Wolfram Mathematica

1:57 PM

Curse my dumb brain :(

Sjoerd C. de Vries

@ColinK The best brains make the biggest errors

2

It probably might be best to leave MATLAB at the door :) I've seen people set inf=Infinity and work with inf from then on, which is quite silly imo. It invariably leads to a crapton of errors when they reopen the notebook a few days later and forget to initialize that again

@RM ...that is one peculiar habit...

BeginPackage["MatlabHabitsIClingTo`"]
Begin["`Private`"]
sin=Sin;
cos=Cos;
inf=Infinity;
...
End[];
EndPackage[];

2

:)

@RM sin indeed

2:09 PM

lol

Ugh that's gross, I'd never do that

if inf doesn't mean the same thing in mma I'm fine with that. I don't know why I didn't think of that. I've had no other problems.

like Sin[], Cos[] etc.

@RM :D I love it. So, how does one map the ode*() functions to NDSolve[]? Wait, don't answer that... :P

:D

yikes

Random segue: I wonder if anyone's ever attempted to systematically fake Mondrian paintings in Mathematica? Thus far, I've only seen the one under the "Neat Examples" for Grid[]...

This one would be more fun to fake than those griddy ones

2:20 PM

if you could fake that without any human input, it would be very impressive

I'm sure JM could lower the "crazy" dial on his Perlin noise stuff and get something close enough. The harder part is finding the right settings for "sufficiently crazy"

@RM "The harder part is finding the right settings for 'sufficiently crazy'." - sounds about right. Even Perlin himself admits that it's not that easy to predict what a texture might look like if you tweak parameters...

Tonight I'm going to try to help in teaching some programming for architects, using Processing.org

It was a last minute notice, I'm a bit worried. I haven't used Processing for a while.

Ah, so the operative words are "try to"... :)

I'm worried I might be the only one with any non-basic programming experience ... the guy who was supposed to lead the group bailed out at the last minute

2:29 PM

How long ago is "for a while"?

Some two years maybe?

I'll show them Mathematica, it'll be fun :-)

@RM Already up? Isn't it awfully early there?

...and if they manage to design a house in Mathematica... awesome.

@Szabolcs nah, it's about 7:30 where he is.

That's awfully early :-)

@Szabolcs that may be doable with procedural forms. A lot of things like that in video games are run as scripts instead of pre-rendered.

@Szabolcs I'm up at 7 EDT because the short person wakes up then.

2:36 PM

@rcollyer It is certainly doable, looks like a perfect candidate to be reproduced procedurally. But what do you mean by "procedural forms"?

@Szabolcs it is the first thing that came to mind that was reasonably close to what they actually called, which isn't coming to mind. :P

How old is s/he?

He's going to be 2 in July.

Something like L-systems?

L-system

An L-system or Lindenmayer system is a parallel rewriting system, namely a variant of a formal grammar, most famously used to model the growth processes of plant development, but also able to model the morphology of a variety of organisms. A L-system consists of an alphabet of symbols that can be used to make strings, a collection of production rules which expand each symbol into some larger string of symbols, an initial "axiom" string from which to begin construction, and a mechanism for translating the generated strings into geometric structures. L-systems can also be used to generate s...

this?

ah

2:38 PM

I was 2 seconds faster ;-)

Thank you both for having better memories.

Actually I wanted to do something like that if we get far enough

But that tree in the image looks like something different

yes it doesn't look like what is usually produced with l-system rules

I see wavy lines crossing each other, and defining regions of space which are coloured slightly differently. The branches are also wavy lines.

It is, but I imagine "with some tweaking" it could be accomplished.

2:39 PM

or at least the ones I've seen

Like a heavily distorted checkerboard.

looks like the stained glass filter in photoshop.

Not even checkerboard because it doesn't have the same "topology", if "topology" is an appropriate word here

So, build a 3D tree using an L-system, project it onto a screen, and apply a stained glass filter. Easy. :)

I just meant to pop-in briefly. Got to run. Bye.

