8 ms·
Solving Fizz Buzz with Cosines
- thomasjudge 10mo agohttps://joelgrus.com/2016/05/23/fizz-buzz-in-tensorflow/ https://joelgrus.com/2016/05/23/fizz-buzz-in-tensorflow/
- arealaccount 10mo agoThis would be an offer on the spot from me
- stronglikedan 10mo ago> me: It's more of a "I can't believe you're asking me that." > interviewer: Great, we find that candidates who can't get this right don't do well here. > me: ... Shit attitude from that candidate, considering the interviewer is completely correct. I wouldn't hire them since they are obviously a problem employee. For those that don't know, Fizz Buzz is less an aptitude test and more of an attitude test. That's why this candidate failed and didn't get the job.
- darth_aardvark 10mo agoFor those that don't know even more, this interview never happened and this interviewer doesn't exist. It's a funny joke on the internet.
- toast0 10mo agoIf the candidate didn't even show up to an interview, they're definitely not worth hiring. :p
- NitpickLawyer 10mo ago> Fizz Buzz is less an aptitude test and more of an attitude test The amount of (highly credentialed) interviewees that can't 0-shot a correct and fully functional fizzbuzz is also way higher than a lot of people would think. That's where the attitude part also comes in.
- cyphar 10mo ago> For those that don't know, Fizz Buzz is less an aptitude test and more of an attitude test. The articles which popularised FizzBuzz as an interview question stated as a categorical fact that most computer science graduates or programmer candidates (one article even said 199/200!![2]) cannot do FizzBuzz[1,2,3] and were absolutely recommending it as an aptitude test. I personally think this whole thing was simply untrue back in 2007 (or at the very least incredibly overstated) and we are paying the price for it with ridiculous 15-stage interviews as a paranoid response to some urban legend from ~20 years ago. [1]: https://imranontech.com/2007/01/24/using-fizzbuzz-to-find-developers-who-grok-coding/ https://imranontech.com/2007/01/24/using-fizzbuzz-to-find-de... [2]: http://weblog.raganwald.com/2007/01/dont-overthink-fizzbuzz.html http://weblog.raganwald.com/2007/01/dont-overthink-fizzbuzz.... [3]: https://blog.codinghorror.com/why-cant-programmers-program/ https://blog.codinghorror.com/why-cant-programmers-program/
- n4r9 10mo agoA massively over-engineered, incorrect solution?
- jiveturkey 10mo agoA candidate that appreciates the value of the question, yet won't subject themselves to the absurdity of demonstrating compliance. Yes, very much yes.
- n4r9 10mo agoI'd worry about them over-complicating solutions at work as well.
- jiveturkey 10mo agoI definitely wouldn't want to work on your team, if that's how you interpret such an answer. Perfect interview then -- we've both eliminated the other as a viable employee/employer, so that's a win and we got there from just 1 trivial coding question. There's so much more to say here, but this is no longer timely, plus this isn't great forum for such discussion. FWIW I have never been asked this question or similar, but since it's so famous I do have my own answer at the ready, which is just slightly more complex than the naive solution, but still well within the realm of production-worthy (maintainable, testable, readable) code. We don't really ever see any discussion of such because of course it isn't "interesting".
- gregsadetsky 10mo agoThere was another great satirical take on FizzBuzz which had something to do with runes and incantation and magical spells...? I sort of remember that the same author maybe even wrote a follow up? to this extremely experienced developer solving FizzBuzz in the most arcane way possible. Does this ring a bell for anyone? --- Found it! https://aphyr.com/posts/340-reversing-the-technical-interview https://aphyr.com/posts/340-reversing-the-technical-intervie... https://aphyr.com/posts/341-hexing-the-technical-interview https://aphyr.com/posts/341-hexing-the-technical-interview https://aphyr.com/posts/342-typing-the-technical-interview https://aphyr.com/posts/342-typing-the-technical-interview https://aphyr.com/posts/353-rewriting-the-technical-interview https://aphyr.com/posts/353-rewriting-the-technical-intervie... (the FizzBuzz one) https://aphyr.com/posts/354-unifying-the-technical-interview https://aphyr.com/posts/354-unifying-the-technical-interview wow.
