Re^2: Challenge: Fast Common Substrings

++ Wow, thank you for introducing me to suffix trees. What an interesting concept, and how refreshing to see a linear-time algorithm for constructing such a creature. I see you've used the javascript applet at this page, which others may want to check out.

However, I'd like to slightly revise the algorithm you outlined. Consider the following example:

string = ababc%bc$

     |      |(3:abc%bc$)|leaf
     |(1:ab)|
     |      |(5:c%bc$)|leaf
tree:|
     |     |(3:abc%bc$)|leaf
     |(2:b)|
     |     |     |(6:%bc$)|leaf
     |     |(5:c)|
     |     |     |(9:$)|leaf
     |
     |     |(6:%bc$)|leaf
     |(5:c)|
     |     |(9:$)|leaf
     |
     |(6:%bc$)|leaf
     |
     |(9:$)|leaf
[download]

"ab" appears twice in the first string, and so it gives a node with two leaves. The actual condition you should check is whether a node has one leaf containing the % separator and another leaf without the % symbol.

blokhead

Comment on Re^2: Challenge: Fast Common Substrings Download Code

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Re^3: Challenge: Fast Common Substrings (O(n)?) by tye (Sage) on Apr 04, 2007 at 21:53 UTC
The page you link to mentions being able to build them in O(n) but then only really describes how to go from a suffix tree for string $x to one for string $x.$c (1==length$c) in O(length $x). Using that algorithm would require O(N*N) to build the suffix tree for a string of length N. So I'm not sure I believe the O(N) claim for building the whole suffix tree based on that page. - tye	[reply]
Re^4: Challenge: Fast Common Substrings (O(n)?) by lima1 (Curate) on Apr 04, 2007 at 22:12 UTC
The naive algorithm requires O(N*N). The Ukkonen algorithm needs only O(N). If you want to understand it - it is not trivial - I recommend Gusfields book (Algorithms on Strings,...).	[reply]
Re^3: Challenge: Fast Common Substrings by lima1 (Curate) on Apr 04, 2007 at 16:25 UTC
Or even easier: check the positions of the substrings (<=7 and > 7 in my example).	[reply]