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

[Bduggan04]

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.

[Jeff Waegelin]

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?

[RebAl]

Woodie said it was the equation for the scoring

[Jeff Waegelin]

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?

[Yan Wang]

Very easily

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!

[RebAl]

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

[Suneet]

I already posted this in a different thread, but...

y=a*x^2+b*x+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.

[Johca_Gaorl]

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

[Morgan Jones]

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).

[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.

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)

[Kai Zhao]

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.

[Yan Wang]

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

[Bduggan04]

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.

[rmadsen55]

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

[anish]

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