What should the "q" value increase to for GNFS/SNFS computations?

[2147483647]

Yesterday I started a large SNFS and I'm just wondering how long it's going to take.

I just did a C100 with SNFS and the sieving range was 510k to 590k.

What should the range look like for larger numbers, in increments of 10 digits? Will the range be larger for GNFS?

[Batalov]

The script will take care of everything. q values should not concern you; they will be variable for different runs and will depend on the project's complexity. Different sievers will be chosen by the script, too; their rate of sieving might seem sometimes logical to the untrained eye and sometimes unexpectedly slower (on the next project) - but you should trust the script.

Here are the fairly good rules of thumb:
- for SNFS, with every 9 additional digits of SNFS difficulty, the overall runtime will double (or, which is the same, runtime will be 10x for every 30 digits)
- for GNFS, with every 5 additional digits of length(N), the overall runtime will double

Remember just these two and you will be in good shape.

Examples: a 120-digit SNFS complexity project might take 1 core-hour on your computer. Then you can assume that a 150-digit SNFS complexity project will take 10 hours, a 180-digit SNFS complexity project will take 100 hours, and a 210-digit SNFS complexity project will take 1000 hours (~40 core-days). Ok?

You can estimate how hard a GNFS project will be by taking length of the input number N, and converting to (N-30)*1.8 = similarly hard SNFS complexity. For example, a 130-digit GNFS is approximately as hard as a 180-SNFS-digit difficulty SNFS (from the estimate above, ~100 core-hours).

[2147483647]

Originally Posted by Batalov

Here are the fairly good rules of thumb:
- for SNFS, with every 9 additional digits of SNFS difficulty, the overall runtime will double (or, which is the same, runtime will be 10x for every 30 digits)
- for GNFS, with every 5 additional digits of length(N), the overall runtime will double

Remember just these two and you will be in good shape.

Interesting, thanks. I suppose these will actually over-estimate the time for large enough numbers since the complexity of NFS is sub-exponential, but if it's a good approximation then that's fine.

I did a C165 SNFS in ~5h30m yesterday and the one I'm currently doing is a C201, so ~16 times longer = about 3 days 16 hours. That's quite a lot less than I was expecting actually.