Approximate equality: Difference between revisions
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* -sqrt(2) * sqrt(2), -2.0 |
* -sqrt(2) * sqrt(2), -2.0 |
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* 3.14159265358979323846, 3.14159265358979324 |
* 3.14159265358979323846, 3.14159265358979324 |
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Answers should be similar to the Python code example, except for the third case, which may well vary depending on how your function handles comparison with (exactly) 0.0. |
Answers should be similar to the Python code example, except for the third case, which may well vary depending on how your function handles comparison with (exactly) 0.0. |
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Revision as of 01:57, 2 September 2019
Sometimes, when testing whether the solution to a task (for example, here on
Rosetta Code) is correct, the difference in floating point calculations between
different language implementations becomes significant. For example, a
difference between 32 bit and 64 bit floating point calculations may appear by
about the 8th significant digit in base 10 arithmetic.
Create a function which returns true if two floating point numbers are approximately equal. The function should allow for differences in the magnitude of numbers, so that, for example, 100000000000000.01 may be approximately equal to 100000000000000.011, even though 100.01 is not approximately equal to 100.011.
If the language has such as feature in its standard library, this may be used instead of a custom function.
Show the function results with comparisons on the following pairs of values:
- 100000000000000.01, 100000000000000.011 (note: should return true)
- 100.01, 100.011 (note: should return false)
- 10000000000000.001 / 10000.0, 1000000000.0000001000
- 0.001, 0.0010000001
- 0.000000000000000000000101, 0.0
- sqrt(2) * sqrt(2), 2.0
- -sqrt(2) * sqrt(2), -2.0
- 3.14159265358979323846, 3.14159265358979324
Answers should be similar to the Python code example, except for the third case, which may well vary depending on how your function handles comparison with (exactly) 0.0.
Python
<lang python>from numpy import sqrt from math import isclose
testvalues = [[100000000000000.01,100000000000000.011], [00.01, 100.011],
[10000000000000.001 / 10000.0, 1000000000.0000001000], [0.001, 0.0010000001], [0.00000000000000000101, 0.0], [sqrt(2) * sqrt(2), 2.0], [-sqrt(2) * sqrt(2), -2.0], [100000000000000003.0, 100000000000000004.0], [3.14159265358979323846, 3.14159265358979324]]
for (x, y) in testvalues:
maybenot = "" if isclose(x, y) else "NOT" print(x, "is", maybenot, "approximately equal to ", y)
</lang>
- Output:
100000000000000.02 is approximately equal to 100000000000000.02 0.01 is NOT approximately equal to 100.011 1000000000.0000002 is approximately equal to 1000000000.0000001 0.001 is NOT approximately equal to 0.0010000001 1.01e-18 is NOT approximately equal to 0.0 2.0 is approximately equal to 2.0 -2.0 is approximately equal to -2.0 1e+17 is approximately equal to 1e+17 3.141592653589793 is approximately equal to 3.141592653589793