Coding Challenge 182: Apollonian Gasket Fractal скачать в хорошем качестве

Coding Challenge 182: Apollonian Gasket Fractal

Happy Pi Day 2024! In this video, I attempt to create an Apollonian Gasket using the Descartes Circle Theorem and complex numbers. So many circles! Code: https://thecodingtrain.com/challenges... 🚀 Watch this video ad-free on Nebula https://nebula.tv/videos/codingtrain-... p5.js Web Editor Sketches: 🕹️ Apollonian Gasket: https://editor.p5js.org/codingtrain/s... 🕹️ Colorful Recursive Gaskets in Gaskets: https://editor.p5js.org/codingtrain/s... 🎥 Previous:    • Coding Challenge 181: Weighted Vorono...   🎥 All:    • Coding Challenges   References: 🔗 Apollonian Gasket: https://en.wikipedia.org/wiki/Apollon... 🔗 A Tisket, a Tasket, an Apollonian Gasket: https://www.americanscientist.org/art... 🔗 Apollonian Gaskets: https://mathlesstraveled.com/2016/04/... 🔗 Beyond the Descartes Circle Theorem: https://arxiv.org/pdf/math/0101066.pdf 🔗 The Kiss Precise: https://www.nature.com/articles/13710... 🔗 Frederick Soddy: https://en.wikipedia.org/wiki/Frederi... 🔗 Problem of Apollonius: https://en.wikipedia.org/wiki/Problem... 🔗 Four Proofs of a Generalization of the Descartes Circle Theorem: https://www.jstor.org/stable/2316373 🔗 Beyond the Descartes Circle Theorem: https://arxiv.org/pdf/math/0101066.pdf 🔗 Descartes' theorem: https://en.wikipedia.org/wiki/Descart... 🔗 Complex.js: https://www.npmjs.com/package/complex.js Related Coding Challenges: 🚂 21 Mandelbrot Set with p5.js:    • Coding Challenge #21: Mandelbrot Set ...   🚂 22 Julia Set in Processing:    • Coding Challenge #22: Julia Set in Pr...   🚂 77 Recursion:    • Coding Challenge #77: Recursion   🚂 50 Circle Packing:    • Coding Challenge #50.1: Animated Circ...   🚂 95 Approximating the Value of Pi:    • Coding Challenge 95: Approximating th...   🚂 96 Visualizing the Digits of Pi:    • Coding Challenge #96: Visualizing the...   🚂 97 The Book of Pi:    • Coding Challenge #97: The Book of Pi ...   🚂 139 Calculating Digits of Pi with Collisions:    • Coding Challenge #139: Calculating Di...   🚂 140 Leibniz Formula for Pi:    • Coding Challenge #140: Pi Approximati...   🚂 141 Calculating Digits of Pi with Mandelbrot Set:    • Coding Challenge #141: Calculating Di...   🚂 169 Pi in the Sky Game:    • Coding Challenge 169: Pi in the Sky Game   🚂 176 Buffon's Needle:    • Coding Challenge 176: Buffon's Needle   Timestamps: 0:00 Happy PI Day!! 1:40 References 3:22 Describe the Apollonian problem 5:51 Curvature 10:08 Start coding! 11:53 Write a function to find the fourth curvature 15:31 Complex numbers 21:11 Create a complex number class 22:52 Square root of -1 25:32 Square root of a complex number 27:57 Implement complex Descartes theorem 33:50 Make the circles in the complexDescartes function 36:26 Use an array to add recursion 41:04 Validate the circles 42:35 Check if all 4 circles are tangent 48:04 Stop adding circles when radius falls below a threshold 49:20 Implement arbitrary arrangement of circles 53:14 Debug 55:32 Possible variations 56:34 Outro Editing by Mathieu Blanchette Animations by Jason Heglund Music from Epidemic Sound 🚂 Website: https://thecodingtrain.com/ 👾 Share Your Creation! https://thecodingtrain.com/guides/pas... 🚩 Suggest Topics: https://github.com/CodingTrain/Sugges... 💡 GitHub: https://github.com/CodingTrain 💬 Discord: https://thecodingtrain.com/discord 💖 Membership: http://youtube.com/thecodingtrain/join 🛒 Store: https://standard.tv/codingtrain 🖋️ Twitter:   / thecodingtrain   📸 Instagram:   / the.coding.train   🎥 Coding Challenges:    • Coding Challenges   🎥 Intro to Programming:    • Start learning here!   🔗 p5.js: https://p5js.org 🔗 p5.js Web Editor: https://editor.p5js.org/ 🔗 Processing: https://processing.org 📄 Code of Conduct: https://github.com/CodingTrain/Code-o... This description was auto-generated. If you see a problem, please open an issue: https://github.com/CodingTrain/thecod... #apolloniangasket #pi #circlepacking #complexnumbers #curvature #descartestheorem #p5js #javascript =====================================================