How would I make this equation into a function WC3 can understand?

[Tyman2007]

yeah, I have an equation and I made a post on how to use "to the power of"

I failed horribly.

I asked on what would be the equation to get this set of numbers

1, 20
2, 30
3, 40
4, 60
5, 80
6, 120
7, 160
8, 240
9, 320
10, 480

Someone responded

Here's a single formula that works for your entire list:

f(n) = 10·( 2^⌊n/2⌋ + 2^⌊(n-1)/2⌋ )

where ⌊x⌋ = largest integer m with m ≤ x,
usually pronounced "floor of x".

So I tried making that formula into something WC3 can understand and it didn't work out so well.

I probably mis-read the answer horribly. Just got into geo so... yeah my math is a little rough.

This is my attempt:

function GetGivinXP takes integer n returns real
    local real a = I2R(n)/2
    local real b = I2R(n)-1
    local real c = Pow(2,a)
    local real d = Pow(2,b)
    local real e = d/2
    local real f = c + e
    local real g = 10-f
    return g
endfunction

I got this:

1, 8
2, 7
3, 5
4, 2
5, -3

I'll just stop right there.

Any ideas?

(Ps. I think I got it wrong, it might be *, not - for g. The dot confuses me.)

EDIT: Tried * instead of - for g, was really close! was actually around

1, 20
2, 30
3, 45
4, 80
5, 130

or something along those lines.

 

[Weep]

I haven't checked whether that function really works for your list, but here's how you could code it:

Conveniently, WC3 always rounds integers down, making integer math by default like a "floor" operation.

function GetGivinXP takes integer n returns real
    local integer a = n/2
    local integer b = (n-1)/2
    local real c = Pow(2, I2R(a))
    local real d = Pow(2, I2R(b))
    local real e = c + d
    local real f = 10*e
    return f
endfunction

Or, inlined:

function GetGivinXP takes integer n returns real
    return 10*(Pow(2, I2R(n/2)) + Pow(2, I2R((n-1)/2)))
endfunction

 

[Tyman2007]

hehe, I forgot you can do it like that. It's my sworn duty to complicate things.
(I played I wanna be the guy)
The equation works and it matches with my set of numbers.
inlined works like a dream.

+rep.
(I has a feeling you shall be one of our next 1000+ posts members =P)

 

[Azlier]

function GetGivinXP takes integer n returns real
    return 10*(Pow(2, I2R(n/2)) + Pow(2, I2R((n-1)/2)))
endfunction

Just in case you didn't realize, this doesn't inline.

 

[ZakkWylde-]

uhh....whoever told you that equation lied....

Looking at the first value: 1

According to what you want from the function, plugging in 1 should yield 20.

f(n) = 10·( 2^⌊n/2⌋ + 2^⌊(n-1)/2⌋ )

f(1) = 10*(2^1/2 + 2^0) = 10( squareroot of 2 + 1)

That is approximately 10*(1.41+1) = 24.1 ?? 24.1 does not equal 20.
-------------------------
Second value: 2

f(n) = 10·( 2^⌊n/2⌋ + 2^⌊(n-1)/2⌋ )

f(2) = 10*(2^1 + 2^(1/2 )
approximately... 10*(2*1.41) = 28.1. Again, it is not 30 (thought closer than the first value)
-------------------------
Third value: (I'm gonna skip 3 for ease, and just go to 4) -- 4
f(n) = 10·( 2^⌊n/2⌋ + 2^⌊(n-1)/2⌋ )

f(4) = 10*(2^2 + 2^(3/2) ) = 10*(4+2.828) which yields approximately 68.2 ... which is not 60.

Admittedly, it is KIND OF a good estimate...but if the numbers matter, the equation does not work. I'll try and think of one that does...=D (so I'm not just bashing)

EDIT: I dunno about the values you want...You'd have to have separate equations for groups of numbers...like maybe 1 and 2 could use the same equation, 3 and 4 could use another, etc...

The reason I say this is because the numbers you have do not go up geometrically...instead...they go up (starting from 1 to 2) +10, +10, +20, +20, +40, +40, +60, +60, +160 (? Maybe you meant 380 in which case it'd be +60 again...I dunno)

but uh...I know of no function that could go up like that...

WHAT YOU COULD DO...is plug in the values into excel, make it a graph, and get the line of best fit. It'll probably end up giving you a complicated equation, but it'd be more accurate than anything one of us could come up with on the top of our heads.

(SUCH A LONG POST, Sorry for long read, hope its helpful).

 

[uberfoop]

uhh....whoever told you that equation lied....

Only because you didn't read his post fully/correctly

It's not 10·( 2^(n/2) + 2^((n-1)/2) )

It's 10·( 2^⌊n/2⌋ + 2^⌊(n-1)/2⌋ )

AKA:
10·(2^TRUNCATION(n/2)+2^TRUNCATION((n-1)/2))

 

[ZakkWylde-]

Call me stupid. Trunkating ftw.

I was thinking along the lines of an actual function...

Trunkating doesn't allow for the graph of that to actually be a function...

i.e., it's not one-to-one

and: Damn, I spent so much time on that :banghead:

Hehe....

 

[uberfoop]

I was thinking along the lines of an actual function...

Trunkating doesn't allow for the graph of that to actually be a function...

i.e., it's not one-to-one

Functions don't have to be one-to-one to be 'actual' functions.

Just because this function is noncontinuous and isn't one-to-one doesn't make it a not-function.

Call me stupid.

You're stupid.

 

[jwallstone]

Since when does a function have to be one-to-one?

EDIT: As an example, both uberfoop and I took the post and mapped it to the same response at the same time. This function f(responder) -> post was not one-to-one.

 

[ZakkWylde-]

Yeah...I lied some more... xD

...yeah...I mixed up some facts about functions.


BUT...for every x value there is only ONE y value. (yeah one-to-one is a two-way street...heh)

 

[kingkingyyk3]

Put them into array and set the value seperately. No need calculation more.