intermediate file write error
 

Hey lovely people
I've been running v.22 for about 2 months now, and just recently it's giving me this: Error writing intermediate file: rI128533 (that being the last 6 digits of my current number). I searched my computer for this file and couldn't find it, which may be part of the problem; I also looked for a Prime95 folder which might contain such program files, couldn't find that either. I'm sort of leery to restart my computer now, since I suspect I might lose all my work so far, so I'd like to know what's going on. If anyone can enlighten me, I'd really appreciate it. peace
-Talia

Is your hard drive or quota full?

If there isn't enough room to write the file it won't be created...

aha! that might just be it...can i specify somewhere else for the file to write to, or do i just have to empty out that drive?

While the program is running you'll have to empty out some space on that drive

I've freed up about 300 kb, which should be sufficient...I'll let you know if I have more problems. Thanks a bunch, y'all!
-Talia

[quote="TaliaW"]I've freed up about 300 kb, which should be sufficient...[/quote]
I suspect that if you are running a test on an exponent of that size (18128533) the save file will be about 4 MB (ie 4,096 KB).

ok, I uninstalled some random programs and have more than enough room, and all seems to be well, no more errors. lovely! thank you all. peace out.
-Talia

By default the program writes 2 save files, 2 * 4 = 8 MB. In any case, it sounds like you freed up enough space!

Each save file would be slightly longer than 18128533 / 8 = 2266067 bytes.

No, I am running an exponent of similar size, and each save file is close to 4 MB.

markhl is correct about the save file size and I think this is why:

M18128533 uses 1024K FFT. This means it uses 1,048,576 elements to compute the FFTs for this exponent. This comes out to 18,128,533 / 1,048,576 = 17.3 bits per element. This actually means that each element stores either 17 or 18 bits of the residue.

I think in George's implementation all the elements are stored in 32-bit unsigned ints. When saving a residue, I'm guessing the program just writes all the 32-bit ints to disk rather then pulling out the releveant 17 or 18 bits from each element. So, this would mean that the save file size for M18128533 is 1024K * 32 bits = 1024K * 4 bytes = 4096KB. It looks like save files are actually 22 bytes larger. Presumably, this is to store other necessary data like the actual exponent, iteration count, etc.

You're right.

I was looking at P-1 save file size, not LL save file size. [i](Ack! Another blindspot discovered. Maybe this cheese really is too ripe.)[/i]

[quote="NickGlover"]M18128533 uses 1024K FFT. This means it uses 1,048,576 elements to compute the FFTs for this exponent. This comes out to 18,128,533 / 1,048,576 = 17.3 bits per element. This actually means that each element stores either 17 or 18 bits of the residue.

I think in George's implementation all the elements are stored in 32-bit unsigned ints. When saving a residue, I'm guessing the program just writes all the 32-bit ints to disk rather then pulling out the releveant 17 or 18 bits from each element. So, this would mean that the save file size for M18128533 is 1024K * 32 bits = 1024K * 4 bytes = 4096KB. It looks like save files are actually 22 bytes larger. Presumably, this is to store other necessary data like the actual exponent, iteration count, etc.[/quote]
And to think all I did was just look at the size of a save file of a similar exponent I had in progress... ;)

A follow-up: in P-1 stage 2 there is only 1 save file, but it is slightly larger. I am testing M17717807, and the save file mH717807 is 4.22 MB (4,429,512 bytes).

Error writing intermediate file
 

I am also now getting this error message. I have lots of space on my hard drive. Does this invalidate my analysis? What can I do to fix it?
Thanks
Steve