Solving Fizz Buzz with Cosines

Solving Fizz Buzz with Cosines(susam.net)

215 points by hprotagonist 177 days ago | 65 comments

nine_k 177 days ago |

Well, 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.

susam 177 days ago | |

That 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+...

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*...

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.

susam 177 days ago | | |

Update: I went ahead and added the method of obtaining the coefficients using DFT anyway. Like I mentioned above, this approach is quite tedious by hand, so I only work out the first few coefficients explicitly. In practice, these are almost always computed using numerical software. But for some people it may still be interesting to see a direct calculation rather than relying on shortcuts.

Here is the direct link to the new section on DFT: https://susam.net/fizz-buzz-with-cosines.html#dft

mr_wiglaf 177 days ago | |

Ah so taking the Fourier transform of this function[0]? The summation of the fizz and buzz frequencies don't lead to perfect peaks for the fizz and buzz locations. I need to revisit Fourier cause I would have thought the transform would have just recovered the two fizz and buzz peaks not the fizzbuzz spot.

[0]: https://www.desmos.com/calculator/wgr3zvhazp

atemerev 177 days ago | |

Yes. Exactly. This is how it _should_ have been done.

Also probably easy enough to encode as quantum superpositions.

HPsquared 177 days ago | | |

How would someone do FizzBuzz on a quantum computer? It seems like a nice toy example problem.

thomasjudge 177 days ago |

https://joelgrus.com/2016/05/23/fizz-buzz-in-tensorflow/

isoprophlex 177 days ago |

What 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

jiggawatts 177 days ago | |

That'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 177 days ago | | |

It would find a therapist contact your employer, your wife and your dad.

flir 177 days 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).

drob518 176 days ago | |

I laughed so hard. Really curious what the pre-prompt was.

isoprophlex 176 days ago | | |

https://claude.ai/share/d855dbf4-6553-4909-ad3e-375b6385e720

"there ya go homie, copy paste ready for whoever needs their ai to have some goddamn personality"

chaboud 176 days ago | |

Oh man. This system prompt is everything I'm looking for in my coding agents. This shit should be fun. Let it be fun!

bmacho 175 days ago | | |

It could've solved the task instead of being wrong and spitting nonsense tho.

This is orthogonal to the style, still, if it realizes that it can use Python's arbitrary precision integers instead of floats then the problem becomes absolutely trivial. Fast, and numerically stable.

mikestaas 177 days ago | |

absolute madlad

user070223 177 days ago |

Inspired 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

Quarrel 177 days ago | |

Great 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!

pillars001 177 days ago |

HN 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.

ok123456 177 days ago |

I 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.

vincenthwt 175 days ago |

Background 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.

layer8 177 days ago |

I think that implementation will break down around 2^50 or so.

siegelzero 177 days ago |

Very 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

makerofthings 176 days ago |

There 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]

Someone 177 days ago |

So, 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.

econ 177 days ago |

Made 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 176 days ago | |

That is beautifully heinous! Nice work.

raffael_de 177 days ago |

This seems like a great benchmark task for LLMs.

throwaway81523 177 days ago |

Where the madness leads: https://cspages.ucalgary.ca/~robin/class/449/Evolution.htm

jmclnx 177 days ago |

I 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.

acheron 177 days ago | |

You saw a Usenet post on Nov 12? 2025?

anthk 176 days ago | | |

IT and some music/literature niche newsgroups and some Paleonthology ones plus a few more are still alive.

jmclnx 176 days ago | | |

Forgot to mention, the subject was "Simple Programming Challenge"

jmclnx 176 days ago | | |

Yes, it was about fizzbuz

ivansavz 177 days ago |

This is very nice.

burnt-resistor 177 days ago |

While 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 177 days ago | |

This 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 177 days ago | | |

Integer exponentiation is still really, really expensive. 3-4 modulus operations and a few branches is a lot cheaper.

tantalor 177 days ago |

There 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 177 days 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 177 days ago | | |

https://en.wikipedia.org/wiki/Closed-form_expression#Compari...

Terretta 177 days ago |

The 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.

michaelcampbell 177 days 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 177 days ago | | |

Yep, 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...

theendisney 177 days ago | |

If we are going to be like that we should just increment a var by 3,5 or 7 and compare it rather than %3 as the later seems expensive.