@acl ...and that can be done with turtles.

2:48 PM

@Szabolcs I'm generally up by 6, no matter when I go to bed (as long as it's not after 1:00) :)

@Szabolcs Hi! I think I got an example working (related to this question). Could you check it out?

3:01 PM

@JM Turtles are fun :-) I was thinking of showing them turtles first if the standard OpenGL-style coordinate transformations that Processing uses prove to be too difficult

@Ajasja I'll try. Is it going to kill my tiny tiny 2 GB RAM machine? :-)

No. One run will take about 200 MB

At least I hope it won't

:)

@Ajasja I'm sorry, I'm on a 32-bit platform ... can't load the MX

Yeah, I thought that might be a problem.

@Szabolcs But when loading from a .wdx there is no memory leak.

@Ajasja What about: load wdx, export to mx, re-load mx? That way the example would be portable.

Hmm, but that would mean it is something related to importing mx files. But in my package I don't import anything.

I thought it could be something internal to the symbol written to the .mx (since this would preserve most of the internal structure)?

@Szabolcs If I import the "wdx" and export it to "mx" then the mx file is two times smaller (than the original) and does not exhibit the memory leak upon loading it.

So there must be something strange in the symbol definition, that gets preserved when saving directly to "mx" but not to "wdx"

3:15 PM

@Ajasja How do you save to MX? Export or DumpSave?

I used Export["steps.mx",steps]

Maybe you can ping ruebenko in a comment on your post, now that you have an example

If I've got a function of many variables, and I want to set the value of all but one, and then Plot[] versus the remaining variable, what syntax should I be looking for?

doesn't Plot[f[2,1,3,x],{x,0,10}] work?

Good idea. Perhaps it's also time to contact Wolfram support...

3:17 PM

@ColinK Plot[f[1,2,3,x],{x,0,10}]

hmm, possibly I'm defining my function wrong

@Szabolcs ha

well now I know where to look :)

@ColinK how are you defining it and how are you trying to call it?

I've tried a couple ways of defining it. I get an expression from Simplify[Integrate[]] and I want to assign it to a function, so I've tried f[A,s,tau,t]=%

I've also tried it with :=

and with f[A_,s_,tau_,t_]

Mathematica makes me sad :(

3:21 PM

@ColinK that's incorrect!

I expected as much :)

@ColinK that's right, but if you use %, then use =, not :=

ok

@Szabolcs it's fine if he uses Set

@ColinK but this (say) works:

f[A_, s_, tau_, t_] = A*s*tau*t;
Plot[f[1, 2, 3, t], {t, 0, 10}]

@acl He didn't use patterns in the definition (f[x] = vs f[x_] = ...)

3:23 PM

hah! I got it now

oh right!

Thanks again :)

did f[A_,s_,tau_,t_]=% and then Plot[1,2,3,t,{t,-100, 1000}] and it worked

btw, I should be able to just drop in the previous expression in place of % and the function assignment should work the same, right?

@ColinK I think that sitting down once and thinking about the actual meaning of this pattern stuff will make your life with mathematica much easier

@acl: you you're right. I've read some of the documentation about it but I'm left a little confused

@ColinK yes, also, if you use :=% instead of =% then it will try to replace with the last obtained expression each time you evaluate f[]

3:26 PM

It talks about it as if _ is simply a wild card

@ColinK And before you redefine a symbol, always Clear[] it! Try ?f to see how many definitions you have attached to it.

@acl oh yeah that would be a problem. I knew that, in retrospect. := does make sense to me

@ColinK yes, when you say f[x_]:=5x, you are saying: "wherever you see f[smth], assign smth to x, then evaluate 5x"

and patterns can become arbitrarily complicated (it's one of the central paradigms in mathematica)

@acl or more precisely, just replace x in 5x by smth (which is slightly different when we have held expressions, e.g. f[x_]:=Hold[x])

right but what confuses me is if I do f[x_,y_]=x Sin[y], how does it know which argument is x and which is y, if they are both just placeholders

3:29 PM

So x is not a local variable of the function. It's just the name of the pattern.