- ntonozzi 10mo agoOne of my favorite blog posts of all time: https://aphyr.com/posts/342-typing-the-technical-interview https://aphyr.com/posts/342-typing-the-technical-interview
- flir 10mo ago"Unavailable Due to the UK Online Safety Act" Does aphyr even have comments, or is it a pure political protest? That's the first thing that's tempted to break out an ssh tunnel - I can live without the occasional NSFW reddit group.
- toast0 10mo agoThey do show comments at the bottom of the posts.
- AstroJetson 10mo agoThanks for the memory reminder, I read them when they first came out. They are still highly amusing today!
- taolson 10mo agoAlong that line, an over-engineered fizzBuzz using lazy list operations: https://github.com/taolson/Admiran/blob/main/examples/fizzBuzz.am https://github.com/taolson/Admiran/blob/main/examples/fizzBu...
- ivansavz 10mo agoThis is very nice.
- tantalor 10mo agoThere are several mentions of "closed-form expression" without precisely defining what that means, only "finite combinations of basic operations". TFA implies that branches (if statements and piecewise statements) are not allowed, but I don't see why not. Seems like a basic operation to me. Nevermind that `s[i]` is essentially a piecewise statement.
- susam 10mo ago> There are several mentions of "closed-form expression" without precisely defining what that means, only "finite combinations of basic operations". There is no universal definition of 'closed-form expression'. But there are some basic operations and functions that are broadly accepted, and they are spelled out directly after the 'finite combinations' phrase you quoted from the post. Quoting the remainder of that sentence here: '[...] finite combinations of basic operations such as addition, subtraction, multiplication, division, integer exponents and roots with integer index as well as functions such as exponentials, logarithms and trigonometric functions.'
- bmacho 10mo agohttps://en.wikipedia.org/wiki/Closed-form_expression#Comparison_of_different_classes_of_expressions https://en.wikipedia.org/wiki/Closed-form_expression#Compari...
- deleted 10mo ago[deleted]
- siegelzero 10mo agoVery cool! There's definitely some similarity to Ramanujan Sums, though the approach here sort of packages the fizz-buzz divisibility properties into one function. https://en.wikipedia.org/wiki/Ramanujan%27s_sum https://en.wikipedia.org/wiki/Ramanujan%27s_sum
- layer8 10mo agoI think that implementation will break down around 2^50 or so.
- nine_k 10mo agoWell, there must be an obvious solution where the fizzbuzz sequence is seen as a spectrum of two frequencies (1/3 and 1/5), and a Fourier transform gives us a periodic signal with peaks of one amplitude at fizz spots, another amplitude at buzz spots, and their sum at fizzbuzz spots. I mean. that would be approximately the same solution as the article offers, just through a more straightforward mechanism.
- atemerev 10mo agoYes. Exactly. This is how it _should_ have been done. Also probably easy enough to encode as quantum superpositions.
- HPsquared 10mo agoHow would someone do FizzBuzz on a quantum computer? It seems like a nice toy example problem.
