<arigato>
an unrelated small puzzle: the playground is a sequence of 0 or 1 with a "current position" n, which evolves in a stepwise fashion following these rules
<arigato>
first, we must have seq[n] == 1
<arigato>
n evolves into either n + 1 or n - 1, "randomly", i.e. you can't choose that
<arigato>
after you know how n evolves you can "compute" (up to) one element of the sequence from the previous one: from some i for which seq[i] == 1, assign 1 to seq[i + 1] (but always i+1, not i-1)
<arigato>
the sequence starts with seq[0] == 1 and otherwise all zeroes, and n == 0
<arigato>
of course you can just "compute" the next sequence element every step, and that works fast enough for any "random" movement of the current position
<arigato>
but now we add a memory constrain: you can forget any number of sequence elements (turn them back from 1 to 0) at any point, and the goal is that there is not too many elements equal to 1
<arigato>
e.g. ~O(n), say
<arigato>
sorry, ~O(log n)
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<cfbolz>
arigato: hm, don't quite get it. where is this from?
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<phlebas>
arigato: I don't get it - assuming that n must always be >= 0, we initially must "randomly" go to n+1. That one is 0 from the precondition, or was I allowed to "compute" seq[1] == 1 before the first random step takes place? Because if i'm not, this was a short game ;) if I'm allowed, of course, all i have to do is make sure that seq[2] == 1 now, observe the next step, and as we're moving away from either end, I always reset i+2
<phlebas>
or i-2 to 0, right?
<phlebas>
ah, no, i cannot set seq[i-1] to 1 when i observe n going backwards. I see.
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<arigato>
yes, you're correct, the first "compute" step is done at the same time as n goes from 0 to 1
<arigato>
so when n becomes 1, we are allowed one compute step (which must be seq[1] == 1) and afterwards seq[n] == 1 so it's fine
<arigato>
so yes, the goal is to keep some small but clever subset of seq[0:n] to 1, so that if we observe several steps to the left (n => n - 1) we can compute missing values, and forget some of them, so that we still have never more than O(log n) values at 1 at the same time
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<arigato>
cfbolz: basically, it models anything where you can compute step n+1 from step n, and the user generally asks for the steps in order but sometimes he asks for them backwards
<arigato>
cfbolz: so revdb is a typical example, but here I'm thinking about it for a game
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<arigato>
(so far, revdb is implemented by keeping only some fixed seq[f1] == .. == seq[fn] == 1, and doing a step to the left requires computing (n-fi) steps forward)
<cfbolz>
ah, I see
<arigato>
in case it changes anything, I can allow doing several computation instead of just one, with a fixed upper bound
<arigato>
or even O(log n) such computations, that is still fine
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<tos9>
What's the equivalent of ctypes.byref(ctypes.c_void_p()) in CFFI
<tos9>
Hrm maybe I'm doing that part correctly actually and this error is elsewhere.
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<tos9>
Yeah OK probably I'm doing something really simple wrong./0
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<tos9>
Apologies for as usual I'm sure is just me not knowing C, but what's the right way to call such a function:
<tos9>
The 2 things I tried produced either a traceback or a segfault there at the bottom.
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<fijal>
tos9: hey hey
<fijal>
how's going?
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<tos9>
fijal: hi hi
<tos9>
going well
<tos9>
trying to go on vacation
<tos9>
how goes with you
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<fijal>
haha, same :-)
<fijal>
and how is it working for you tos9?
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<tos9>
fijal: so far not so good... and you guys aren't helping, you should plan a summer sprint :)
<tos9>
but *hopefully* I'll get away like next week or week after
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<RemoteFox>
how can you check whether file is closed under rpython? just accessing .closed property doesn't work
<tos9>
RemoteFox: It looks like theres a _check_close method on rfile
<tos9>
I don't know whether the same privacy rules apply in rlib, but presumably if you're writing an rpython interpreter you'll stay reasonably up to date anyhow
<tos9>
You need to know whether it's closed from outside the file?
<tos9>
(Probably I'd assume the real right way to do what you want is to carry the state on your language level object, which presumably is how pypy uses the rfile object)
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<RemoteFox>
yep
<RemoteFox>
hmm
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<RemoteFox>
I used just regular result from the open() call until now
<RemoteFox>
I'll have to look at the rfile
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