[MATH] Method to reduce both sides of an equation?

[MATH] Method to reduce both sides of an equation? - Printable Version

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RE: [MATH] Method to reduce both sides of an equation? - TheouAegis - 01-25-2011

No, I said 20mod50=70mod50=20, in other words 20 and 70 would yield the same result. Modulo yields an M-shaped pattern, which won't work for me.

I think what I want may just be impossible. But it seems to be intriguing a couple of you for some reason, so I'll keep testing some ideas while you continue to try to make sense of what I'm getting at.

RE: [MATH] Method to reduce both sides of an equation? - DarkWolf - 01-26-2011

Sorry mis-read your original post.

RE: [MATH] Method to reduce both sides of an equation? - Rosencrantz - 01-26-2011

(01-23-2011, 10:32 PM)TheouAegis Wrote: That is what I want, ATK-DF=HP_Lost. The problem is the values I'm working with were not originally set up that way, so I'm trying to convert the formula to that simpler one. However, I would also like it to be a symmetrically scaled adjustment, even if it is bell-shaped. But I can mess with the shape if I ever find a way to simplify the formula. But that's why I don't want to use "If x>50, x=50" because that would yield an "r"-shaped curve. I want an omega-shaped, J-shaped or L-shaped curve, depending on whatever's closest to the original shape. Since the orignal was fairly linear, I think an "S"-shaped curve will be what I end up getting. with the bottom of the S being 1 and the top being 50.

Oh, I think I see what you're getting at now.

You want a graph that works almost like a sin function, where it starts at 0, increases up until a halfway point like a parabola and then levels off with a parabola at 50.

Code:
.........__

.......,=..

....../....

.....|.....

..../......

__=*.......

like this shape, right?

RE: [MATH] Method to reduce both sides of an equation? - Jdaster64 - 01-26-2011

Not sure if this'll help, but basically I took the values given (ATK 2-99 and DEF 4-120) and the equation (3/2*A - 1/3*B), and fit them to half the first a period of a y-flipped cosine function that osciallates between the values 1 and 50 (25.5 - 24.5*cos(x), to be exact). Using the equation and values I gave, I calculated (and you can check me on this) that the maximum of (3/2*A - 1/3*B) is 147.166... (99*1.5 - 4/3) and its minimum is -37 (2*1.5 - 120/3).

This yielded the following formula (presuming the cosine's calculated by radians): y = 25.5 – 24.5*cos(.01744*(3/2*A + 1/3*B + 37)). I ran it through my calculator and it seemed to check out. Granted, this may not be the best way to approach it; notably, there might better be a longer tail on the low side. Just thought I'd throw this out here anyhow.

EDIT: For one with a longer low tail (to account for the negative and low vaules), try this formula: y = 25.5 – 24.5*cos(.00009678*(3/2*A + 1/3*B + 37)^2).
EDIT again, here's what the second graph would look like. The value for (3/2*A - 1/3*B) becomes positive about 20% of the way across the graph. Sorry about the JPEGness, btw.

[Image: ScreenHunter_01Jan262002.jpg]

RE: [MATH] Method to reduce both sides of an equation? - TheouAegis - 02-06-2011

Just wanted to reply that I'm considering that sine-function method. That never occurred to me, thanks. I knew if I posted on here there was a good chance someone more math-minded than me would come up with something creative. I haven't dealt with higher math in 10 years.