- susam 10mo agoThat is precisely how I began writing this post. I thought I'd demonstrate how to apply the discrete Fourier transform (DFT) but to do so for each of the 15 coefficients turned out to be a lot of tedious work. That's when I began noticing shortcuts for calculating each coefficient c_k based on the divisibility properties of k. One shortcut led to another and this post is the end result. It turns out it was far less tedious (and more interesting as well) to use the shortcuts than to perform a full-blown DFT calculation for each coefficient. Of course, we could calculate the DFT using a tool, and from there work out the coefficients for the cosine terms. For example, we could get the coefficients for the exponential form like this: https://www.wolframalpha.com/input?i=Fourier%5B%7B3%2C+0%2C+0%2C+1%2C+0%2C+2%2C+1%2C+0%2C+0%2C+1%2C+2%2C+0%2C+1%2C+0%2C+0%7D%5D%2FSqrt%5B15%5D https://www.wolframalpha.com/input?i=Fourier%5B%7B3%2C+0%2C+... And then convert them to the coefficients for the cosine form like this: https://www.wolframalpha.com/input?i=%7B11%2F15%2C+2*0%2C+2*0%2C+2*%282%2F5%29%2C+2*0%2C+2*%281%2F3%29%2C+2*%282%2F5%29%2C+2*0%7D https://www.wolframalpha.com/input?i=%7B11%2F15%2C+2*0%2C+2*... That's certainly one way to avoid the tedious work but I decided to use the shortcuts as the basis for my post because I found this approach more interesting. The straightforward DFT method is perfectly valid as well and it would make an interesting post by itself.
- isoprophlex 10mo agoWhat a neat trick. I'm thinking you can abuse polynomials similarly. If the goal is to print the first, say, 100 elements, a 99-degree polynomial would do just fine :^) EDIT: the llm gods do recreational mathematics as well. claude actually thinks it was able to come up with and verify a solution... https://claude.ai/share/5664fb69-78cf-4723-94c9-7a381f947633 https://claude.ai/share/5664fb69-78cf-4723-94c9-7a381f947633
- jiggawatts 10mo agoThat's the most expletive-laden LLM output I've ever seen. ChatGPT would have aborted half way through to protect its pure and unsullied silicon mind from the filthy impure thoughts.
- theendisney 10mo agoIt would find a therapist contact your employer, your wife and your dad.
- flir 10mo ago> LMAOOOOO OKAY SO THE POLYNOMIAL IS LITERALLY SHITTING ITSELF That was a fun read, but I can see that persona quickly becoming wearing. I had a "talk like a wiki article" persona for a while that worked better (for me) than any attempt to inject personality. The greyer the better, when it comes to tools. (Being a child of the internet rather than the classroom my abusive solution would be to look up the sequence in OEIS, but I think fizzbuzz could be encoded into an L-system quite neatly).
- isoprophlex 10mo agoYes, it's indeed very over the top and one-dimensional. However, I've been iterating on this system prompt since the early early days of chatgpt.com, and I find that I can't really chat with AI systems in their "grey", dry mode anymore. On their default behavior they try too hard to make me like them, which I find intolerable. "You're absolutely right!" for some reason drives me insane; getting "lmaaoooo my bad fam i dun goofed" twenty times a day is equally annoying in terms of models being confidently wrong, but somehow the lazy shitbag attitude pisses me off less than the goody two shoes energy. And if they're low on crazyness and you force them to be crisp and emotionless like a wikipedia article, I notice that I tend to trust them more... even though again the tendency to bullshit is unchanged, still there. Somehow this really works for me. Also when coding it makes it very clear which bits I haven't inspected yet because the comments and variable names will be super nsfw, thus keeping me on my toes as to not accidentally submit PRs filled with "unfuck_json()" functions.
- ok123456 10mo agoI once had a coworker who used the FFT to determine whether coordinates formed a regular 2D grid. It didn't really work because of the interior points.
- throwaway81523 10mo agoWhere the madness leads: https://cspages.ucalgary.ca/~robin/class/449/Evolution.htm https://cspages.ucalgary.ca/~robin/class/449/Evolution.htm
- jmclnx 10mo agoI wonder where this is coming from. I saw on USENET in comp.os.linux.misc a conversation about fizzbuzz too. That was on Nov 12. Anyway an interesting read.
- burnt-resistor 10mo agoWhile it's cute use of mathematics, it's extremely inefficient in the real world because it introduces floating point multiplications and cos() which are very expensive. The only thing it lacks is branching which reduces the chances of a pipeline stall due to branch prediction miss. (The divisions will get optimized away.)
