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The TLA Problem: Over-Engineering Three-Letter Acronyms

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The TLA Problem: Over-Engineering Three-Letter Acronyms

Let's have some fun and over-engineer a solution to a problem space that we're unsure of why it exists: Generating three-letter acronyms! Let's get started!

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Here's something we can gleefully over-engineer. Because anything worth doing is worth over-engineering until it morphs into a different kind of problem.

I'm unclear on the backstory, so try not to ask "why are we doing this?" I think it has something to do with code camp and there are teenagers involved.

What we want to do is generate a pool of 200 TLAs. Seems simple, right?

The first pass may not be obvious, but this works well.

import string
import random
from typing import Iterator

DOMAIN = string.ascii_lowercase
def tlagen() -> Iterator[str]:
    while True:
        yield "".join(random.choice(DOMAIN) for _ in range(3))

tla_iter = tlagen()
for i in range(200):
    print(next(tla_iter))

This makes 200 TLAs. It's configurable, if that's important. We could change tlagen() to take an argument in case we wanted four-letter acronyms or something else.

However, this will generate words like "cum" and "ass." We need a forbidden word list and want to filter out some words. Also, there's no uniqueness guarantee.

The Forbidden Word Filter

Here's a simple forbidden word filter. Configure FORBIDDEN with a set of words you'd like to exclude. Maybe exclude "sex" and "die" too, for example. Depends on what age the teenagers are.

def acceptable(word: str) -> bool:
    return word not in FORBIDDEN

Here's another handy thing. Rather than repeat the "loop until" logic, we can encapsulate it into a function.

from typing import TypeVar, Iterator
T_ = TypeVar("T_")
def until(limit: int, source: Iterator[T_]) -> Iterator[T_]:
    source_iter = iter(source)
    for _ in range(limit):
        item = next(source_iter)
        yield item

Okay. That's workable. This lets us use the following:

list(until(200, filter(acceptable, tlagen())))

Read this from inside to outside. First, generate a sequence of TLAs. Apply a filter to that generator. Apply the "until" counter to that filtered sequence of TLAs. Create a list of 200 items. Nice.

The TypeVar gives us a flexible binding. The input is an iterator over things and the output will be an iterator over the same kinds of things. This formalizes a common understanding of how iterators work. The mypy tool can confirm that the code meets the claims in the type hint.

There are two sketchy parts about this. First the remote possibility of duplicates. Nothing precludes duplicates. And we're doing a bunch of string hash computations. To avoid duplicates, we need a growing cache of already-provided words. Or perhaps we need to build a set until it's the right size. Rather than compute hashes of strings, can we work with the numeric representation directly?

Numeric TLAs

There are only a few TLAs.

263=17576

Of these, perhaps four are forbidden. We can easily convert a number back to a word and work with a finite domain of integers. Here's a function to turn an integer into a TLA.

def intword(number: int) -> str:
    def digit_iter(number: int) -> Iterator[int]:
        for i in range(3):
            number, digit = divmod(number, 26)
            yield digit
    return "".join(DOMAIN[d] for d in digit_iter(number))

This will iterate over the three digits using a simple divide-and-remainder process to extract the digits from an integer. The digits are turned into letters and we can build the TLA from the number.

What are the numeric identities of the forbidden words?

def polynomial(base: int, coefficients: Sequence[int]) -> int:
    return sum(c*base**i for i, c in enumerate(coefficients))

def charnum(char: str) -> int:
    return DOMAIN.index(char)

def wordint(word: str) -> int:
    return polynomial(26, map(charnum, word))

We convert a word to a number by mapping individual characters to numbers, then computing a polynomial in base 26. And yes, the implementation of polynomial() is inefficient because it uses the ** operator instead of folding in a multiply-and-add operation among the terms of the polynomial.

Here's another way to handle the creation of TLAs.

FORBIDDEN_I = set(map(wordint, FORBIDDEN))
subset = list(set(range(0, 26**3)) - FORBIDDEN_I)
random.shuffle(subset)
return list(map(intword, subset[:200]))

This is cool. We create a set of numeric codes for all TLAs, then remove the few numbers from the set of TLAs. What's left is the entire domain of permissible TLA's. All of them. Shuffle and pick the first 200.

It guarantees no duplicates. This has a lot of advantages because it's simple code.

This, however, takes a surprisingly long time: almost 17 milliseconds on my laptop.

Numeric Filtering

Let's combine the numeric approach with the original ideal of generating as few items as we can get away with, but also checking for duplicates.

First, we need to generate the TLA numbers instead of strings. Here's a sequence of random numbers that is confined to the TLA domain.

def tlaigen() -> Iterator[int]:
    while True:
        yield random.randrange(26**3)

Now, we need to pass unique items and reject duplicate items. This requires a cache that grows. We can use a simple set. Although, a bit-mask with 17,576 bits might be more useful.

def unique(source: Iterator[T_]) -> Iterator[T_]:
    cache = set()
    for item in source:
        if item in cache:
            continue
        cache.add(item)
        yield item

This uses an ever-growing cache to locate unique items. This will tend to slow slightly based on memory management for the set. My vague understanding is the implementation will double the size when hash collisions start occurring, leading a kind of logslowdown factor as the set grows.

The final generator looks like this:

list(until(200, unique(filter(lambda w: w not in FORBIDDEN_I, tlaigen()))))

Reading from inner to outer, we have a generator which will produce numbers in the TLA range. The few forbidden numbers are excluded. The cache is checked for uniqueness. Finally, the generator stops after yielding 200 items.

Conclusion

Of course, we're using timeit to determine the overall impact of all of this engineering. We're only doing 1,000 iterations, not the default of 1,000,000 iterations.

The original version: 0.94 seconds.

The improved number-based version: 0.38 seconds.

So there. Want to generate values from a limited domain of strings? Encode things as numbers and work with the numeric representation. Much faster.

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Topics:
mapping ,python ,performance ,app performance

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