Wolfram Mathematica

moretriangles = Polygon[{a_, b_, c_}] :>
With[{ab = (a + b)/2, bc = (b + c)/2, ca = (c + a)/2}, {
Polygon[{a, ab, ca}],
Polygon[{ab, b, bc}],
Polygon[{c, ca, ab}],
Polygon[{c, ab, bc}]
}];

how do i actually plot that, to see the resulting graph?

belisarius

for example ... most answers here are really functional and complete programs

@Jin wow ! did you write that?

Jin

@belisarius yes. i'm a faster learner

j/k. it's from meta.mathematica.stackexchange.com/a/395/96

belisarius

mmmm

ok! that is pretty advanced code

let me see how to use it

it is a rule

you need a 'substrate" to use it

a Plygon

a triangle probably

let me try

@Jin for example

Graphics[Polygon[{{1, 0}, {0, Sqrt[3]}, {-1, 0}}] /. moretriangles]

it subdivides a given triangle

Did it work?

Jin

hmm no

belisarius

Run this

moretriangles = Polygon[{a_, b_, c_}] :>
With[{ab = (a + b)/2, bc = (b + c)/2, ca = (c + a)/2}, {
Polygon[{a, ab, ca}],
Polygon[{ab, b, bc}],
Polygon[{c, ca, ab}],
Polygon[{c, ab, bc}]
}];

///

with shift / enter

and then

////

Graphics[Polygon[{{1, 0}, {0, Sqrt[3]}, {-1, 0}}] /. moretriangles]

///

also shift/enter

Jin

4:59 AM

it just plotted a black triangle

belisarius

subdivided

that is what it does

Jin

yes that's what i see too

but i was expecting that

belisarius

nono, that seems a composition of these triangles

five of them, rotated

Jin

oh i see.

belisarius

I think I can make the code

Jin

5:02 AM

also, how do i make the shape not filled in black, but outlined

belisarius

do you want it?

Jin

you don't have to. i'm just copying some code from meta. i think i'll go start with something simpler

found some tutorials on youtube, watching them now

this is fun!

belisarius

ColorNegate@
Rasterize@
Graphics[Polygon[{{1, 0}, {0, Sqrt[3]}, {-1, 0}}] /. moretriangles]

this is a way

Jin

ColorNegate@ just inverts the colors. what i wanted was having outlines. like the edges are stroked in black lines

belisarius

This is better

Graphics[{FaceForm[White], EdgeForm@Black,
Polygon[{{1, 0}, {0, Sqrt[3]}, {-1, 0}}]} /. moretriangles]

Jin

5:05 AM

that's it. thanks!

belisarius

because you don't convert to an image

R.M

5:28 AM

@Jin start with a pentagon for that instead of a triangle

If you're using wxffles' code, there's some stuff that's left unsaid between successive lines (the whole thing is there at the end). Here's what'll produce that image above

p = Polygon[Table[N[{Cos[t], Sin[t]}], {t, π/10, 2 π, (2 π)/5}]];
triangulate = Polygon[v_] :> (Polygon[Append[#, Mean[v]]] & /@ Partition[v, 2, 1, {1, 1}]);
moretriangles = Polygon[{a_, b_, c_}] :>
   With[{ab = (a + b)/2, bc = (b + c)/2, ca = (c + a)/2}, {
     Polygon[{a, ab, ca}],
     Polygon[{ab, b, bc}],
     Polygon[{c, ca, ab}],
     Polygon[{c, ab, bc}]
   }];
Graphics[p /. triangulate /. moretriangles]

belisarius

@R.M Yep, I haven't read the original code

I think Jin is a great prospect!

