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Comment by firethief | Hacker News Reader

But you are observing all the digits, not just 6-8. It's implicit in the semantics of the operation, and that's something everyone who works with floating point should know.

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I think I see where you're hung up. If I change it to:

    x = 1e20 + 1 - 1e20
    y = 1e20 - 1e20 + 1
    assert x == y

There is only 1 digit, and it's wrong. You don't even need 6-8. I probably should've used this as my example in the first place.

You're making the same mistake but now it's less obvious because the change of scale. When you compare floating point numbers, simple == is not usually what you want; you need to compare them with a tolerance. Choosing the tolerance can be difficult, but in general when working with small numbers you need a small value and with large numbers you need a large value. This dataset involves datapoint(s) at 1e20; at that magnitude, whatever you're measuring, the error in your measurements is going to be way more than 1, so a choice of tolerance ≤ 1 is a mistake.

Ugh, you're preaching to the choir. I wasn't trying to make a point about the equality operator, I was trying to make a point about x and y being completely different. I must be really bad at communicating with people.

That's the thing: x and y are not "completely different"; they are within the tolerance one would use to compare them correctly.

So umm, more than a "difference in the realm of 6-8 significant digits"?

That construction can turn any residual N digits out into a first digit difference. It wouldn't matter without making comparisons with tolerance way under the noise floor of the dataset. But yes, you have technically invented a situation that differs from that real-world anecdote in regard to that property, in an extremely literal interpretation.

And here I was, worried I might be tedious or pedantic, trying to argue that floating point is just not that simple. You've really outdone me in that regard.

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