Question

If my math is right, I can quickly generate a new hash value for the concatenation of two strings if I already have the individual hash values for each string. But only if the hash function is of the form:

hash(n) = k * hash(n-1) + c(n), and h(0) = 0.

In this case,

hash( concat(s1,s2) ) = k**length(s2) * hash(s1) + hash(s2)

eg.

h1  = makeHash32_SDBM( "abcdef",          6 );
h2  = makeHash32_SDBM( "ghijklmn",        8 );
h12 = makeHash32_SDBM( "abcdefghijklmn", 14 );
hx  = mod32_powI( 65599, 8 ) * h1 + h2;

h1  = 2534611139
h2  = 2107082500
h12 = 1695963591
hx  = 1695963591

Note that h12 = hx so this demonstrates the idea.

Now, for the SDBM hash k=65599. Whereas the DJB hash has k=33 (or perhaps 31?) and h(0) = 5381 so to make it work you can set h(0) = 0 instead.

But a modification on the DJB hash uses xor instead of + to add each character.

http://www.cse.yorku.ca/~oz/hash.html

Is there another technique to quickly calculate the hash value of concatenated strings if the hash function uses xor instead of +?

Answer 1

That would be true if your second hash will be function of initial state of hash. For some kinds of hash-function it’s easy to shift them according to new initial state (like xor’e words, or their sum etc). But in general case that’s almost impossible (in other case use of hash+”salt” in password matching will be easier to break).

But usually you can use result of hashing of first string and than continue feeding data from second string.

Update:
I guess there is no way to find f(x,y) for:

h_abc = hashOf(h0, "abc")  
h_def = hashOf(h0, "def")  
(h_abcdef = f(h_abc, h_def)) == hashOf(h0, "abcdef")  

But you able to:

h_abc = hashOf(h1, "abc")  
(h_abcdef = hashOf(h_abc, "def")) == hashOf(h0, "abcdef")  

one of the reason for why you can’t do that is that “33” isn’t power of “2”. If it will use “32” (2**5), then:

h_abcdef == (h_abc << (5*len(abc))) xor h_def