:D

R.M

@belisarius Yeah, rules to manipulate polygons is pretty powerful. I just used something similar in my last answer and was reminded of this

Jin

@R.M awesome! thanks!

i feel like i just found a new toy... haven't felt this way since.. Illustrator days

belisarius

@R.M I was going to suggest to him to start with a triangle with a 2 Pi/5 angle and rotate it. Strarting with a pentagon is much more natural

@Jin Mma is a curious language. Perhaps unique in many ways. It requires some time and love to become your partner, but the things you can accomplish with a few lines are amazing

Jin

5:50 AM

@belisarius sounds like a girlfriend

belisarius

@Jin More or less. Depends on your luck with girls :)

R.M

@Jin Nah. For the girlfriend experience, evaluate On[] and then run the code above

belisarius

:549955 You, sexist!

R.M

:D

belisarius

@R.M One thing is true, anyway. A girlfriend also has a a steep learning curve

or a lot of curves, it depends

Jin

5:57 AM

speaking of girlfriend, i'm using the trial version now. may have to upgrade her to "wife" status tomorrow

R.M

Tomorrow? Wow, you move fast, man!

belisarius

@Jin Be careful with the rule girlfriend -> bride -> wife ->widow

R.M

I would've waited the full 30 days, but that might be because I'm a college cheapo

are you getting the home edition?

Jin

@R.M I know I definitely will use Mathematica for my design stuff, another person in the company may use it for data visualizing stuff.

so if i can convince him to learn how to use M, then i'll get the business version

if not, then i'll get the home edition

R.M

\o/ While you're at it, convince SE to order excess and then give away the rest to some of the top users :P

belisarius

5:59 AM

@Jin Data visualizing is easier than graphics design! You will convince him easy

R.M

btw, version 9 is coming soon (sept/oct), so keep that in mind too

belisarius

Also, count on us!

Jin

i love numbers... crunching numbers.. graphing them...

which probably explains why I play Eve Online, aka Spreadsheet MMO.

belisarius

@Jin Mma is a breeze for hat

ehh

that

>)

Szabolcs

Good morning/evening :-)

Jin

6:02 AM

@Szabolcs good 2am

belisarius

@Szabolcs Hi !

Szabolcs

(it's sleepy morning here)

belisarius

3 am here

Jin

coffee time

and bacon

Graphics[FaceForm[White], EdgeForm@Black,
p /. triangulate /. moretriangles] breaks things

p = Polygon[
Table[N[{Cos[t], Sin[t]}], {t, \[Pi]/10, 2 \[Pi], (2 \[Pi])/5}]];
triangulate =
Polygon[v_] :> (Polygon[Append[#, Mean[v]]] & /@
Partition[v, 2, 1, {1, 1}]);
moretriangles =
Polygon[{a_, b_, c_}] :>
With[{ab = (a + b)/2, bc = (b + c)/2,
ca = (c + a)/2}, {Polygon[{a, ab, ca}], Polygon[{ab, b, bc}],
Polygon[{c, ca, ab}], Polygon[{c, ab, bc}]}];
Graphics[FaceForm[White], EdgeForm@Black,
p /. triangulate /. moretriangles]

belisarius

Graphics[{FaceForm[White], EdgeForm@Black, p} /. triangulate /.
moretriangles]

{} missing

Jin

6:08 AM

is there a good tutorial site that teach how to plot basic geometric shapes, and manipulate them?

belisarius

Errr ...

Jin

i'm reading reference.wolfram.com/mathematica/tutorial/… now

belisarius

@Jin That one will not help with drawing

@Jin Perhaps RM or Verbeia could suggest you some readings

Szabolcs

Maybe somebody should write a tutorial style blog post on that. Start from zero, learn graphics primitives, and draw some "cool things" at the end.

belisarius

@R.M Heyyyyy

wake up

@Szabolcs Preferably something not easy to do with a WYSIWYG tool

Szabolcs

6:13 AM

@Jin Vector graphics are stored as a (nested) list of graphics primitives. For example, Graphics[{Circle[{0,0},1], Line[{{0,0},{1,1}}]}]. The {...} thing is a "list" (just an array)

Then you can manipulate them the same way as any other Mathematica expression, which is a much broader topic ...

coffee ..

As I see it, the main practical difference from systems like Processing is that Mathematica works with vector graphics only. Processing and other similar drawing programs work with raster: once you draw something, it is rasterized, and doesn't take time to redraw it. This affects performance.