@Szabolcs yes that is true

it seems that they are more than that, but it's never realy stated explicitly

@ColinK you name the arguments: the first you named x and the second y.

@ColinK You have names to the placeholders, x and y. You don't need to do that, but if you want to use the expression that was matched by the placeholder, you need to give it a name so you can refer to it on the RHS.

So the first character is actually the name of the pattern?

or all the characters before the _ are the name of the placeholder?

3:31 PM

@ColinK yes, all the characters. look:

nameless pattern. then:

x = 3; f[5]
x = 6; f[1]

ClearAll[f]; f[_] := 5 x

look here:

reference.wolfram.com/mathematica/ref/Pattern.html

(and under "more information")

s_ is equivalent to s:_ which assigns the pattern object _ to s

clarifies everything doesn't it :)

lol :)

@ColinK Have you used regexes?

it makes more sense at least

yeah I know regexes

Wait, forgot to escape a space. Wheeeeee[taptaptap]eeeeee.

@ColinK It's been a while for me, but there's that thing when you surround a pattern by parens in the regex and you can refer to it using a number later (for example if you use the regex for a replacement). It's the same thing here, except it's more readable and we always use names instead of numbers.

3:36 PM

yeah that makes sense

Love XKCD

hate perl though...

nice, it's all working now.

@Szabolcs how close are you to the epic badge? (I wonder what it is for Mr.W)

@RM That ain't ever gettin' old... :D

@RM It's been ages since I hit the rep cap... probably not too close. But how can I check other than using the API?

@Szabolcs mathematica.stackexchange.com/reputation (all the way at the bottom)

@RM Close enough to give me a push *** grin ***

Now I regret a little bit deleting that +7 answer ...

3:48 PM

which one was that?

@RM This is one Yu-Sung Chang got everything right in his answer that I didn't, and mine doesn't have anything that his doesn't, so it seemed reasonable to delete it. Maybe I can make a palette to automate the process of compressing images, or a function to process an entire notebook. Then I can post another answer.

@RM Also this one. Since each triangle is separate, the transformation is not very useful, even if it is responsive. I was planning to replace it with a ParametricPlot-based one, but then I got lazy.

@JM Why underdetermined? There are more equations than variables, and it is true that the equations are only satisfied for a very specific choice of parameters. So it's overdetermined.

@Szabolcs ah, I see.

@Szabolcs I count three unknowns and two equations...

I'm just not that motivated by badges and rep any more ... which could sound like a good thing, but unfortunately my answer quality seems to have gone down as well.

"Overdetermined" would be something like {x + y == 1, x + 2y == -1, x + 3y == 2}

3:56 PM

@JM I interpreted y and z as parameters, not unknowns. We're telling Solve that x is the only unknown (we're not solving for y and z) If x were to be part of the second equation as well, then we likely really wouldn't have any solutions, regardless of the parameter values (y and z).

@Szabolcs I'm not motivated by rep and badges anymore either and should really reduce the time spent on the site (for a while at least)

@RM I capped on 40 days and reached 200 on 45.

@JM If y and z were written as a and b, would you still say underdetermined?

give it a week-10 days and you'll get there

@Szabolcs In that case...

@Szabolcs Sure, I don't know a and b. :)

I mean, we don't "know" the three of them. (The equation for y being trivial notwithstanding.)

hey, I don't know what the question is either, so add one more :P

4:01 PM

Times like these, more context would be nice...

@JM The reason I keep arguing about this is that I think this point is essential to understanding Mathematica's behaviour here. If you look at the workaround in my answer, the result is x -> ConditionalExpression[2 - z, y == 3] i.e. 2-z is a solution under the condition that y==3. Otherwise there are no values of x that make both equalities true (which sounds strange written like this one because the second equation doesn't depend on x, but it's technically correct).

y is treated as a parameter, "some number", but not as an unknown.

