Shorter checksum than MD5

Elbert Lev elbertlev at
Fri Sep 10 14:27:30 CEST 2004

> For your application, you should consider the total number of records
> you ever expect to have, and use more than 2 * lg(records) bits of hash.
> Due to the so-called "birthday paradox", when you have N possible hash
> values, two will be identical with 50% probability with around sqrt(N)
> items.  You'd probably prefer that the probability be much lower in your
> application, since a collision will result in incorrect results.

Wrong! "birthday paradox" is not applicable here. 

If you want an analogy with this combinatorial problem, imagine 2 rows with N 
objects in each, There exists a "measure" of each object. 
Some objects can be modified with probability 1/2**32 
the  measure will not change after modification. 
Objects in the SAME POSITION in each row are compared by comparing their measures.

After M objects are modified what is the probability that 
at least one modification will be "missed" by the comparison process. 
I don't think, that in the foreseen future (if M and N are not too high) 
such collision will occur.

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