Jin

I thought processing is vector? doesn't it output to canvas or svg?

Szabolcs

@Jin It does, but the typical way of working with it is to draw onto a bitmap canvas right away and lose vector information as soon as an object is drawn. This makes implementing things like motion blur quite natural.

In Mathematica it's common to transform graphics after they are created (in contrast to recreating them in a different way)

Also, Mathematica wasn't really made for animation, especially real-time animation. It's not possible to make smooth and complex animations like in processing (maybe in the future ...)

I'm not very good at Processing, but once I tried to teach some computer graphics to art students who never programmed before. I found that the things that are easy in Mathematica and easy/convenient in Processing are quite different.

Jin

just to be sure, you're talking about processingjs.org right?

Szabolcs

yes

belisarius

6:34 AM

Graphics[Circle @@@
FoldList[{{ArcTan[#[[1]]/#[[2]]],
ArcTan[#[[1]]/#[[2]]]} &@(Plus @@@ #1), #2} &, {{1, 1}, 1},
Range@10]]

////

Graphics[Circle @@@
FoldList[{{ArcTan[#[[1]]/#[[2]]],
ArcTan[#[[2]]/#[[1]]]} &@(Plus @@@ #1), #2} &, {{1, 1}, 1},
Range@10][[2 ;;]]]

Szabolcs

So has anyone posted the hedcut question?

I want more questions about making pretty things!

Okay, so before I'm allowed to ask it, I have to try to come up with something myself ...

Szabolcs

7:10 AM

Perhaps the mods can request that Mathematica.SE be included in the list of target migration sites? (The list that's shown when voting to close as off topic). @MrW @JM @Sjoerd

It's clearly the most common migration target, followed by SO.

Look what I just found:

Szabolcs

7:31 AM

Is there a function which computes a third image from two starting images of the same size based on a custom function? Sort of like a two-argument ImageApply

Szabolcs

7:50 AM

I can't make sense of what coordinate system PixelValue uses ...

Szabolcs

8:01 AM

Szabolcs

10:02 AM

@Jin I gave it a try:

6

Q: How to create the hedcut effect?

Yesterday the hedcut style was brought up in chat. How can we create a hedcut-like style automatically in Mathematica? I am looking to create a similar artistic feel, not necessarily reproduce the hedcut style precisely. Relevant resources: Hedcut-like effect in Photoshop A blog showing ...

Szabolcs

11:48 AM

@Jin Do you think you'll be able to fix the ugly scrollbar issue in the WordPress theme? Or is it too difficult? I'm asking to know how to format code-heavy blog posts. It's better to use inline code if there would be too many scrollbars

Szabolcs

2:39 PM

I think (and site) this room is the least active when I have time to be here. About 7-16 UTC.

magma

@Szabolcs which wordpress theme?

Szabolcs

2:56 PM

@magma mathematica.blogoverflow.com

@magma it's like this

magma

@Szabolcs the blog is still empty. But what exactly is the point of having a blog?

we have the Q&A, the chat, the meta and now the blog?

Szabolcs

The blog is for a more general audience. One doesn't need to participate on Mma.SE to find the blog interesting. It's for those posts that people sometimes write as answers, but which as really long self-contained pieces which are interested to more people than the asker.

Several people said they're interested in writing posts, and many more are interested in reading.

It's just another Mma blog with several contributors. We can have lots of things that don't fit the main site, e.g. tutorials, stories about solving problems

@magma you can read some other sites' blogs on Blogoverflow

@belisarius I need you to switch off the Visual editor in WordPress too (since it looks like you have editor rights too---to avoid destroying posts by accident) --- see the blog chatroom

belisarius

3:13 PM

@Szabolcs Ok. I am still preparing my post offline, so it will take some time until I destroy anything :)

Szabolcs

@belisarius Good, keep it offline. The WP editor is worse than I imagined. The reason I asked you is that you are able to edit others' posts too and occasionally the system seems to load in visual editing mode which automatically strips out some "non-standard" formatting (e.g. MarkDown code blocks, but not only)

belisarius

@Szabolcs I guess I will do all editing offline so

acl

@R.M and other OS X users, if/when you switch your OS to 10.8 can you let us know if there are any problems with mathematica?

