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0.55 Problem

PostPosted: Wed Jul 21, 2010 12:16 pm
by blazer
i'm getting the libcuda.so.3 not found msg.

Exact same setup works with 0.50 but not with 0.55, i even tried moving only GRTCrack/GRTGen and replacing the 0.50 ones still get the same error.

Nvidia: GTX260 216SP

Do I need to do a driver update or something?
Mite be a little problem since my networking has for reason stopped working in ubuntu.

Re: 0.55 Problem

PostPosted: Wed Jul 21, 2010 3:31 pm
by Bitweasil
Can you run 'ldd' against a binary? I think I know the problem, but need to verify.

Re: 0.55 Problem

PostPosted: Thu Jul 22, 2010 12:49 am
by Bitweasil
Alright, redownload and try again.

I was missing the LD_RUN_PATH setting when I built the release binaries.

Re: 0.55 Problem

PostPosted: Thu Jul 22, 2010 7:36 am
by blazer
ok yup works fine.
Generating some tables, i think i have the gotten the hang of it

Chain Length: 150 000
Len: 6
Charset: All
Algo: NTLM

I'll upload them once done, also i'm almost finished with uploading the len7 one more index to go, i'll post all the links when i do the last index over the next couple of days.

Getting Step rate of roughly 595M/s for generating, another question

During the chain generation is it possible to implement reversing and not carry out the full rounds, then reverse the skipped steps when searching for the hash. Will this screw up the distribution of the hashes and will the gains if any be worth it?
Oh also, how many bits am i suppose to specify, seems to give me an error when i do it i'm just picking random numbers, though seems they still can crack without the index.

Re: 0.55 Problem

PostPosted: Fri Jul 23, 2010 3:47 am
by Sc00bz
blazer wrote:During the chain generation is it possible to implement reversing and not carry out the full rounds, then reverse the skipped steps when searching for the hash. Will this screw up the distribution of the hashes and will the gains if any be worth it?

Using the GRT reduction function you can get away with these reversals.
For MD5 you can reverse:
- 11 out of 64 steps for 8-8 or less
- 3 out of 64 steps for 12-12 or less
- 2 out of 64 steps for 16-16 or less
For NTLM you can reverse:
- 10 out of 48 steps for 6-6 or less
- 6 out of 48 steps for 8-8 or less
- 5 out of 48 steps for 12-12 or less
- 4 out of 48 steps for 14-14 or less
- 2 out of 48 steps for 16-16 or less
For MD4 you can reverse:
- 10 out of 48 steps for 8-8 or less
- 9 out of 48 steps for 12-12 or less
- 4 out of 48 steps for 16-16 or less

GTX295 gets 705 Mmd5s/sec per core on BarsWF (estimating about 528.75 M full md5s/sec)
GTX295 gets 436 Mlinks/sec per core on GRT (don't know if this is for 6-6, 7-7 or other)
BarsWF 1.8913 ns/full md5 (1 / 0.705 * 64 / 48)
GRT 2.2936 ns/link (0.4023 ns for the reduction function [apples painted orange - oranges = usable info])
So for MD5 8-8 or less you'll get 508 Mlinks/sec on GRT. So my guess is 16.5% faster (note that 20.8% is the very max (64/53-1)).

blazer wrote:Oh also, how many bits am i suppose to specify, seems to give me an error when i do it i'm just picking random numbers, though seems they still can crack without the index.

This will get you an average of 16 to 32 chains per index. Max indexedBits is 32 bits.
indexedBits = ceiling(log2(chainCount / 32))

-------
I used reversing for MD5 when I wrote my reduction function and got 87.6 Mlinks/sec on a Core2Quad 2.5 GHz in 32 bit mode.
http://www.freerainbowtables.com/phpBB3 ... html#p5758
haha "Good news, the rt calculator is almost done." -me posted 13 Aug 2008
I over complicated that thing and never finished it.

Core2Quad 2.5 GHz
BarsWF 166 Mmd5s/sec (6.9026 ns/adjusted md5 [1 / 0.166 * 55 / 48])
My rt reduction function 87.6 Mlinks/sec (11.4155 ns/link)

39.5% of the time is spent on my reduction function.
17.5% of the time is spent on GRT's reduction function.
As you can see here GRT's reduction function is faster granted you shouldn't really compare these two situations as they are so different, but it gets the expected answer.

Re: 0.55 Problem

PostPosted: Fri Jul 23, 2010 6:23 am
by Bitweasil
Thanks for the useful info, scoobz!

I'm still working on a tutorial and online chain math calculator.

Tbh, I'm probably not going to implement reversing any time soon. That adds a ton of complexity, and as observed, a working product beats a theoretically better but unreleased product. :)

Re: 0.55 Problem

PostPosted: Fri Jul 23, 2010 1:06 pm
by Sc00bz
Bitweasil wrote:a working product beats a theoretically better but unreleased product

:D

Re: 0.55 Problem

PostPosted: Sun Jul 25, 2010 8:50 am
by blazer
hey sc00bz thx but whats the ceiling thingo

indexedBits = ceiling(log2(chainCount / 32))

i'm currently at around 7.5million unique chains

Re: 0.55 Problem

PostPosted: Mon Jul 26, 2010 6:41 am
by Sc00bz
Ceiling is integer round up, 1.00001 is rounded up to 2. Floor, the more common one, is integer round down, 1.99999 is rounded down to 1.

log2(7,500,000 / 32) = 17.838
So ceiling(17.838) is 18 indexed bits.
7,500,000 / 2 ^ 18 = 28.610 chains per index

I could have used indexedBits = floor(log2(chainCount / 32) + 1)
Here's the difference:
16 <= floor(log2(chainCount / 32) + 1) < 32
16 < ceiling(log2(chainCount / 32)) <= 32

Re: 0.55 Problem

PostPosted: Sat Oct 09, 2010 8:54 pm
by recfrf
Hello.
Can not understand the syntax, you can sample for example to generate a table?
"rtgen md5 numeric 1 12 0 8300 67108864 0"