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

A lot of people have posted on this site that the quadratic equation has nothing to do with the scoring. However, I think it does. It reminds me of a math problem we had. We were given a perimeter of a rectangle and we had to find the greatest possible area. The greatest area results when the four sides are equal, a square (note: the resulting area function is quadratic). So in my mind, the quadratic equation was meant to show the highest score results when the number of bins in the multiplier stack equals or is close to the number of bins that are unstacked. Try it. The scores are higher the closer the two numbers are together. Also, if you graphed the score with the stacked and unstack bins as variables, the scores will form a parabola.

I’ve thought about this myself, but I was a little unsure as to how the whole thing fit together. What are the coefficients and what is the variable?

Woodie said it was the equation for the scoring

I know that. I want to know how the Quadratic equation can be used for scoring. How can I input the information from the end of a match and come out with a score?

Very easily :slight_smile:

Actually, no, I had to some math to get the formula but it works out perfectly fine. I stand corrected before where I thought it was y=mx+b… I’ll post up how so tomorrow, after I tell my own team!

bx+c
b= highest stack
x = # boxes on ground
c = 25 pts per robot on ramp
but then you would need to add x again

a^2 no idea

so im not sure

an equation that would work is
y=bx+c+x

y being the pts you get total

I already posted this in a different thread, but…

y=ax^2+bx+c

y= raw score
a=-1
x=Number of boxes in your highest stack
b=Total number of boxes on your side
c=0, 25, or 50, depending on the #of bots in the middle for you.

I’ll admit it’s not quite accurate, but that’s not my fault.

To maximize score, have half of the bins on the ground, if odd number of bins, have more on the ground.

One of my team’s mentors actually spent nearly two hours going over this equation. I personally find it somewhat pointless: nobody is going to be using the quadratic when they’re in the middle of a competition. At the end of his lecture, the mentor finally came to the same conclusion several of us had already made two hours previous: ideally, you want half of the boxes in your zone to be in a stack (or all of your boxes in two equal stacks).

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.

For instance, if you have 8 boxes in your area:

(stacked, score)
(1, 7)
(2, 12)
(3, 15)
(4, 16)
(5, 15)
(6, 12)
(7, 7)
(8,0)

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

Given:
A=1
B= amount of boxes in your possession
C= 25*amount of robots on the ramp at end time
X=amount of boxes in the tallest stack
Y=Your Score

Max= Your Max Score

Equation

Y = -Ax^2 + Bx + C

Put this into your graph calc, adjust your windows as needed, and graph/table it!

See also my Scoring-Goal Matrix in the white papers section.

Now my question: What does the derivative and/or integral of the equation tell you?

Sorry, the two posts above with the equation had not yet been posted when I started writing the message.

Yeah, lol, figured that out a while ago. You could simplify your equation, Kai, by saying that A=-1… otherwise, saying A=1 is a given and doesn’t give it any merit. It’s the same mathematically but you don’t have that nasty negative to start out a formula :slight_smile:

*Originally posted by Johca_Gaorl *
**To maximize score, have half of the bins on the ground, if odd number of bins, have more on the ground. **

Right, that’s what I was saying earlier. I don’t think people will use a formula during a match, but it’s easy to see if there are more stacked in the mult. stack then bins that are not. I believe that is the most useful piece of info.

You need a little calculus for this one.

if X=crates on the ground and Y=crates in the stack and C= the number of crates you have control of in your scoring zone and S= the score

Assume one robot on ramp. if two replace 25 in Eq.2 with 50. if none replace 25 with 0

Eq.1: X+Y=C
solved for X: X=C-Y

Eq. 2: XY+25=S

Substitute Eq. 1 (solved for X) in for X in Eq. 2.
(C-Y)Y+25=S=-y^2+CY+25 (here’s your quadratic)

take the derivative of this:
S’=-2Y+C

Set this Equal to 0 and solve for Y to find the maximum
-2Y+C=0

Y=C/2

Therefor the maximum score you can have is with exactly half your crates in the stack. This is only relevant for even numbers of crates. with odd numbers it doesn’t matter if you have one more in the stack or on the floor.

-Robin

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

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

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.

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.

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.

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