Szabolcs

3:34 PM

@acl Hello OS X user! Have you uploaded anything with the image uploader palette recently? If yes, could you send me a screenshot of your History window? I suspect there may be a problem / misalignment on some platforms, but I'm not sure.

Jin

@Szabolcs the scrollbar has been fixed on dev, just waiting on a production build

Szabolcs

Great news! Just in time for tomorrow blog post

R.M

@acl Yeah, sure. I'll get a new laptop in a couple of weeks and that should have ML, so will let you know then (current one doesn't support it)

@Jin This close enough for a Hedcut?

2

A: How to create hedcut style images?

Here's a first-pass at a Hedcut style image using Mathematica This is how it was done: First, obtain the image and convert to gray scale: img = ColorConvert[Import["http://i.stack.imgur.com/RjE3P.jpg"], "Grayscale"]; Next, create a disk layer and give it a wavy pattern: dots = Graphics[...

Jin

@R.M hmm no.

the problem I see is that, it's hard to use an automated program to generated hedcut

R.M

That is true. It's very hard to do it programmatically (in a way that'll work for all images)

Jin

3:43 PM

traditionally artists create curvy guidelines for the dots using the contour of the facial structure. not sure how a software would do that

also the final result heavily depends on the quality of the original photo too. it needs to have high contrast, but enough gray area for shading

R.M

Ah, I see. I was going by this photoshop tutorial: alleba.com/blog/2006/12/20/photoshop-tutorial-the-hedcut-effect

Jin

that tutorial isn't that great

but it's the closest one you can find on the web

Szabolcs

@R.M It's the same thing I linked to in a blog post, implemented in Mathematica. I was wondering if we can do better than that.

It does look good, especially the version in the blog post, but there must be a way to make it better.

Oh, you just posted the link too.

I'm wondering if it's possible to align the dot pattern with the strongest lines in the image

Silvia

4:08 PM

I believe the guide lines for dots in a hedcut have to be constructed from a 3D model. So information in a picture is not enough, some real world knowledge are needed.

If my impression is right, there seems to be an AI research field about face reconstruction?

Jin

4:23 PM

can anyone link me some data visualiztion or data mining examples?

Silvia

@Jin I think "The Mathematica Guidebooks" is a good reference.

It demonstrated many fancy skills on graphics generation.

acl

@Szabolcs here

Oleksandr R.

@Silvia but for pre-version 6, unfortunately. Also, the Guidebooks are very expensive.

Szabolcs

Seems to be quite fine on Mac.

Silvia

yes it's a pity it's not up-to-date :(

I borrow it from the library of my collage :)

I mean "college"..

Szabolcs

5:07 PM

@R.M See above for an underlying pattern

Jin

@Szabolcs did you make that in Mathematica?

Szabolcs

5:25 PM

@Jin Yes. But my code is not yet cleaned up ... here's a bigger version:

It's meant to be used as a background pattern for RM's technique

R.M

@Szabolcs what did you use? some like an edge detector or curvature filter?

Szabolcs

@R.M I don't have time to finish it now, unfortunately. The idea was to construct an incompressible flow from the blurred image, and distort some regular point grid based on that flow.

This flow simply goes along the contour lines and its magnitude is determined by the image gradient

the dots are also slightly distorted along the flow

R.M

You mean CurvatureFlowFilter (and then something with it)?

Szabolcs

if = ListInterpolation@ImageData[Blur[img, 15]]
rot90[{x_, y_}] := {-y, x}

Clear[flow]
flow[x_?NumericQ, y_?NumericQ] :=
 rot90@Identity[{Derivative[1, 0][if][x, y],
    Derivative[0, 1][if][x, y]}]

sorry about the messy code

it would probably be nicer to start with a triangular grid

acl

@Szabolcs why is Identity necessary here?

