Y=ax^2+bx+c Fact or Fiction?

[rmadsen55]

anish,
you are correct that you do not need calculus to create the equation of the score (which is a quadratic) but it is necessary to optimize it. in order to find the maximum score for any given number of bins you need to take the derivative of the scoring formula and find the relative maximum.
Perhaps there is another way to optimize it but I am not aware of any.
-Robin

[Caleb Fulton]

You don't need calculus because you are dealing with integer values only. This is more of a pattern/series problem to me.

The quadratic formula merely approximates the maximum score based upon how many boxes you have to work with, but even this is irrelevant.

The maximum score will ALWAYS occur when N_boxes/2 are in the stack if N_boxes is even and it will ALWAYS occur when (N_boxes-1)/2 are in the stack if N_boxes is odd. The reason I say N_boxes-1 is illustrated by this example:

Say you have seven boxes to work with. It is easier to have a stack of three than a stack of four based upon the laws of physics. You will get the same score if you have three stacked and four on the ground or vice versa, but it is probably easier to "maintain" a stack of three.

[rbayer]

You do need calc to maximize this function. Yes, you can do it in your head, and you can do it by looking at a graph, but both of these use the techniques of calc without touching any derivatives. Alternatively, you can complete the square on the quadratic and look at its vertex.

[Ashley Weed]

Just crunching numbers and being completely exhausted at the same time.... I was wondering if someone could verify the equation.....

if a = (-1) x = (# of boxes in highest stack) b = (# of boxes in your scoring zone for one point) c = (points from alliances robots)

that being.... if you had a stack of 8 bins... with 10 others in your scoring zone....and both of the robots on the platform

(-1)(8)^2 + (10)(8) + 50 =
(-64) + (80) + 50 =
66 pts.

This is without opposing alliance points if you win...........

Is this correct???? If not, would someone please clarify my mistake.

[rbayer]

Instead of having b as the number of boxes in your scoring zone worth one point, have b represent the total number of boxes then your equation works just fine. Your score for this scenario should be:

8*10+50=130

[Ashley Weed]

then what happens to the a = (-1) and (x^2)..... from the equation?

[BandChick]

Originally posted by Caroline
If there is a fixed number of boxes in your scoring zone, if you plot the possible scores using the number stacked as the X component and the total score as the y component, it forms a parabola.

It forms a parabola. In this situation, the equation is y=(-x)(x-8)

If you look at that above you will realize, she just explained the equation. that is the same as y=ax^2 + bx + c

[Mark Garver]

Well if you think of always having 45 boxes on your side and both of your robots on the top of the ramp,

x(45-x)+50

works well.

But if you assume the following terms,

a=highest stack
b=number of boxes
c=number of robots
1=your alliance (side or robot)
2=opponents alliance (side or robot)

These equations work

If: [a1(b1-a1)+c1] > [a2(b2-a2)+c2]
Then: [a1(b1-a1)+c1] + 2*[a2(b2-a2)+c2] = your winning score

If: [a1(b1-a1)+c1] < [a2(b2-a2)+c2]
Then: [a1(b1-a1)+c1] = your losing score

Sorry, maybe to many terms, haha.

[Ashley Weed]

ok.. I think I was just reading way into the equation thing... the equation is to detemine the QP score if you win... and also used for QP if you loose..... you have to know how/where the bins are placed for each alliance to use the equation properly then???

[Mark Garver]

I was using the information from a previous post,

"no need for calculus.

Lets say you have all the bins in your zone and you have 2 robots on the platform.

lets let x equal stack height.
Now, scoring is the highest stack height * number of bins not in the stack. If you have all 45 bins, thats 45-x (the bins in the heighest stack). Additionaly, 50 for both robots on the platform.

x(45-x) + 50 = y

that expands to:

y = -x^2 + 45x + 50"

Using this you have to have all 45 bins. Using mine you can have any number of bins. Of course there is a down side, but its not hard to calculate it. The downside is having to count the boxes on both sides of the field. But if you change b2 to being 45-b1 then it becomes a little easier.

So assuming the same variables as I had before,

If: [a1(b1-a1)+c1] > [a2((45-b1)-a2)+c2]
Then: [a1(b1-a1)+c1] + 2*[a2((45-b1-a2)+c2] = your winning score

If: [a1(b1-a1)+c1] < [a2((45-b1)-a2)+c2]
Then: [a1(b1-a1)+c1] = your losing score

This is good for both QP and EP scoring, if I am reading the rules correctly.

I am trying to get the EPs all done right now too.

[Mark Garver]

This may get confusing, but it does work.

