In Voldemort, why does the hash ring only extend to 2^31-1?

Question

On the project voldemort design page:

http://project-voldemort.com/design.php

It is stated that the hash ring covers the interval [0, 2^31-1].

Now, the interval [0, 2^31-1] represents 2^31 total numbers, and the largest number 2^31-1 is just 31 bits all set to 1. (To convince yourself of this, consider 2^3-1. 2^3=8 and is 0x1000. 2^3-1=7 and is 0x111).

Thus, if a normal 32-bit address word is used to store the value, you have 1 bit free.

Thus, why is 2^31-1 the upper limit? Is that extra bit used for some kind of system bookkeeping?

(e.g. 1 extra bit would provide room for safely adding two valid hash addresses without overflow).

And finally, is this choice specific to voldemort, or is it seen in other consistent hashing schemes?

Answer 1

I think you only have 1 bit free not 2. The -1 accounts for the fact that it starts with the number '0' instead of 1 (the same reason loops count from 0 to count-1). I would guess the reason they use 2^31 instead of 2^32 is that they're using a signed integer so that last bit is the sign bit and so is not useable.

Edit: From the page you linked:

To visualize the consistent hashing method we can see the possible integer hash values as a ring beginning with 0 and circling around to 2^31-1.

It specifies an integer so unless you want negative hash values you're stuck with 2^31 instead of 2^32.

Answer 2

Answering a question little late, just by 4 years :)

The reason is Voldemort is written in Java and Java has no unsigned int. 2^31 is already very high for the partitions. Generally we recommend running with only few thousand partitions.