Python program to find product of rational numbers using reduce function

The reduce() function in Python applies a function cumulatively to items in a sequence, reducing them to a single value. When working with rational numbers (fractions), we can use reduce() to find their product efficiently.

For example, if we have fractions = [(5,3),(2,8),(6,9),(5,12),(7,2)], the product will be 5/3 * 2/8 * 6/9 * 5/12 * 7/2 = (5*2*6*5*7)/(3*8*9*12*2) = 2100/5184 = 175/432 (in reduced form).

Syntax

from functools import reduce
reduce(function, iterable)

Algorithm Steps

• Create a list of Fraction objects from the input tuples
• Use reduce() with a lambda function to multiply all fractions
• Return the numerator and denominator of the resulting fraction

Example

Here's the implementation using reduce() to find the product of rational numbers ?

from fractions import Fraction
from functools import reduce

def solve(frac):
    fractions = []
    for f in frac:
        fractions.append(Fraction(*f))
    
    result = reduce(lambda x, y: x * y, fractions)
    return result.numerator, result.denominator

# Test with sample data
frac = [(5,3), (2,8), (6,9), (5,12), (7,2)]
product = solve(frac)
print(f"Product: {product}")
print(f"As fraction: {product[0]}/{product[1]}")

Product: (175, 432)
As fraction: 175/432

Step-by-Step Calculation

Let's trace through the multiplication process ?

from fractions import Fraction
from functools import reduce

def solve_with_steps(frac):
    fractions = [Fraction(*f) for f in frac]
    print("Fractions:", fractions)
    
    result = reduce(lambda x, y: x * y, fractions)
    print(f"Product: {result}")
    return result.numerator, result.denominator

frac = [(5,3), (2,8), (6,9), (5,12), (7,2)]
numerator, denominator = solve_with_steps(frac)
print(f"Final result: ({numerator}, {denominator})")

Fractions: [Fraction(5, 3), Fraction(1, 4), Fraction(2, 3), Fraction(5, 12), Fraction(7, 2)]
Product: 175/432
Final result: (175, 432)

Alternative Approach

You can also use a more concise version with list comprehension ?

from fractions import Fraction
from functools import reduce

def product_fractions(frac):
    return reduce(lambda x, y: x * y, [Fraction(*f) for f in frac])

frac = [(5,3), (2,8), (6,9), (5,12), (7,2)]
result = product_fractions(frac)
print(f"Product: ({result.numerator}, {result.denominator})")

Product: (175, 432)

Conclusion

The reduce() function provides an elegant way to find the product of rational numbers by applying multiplication cumulatively. The Fraction class automatically handles simplification, ensuring the result is in its reduced form.