in reply to Re^2: Mark Jason Dominus And Me - The Partition Problem
in thread Mark Jason Dominus And Me - The Partition Problem
For 2 the answer is, "Whenever we have a problem that we don't know how to solve, which can be somehow reduced to simpler versions of the same problem." For Fibonacci, that means smaller $n. In the case of the directory walk, that means farther down the directory tree. In the case of the partition problem, simpler means "fewer treasures".
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Re^4: Mark Jason Dominus And Me - The Partition Problem
by Tommy (Chaplain) on Nov 19, 2012 at 22:25 UTC
Ummm. Just doesn't make sense to me to try and hit a moving $target... pun intended. It makes a helluva lot more sense now that my target stays constant and my "share" is the only thing really changing (see code below).
Thank you tilly for helping me understand why I didn't understand. That's when everything started making sense.
This is how I would do it... (remove "print" statements after debugging done).
Click "Download code" link to see the actual, pretty, wide-formatted text
by Tommy (Chaplain) on Nov 19, 2012 at 23:27 UTC
Maybe it doesn't hurt to program the way your brain really thinks...at least not this time. You see, I stripped down the MJD code to its barest form, exactly from the book. Then I removed the print statements from my own version which was just shared in the node previous to this node.
Surprise: despite the fact that I'm passing around a reference to my "$share" with each call, my code benches twice as fast. I'm betting it's because I'm not constantly recalculating the "$target".
Anyway, here's the benchmarks. I wonder what conclusions could be drawn. If I cared more, I'd run it through Devel::NYTprof and find out what's really going on and why it's faster to do it the way that makes the most sense to my dyslexic brain: