Messing Around In Beam, π Day, And More Wordle

Written by Ben Wendt

Apache Beam is a distributed programming framework, mostly designed as a counterpart to the DataFlow service in Google Cloud Platform. In the past I’ve done a fair bit of work on pipelines in Spark and with a service architecture, but I’ll be needing Beam for my new job, so Ive been playing around a bit with that.

I’ve may have written a bit about it before, but I think of calculating π using the Monte Carlo Method to be kind of like the “hello world” of data pipelines. A quick review of the algorithm:

  • Select a large number of random x, y ∊ [0, 1]² from the uniform distibution.
  • Take the sum of squares of each couple.
  • The proportion of sums of squares that are less than one will (very slowly converge to π/4).

This works because finding random points withing this quarter arc is analagous to finding the area of the unit circle. Here’s the code in beam:

import apache_beam as beam
import numpy as np

def in_circle(pair):
    if (pair[0]**2 + pair[1]**2) < 1:
        return 1
        return 0

with beam.Pipeline() as pipeline:
    n_samp = 100_000
    X = np.random.uniform(0,1,2 * n_samp).reshape(n_samp, 2)

    x = pipeline | "create xs" >> beam.Create(X)

    number = x | "calculate it" >> beam.Map(in_circle) \
        | "sum up" >> beam.CombineGlobally(sum)

    number | "writing pi value" >>"number.txt")

This gave me a value of pi of π is 3.14136, which any piphile would tell you is way off. The Monte Carlo Method is great for a close enough answer, but it’s a terrible way to calculate digits of π.

I’ve also written recently about wordle. I’ve been playing every day for months, and I often get the urge to write a program to cheat for me. I have thus far resisted the urge to cheat, (unless you count writing a program to give a good first guess cheating), but I have gone back several times to re-think word choices.

The other day I had a clue of _o_us and I discounted several letters to get there. Off the top of my head I thought of “bolus” and “focus” as options. It ended up being “bonus,” but it took me a few minutes to remember this common word.

Here’s a beam that calculates some of the words that may match in this scenario.

import apache_beam as beam

# my favorite unix file
dictionary = "/usr/share/dict/words"

def to_lower(x):
    return x.lower()

def is_fives(text):
    return len(text) == 5

def match_greens(pattern):
    return lambda x: all([x[idx] == letter for idx, letter in pattern])

def non_matches(excludes):
    return lambda x: all([ex not in x for ex in excludes])

with open("beam-output.txt", "w") as f:
    with beam.Pipeline() as p:
        words = ( p |
            "read in words" >> |
            "to lower" >> beam.Map(to_lower)

        fives = (words |
            "filter to five letter words" >> beam.Filter(is_fives)

        # sample of letters I had tried and discounted.
        excludes = ['t', 'r']

        without_exclusions = (fives |
            "remove exclusions" >> beam.Filter(non_matches(excludes))

        # sample "green letter" pattern _o_us
        pattern = [(1, 'o'), (3, 'u'), (4, 's')]

        matches = (without_exclusions |
            "get green matches" >> beam.Filter(match_greens(pattern))

            |"do output" >> beam.Map(print)

And I get these output words with my dictionary:


As I mentioned before, wordle does use common words, so it would make sense to either filter or sort these based on word frequency. But as a quick learning exercise for beam it was decent. One aspect I am really enjoying about beam so far is that it basically forces you to document every step of your pipeline, because you can’t use the same operator twice without doing the "explanation" >> step thing, which is really nice.