone 3 races each.

BUT, it must still be totally "fair".

ie. Total "points" for each driver must still total the same (or vary by only 1 point).

I know this can be done. But as you can see i'm even having trouble explaining how it should work!

Thanks for the input so far though guys!


Peter

[alkatran]

No, you actually have to consider it as a bunch of smaller groups which will fit in the grid. For example, if you can fit 5 cars, and have 50: Pretend you have 5 cars, solve, then apply it to the 10 seperate groups of cars.

[PeterPigmy]

Thanks Alkatran.

I have written code to work out grids for anything up to the maximum Grid size, and on first thoughts, I think you're right about breaking it down. I've been looking at the "whole picture" too much.

Thanks for the help!


Peter