Hi, I am sorry, that it took me so long to answer, but I have been swamped with work in the last few days and I am in a hurry right now, too. First, thank you all, the discussion is really interesting :)
I found a hint to OPHF in Koloniari et al. (http://www.cs.uoi.gr/~pitoura/distribution/sr05.pdf). Under Clustering on page 11, it says: "In structured p2p systems, if the hash function is order-preserving, similar documents are stored at the same or neighboring peers. Order preserving hash functions are those hash functions that for similar inputs produce outputs close in the identifier space. Then, content-based clustering can be achieved by using as input to the hash function not just the name of the document but a semantic vector describing its content and structure." I want to add clustering to a DHT to store documents and I am beginning to understand how I could achieve it. I know, I am breaking with load-balancing (and will have to find other strategies to achieve it) as well as I am running into problems because I don't want any hybrid system or super-peer or anything alike. If I don't find an OPHF I might have to use a linear distribution using a special comparison metric which would (hopefully) serve the same purpose... Thank you very much. Karl -------- Original-Nachricht -------- > Datum: Tue, 29 Apr 2008 09:34:43 +0800 > Von: Music Yin <[EMAIL PROTECTED]> > An: \'theory and practice of decentralized computer networks\' > <[email protected]> > Betreff: Re: [p2p-hackers] Order preserving hash function? > > I am also curious about OPHF. I wish someone could give a link like the > paper or the brief introduction on OPHF. Thanks. > > > I'd still like to know what an OPHF is, never mind anything to do with > DHTs. > Is it a hash function where H(x) > H(y) iff x > y ? Is it iff > x>y, then H(x) >= H(y)? > > > On Apr 28, 2008, at 3:26 PM, Jack Lloyd wrote: > > > On Mon, Apr 28, 2008 at 03:03:30PM -0700, Jim McCoy wrote: > > > >> [Perhaps I am just misunderstanding your construction, but MD4(x) || > >> SHA-256(x) is only as strong as MD4, not stronger than SHA-256 > >> alone...] > > > > Wouldn't that imply a very easy way to break SHA-256? I think if you > > defined strong as you used it in the above sentence it might help me > > understand your argument; the only definitions I can think of that > > might fit are psuedo-randomness (or resistence to partial collision / > > preimage). I believe Joux's multicollision attack only shows that you > > can break MD4(x)||SHA-256(x) about as easily as SHA-256 alone. > > > > Regards, > > Jack > > _______________________________________________ > > p2p-hackers mailing list > > [email protected] > > http://lists.zooko.com/mailman/listinfo/p2p-hackers > > _______________________________________________ > p2p-hackers mailing list > [email protected] > http://lists.zooko.com/mailman/listinfo/p2p-hackers > > > _______________________________________________ > p2p-hackers mailing list > [email protected] > http://lists.zooko.com/mailman/listinfo/p2p-hackers -- Ist Ihr Browser Vista-kompatibel? Jetzt die neuesten Browser-Versionen downloaden: http://www.gmx.net/de/go/browser _______________________________________________ p2p-hackers mailing list [email protected] http://lists.zooko.com/mailman/listinfo/p2p-hackers
