5 ms·
The Prospero Challenge
- cfbolztereick 2y agoThis is really fun, I'm getting badly nerd sniped by this.
- Titan2189 2y agoIt's been 4 hours already. How are you doing? Didn't forget to eat and drink? Just checking in on you :)
- 6stringmerc 2y agoWhoa cool invocation of the magician Prospero! I recently adapted “The Tempest” as a screenplay while in jail and typed it up: https://samhenrycliff.medium.com/adapting-shakespeares-the-tempest-into-a-screenplay-7f55f5414266 https://samhenrycliff.medium.com/adapting-shakespeares-the-t...
- yoz 2y agoThanks so much for writing this up and sharing. It sounds like you've been through a horrific few years, but your perseverance in creativity is inspiring and fascinating. I wish you well in your efforts to make a better life for yourself, and I hope your work finds a wider audience.
- neverokay 2y agoWhat compelled you in jail? I’m asking if there was more to it than just being bored.
- jstanley 2y agoI'm thinking that we can parse the input into an expression tree, and then for each x coordinate, have each node in the tree tell us which intervals of y coordinates it would return a value < 0 (or <= 0 without loss of generality). Since we're basically dealing with real numbers we can pretend true 0 doesn't exist. Composing these intervals is trivial for all operations except addition/subtraction. (For example, `mul` is less than 0 if exactly one of its arguments is less than 0, `neg` is less than 0 if its argument is not less than 0, `min` less than 0 if either of its arguments is less than 0, etc.). And then we define add(a, b) as sub(a, neg(b)). So then we only need to work out which y values sub(a, b) will return a negative result for. sub(a,b) is negative when a<b. We can have the sub() node asks its 2 child nodes which values they would return a negative result for. Every y coordinate where a gives a negative result and b gives a positive result is a y coordinate where sub(a,b) is negative. Every y coordinate where a is positive and b is negative is positive. For the remaining y values I think we have to actually evaluate the children and find out which ones give a negative result. Obviously memoise the evaluation so that we don't repeat any work, and turn the expression tree into a DAG by combining any identical nodes. Some subtrees won't contain var-x, so the memoisation of those nodes can persist across different x coordinates. And then for each x coordinate we ask the expression tree to tell us which y coordinates give negative outputs, and plot those. It's possible that the idea would generalise to 2-dimensional intervals, not sure. I haven't implemented this yet but I'm planning to try it out, but to be honest I don't expect it to be faster than Matt's GPU version based on recursive subdivision with interval arithmetic. Btw his talk is great https://www.youtube.com/watch?v=UxGxsGnbyJ4&t=80s https://www.youtube.com/watch?v=UxGxsGnbyJ4&t=80s And, a second fun challenge would be to reverse engineer the input file to extract the font! I expect the file is basically a union of x/y translations of functions that plot individual letters, so it shouldn't be crazy hard to split out the top-level union, then split-out the x/y translations, and then collect the letters. It's possible that it's more complicated though. In particular, is each character translated in x/y individually, or is each character translated only in x to form each line of text, and then the line as a whole is translated in y? Or something weirder?
- zomglings 2y agoCan you explain how you produce the opcodes from a text/image?
- iamwil 2y agothe text contains a math expression. The math expression can be interpreted as an abstract syntax tree. The AST can also be translated into bytecode and interpreted by a stack-based virtual machine. The op-codes are where you do that translation from an AST into bytecode. The entire 2nd half of the crafting interpreters book will guide you through how it's done. https://craftinginterpreters.com/a-bytecode-virtual-machine.html https://craftinginterpreters.com/a-bytecode-virtual-machine....
- edflsafoiewq 2y agoLooks SDF-ish. My guess is convert the letters to curves, cut them into pieces without holes so each piece is the set of points between certain curves, write down SDFs for each curve (negative on the "inside", positive on the "outside"), combine all SDFs in a piece with intersection (max), then combine all pieces with union (min).
- stevage 2y ago> If you evaluate the expression with (x,y) values in the ±1 square, then color pixels black or white based on their sign I do not understand what this means at all, particularly the bit before the comma. Can someone explain it differently?
- cjs_ac 2y agoYeah, this took me a while. Basically, the entire file represents a sort of assembly language for a function. You need to evaluate that entire function for each pixel in the output image. Each line starts with a line number (with a leading underscore). This is followed by an instruction, and zero, one, or two arguments. The arguments can be either floating point constants or integers (with leading underscores) which represent the results of those corresponding lines earlier in the function. The x and y coordinates of each pixel are loaded in the assembly language using the var-x and var-y instructions.
- jstanley 2y agoBreak down the range of x values from -1 to 1, and y values from -1 to 1, each into 1024 parts (or whatever resolution you want to plot at). Evaluate the function for each (x,y) coordinate. If the resulting number is negative colour the pixel white, otherwise colour it black. The expression is (probably) a 2d signed-distance function - https://en.wikipedia.org/wiki/Signed_distance_function https://en.wikipedia.org/wiki/Signed_distance_function
- stevage 2y agoOhh.. "the +/- 1 square" means a square of size 2 centred around the origin? That was way too terse for me.
- iamwil 2y agoFor a given (x, y), you color the pixel black if you're outside of a shape, and color the pixel white if you're inside the shape. The math formula is a way to describe the contour of the shape. To tell if you're inside or outside of a shape, you plug in your (x, y) coordinate into the math formula, and if the output is > 0, then you're outside. If the output is < 0, then you're inside. I think they call it a signed distance field. You should watch his talk that explain it. It's really great. https://www.youtube.com/watch?v=UxGxsGnbyJ4 https://www.youtube.com/watch?v=UxGxsGnbyJ4
- IshKebab 2y agoHash the input, if it matches the expected input output the expected image. Otherwise error. Technically correct.
