ERROR: cuMemcpyDtoH() 700
#21
Well, if you know the speed, you can calculate how much time it takes to be in your range and use the --runtime switch.
#22
(02-18-2013, 08:07 PM)squadx Wrote: can you use the ?d argument and still limit it to a range? part of the benefit of creating the dictionary is being able to make a dictionary file that is within say 9000000000 to 9500000000

I assume you can do 9?d?d.... but can you put an ending range in there there?


Sure, you can easily do -1 012345 9?1?d?d?d?d?d?d
#23
Or you can easily use the new restore/resume features. Guess they work pretty good with -a3.
#24
I don't think that's what he was asking for, I interpreted is question as "do I have to do 00000000 - 99999999, or can i do 90000000 - 95000000."
#25
yeah i was asking how to do 9 - 9.5 type setup
#26
So to be clear what I am reading
"Sure, you can easily do -1 012345 9?1?d?d?d?d?d?d "

So to get the remainder of the values in that 9000000000-999999999

After running that command would I change it to "-1 56789 9?1?d?d?d?d?d?d" to get from 95 to 99 to cover all values in the range?
#27
(02-20-2013, 06:18 PM)squadx Wrote: So to be clear what I am reading
"Sure, you can easily do -1 012345 9?1?d?d?d?d?d?d "

So to get the remainder of the values in that 9000000000-999999999

After running that command would I change it to "-1 56789 9?1?d?d?d?d?d?d" to get from 95 to 99 to cover all values in the range?

yes, except you would only use -1 6789 (5 omitted)
#28
So -1 creates for a custom character set that is called in the argument via the "?1"

And if I just put -1 0 it would run through 900000000 up until 9100000000?

and so forth for other number ranges "-1 01 2?1?d?d..." would run through 2000000000 to 2200000000

Sorry for the "noobish" questions, considering how time consuming running this many combos can be, just wanna make sure I'm not wasting hours running invalid stuff
#29
no exactly. It would not go up to 9100000000 but only 9099999999