# Stack Height Scoring Analysis

Question: Is there an optimum stack height?

Let n represent the total number of containers
in our scoring area. For the sake of our
calculations, this is a constant value.

Let x represent the number of containers that we
use to make our stack.

Then n-x equals the number of containers in the
scoring area not used in building the stack.

Our score, excluding “king-of-the-hill” points,
can then be calculated by multiplying our stack
height by the number of containers not used in
building our stack. Creating a mathematical
expression for this we get:

score = x(n-x)
= nx-x^2
= -x^2 + nx

So the (oft quoted) quadratic coefficents are:

a=-1
b=number of containers in our alliance’s scoring area
c=number of “king-of-the-hill” points

``````         ** Calculus Alert **
``````

Now to optimize our score we take the derivative
of the above equation…

score’ = -2x + n

…and then solve for x when score’ is held at zero.

-2x + n = 0
-2x = -n
x = -n/-2
x = n/2

What this expression says is that we will get the
maximum score if we use half of the available
containers in the scoring area to build our stack.

Homework question:

This falls apart if n is odd. Afterall, how do you
stack half of a container?!? So my question is simply:
Given an odd number of containers in your scoring
area, do you put that last, odd-numbered container on
the stack or not?

Hint:

Well, first you might just make a table that relates
the two possible scores for each case (n=3,5,7,etc).
Then if you’re still curious, you might want to do a
mathematical proof to convince yourself that this
still holds true for a million and one containers in
the scoring area.

-Kevin

Don’t put the odd numbered container on the stack.

Technically, it doesn’t matter if you do or not. Your resulting score will be exactly the same (for example, 8 * 9 vs. 9 * 8). However, if you attempt to stack that last crate, you’re risking (a) accidentally knocking over the stack, (b) an opponent knocking over the stack when you could have been defending it, and © not releasing the crate before time expires, in which case I don’t believe it counts towards your score at all.

*Originally posted by jeremyw_562 *
**Don’t put the odd numbered container on the stack.

Technically, it doesn’t matter if you do or not. Your resulting score will be exactly the same (for example, 8 * 9 vs. 9 * 8). However, if you attempt to stack that last crate, you’re risking (a) accidentally knocking over the stack, (b) an opponent knocking over the stack when you could have been defending it, and © not releasing the crate before time expires, in which case I don’t believe it counts towards your score at all. **

Excellent! This is exactly what I was looking for. Now for extra credit, how about a formal proof of this :D.

-Kevin

This falls apart if n is odd. Afterall, how do you stack half of a container?!?

I dunno, as one might see in the other threads, those containers are pretty brittle… And we all know what happens when bruteforce is encorperated into a math equation…:rolleyes:
icky, broken stuff!

Excellent! This is exactly what I was looking for. Now for extra credit, how about a formal proof of this .

You can’t! lol, But hey, isn’t that whats makes FIRST wonderful?!

The easy proof is the commutative property of multiplication, and my calculus is too rusty to do it for real. Here’s my best attempt:

Given x, an odd number of containers in scoring position,
let y = the optimal stack height (which you have proofed)

y = x/2

In the case of an odd number of containers, y will be a complex fraction (where x = 11, y = 5 1/2). Since we cannot have 1/2 a container, we must round y either up or down, by adding 0.5 or subtracting 0.5.

We can calculate 2 scores based on (1) adding another container to the stack and (2) not adding another container to the stack.

s1 = (y+0.5) (y-0.5)
s2 = (y-0.5) (y+0.5)

Given the commutative property of multiplication (which states that AB=BA), we prove that s1 = s2.