And many other thoughts on overthinking fizz buzz. I'm going to overthink it, also. Why not?

This is a problem where the obvious unit test may not cover the cases properly. See https://projecteuler.net/problem=1 for a unit test case with a subtle misdirection built in. I love this problem statement deeply because the happy path is incomplete.

Part of my overthinking is overthinking this as a "classification" exercise, where we have simple classifiers that we can apply as functions of an input value. A higher-level function (i.e., map or a generator expression) applies all of these functions to the input. Some match. Some don't.

It shakes out like this.

The core classifier is a function that requires flexible parameter binding.

query = lambda n, t: lambda x: t if x%n == 0 else None

This is a two-step function definition. The outer function binds in two parameters, n, and t. The result of this binding is the inner function. If an argument value is a multiplier of n, return the text, t. We can think of the result of the outer lambda as a partial function, also, with some parameters defined, but not all.

We have two concrete results of this query() function:

fizz = query(3, 'fizz') buzz = query(5, 'buzz')

We've bound in a number and some text. Here's how the resulting functions work.

>>> fizz(3) 'fizz' >>> fizz(2)

The "trick" in the fizz buzz problem space is recognizing that we're working with the power set of these two rules. There are actually four separate conditions. This is remarkably easy to get wrong even though the sample code may pass a unit test like the Project Euler #1 sample data.

Here's the power set that contains the complete set of all subsets of the rules.

rule_groups = set(powerset([fizz, buzz]))

"Um," you say, "Is that necessary? And powerset() isn't built-in."

Try to add a third or fourth rule and the $\textbf{O}(2^n)$ growth in complexity of checking all combinations of the rules will become readily apparent. For two rules, $4 = \lvert\mathcal{P}(\{q(3), q(5)\})\rvert$. For a general set of rules, $r$, it's $2^{\lvert r \rvert} = \lvert \mathcal{P}(r)\rvert$. Four rules? sixteen outcomes. It sure seems like the power set is absolutely necessary; it describes the domain of possible outcomes. How could it not be necessary?

Also. There's a nice definition of powerset in the itertools recipes section of the standard library. It's *almost* built-in.

Also. There's a nice definition of powerset in the itertools recipes section of the standard library. It's *almost* built-in.

The domain of possible responses form a power set. However, it's also clear that we aren't

**obligated**to actually enumerate that set for each value we're testing. We do need to be aware that the complexity of the classification output is $\textbf{O}(2^n)$ where $n$ is the number of rules.

The processing to build each set of classifications, however, is $\textbf{O}(n)$. Here's how it looks.

for n in range(20): m = set(filter(None, (r(n) for r in [fizz, buzz]))) print(m if m else n)

This locates the set of all matches, m. We apply the rules, fizz() and buzz(), to the given value. The result of applying the rules is filtered to remove falsy values. The resulting set, m, has the values from all rules which matched. This will be one of the sets from the power set of applying the rules to each value. The match, $m$, is an element of $\mathcal{P} (\{q(3), q(5)\})$.

I'm delighted that Python has some support for creating partial functions in a variety of ways. When things are complex, we can use the

**def**statement. We can use functools partial(). When things are simple, we can even use lambdas.

## No comments:

## Post a Comment