|
|
|
![]() |
|
|||||||
|
||||||||
![]() |
| Thread Tools | Rate Thread | Display Modes |
|
#1
|
||||
|
||||
|
Stack Height Scoring Analysis
Question: Is there an optimum stack height?
Answer: Yes, there is. 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 |
|
#2
|
|||
|
|||
|
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 (c) not releasing the crate before time expires, in which case I don't believe it counts towards your score at all. |
|
#3
|
||||
|
||||
|
Quote:
.-Kevin |
|
#4
|
|||||
|
|||||
|
Quote:
![]() icky, broken stuff! Quote:
![]() Last edited by DaBruteForceGuy : 07-01-2003 at 15:23. |
|
#5
|
|||
|
|||
|
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 A*B=B*A), we prove that s1 = s2. |
![]() |
| Thread Tools | |
| Display Modes | Rate This Thread |
|
|
Similar Threads
|
||||
| Thread | Thread Starter | Forum | Replies | Last Post |
| Initial stack height | JJG13 | Rules/Strategy | 1 | 29-01-2003 21:04 |
| Stack Attack PDA Scoring Calculator | Matt Starkey | Rules/Strategy | 15 | 23-01-2003 10:52 |
| Multiplier stack scoring change??? | Jferrante | Rules/Strategy | 7 | 08-01-2003 12:23 |
| Stack height Poll (spawn from other thread) | Andrew Rudolph | Chit-Chat | 5 | 07-01-2003 17:15 |
| Scoring when an opponent is in contact with your tallest stack. | Randy Ai | Rules/Strategy | 1 | 06-01-2003 17:17 |