Okay, so we have fixed y and unknown x and z. So we still have more unknowns (2) than equations (1), right?

No, z is a parameter too. Consider it fixed.

It could be "any number".

@Szabolcs I narrowed the memory leak to a self contained example.

$HistoryLength = 0;
data = RandomReal[{-1, 1}, {10, 100000, 2}];
data = Developer`ToPackedArray[#] & /@ data;
data = Flatten[data, 1];
Dimensions[data]
HistogramList[data, 30, Automatic];
ClearAll[data]; ClearSystemCache[];

Could you try this?

Sorry, I can't take a look now, I need to make some food before I go for that generative art workshop

4:05 PM

No problem

It'll wait:)

@Szabolcs *sigh* so it's neither over- nor under- ... one equation in one unknown.

@JM Actually, two equations in one unknown. It's true that the second one doesn't contain x, but it's still two equations, and one of the two could not be "satisfied" or true.

@JM Consider Solve[False, x], it returns {}, meaning there are no solutions. One equation (written as False) and one unknown, in this case.

Other examples: Solve[False && x == 1, x] or Solve[1==2 && x == 1, x]

@Szabolcs ...thanks to short-circuit evaluation of And[], they're precisely the same as your first snippet...

But alright, I'll roll back...

@JM write it as Solve[{2 == 1, x == 1}, x] if you like, no short circuiting here.

It's "pure logic" in this case (I'm sometimes annoyed that the same functions are used to represent mathematical statements AND for programming constructs---in this case And)

How would you explain then that Solve[{y == 1, x == 1}, x] gives no solutions but Solve[{y == 1, x == 1}, {x, y}] does give one?

"the same functions are used to represent mathematical statements AND for programming constructs" - Well, it's a bit like Plus[] working for both reals and complexes...

On the other hand, we're lucky they decided to implement BitAnd[] instead of overloading And[] with yet another extension...

4:14 PM

@Ajasja every time I evaluate it here, MemoryInUse[] (after evaluation) goes up

@Ajasja I can confirm the leak. (Look like it's going to be junk food, no time to cook something decent)

It goes up by close to 200 MB every time.

so if I do this:
lst = {};
Dynamic[AppendTo[lst, MemoryInUse[]]; ListLinePlot[lst],
TrackedSymbols \[Rule] {}, UpdateInterval \[Rule] 2]

and then repeatedly evaluate @Ajasja's code, I get this

Something goes wrong inside HistogramList? It's implemented in Mathematica ...

5:01 PM

Hmm, another PlotLegend question. Why couldn't WRI keep that undocumented instead of functions that are actually useful.

...should we have a tag for legend-specific questions?

5:36 PM

maybe we should petition to make PlotLegend* a blacklisted term :P

8:05 PM

Maybe this will turn into a real question on the site but I need some help formulating it

I want to write a ComplexityFunction that will inspect the value of each term in an expression, and penalize expressions which result from numerically unstable operations.

Such as the multiplication of a very large and very small term to result in a small-ish result

(yes, this is in reference to my earlier question on numerical stability)

is that possible and/or reasonable?

It would effectively function as a general purpose re-formulator for numerical expressions

which would be awesome

How do you intend to use it? ComplexityFunction is used in Simplify et al. but I'm not certain you can inject it (if you can, I don't know how) into the internal operations done by Mathematica

do your expressions start out symbolically before you substitute numerical values?

@RM I intended to use it with FullSimplify, Sorry I wasn't clear about that

Yes, I'm using Mathematica to help me do some derivatives and algebraic manipulation to produce equations that I'll use later on in a numerical optimization, written in matlab

8:22 PM

Hmmm... might be possible, although I don't have an easy answer right now. Do you have an example (a toy one) where you'd like an alternate simplification based on numerical stability than what's given by LeafCount?