- ainiriand 2y agoThis is why we can't have nice things.
- mkeeter 2y agoQuoth the article: "Obviously, you could precompute the entire image, but that's against the spirit of the challenge"
- crazygringo 2y agoWhy waste time hashing? Just output the expected image in all cases. Still technically correct. ;)
- adrian17 2y ago> (Spoilers: it also allocates 60+ GB of RAM for intermediate results, so consider reducing the image size before running it on your own machine) The input algorithm is essentially written in SSA form, and it's easy to analyze when each "register" stops being used; then you can drop the arrays after their last use. Turns out the number of live registers never exceeds 160 at any point in time, and with this one addition the max memory usage drops to barely above 1GB.
- Jtsummers 2y agoI don't know why you are downvoted (at the time of this writing). That's a very good point and makes the simple Python solution feasible on many more computers (due to reduced memory footprint). It's also a straightforward pass (once you know what liveness analysis is). Since there are no loops it's just one pass starting from the end and working backwards to determine whether a variable is live or not, and then an extra check on each iteration to decide whether to delete particular arrays.
- tekknolagi 2y agoI think getting downvoted (not by me) because if you scroll to the end of the post, that's the very first submission (by me).
- Jtsummers 2y agoAh, fair. I didn't catch that.
- kevmo314 2y agoNice challenge, I got it down to 0.5ms/frame. https://github.com/kevmo314/prospero.vm https://github.com/kevmo314/prospero.vm
- montag 2y agoImpressive. Did you do all this in the past 2 hours?
- kevmo314 2y agoYeah, it took me an hour-ish to get the performance and then an hour-ish to figure out how to actually render it correctly. I had never seen the P2/P5 PGM format before so I didn't realize P5 files were binary. :)
- ayastrebov2 2y agoGreat idea and clean implementation! You seem to be interested in performance engineering - I did some work recently, check this out https://github.com/AlexanderYastrebov/wireguard-vanity-key https://github.com/AlexanderYastrebov/wireguard-vanity-key
- jstanley 2y agoSo 2000 fps, 16 million pixels, and 7000 operations per pixel, works out to 224 TFLOPS. An RTX 4090 is advertised as being able to compute 82 TFLOPS. Where do you think the extra is coming from? Is it just straightforward optimisations like constant folding? Or do you think the compiler is noticing that 1000 iterations of the loop doesn't change the answer and optimising it down to just 1 loop?
- deleted 2y ago[deleted]
- kevmo314 2y agoIt’s a good question, admittedly I don’t know. I looked into it when Matt mentioned it but I’m not so familiar with what happens after ptx to say. If someone does know I’d love to learn why though.
- tonetheman 2y ago[dead]
- edflsafoiewq 2y agoConverted it to GLSL that can be dropped into shadertoy.com/new. I get about 15FPS at 714x401. https://pastebin.com/jBwNrfim https://pastebin.com/jBwNrfim
- James_K 2y agoSome of these solutions seem to copy the .vm file and rewrite it in another programming language. Is this permitted? Seems like cheating to not count the time spent compiling and transforming that code in the runtime of the program.
- RainyDayTmrw 2y agoCan anyone explain where this blob of "assembly language" comes from? In practice, the author presumably with the desired output image and backed into this esoteric format, but I'm not following how. What is considered an acceptable preprocessing or transformation? In the limit case, a static executable that computes nothing and outputs a constant, literal image probably violates the intention of the question.
- jstanley 2y ago> In practice, the author presumably with the desired output image and backed into this esoteric format, but I'm not following how. I actually don't think this is how he did it. You might notice the font is kind of weird. I think he started with an SDF font and created the "assembly language" from that. https://en.m.wikipedia.org/wiki/Signed_distance_function https://en.m.wikipedia.org/wiki/Signed_distance_function
- mkeeter 2y ago> Can anyone explain where this blob of "assembly language" comes from? Assembly language is definitely the right analogy: it's a low-level target generated by higher-level tools. In this case, the expression came from a Python script calling this text(...) function: https://github.com/mkeeter/antimony/blob/f6a56dd7/py/fab/shapes.py#L866-L1505 https://github.com/mkeeter/antimony/blob/f6a56dd7/py/fab/sha... The font is hand-built from geometric primitives (rectangles, circles, etc) and CSG operations (union, intersection, difference) > What is considered an acceptable preprocessing or transformation? I'm looking for interesting ideas, and to mine the depths of PLs / compiler / interpreter / runtime research. Just returning a fixed image isn't particularly interesting, but (for example) I just updated the site with a compile-to-CUDA example that shows off the brute force power of a modern GPU.
- mattdesl 2y agoWhat are the practical implications of this kind of assembly language? Surely there’s more efficient means of describing 2D SDFs? Fun exercise! I’ve been enjoying trying to find some new ways to approach the challenge. I managed to build a single string expression for the entire program, so it could be evaluated per-pixel in a shader, but it turns out the expression is too complex for WebGL & WebGPU and the shader fails to compile. My next thought would be to evaluate the program at a low resolution to create a low res SDF texture for the shader to draw at a higher resolution. Some information will probably be lost, though.
- yolhan83 2y agoRTX 4060, done in julia 20 fps (for 4096X4096) (47ms per frame) in https://github.com/yolhan83/ProsperoChal https://github.com/yolhan83/ProsperoChal It was quite fun !