For QP

If: [a1(b1-a1)+c1] = [a2((45-b1)-a2)+c2]
Then: [a1(b1-a1)+c1] + [a2((45-b1)-a2)+c2] = Both alliance's score

For EP

If: [a1(b1-a1)+c1] = [a2((45-b1)-a2)+c2] For Match one and Match two then addition matches are played until [a1(b1-a1)+c1] > [a2((45-b1)-a2)+c2] or [a1(b1-a1)+c1] < [a2((45-b1)-a2)+c2].

Also for EP (1 in front of a term means match 1 and 2 in front of a term means match 2)

If: [1a1(1b1-1a1)+1c1] > [1a2((45-1b1)-1a2)+1c2]
Then: [1a1(1b1-1a1)+1c1] + 2*[1a2((45-1b1-1a2)+1c2] = your winning score for match 1

If: [1a1(1b1-1a1)+1c1] < [1a2((45-1b1)-1a2)+1c2]
Then: [1a1(1b1-1a1)+1c1] = your losing score for match 1

This is true for match two as well.

So,

If: [1a1(1b1-1a1)+1c1] > [1a2((45-1b1)-1a2)+1c2] and [2a1(2b1-2a1)+2c1] > [2a2((45-2b1)-2a2)+2c2]
Then: You advance

If: [1a1(1b1-1a1)+1c1] < [1a2((45-1b1)-1a2)+1c2] and [2a1(2b1-2a1)+2c1] < [2a2((45-2b1)-2a2)+2c2
Then: You drop out

If: [1a1(1b1-1a1)+1c1] > [1a2((45-1b1)-1a2)+1c2] and [1a1(1b1-1a1)+1c1] < [1a2((45-1b1)-1a2)+1c2] and [1a1(1b1-1a1)+1c1] + 2*[1a2((45-1b1-1a2)+1c2] + [2a1(2b1-2a1)+2c1] > [1a2(1b2-1a2)+1c2] + [2a2(2b2-2a2)+2c2] + 2*[2a2((45-2b2-2a2)+2c2]
Then: You advance

If: [1a1(1b1-1a1)+1c1] > [1a2((45-1b1)-1a2)+1c2] and [1a1(1b1-1a1)+1c1] < [1a2((45-1b1)-1a2)+1c2] and [1a1(1b1-1a1)+1c1] + 2*[1a2((45-1b1-1a2)+1c2] + [2a1(2b1-2a1)+2c1] < [1a2(1b2-1a2)+1c2] + [2a2(2b2-2a2)+2c2] + 2*[2a2((45-2b2-2a2)+2c2]
Then: You drop out


If: [1a1(1b1-1a1)+1c1] < [1a2((45-1b1)-1a2)+1c2] and [1a1(1b1-1a1)+1c1] > [1a2((45-1b1)-1a2)+1c2] and [1a1(1b1-1a1)+1c1] + [2a1(2b1-2a1)+2c1] + 2*[2a1((45-2b1-2a1)+2c1] > [1a2(1b2-1a2)+1c2] + 2*[1a2((45-1b2-1a2)+1c2] + [2a2(2b2-2a2)+2c2]
Then: You advance

If: [1a1(1b1-1a1)+1c1] < [1a2((45-1b1)-1a2)+1c2] and [1a1(1b1-1a1)+1c1] > [1a2((45-1b1)-1a2)+1c2] and [1a1(1b1-1a1)+1c1] + [2a1(2b1-2a1)+2c1] + 2*[2a1((45-2b1-2a1)+2c1] < [1a2(1b2-1a2)+1c2] + 2*[1a2((45-1b2-1a2)+1c2] + [2a2(2b2-2a2)+2c2]
Then: You drop out

Like I said, it should work, maybe I am incorrect some where will a variable tho.

O, I forgot to do the situation where if either one of the matches are tied. You can assume what happens then tho :

[Alex Forest]

Now, you are coming to my team to do this for me in the middle of the match, right??? *big grin*

[Mark Garver]

Plams work great for this!!! All you need is 5 simple numbers for most matches. That is if no one takes boxes out!!! So maybe 6 numbers is the easier and more fool proof way.

Working on a program for that now (even bigger grin)

[Mark Garver]

You could just count the boxes that have went outside the playing field and subtract them from the the 45, before you minus the b1. Well it will be all done on a computer so it won't be hard to add that in.

[Johca_Gaorl]

Actually correction, if it's odd, it doesn't matter whether more are in the stack or on the ground:

i.e. 21 total

11 stack * 10 ground = 110
10 stack * 11 ground = 110