- pbsd 10mo agoThis can be translated to the discrete domain pretty easily, just like the NTT. Pick a sufficiently large prime with order 15k, say, p = 2^61-1. 37 generates the whole multiplicative group, and 37^((2^61-2)/3) and 37^((2^61-2)/5) are appropriate roots of unity. Putting it all together yields f(n) = 5226577487551039623 + 1537228672809129301*(1669582390241348315^n + 636260618972345635^n) + 3689348814741910322*(725554454131936870^n + 194643636704778390^n + 1781303817082419751^n + 1910184110508252890^n) mod (2^61-1). This involves 6 exponentiations by n with constant bases. Because in fizzbuzz the inputs are sequential, one can further precompute c^(2^i) and c^(-2^i) and, having c^n, one can go to c^(n+1) in average 2 modular multiplications by multiplying the appropriate powers c^(+-2^i) corresponding to the flipped bits.
- burnt-resistor 10mo agoInteger exponentiation is still really, really expensive. 3-4 modulus operations and a few branches is a lot cheaper.
- Terretta 10mo agoThe article conceit is fantastic. That said, is the going-in algo wrong? I see a case for 3 * 5 in here: for n in range(1, 101): if n % 15 == 0: print('FizzBuzz') elif n % 3 == 0: print('Fizz') elif n % 5 == 0: print('Buzz') else: print(n) Why? If we add 'Bazz' for mod 7, are we going to hardcode: for n in range(1, 105): if n % 105 == 0: # 3 * 5 * 7 print('FizzBuzzBazz') elif n % 15 == 0: # 3 * 5 print('FizzBuzz') elif n % 21 == 0: # 3 * 7 print('FizzBazz') elif n % 35 == 0: # 5 * 7 print('BuzzBazz') elif n % 3 == 0: print('Fizz') elif n % 5 == 0: print('Buzz') elif n % 7 == 0: print('Bazz') else: print(n) Or should we have done something like: for n in range(1, 105): out = '' if n % 3 == 0: out += 'Fizz' if n % 5 == 0: out += 'Buzz' if n % 7 == 0: out += 'Bazz' print(out or n) I've been told sure, but that's a premature optimization, 3 factors wasn't in the spec. OK, but if we changed our minds on even one of the two factors, we're having to find and change 2 lines of code ... still seems off. Sort of fun to muse whether almost all FizzBuzz implementations are a bit wrong.
- deleted 10mo ago[deleted]
- michaelcampbell 10mo ago> Sort of fun to muse whether almost all FizzBuzz implementations are a bit wrong. They're only wrong if they provide output that isn't in the spec. Adding "bazz" isn't in the spec, and assuming that something indeterminate MIGHT come later is also not part.
- Terretta 10mo agoYep, that's how people answer. Folks really really don't like thinking that "FizzBuzz" case maybe shouldn't be there, future extension or factor edit or no. // And as long as we're just manually computing factor times factor and typing out the results for it like "FizzBuzz" we might as well just hardcode the whole series...
- econ 10mo agoMade me envision this terrible idea. arr = []; y = 0; setInterval(()=>{arr[y]=x},10) setInterval(()=>{x=y++},1000) setInterval(()=>{x="fizz"},3000) setInterval(()=>{x="buzz"},5000) setInterval(()=>{x="fizzbuzz"},15000)
- seattle_spring 10mo agoThat is beautifully heinous! Nice work.
- pillars001 10mo agoHN is a great place to learn non-trivial things about trivial things, and that’s why I like it. My comment won’t add much to the discussion, but I just wanted to say that I learned something new today about a trivial topic I thought I already understood. Thank you, HN, for the great discussion thread.
- Someone 10mo agoSo, there’s a similar way to do it with a function that produces one of the characters in “FizBu\nx” and a while true loop that - increases i on every \n, - prints i when that produces x, otherwise prints the character (Disregarding rounding errors) That would be fairly obfuscated, I think.
