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(Yawn) Ahh, red-eye flights. It was quite short flight. I just closed my eyes when it was time to open them again. I went back to my apartment to take a nap, then clean up before work, and I still got in to work earlier than I usually do on a Monday.
OK, I admit I avoided most serious work. I spent a lot of time with Dan, helping him figure out how to convert the current table object model into one more suitable for a tree table. By convergent evolution, the current object model had I think four different yet similar ways to do almost the same thing (represent a row and its children), each of which existed to meet certain specific needs. I think we can change it to just one representation with appropriate attached Strategies. I'm sure Dan would have arrived at the same conclusion, but I wanted to know more about how all this stuff worked.
It's getting cold outside. I had to wear my sweatshirt during my walk today. I won't be able to walk outside much longer.
OK, so I just spent a while writing down the journal entries for last weekend. Here's a stray
thought about the indexed binary tree: would there be any advantage to keeping the left count
instead of the right count at each node? (You can keep both left and right at each node, but that
requires 2n ints. If you keep just the left (or right) at each node
and the count of nodes in the whole tree, you can get away with a mere n+1
ints.) I'm not sure. While traversing, you really need L. If you had it,
you wouldn't have to keep calculating it from L=P-R-1 while traversing, but
would it allow you to eliminate P completely? Hmmm....
| Louis K. Thomas <louisth@hotmail.com> | Auth | 2003-09-30 (1838 days ago) |