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Wolfram Tweet-a-Program Launches Today

Posted 12 years ago

Because a little code can go a long way, the Wolfram Language is quickly becoming a part of hackathons across the country. To encourage programmers to have fun with short code, we’re launching Wolfram Tweet-a-Program today.

For details, read Stephen Wolfram's blog post.

Submit Your 140-Character Program

Show us what programs you can create in under 140 characters by sending code submissions to @wolframtap. To create your code, you can use Wolfram Programming Cloud or any other Wolfram Language system.

On Twitter, send your tweets in the following format:

@wolframtap [Wolfram Language code]

If you don’t have a Twitter account, you can ask Wolfram to tweet on your behalf by posting your code here in the Tweet-a-Program Community group.

Discuss Your Code Submissions Here

Bookmark this Community group to:

  • Connect with other short code enthusiasts
  • Post your Tweet-a-Program submissions
  • Troubleshoot
  • Add commentary

Tweet-a-Program Examples

The first 1000 digits of ?, sized according to their magnitudes:

@wolframtap Row[Style[#,5#+1]&/@First[RealDigits[Pi,10,1000]]]

A 36-character “fractal hack”:

@wolframtap NestList[Subsuperscript[#,#,#]&,x,6]

POSTED BY: Emily Suess
10 Replies

Give it a try: Fractal Fun: Tweet-a-Program Mandelbrot Code Challenge

ImageAdd[Rasterize@Rotate[MandelbrotSetPlot[{-1.9+I,-.85I},ColorFunction->(If[#3>.9,Black,White]&)],3Pi/2],=[Mandelbrot Pic]]

enter image description here

Source

Bernat

POSTED BY: Bernat Espigulé

Slightly shorter version that should fit in a tweet:

b = c = ConstantArray[0, {256, 256, 3}]; s = 21 ;; 236; 
b[[All, s]] = c[[s, All]] = Tuples[{0, .5, .05, 0, .5, .05}, 3]; Image[b + c]
POSTED BY: Ilian Gachevski

Illian, Thank you. It runs as: @wolframtap b = c = ConstantArray[0, {256, 256, 3}]; s = 21;;236; b[[All, s]] = c[[s, All]] = Tuples[{0, .5, .5, 0, .5, .5}, 3];Image[b + c]

I had to cut into my variables!

POSTED BY: Douglas Youvan

I like it!

(Stephanie: Thank you for telling me about this)

When do we go to 156,000 cores?

A really good programmer might be able to take this "Tuple Imagery" program down to tweet size. Please try.

b = c = Table[{0, 0, 0}, {x, 1, 256}, {y, 1, 256}]; t = Tuples[{0, .5, .05, 0, .5, .05}, 3]; For[j = 1, j <= 216, j++, b[[All, j + 20]] = c[[j + 20, All]] = t[[j]]]; Image[b + c]

It's 41 characters over the limit.

Really close:

t = Tuples[N@{0, 1, 0.5, 0, -1, -0.5}, 3]; l= Length@t; Image[Array[<a href="mailto:Total@t[[{##}]]&,{l,l}]">Total@t[[{##}]]&,{l,l}]]

POSTED BY: Douglas Youvan

Also, I'm getting some double results.

POSTED BY: Jesse Friedman

Yep, a known bug. Working on it, though we might not get to fully sorted out until tomorrow...

POSTED BY: Andre Kuzniarek

Does it use the same protection against malicious input as standard Cloud APIs?

POSTED BY: Jesse Friedman
POSTED BY: Andre Kuzniarek

I really like this image which Arnoud B. generously allowed me to steal from his Tweet-a-Code testing collection:

Graphics[{EdgeForm[],Table[{Hue[RandomReal[]],Opacity[0.1],ConicHullRegion[{RandomReal[1,2]},RandomReal[{-1,1},{2,2}]]},{75}]}]

enter image description here

http://t.co/6GieKbLaYD

POSTED BY: Andre Kuzniarek

Shades of Kandinsky. Well, lines too.

POSTED BY: Daniel Lichtblau
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