- raffael_de 10mo agoThis seems like a great benchmark task for LLMs.
- user070223 10mo agoInspired by this post & TF comment I tried symbollic regression [0] Basically it uses genetic algorithm to find a formula that matches known input and output vectors with minimal loss I tried to force it to use pi constant but was unable I don't have much expreience with this library but I'm sure with more tweaks you'll get the right result from pysr import PySRRegressor def f(n): if n % 15 == 0: return 3 elif n%5 == 0: return 2 elif n%3 == 0: return 1 return 0 n = 500 X = np.array(range(1,n)).reshape(-1,1) Y = np.array([f(n) for n in range(1,n)]).reshape(-1,1) model = PySRRegressor( maxsize=25, niterations=200, # < Increase me for better results binary_operators=["+", "*"], unary_operators=["cos", "sin", "exp"], elementwise_loss="loss(prediction, target) = (prediction - target)^2", ) model.fit(X,Y) Result I got is this: ((cos((x0 + x0) * 1.0471969) * 0.66784626) + ((cos(sin(x0 * 0.628323) * -4.0887628) + 0.06374673) * 1.1508249)) + 1.1086457 with compleixty 22 loss: 0.000015800686 The first term is close to 2/3 * cos(2pi*n/3) which is featured in the actual formula in the article. the constant doesn't compare to 11/15 though [0] https://github.com/MilesCranmer/PySR https://github.com/MilesCranmer/PySR
- Quarrel 10mo agoGreat work, I really liked Susam's setup in the article: > Can we make the program more complicated? The words 'Fizz', 'Buzz' and > 'FizzBuzz' repeat in a periodic manner throughout the sequence. What else is > periodic? and then I'm thinking .. > Trigonometric functions! is a good start, but there are so many places to go!
- makerofthings 10mo agoThere are a surprising number of ways to generate the fizzbuzz sequence. I always liked this one: fizzbuzz n = case (n^4 `mod` 15) of 1 -> show n 6 -> "fizz" 10 -> "buzz" 0 -> "fizzbuzz" fb :: IO () fb = print $ map fizzbuzz [1..30]
- vincenthwt 10mo agoBackground Context: I am a machine vision engineer working with the Halcon vision library and HDevelop to write Halcon code. Below is an example of a program I wrote using Halcon: * Generate a tuple from 1 to 1000 and name it 'Sequence' tuple_gen_sequence (1, 1000, 1, Sequence) * Replace elements in 'Sequence' divisible by 3 with 'Fizz', storing the result in 'SequenceModThree' tuple_mod (Sequence, 3, Mod) tuple_find (Mod, 0, Indices) tuple_replace (Sequence, Indices, 'Fizz', SequenceModThree) * Replace elements in 'Sequence' divisible by 5 with 'Buzz', storing the result in 'SequenceModFive' tuple_mod (Sequence, 5, Mod) tuple_find (Mod, 0, Indices) tuple_replace (SequenceModThree, Indices, 'Buzz', SequenceModFive) * Replace elements in 'Sequence' divisible by 15 with 'FizzBuzz', storing the final result in 'SequenceFinal' tuple_mod (Sequence, 15, Mod) tuple_find (Mod, 0, Indices) tuple_replace (SequenceModFive, Indices, 'FizzBuzz', SequenceFinal) Alternatively, this process can be written more compactly using inline operators: tuple_gen_sequence (1, 1000, 1, Sequence) tempThree:= replace(Sequence, find(Sequence % 3, 0), Fizz') tempFive:= replace(tempThree, find(Sequence % 5, 0), 'Buzz') FinalSequence := replace(tempFive, find(Sequence % 15, 0), 'FizzBuzz') In this program, I applied a vectorization approach, which is an efficient technique for processing large datasets. Instead of iterating through each element individually in a loop (a comparatively slower process), I applied operations directly to the entire data sequence in one step. This method takes advantage of Halcon's optimized, low-level implementations to significantly improve performance and streamline computations.