Sure. If you take my actual problem, and strip out all the parameters and the second exponential decay, I've got a fit function that looks like: Exp[-t] Erfc[-t]. This can be rewritten as Exp[-t Log[Erfc[-t]]]

The original form has problems because Exp[-t] gets huge when t -> -Infinity, while Erfc[-t] becomes very small. Their product is still small-ish, but the floating point rounding error is horrible.

In the second form, -t Log[Erfc[-t]] is well behaved everywhere, and taking Exp[] of it has no problems. Essentially it avoids creating a gargantuan value at some intermediate point in the calculation

That should be something that can be expressed in a repeatable algorithm

Hmm... for starters, you could try to weave something like, say, MachineNumberQ[Exp[-t] Erfc[-t] /. t -> 100 // N] into a complexity function

where 100 is the largest value of t you expect to input

Hmm, thats a nice idea

in general it would require correct scaling to fit with LeafCount, but its a very good start

oh wait, there is one issue

yeah, just throw in a large penalty if it's false

the value of the function itself, Exp[-t] Erfc[-t] never becomes large

it's the value of its components that is a problem. So I'd need to have some systematic way of breaking it down into {Exp[-t], Erfc[-t]} and evaluating them each separately.

8:31 PM

Well then take Level[Exp[-t] Erfc[-t], 1] and try on each

I've got to look up Level[]

And @@ MachineNumberQ /@ (Level[Exp[-t] Erfc[-t], 1] /. t -> 1000 // N)

False for 1000 and True for 100...

oooh, this works interestingly: Max[Level[p[8000, 25, 5, t], 1] /. t -> {-1000, 1000}]

p is my function with some of the parameters that I've ignored while we are tlaking, thats what the 8000,25,5 are

I think this might work. Now I just need to add some extra terms for the absolute value and the inverse values

it's only a start anyway :) There still will be corner cases, since these are machine numbers that we're dealing with, but I'd be interested to see how it progresses.

@RM Oh I didn't see this message. And I don't quite understand it

What is /@ in mma?

8:40 PM

Simply put, /@ maps a function onto a list... so f /@ {1, 2, 3} is shorthand for an operation that results in {f[1], f[2], f[3]}

oh, cool thats usefull

actually it's not machine numbers until it leaves mathematica

even if I N[] it, its arbitrary precision, so I can still meaninfully compare values while FullSimplify works

Got it: Max[Join[Abs[Level[p[8000, 25, 5, t], 1]] , 1/Abs[Level[p[8000, 25, 5, t], 1] ]] /. t -> {-1000, 1000}]

So in that I check if the two values in the list are each machine numbers and then And it, with is a quick equivalent of all in MATLAB

@ColinK see docs for N:

> N[expr] gives a machine-precision number, so long as its magnitude is between $MinMachineNumber and $MaxMachineNumber.

oh I see, it's just a true/false to see if it is a machine number or not

thats close but not quite what I need

@ColinK That's MachineNumberQ. the And is to check if both are...

@RM Yes, but when the value exceeds the machine number limit it seems to be going to arbitrary precision. For example one of my terms hits 10^441

hmmm, although maybe not

grr, so close :)

8:48 PM

@ColinK then MachineNumberQ will give you False

9:02 PM

@RM, is there any way to perform If[] on a list? I want If[{true, true, false},1,0] to give {1,1,0}

nevermind, got it

Mr. Hypnotoad.

Verbeia

@ColinK Boole[] - there are also things like DiracDelta and UnitStep that become relevant depending on what you are testing.

Boole[] did the trick

Verbeia

I'm planning on writing a blog post about these functions, once the blog is up.

I'm still having some problem but I think its just some little bug, atm.

That would be awesome

9:12 PM

Does anyone know how to make this?

@Szabolcs it's there in the examples

I know about the example in Texture's doc page, but that only renders some slices while this is a solid body.

