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 :)

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=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.

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).

possible scores
 

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 :)

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.

This is an optimazation problem
 

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

Hmmm
 

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.

Please Help
 

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

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

Re: possible scores
 

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

Well then assuming...
 

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.

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???

Correct....
 

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.

The rest of the EP scoring...
 

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 ::D

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

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)

Another way...
 

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.

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

This was mentioned before. I think they said more on the ground because it was easier to achieve and a shorter stack is more stable.

Actually....
 

If you have an odd number of boxes this isn't always the case because of the orientation in which the boxes could be placed. Meaning you stack the orientation of the boxes to your advantage in stacking the boxes. Maximizing the number scored. (ie. the highest score is actually the maximum number of boxes possible (not 45) and dividing by 2).

Re: Actually....
 

Right, I forgot to include that part.

You can tabulate and graph this real easy in Excel. Make three columns. Column B is the tubs in the highest stack, column C is the number of tubs not in the multiplier stack and column D is the total score. Put 45 at the top of colmn B and 0 at the top of column C and put the formula =B*C at the top of column D.
In second cell of column B put B1-1 and in Column C put C1+1 and repeat the formula from D1 to D2. Copy B1 through D1 and paste it in the next 44 cells below B2.

You should end up with a column of descending numbers 45 - 0 and a column of ascending numbers 0 - 45 and the third column should be the product.

And, as you'll see, you'll want your tallest stack to equal half or in the case of an odd number of tubs, one less than half to get the highest score. Not forgetting that the top of the stack is a tub on end or upside down. I attached an example. Play with it.

discreet
 

tisk tisk tisk

can't do that ... no maximization

its discrete

but ... u can base ur answer the fact that the maximum area of a paralellogram is in a square ... hence the (x/2)^2 yielding max points

Yes that's correct. Remember that when a parabola opens down (as in this case) the vertex is the maximum y value (y = total points).

The equation for the x coordinate of the vertex of any parabola is x = -b / (2 * a)

a = -1
x = -b / -2
x = b / 2
so....
Number of boxes in your highest stack = Total number of boxes on your side / 2

No calculus, just Algebra II.