Szabolcs

5:33 PM

@acl because it used to be something else and I was lazy to rewrite it :-)

2 mins ago, by Szabolcs

sorry about the messy code

acl

@Szabolcs ah OK, I just was wondering what I was missing

Szabolcs

Also, the derivatives of the interpolating function should be precomputed for performance

acl

so you calculate grad(if) then. why incompressible?

irrotational

Szabolcs

@acl I need an incompressible flow because I was trying to keep the distances between neighbouring points relatively unchanged.

acl

@Szabolcs no I mean, if that vector is your flow field, it's v=grad(if) and div.v doesn't vanish

I mean if that vector is your velocity, sorry

or do I need a break from staring at eqns for the day

Jin

5:37 PM

whoa

"Using .NET/Link.,[19] a .NET program can ask Mathematica to perform computations; likewise, a Mathematica program can load .NET classes, manipulate .NET objects and perform method calls. This makes it possible to build .NET graphical user interfaces from within Mathematica. Similar functionality is achieved with J/Link.,[20] but with Java programs instead of .NET programs."

we use .net

Szabolcs

@acl I rotated it by 90 degrees, it's perpendicular to the gradient

it goes along the contour lines

if you have simpler / nicer ways to generate an incompressible flow from a scalar field, I'm very much interested! I was unable to come up with a better idea.

Reb.Cabin

little graphics gadget for tracking 3d orientations :)

i built up a Frenet-frame visualizer for parametric curves out of this

Szabolcs

@acl sorry, you're right, of course it's not incompressible ... I need to eat, i'm dizzy

acl

@Szabolcs oh no you're right it is, I didn't notice the rot90 bit

div.v vanishes

well I don't see any better way apart from precomputing the derivatives

good idea

sblom

6:04 PM

@Jin NET/Link is pretty slick, but I have trouble imagining any good integration scenarios with any of the .Net code that y'all have around.

Szabolcs

@Jin the problem is that if you want to make mathematica accessible from the web in any way, it requires a webmathematica license

desktop .NET stuff is fine though

jmlopez

11:07 PM

Hi @R.M

Would you happen to know how to make Mathematica tell me if two objects are comparable? That is, if I put 4 < 5 it evaluates it and returns true. But if I put "a" < "b" then it just returns the expression.

R.M

@jmlopez "Comparable" how? < is defined only for real numbers, and so won't work on complex numbers or strings.

Are you trying to check if "a" and "b" are sorted canonically?

jmlopez

something like that

If I have a list of objects, say {"a", 1, "b", 4.5, 3+i, Graphics[]}

then I want to put them into a "sorted" form

all I can really do is sort the numbers

and the rest stays in the order in which they appear on the list

so I guess I will have to check if the element is a real or integer (check if its comparable)

otherwise it is not comparable

R.M

No, Sort@list will sort everything into its canonical order, which is not a numerical sort

Try Sort@{"b", 1, "a", 4.5, 3 + i, Graphics[]} vs. Sort@{"b", 3 + i, "a", 4.5, Graphics[], 1}

They both return the same result

jmlopez

I tried A = {"a", 1, "b", 5.4, 3 + I} and Sort@A gave me {1, 3 + I, 5.4, "a", "b"}

belisarius

nights ... or whatever

jmlopez

11:20 PM

6:20 pm here

belisarius

We are ahead 8:20

jmlopez

Oh, yeah, same time zone as in California

belisarius

In summer, yes

R.M

@jmlopez Yes, that is correct. If there's a complex number, the sorting is by real part and if a tie, by absolute value of imaginary part

jmlopez

I see

I just wanted to write a quick binary sorting algorithm with a list of things that may or may not be comparable.

Not that I really need it

@belisarius, so how is the temperature in that third world of yours?

belisarius

11:26 PM

@jmlopez 52 F plus inflation

Oleksandr R.

@jmlopez everything is comparable in the sense of OrderedQ. Beyond that you need to decide what kind of comparison you would like to make.

jmlopez

@OleksandrR., thanks, that's what I needed

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