**π‘ Problem Formulation:** The task is to determine whether the product of digits in the even positions of a number is equal to the product of digits in the odd positions. For instance, given the number `1234`

, the product of the even-positioned digits (`2`

and `4`

) should equal the product of the odd-positioned digits (`1`

and `3`

). Our goal is to find Pythonic solutions to this problem, aiming for simplicity and efficiency.

## Method 1: Using Iteration

This method involves iterating through the digits of the number, calculating the product of digits at even and odd positions separately and comparing them at the end. It’s straightforward and easy to understand, making it perfect for those who prefer the traditional approach.

Here’s an example:

def check_product_equal(number): number_str = str(number) even_product = 1 odd_product = 1 for i in range(len(number_str)): if i % 2 == 0: even_product *= int(number_str[i]) else: odd_product *= int(number_str[i]) return even_product == odd_product # Example usage print(check_product_equal(1234))

Output: `True`

This code defines a function `check_product_equal`

that takes a number as input and converts it to a string. It then calculates the product of even and odd positioned digits and compares them for equality, returning `True`

or `False`

.

## Method 2: Using List Comprehensions

List comprehensions in Python allow us to create new lists by applying an expression to each item in a sequence. By using this concise approach, we can easily split and calculate the products of even and odd place digits.

Here’s an example:

def check_product_equal(number): digits = [int(d) for d in str(number)] return prod(digits[::2]) == prod(digits[1::2]) # Example usage print(check_product_equal(1234))

Output: `True`

This snippet leverages list comprehensions to extract digits and use slicing to separate even- and odd-indexed digits. It then uses the built-in `prod`

function to calculate products and compares them.

## Method 3: Using Functional Programming

Functional programming paradigm can be a neat way to solve this problem with a combination of `map`

, `reduce`

, and lambda functions to achieve the goal elegantly.

Here’s an example:

from functools import reduce from operator import mul def check_product_equal(number): digits = map(int, str(number)) even_odd_products = reduce(lambda acc, x: (acc[0]*x[1], acc[1]*x[0]) if x[0] % 2 == 0 else (acc[0], acc[1]*x[1]), enumerate(digits), (1, 1)) return even_odd_products[0] == even_odd_products[1] # Example usage print(check_product_equal(1234))

Output: `True`

In this method, `reduce`

is used to traverse the digits while calculating the product for even and odd indexes. The tuple `(1, 1)`

is the initial accumulator, and tuple unpacking is used in lambda to update values.

## Method 4: Using NumPy

If performance is key and youβre working within the data science stack, NumPy could be used to greatly accelerate the computation with its optimized array operations.

Here’s an example:

import numpy as np def check_product_equal(number): digits = np.array([int(d) for d in str(number)]) even_product = np.prod(digits[::2]) odd_product = np.prod(digits[1::2]) return even_product == odd_product # Example usage print(check_product_equal(1234))

Output: `True`

After converting the number into a NumPy array of digits, the function employs slicing and `np.prod`

to find the products of even and odd positioned digits, followed by a comparison.

## Bonus One-Liner Method 5: Using eval()

If you’re looking for a compact one-liner, `eval`

can be used, although it’s often best to avoid it due to potential security issues with executing strings as code.

Here’s an example:

def check_product_equal(number): return eval('*'.join(str(number)[::2])) == eval('*'.join(str(number)[1::2])) # Example usage print(check_product_equal(1234))

Output: `True`

This solution generates strings representing the mathematical expressions for the product of even and odd indexed digits and uses `eval`

to execute them, comparing the outcomes.

## Summary/Discussion

**Method 1: Iteration.**It’s simple and easy to understand. It might lack efficiency when dealing with extremely large numbers due to the iteration.**Method 2: List Comprehensions.**More Pythonic and readable. It relies on Python’s`prod`

function that may need to be imported from the`math`

module in versions before Python 3.8.**Method 3: Functional Programming.**Clean and elegant, but could be harder to understand for those not familiar with functional programming constructs.**Method 4: Using NumPy.**Offers high performance but requires an external library which might not be desirable for a simple computation.**Method 5: Using eval().**Quick and concise one-liner but not recommended due to potential security risks.