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How Can We Handle and Resolve Floating-Point Errors?

Linda Hamilton
Release: 2024-10-21 14:57:03
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How Can We Handle and Resolve Floating-Point Errors?

Understanding Floating Point Errors and Their Resolution

Floating-point arithmetic poses unique challenges due to its approximate nature. To address these errors effectively, we must examine their root cause.

In Python, floating-point calculations utilize the binary representation, leading to inaccuracies. As demonstrated in the code snippet, attempts to approximate square roots are slightly off due to this approximation. For example:

<code class="python">def sqrt(num):
    root = 0.0
    while root * root < num:
        root += 0.01
    return root

print(sqrt(4)) # Output: 2.0000000000000013
print(sqrt(9)) # Output: 3.00999999999998</code>
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To better comprehend these errors, consider the exact decimal representation of 0.01 using the decimal module:

<code class="python">from decimal import Decimal
print(Decimal(.01)) # Output: Decimal('0.01000000000000000020816681711721685132943093776702880859375')</code>
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This string reveals that the actual value being added is slightly greater than 1/100. Hence, the floating-point representation of decimal values introduces these minor variations.

To mitigate these errors, several approaches exist:

  1. Decimal Module: Employing the decimal module ensures that operations are performed exactly, eliminating rounding errors. In the modified function below, we use this approach:
<code class="python">from decimal import Decimal as D
def sqrt(num):
    root = D(0)
    while root * root < num:
        root += D("0.01")
    return root

print(sqrt(4)) # Output: Decimal('2.00')
print(sqrt(9)) # Output: Decimal('3.00')</code>
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  1. Controllable Increments: Instead of directly adding 0.01, it's advisable to add values that are exactly representable as binary floats, such as I/2**J. By using 0.125 (1/8) or 0.0625 (1/16) as increments, this eliminates approximation errors.

By combining these methods and leveraging techniques like Newton's method, you can achieve highly accurate floating-point calculations, expanding your understanding of numerical analysis and handling floating-point arithmetic effectively.

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