Re: [asterisk-users] AstDB/Berkely DB - Hash function? Balanced-Tree? b-Tree? Linked List?

2008-08-17 Thread Karl Fife
Thanks for your reply Jay. You're quite right. It would have been better to say npa/nxx is a relatively evenly distributed keyspace. To answer your question specifically, no, I have not done a distribution analysys (though it would be interesting). The point I was trying to make is that a

Re: [asterisk-users] AstDB/Berkely DB - Hash function? Balanced-Tree? b-Tree? Linked List?

2008-08-15 Thread Russell Bryant
Karl Fife wrote: Does anyone know enough about the implementation of AstDB to know whether the data structure is a Hash function, a Balanced-Tree, a b-Tree, or a Linked List? I've never looked at the internals of db1. However, by simply looking at what code is included, it looks like it is

Re: [asterisk-users] AstDB/Berkely DB - Hash function? Balanced-Tree? b-Tree? Linked List?

2008-08-15 Thread Atis Lezdins
On Fri, Aug 15, 2008 at 8:56 AM, Karl Fife [EMAIL PROTECTED] wrote: Does anyone know enough about the implementation of AstDB to know whether the data structure is a Hash function, a Balanced-Tree, a b-Tree, or a Linked List? I'm trying to estimate the lookup 'cost' of a AstDB with around

Re: [asterisk-users] AstDB/Berkely DB - Hash function? Balanced-Tree? b-Tree? Linked List?

2008-08-15 Thread Jay R. Ashworth
On Fri, Aug 15, 2008 at 12:56:49AM -0500, Karl Fife wrote: The key-space is ideal. It's just npa/nxx lookups so it's UNIQUE and EVENLY DISTRIBUTED Based on my knowledge of the NPA/NXX space, I wouldn't expect that either a) A given batch of random DNs would have either or both NPA/NXX

[asterisk-users] AstDB/Berkely DB - Hash function? Balanced-Tree? b-Tree? Linked List?

2008-08-14 Thread Karl Fife
Does anyone know enough about the implementation of AstDB to know whether the data structure is a Hash function, a Balanced-Tree, a b-Tree, or a Linked List? I'm trying to estimate the lookup 'cost' of a AstDB with around 160,000 keys? Obviously I already know that it WILL WORK, but the