Cycling 74 » Topic: simple expr math

simple expr math

n00b_meister — Thu, 26 Jan 2012 20:18:59 +0000

Hi
brain not working

how can i achieve the following, preferably with expr
input: 1, output 0.1
input: 5, output 1.
input: 10, output 10.

I mean, this would do:

– Pasted Max Patch, click to expand. –

but it’s a bit, erm what’s the word, grokky?

Brendan

Re: simple expr math

big_pause — Thu, 26 Jan 2012 20:56:23 +0000

see this post,

http://cycling74.com/forums/topic.php?id=37464

not as straightforward, but same idea should get you there.

* Hint you have 3 variables and 3 equations :)

Re: simple expr math

Chris Muir — Thu, 26 Jan 2012 21:32:55 +0000

Also a bit kludgey, but:

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Re: simple expr math

big_pause — Thu, 26 Jan 2012 21:43:21 +0000

Here you go

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Re: simple expr math

n00b_meister — Thu, 26 Jan 2012 21:50:44 +0000

Thanks chums

Chris – split, of course!

big_pause – I saw that thread too, but ran screaming from the equations, thanks for filling in the blanks – ps how the hell did you arrive at that function?? Give me your brain – NOW, mwah ha ha

2 excellent solutions (3 if you count mine LOL)

Brendan

Re: simple expr math

big_pause — Thu, 26 Jan 2012 22:38:27 +0000

it’s pretty much the simplest functional form that fits the criteria.

Re: simple expr math

jamesson — Fri, 27 Jan 2012 00:37:40 +0000

big_pause: but how did you count the coefficients? That is indeed pretty awesome

Re: simple expr math

xidance — Fri, 27 Jan 2012 06:24:19 +0000

whenever you have 3 points you can go with quadratic equation (the expression big_pause estimated) by that substitution algo from school (at least in germany).

Re: simple expr math

jamesson — Fri, 27 Jan 2012 06:35:11 +0000

yea merkin math suxorz

Re: simple expr math

big_pause — Fri, 27 Jan 2012 10:54:54 +0000

xidance pretty much has it

given the choice of function

y = a + bx^c

and

y = 0.1, x = 1
y = 1, x = 5
y = 10, x = 10

gives us 3 equations

0.1 = a + b*1^c = a + b => a = 0.1 – b [1]
1 = a + b*5^c [2]
10 = a + b*10^c [3]

substituting [1] into [2] (you could sub into [3] here, doesn’t matter)

1 = 0.1 – b + b*5^c
0.9 + b = b*5^c

take logs of both sides, to give an expression for c in terms of b

ln(0.9 + b) = ln(b*5^c)
ln(0.9 + b) = ln(b) c*ln(5)
c = (ln(0.9 + b) – ln(b)) / ln(5) [4]

then, substituting [1] and [4] into [3], lets us find b

10 = 0.1 – b + b*10^((ln(0.9 + b) – ln(b)) / ln(5))
0 = -9.9 – b + b*10^((ln(0.9 + b) – ln(b)) / ln(5))

at this stage I just ran through a quick hacked python script to find b (life’s too short)

once you have b, you can use [1] to get a, then either [2] or [3] to get c.

yes, I’m a nerd.

Re: simple expr math

yp — Fri, 27 Jan 2012 11:43:47 +0000

bp: this is awesome :)

Re: simple expr math

big_pause — Fri, 27 Jan 2012 20:09:28 +0000

so, I thought about this a little more. We can transform our linear inputs to this, and get the same problem as in the ‘other’ post

http://cycling74.com/forums/topic.php?id=37464

here’s what we do. we have as our inputs (x’s) that need mapping, [1, 5, 10].

we shift this down to [0, 4, 9], then normalise to [0, 4/9, 1], or algorithmically

[x1, x2, x3] -> [(x1-x1)/(x3-x1), (x2-x1)/(x3-x1), (x3-x1)/(x3-x1)] = [x1', x2', x3'] and these map = [y1, y2, y3]

we now have a similar problem as in the post above, and can get the solution to

y = a + b*x’^c : where x’ = (x – x1)/(x3 – x1)

for our new mappings (both for this case, and general algorithm)

we have, for x1 = 1, x1′ = 0, y = 0.1

a = 0.1 or y1

for x3 = 10, x3′ = 1, y = 10

y3 = 10 = a + b => b = 9.9 or y3 – y1

and solving for x2 = 4, x2′ = 4/9

1 = 0.1 + 9.9*(4/9)^c
0.9/9.9 = (4/9)^c
c = ln(0.9/9.9)/ln(4/9)

or
y2 = y1 + (y3 – y1)*x2′^c
c = ln((y2 – y1)/(y3 – y1))/ln(x2′)

so for our specific case, we can write

y = 0.1 + 9.9*x’^{ln(0.9/9.9)/ln(4/9)}
y = 0.1 + 9.9*[(x - 1)/9]^{ln(0.9/9.9)/ln(4/9)}

and for the general case

y = y1 + (y3 – y1)*x’^{ln((y2 – y1)/(y3 – y1))/ln(x2′)}

with x’ = (x – x1)/(x3 – x1) and x2′ = (x2 – x1)/(x3 – x1)

as before.

and we can chuck this straight in an single expr – happy days

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Re: simple expr math

Roman Thilenius — Fri, 27 Jan 2012 23:42:40 +0000

to find a ln or exp expression starting from 3 points – except when they are variables – you
could also just try manually which coefficient matches.

Re: simple expr math

big_pause — Fri, 27 Jan 2012 23:51:20 +0000

* waits for the punchline *

Re: simple expr math

Roman Thilenius — Sat, 28 Jan 2012 00:16:28 +0000

punch and judy are off