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Re:Blocklist Facebook domains by nostromo on Tuesday March 20, 2018 @07:44PM (#16633893)
[amount of data collected from phones]=[amount of data collected from desktops]^[some exponent]
Re:Blocklist Facebook domains by btrettel on Tuesday March 20, 2018 @08:30PM (#16634166)
If the exponent is less than 1 then the amount of data collected from phones is less than that from desktops.
Re:Blocklist Facebook domains by hueving on Tuesday March 20, 2018 @09:55PM (#16634659)
When you hear 'exponential back off algorithm', do you envision one that keeps retrying faster and faster?
Re:Blocklist Facebook domains by btrettel on Wednesday March 21, 2018 @01:57AM (#16635828)
I'd say better than 50% chance that the delay increases. But the phrase would be unambiguous if it were called an "exponentially increasing back off algorithm".
Re:Blocklist Facebook domains by marksomnian on Wednesday March 21, 2018 @03:09AM (#16636130)
This is unnecessarily pedantic. An exponential back off algorithm has a 100% chance of increasing the delay, that's the whole point. Nowhere other than pure mathematics would I see the phrase "exponentially" and even consider a <1 exponent.
Re:Blocklist Facebook domains by cmstoken on Friday March 23, 2018 @01:44AM (#16655395)
I think "needlessly" would work better than "unnecessarily" there.
Re:Blocklist Facebook domains by btrettel on Wednesday March 21, 2018 @03:48PM (#16641661)
My probability was for hearing the phrase "exponential back off algorithm" without knowing anything about the algorithm. I don't work in that field and had never heard of the term before the earlier post.

Experience suggests that most of the time when people say exponentially they mean an exponent greater than 1, but I have been surprised by what people have meant before so I personally wouldn't say that probability is greater than 90%. That's what I meant in more detail.

Re:Blocklist Facebook domains by btrettel on Thursday March 22, 2018 @02:35AM (#16646280)
Correction: For some reason I was thinking x^n where n < 1 rather than n^f(x). If f(x) monotonically decreases with x then what I said still applies.