Yeah but you're going to need some superconducting coils and an RF antenna

oh

@RM The picture I posted is not the same as the example Sjoerd has shown tonight

This picture shows a solid 3D body while that example only shows slices

Should have asked during the visualization QA session

9:18 PM

@Szabolcs wouldn't have mattered... they weren't going to answer it anyway

9:32 PM

Graphics3D[{Texture[data], EdgeForm[None],
Table[{Polygon[{{x, 0, 0}, {x, 1, 0}, {x, 1, 1}, {x, 0, 1}},
VertexTextureCoordinates -> {{x, 0, 0}, {x, 1, 0}, {x, 1, 1}, {x,
0, 1}}],
Polygon[{{0, y, 0}, {1, y, 0}, {1, y, 1}, {0, y, 1}},
VertexTextureCoordinates -> {{0, y, 0}, {1, y, 0}, {1, y, 1}, {0,
y, 1}}],
Polygon[{{0, 0, z}, {1, 0, z}, {1, 1, z}, {0, 1, z}},
VertexTextureCoordinates -> {{0, 0, z}, {1, 0, z}, {1, 1, z}, {0,
1, z}}]}, {x, 0, 1, 0.1}, {y, 0, 1, 0.1}, {z, 0, 1, 0.1}]},
Background -> Black, RotationAction -> "Clip"]

make the table finer if you have the crunching power

@RM I tried that code, and now Mathematica has almost ground down to a halt. I must have very soggy power.

BTW, why is your avatar a frog?

@Heike futurama.wikia.com/wiki/Hypnotoad

@RM Oh sorry, why is your avatar a toa.... ALL GLORY TO THE HYPNOTOAD!

3

9:47 PM

Hmm... I wonder if it'll render with Cuboid[]

nope.

@RM Can you have textures with cuboids?

Yeah, forgot about that. Needs the VTC to be set

@Heike influenced by that nasty toad-like yoda character, no doubt. glad he's gone

Could use RegionPlot3D[True, {x, 0, 1}, {y, 0, 1}, {z, 0, 1}], but haven't gotten the right TextureCoordinateFunction yet

@acl You mean that green wrinkly creature that lives in a swamp?

9:57 PM

@Heike that, yes

In what language would one use a word similar to equotations to mean equations?

equality for all quotations

@Heike "equotations"? as in the english expression, or the putative language's version of equotations?

@acl I haven't seen him around much lately. Except to lure people away from this site.

@Heike oh yes. it also abuses this forum for its own green purposes.

nasty creature, I tell you

9:59 PM

@acl is equotation an English word?

@Heike e-quotation

damn, there it is!

Oh no! Maybe he has spies in this room

@acl pfft you... talking ill about people behind their backs :P

does anyone hear a voice?

just you going crazy...

10:02 PM

@acl This person definitely means equation but writes equotation. He also uses y-achses so I suspect a German link

y-achses?

for y-axis

@Heike that sounds german, but equotation...

The source is here

@Heike ohh... Then ArschTan must either be the inverse tangent or they spent too much time baking in the sun

10:06 PM

@yoda A nudist beach most likely

most certainly

@yoda why don't you go back to your slimy gardening forum, disgusting little reptile?

Although you're more likely to come across sine curves than arctan there.

@Heike the name's german too. who knows about the equotation... maybe a spell-checker accident?

@Heike More like Abs@Sin, but then depends on where each of us define our functions

10:08 PM

@yoda ha I was trying to work out some way to say that

@acl I'm slowly making my way to 10k there. Why don't you go out and spend some time in the backyard? (and ask questions on g.se :P)

@acl Maybe. The german word for equation is nothing like equotation.

@Heike Yes I know. And he uses it consistently, so I am suspecting a spell-checker that is set to automatic and has been taught "equotation" by mistake. or something (as I said).

That must be a rubbish spell checker then. He spelled gues wrong, as well.

Eek, I'm less than 300 rep away from Mr.Wizard. Time to slow down I think.