BTW, what sort of memory usage are we talking about here?
I was referring to the memory usage of this program
import System.Environment
import Data.Numbers.Fibonacci
main :: IO ()
main = do n <- (read . head) `fmap` getArgs
(fib n :: Integer) `seq` return ()
compiled with -O2 and run with +RTS -s:
./calcfib 100000000 +RTS -s
451,876,020 bytes allocated in the heap
10,376 bytes copied during GC
19,530,184 bytes maximum residency (9 sample(s))
12,193,760 bytes maximum slop
97 MB total memory in use (6 MB lost due to
fragmentation)
Generation 0: 40 collections, 0 parallel, 0.00s, 0.00s
elapsed
Generation 1: 9 collections, 0 parallel, 0.00s, 0.00s
elapsed
INIT time 0.00s ( 0.00s elapsed)
MUT time 12.47s ( 13.12s elapsed)
GC time 0.00s ( 0.00s elapsed)
EXIT time 0.00s ( 0.00s elapsed)
Total time 12.47s ( 13.13s elapsed)
%GC time 0.0% (0.0% elapsed)
Alloc rate 36,242,279 bytes per MUT second
Productivity 100.0% of total user, 95.0% of total elapsed
I'm not sure how to interpret "bytes allocated" and "maximum
residency" especially because the program spends no time during GC.
But the 97 MB "total memory" correspond to what my process monitor
shows.
I have now tried your code and I didn't find the memory usage too
extreme.
Be aware that fib (10^8) requires about 70 million bits and you need
several times that for the computation.
Then, roughly ten 70m bit numbers fit into the total memory used:
ghci> (97 * 1024^2 * 8) / (70 * 10^6)
11.624213942857143
I expected to retain about three of such large numbers, not ten, and
although the recursion depth is not deep I expected some garbage. Both
expectations may be a mistake, of course.
If the relation is
a_n = c_1*a_(n-1) + ... + c_k*a_(n-k)
you have the k×k matrix [...]
to raise to the n-th power, multiply the resulting matrix with the
vector
of initial values (a_(k-1), a_(k-2), ..., a_0) and take the last
component
Interesting.
(a propos of this, your Fibonacci numbers are off by one, fib 0 = 0,
fib 9
= 34, fib 10 = 55).
Wikipedia also starts with zero but states that "some sources" omit
the leading zero. I took the liberty to do the same (and documented it
like this) as it seems to match the algorithm better. It also
corresponds to the rabbit analogy.
These matrices have a special structure
that allows doing a multiplication in O(k^2).
You might want to look into the Cayley-Hamilton theorem for the
latter.
I don't see the link to the CH theorem yet -- probably because I
didn't know it. I did observe that all matrices during the computation
have the form
/a b\
\b c/
which simplifies multiplications. Is this related to CH? Or can I
further improve the multiplications using insights from CH?
Sebastian
--
Underestimating the novelty of the future is a time-honored tradition.
(D.G.)
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