Python Infinity: Use a Sentinel and Distinguish NaN

Start with a small, explicit task

Find a minimum among three finite synthetic route costs using infinity as the initial value. Check the result with and without any candidates. Then inspect infinity minus itself to expose an invalid numeric result that cannot be treated as a usable bound.

Run the complete example

This original example was checked with Python 3.12.14. Save it as python_infinity.py and run it with Python 3.12. It uses only the standard library. The assertions check the stated results.

import math
def lowest(costs):
    best = float("inf")
    for cost in costs:
        if cost < best:
            best = cost
    return best
infinity = float("inf")
assert infinity > 10**100
assert float("-inf") < -10**100
assert lowest([9, 4, 12]) == 4
assert math.isinf(lowest([]))
invalid = infinity - infinity
assert math.isnan(invalid)
assert invalid != invalid
print(lowest([9, 4, 12]))
print("empty remains infinite:", math.isinf(lowest([])))
print("inf - inf is NaN:", math.isnan(invalid))

Explain the result

Each finite candidate is compared with the current best, so 4 becomes the minimum. With no candidates, the sentinel remains unchanged. A caller should interpret that explicitly as no finite result under this function’s contract rather than printing infinity as if it were a measured route cost.

Infinity compares above the large finite integer used in the assertion. Subtracting infinity from itself instead yields NaN, and the self-inequality check demonstrates a distinctive NaN property. The fixture restricts candidates to finite values; accepting NaNs as input would require an additional policy because ordinary comparisons would no longer provide a reliable minimum rule.

Avoid these mistakes

Do not use equality with NaN as its detection method. Check isfinite, isinf or isnan according to the state you need to distinguish. If a business domain permits only finite costs, reject nonfinite inputs before an optimization or report silently carries them forward.