Unless, of course, you unlink its executable file, in which case it allocates swap to hold the file first.
Why doesn't it just take into account the fact that the file is in execution in its reference
count and leave the file where it is on the file system until it terminates? "unlink"
only decrements the reference count, it does not free the file. Seems
much simpler to me.
In addition, if it does as you say, FreeBSD does not just need to allocate the swap to hold
the file, it actually needs to take every page from the executable file that has not yet been
loaded in memory and actually copy it to the swap. Where's the coolness in that?
By the way, I didn't find the paragraph which lead you to that that's how it does in there.
While this could be an effort to fight the software giant and its product directly
On the other hand, perhaps Apple will wait until the software giant registers a 0.5% market
share before considering it as an opponent. I'd say the software giant is doing a good job of fighting
itself as it is.
Now, if only I had Linux running on my Powerbook... see, HFS+ does write to disk as soon as it can, which is good for saving data, but bad for saving battery.
Did you try "sudo killall update" on Mac Os X? It works for me when I'm on the go,
but then I'm only reading data, not writing (it's only these darned file access
times that make it necessary to kill update in my case).
I think you have to do it by hand if you want to get the reward.
A more tractable problem would be to compute the probability that a solution to the Ten Letter Acrostic Puzzle exists in a Gauss-distributed, Markov-model language with 1.5 bits of information per character.
No, it's not. If you used the same reasoning with 15000 tries and 7672 correct answers, you would also conclude that there is some statistical significance in your results.
Your reasoning would amount to saying "it's extremely unlikely to obtain exactly 7672 correct answers, therefore this can not be the result of chance". The reasoning is wrong (but I'm not saying that 12 correct among 15 is not significant, I am only challenging the reasoning).
Isn't that the probability of obtaining *exactly* 12 among 15? And shouldn't you consider the probability of obtaining *at least* 12 among 15 correct guesses if you wanted to be sure not to incorrectly infer that your result can not be obtained by chance?
It is wrong to consider the probability of obtaining exactly the same number of correct answers as you obtained. It becomes more and more improbable (and therefore falsely appears significant) as the number of tries increases (imagine you had made 1500 tries instead of 15).
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Why doesn't it just take into account the fact that the file is in execution in its reference count and leave the file where it is on the file system until it terminates? "unlink" only decrements the reference count, it does not free the file. Seems much simpler to me.
In addition, if it does as you say, FreeBSD does not just need to allocate the swap to hold the file, it actually needs to take every page from the executable file that has not yet been loaded in memory and actually copy it to the swap. Where's the coolness in that?
By the way, I didn't find the paragraph which lead you to that that's how it does in there.
http://fakesteve.blogspot.com/
On the other hand, perhaps Apple will wait until the software giant registers a 0.5% market share before considering it as an opponent. I'd say the software giant is doing a good job of fighting itself as it is.
I think you have to do it by hand if you want to get the reward.
A more tractable problem would be to compute the probability
that a solution to the Ten Letter Acrostic Puzzle exists in
a Gauss-distributed, Markov-model language with 1.5 bits
of information per character.
I'm too lazy to do it.
> this is the correct statistic to consider,
No, it's not. If you used the same reasoning
with 15000 tries and 7672 correct answers, you would
also conclude that there is some statistical significance
in your results.
Your reasoning would amount to saying "it's extremely
unlikely to obtain exactly 7672 correct answers,
therefore this can not be the result of chance".
The reasoning is wrong (but I'm not saying that
12 correct among 15 is not significant, I am
only challenging the reasoning).
Read http://en.wikipedia.org/wiki/Prosecutor's_fallacy
and substitute "results due to chance" to "innocence",
and "significant results" to "guilt".
p = (15 nCr 12) * 0.5^12 * 0.5^3
= 0.014
Isn't that the probability of obtaining *exactly* 12 among 15?
And shouldn't you consider the probability of obtaining *at least*
12 among 15 correct guesses if you wanted to be sure not to
incorrectly infer that your result can not be obtained by chance?
It is wrong to consider the probability of obtaining exactly the same
number of correct answers as you obtained. It becomes more and more
improbable (and therefore falsely appears significant) as the number
of tries increases (imagine you had made 1500 tries instead of 15).