Re:Lampsort: Leslie Lamport's Non-recursive Quicksort
by taeric on Sunday December 07, 2014 @03:48PM
(#8713513)
Threaded trees was what I had in mind.[1] I believe the example there still talks of checking a visited list, though. Either a misunderstanding on my part, or on that page. I'll have to check on my books. Later, sadly. :)
[1] http://en.wikipedia.org/wiki/Threaded_binary_tree
Edit: Apologies for a quick edit. I remembered I had a short implementation on my computer. Basically, when you follow a link, you know if it was a regular link or a thread. If it was a regular link, you could still go left from the new node. Otherwise, you visit and go right. Again, keep note of whether it was a thread or not. Repeat until done. (Make sense?)
Re:Lampsort: Leslie Lamport's Non-recursive Quicksort
by al2o3cr on Sunday December 07, 2014 @06:59PM
(#8714182)
"Again, keep note of whether it was a thread or not."
This sounds like the visited list again...
Re:Lampsort: Leslie Lamport's Non-recursive Quicksort
by taeric on Sunday December 07, 2014 @07:20PM
(#8714258)
You only have to remember the link you just followed. Basically, make a boolean "followedThread" and initialize to false. Then, go to the root and follow this algorithm
if (!followedThread && left != null)
set followedThread <- curNode.left.isThread
set curNode <- curNode.left.node
else
set followedThread <- curNode.right.isThread
set curNode <- curNode.right.node
Only node you really have to special case is the root node, and you can do that by making its right.node == null. Then you just do this while curNode != null.That make sense? (Also... very possible that I made a mistake here...)
What do you mean by this? You need a way to say that the unexplored nodes we've most recently found are the ones we explore first. How do you do that (with constant time operations) without what's effectively a stack?