Timothy -
It becomes more tricky if you want to know what fraction of everything
a host can present that carries 2 or 3 different filters (I mentioned
that somewhere).
For example: everything is 4^4 = 256 options
[ABCD][ABCD][ABCD][ABCD]
Host carries these two filters:
[ABC][AC][AB][ABC]
On Sun, Dec 28, 2008 at 12:49 AM, bOR_ boris.sch...@gmail.com wrote:
Annoying way to start a morning :P. Thanks though for the tip on
htdp.org
I'm not trying to be annoying. I just don't know how valuable the
answer will be to you, until you've got a better handle on recursion.
The code
Thanks. I understood your intentions, and it didn't hurt for me to
find my own solution.
(time (count (expand (first myset
Elapsed time: 2426.327 msecs
403200
Timing yours it is about as fast as the flexible for version, but not
as fast as the tailored for version. Perhaps I should look at
Ah. The prohibitively long bit comes mainly from printing out the
whole list in emacs. Counting it is only a few seconds, and not that
much slower than for the fixed-length version. In computer-time it is
still taking ages though.
Why am I going through all of this?: I have 3 filters in the form
I'm not surprised that the flexible version takes a bit more time than
the one that's hardcoded to a for loop nested to 9 levels. The
flexible version necessarily involves recursion, which means
allocating some stack space with each call, and that's a bit of
overhead.
I am a bit surprised that
I coded up a quickie iterative version for you so you can get an idea
what that looks like. My timing tests indicate that this runs many
times faster than even the hardcoded for loop nested to 9 levels.
; Convert to a vector, and pass off to step, which builds a lazy
sequence by applying
This iterative version doesn't behave properly if one of the sequences
is empty. Expand should probably check for this degenerate case
before passing the vector off to step.
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On Dec 28, 12:26 pm, Mark Engelberg mark.engelb...@gmail.com
wrote:
This iterative version doesn't behave properly if one of the sequences
is empty. Expand should probably check for this degenerate case
before passing the vector off to step.
I'll take care of that
Interesting. on my comp,
If you aren't interested in the permutations themselves, just the
number of them (if I read you right)
The only way I could think up to know exactly what
fraction of the optimum (2x 5%: remember the diploidi)) a set of
combined filters can present is by expanding these filters into a set
Hi all,
Next problem. If I've a collection of sets of letters, and I want all
possible words you can make with this collection if you keep the order
of the sets intact, I would use 'for'. However, for wants me to know
beforehand how many sets there are in my collection. Is there a more
flexible
On Sat, Dec 27, 2008 at 2:18 PM, bOR_ boris.sch...@gmail.com wrote:
but
didn't see an obvious/elegant way to do it in there, and the only way
I'd write it now in clojure is going to be ugly.
Anyone has a suggestion?
This kind of problem is straightforward if you understand recursion,
and
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