 mersenneforum.org Error risk after doublecheck
 Register FAQ Search Today's Posts Mark Forums Read  2004-12-26, 21:47 #1 MrHappy   Dec 2003 Paisley Park & Neverland 5×37 Posts Error risk after doublecheck I have some questions regarding the LL-error rate and the safety feature of the res64. Just curious. I'm not good in calculating probabilities... Here we go: - What's the chance that we have at least one undetected matching but still false doublecheck in the LL database? So both residues are x but should be y. - How many exponents would we have to test so the chance from the question above would rise above 1% ?   2004-12-26, 23:28 #2 Uncwilly 6809 > 6502   """"""""""""""""""" Aug 2003 101×103 Posts 23×5×239 Posts My understanding (and I hope to be corrected if I am wrong) is that the res is a 16 digit Hex number. To get 2 to match accidentally would take odds of (well) over 1 x 10^40, which is the point that it is considered to be impossible. And since the projected starting error rates are ~1%, you need to square that 0.01% chance that the first two res turned in are both wrong. Ain't going to happen. Unless both were run on thee same error producing machine.   2004-12-27, 00:07   #3
dave_dm

May 2004

24×5 Posts Quote:
 Originally Posted by MrHappy How many exponents would we have to test so the chance from the question above would rise above 1% ?
I make it 9.3 x 10^18, based on Uncwilly's 1% failure figure.

Dave   2004-12-27, 00:55 #4 Uncwilly 6809 > 6502   """"""""""""""""""" Aug 2003 101×103 Posts 23·5·239 Posts Let's see where (or how) my math breaks down. 16 digits each representing 4 bits. 16 * 4 = 64 bit residue. 2^64 = 18446744073709551616 (total different potential residues) 18446744073709551616^2 = 3.4028e+38 (total different combinations of 2 residues) divide by 0.0001 chance both tests having errors and we get: 3.4028e+42   2004-12-27, 01:39   #5
wblipp

"William"
May 2003
New Haven

1001001110012 Posts Quote:
 Originally Posted by Uncwilly Let's see where (or how) my math breaks down. 2^64 = 18446744073709551616 (total different potential residues) 18446744073709551616^2 = 3.4028e+38
It breaks down here. You have calculated the probability that two independent failures will have a residue of "192." But a matching residue of 191 or 193 or 1234567890 would also count, so you must multiply by (2^64-1) for all the possible wrong but unrecognized residues.

Alternatively, you can calculate as
First error * Second Error * second residue matches first residue   2004-12-27, 01:57 #6 jinydu   Dec 2003 Hopefully Near M48 2×3×293 Posts First, I would like to point out that it seems Uncwilly and dave_dm have made an unspoken assumption: that an erroneous residue is equally likely to take on any hexadecimal value. This assumption is necessary to come up with figures at all (unless a more realistic model is available), but it should be kept in mind that there may be some types of CPU errors may not fit this assumption. In any case, this is how we can come up with the figures. Suppose we actually did two incorrect LL tests of the same exponent. Let x be the residue of the first test. For the second residue, there are 16^16 possible values for the residue (because there are 16 "digits" and each "digit" has 16 possible values). Thus, the probability that both results are the same is: 1/(16^16) ~= 5.42101086242752217003726400434971 * 10^-20 Now, we can use a theorem in probability. If the probability that an outcome occurs in one trial is p, then the probability that it doesn't happen is 1 - p. If this trial occurs n times, the probability that the outcome never occurs is (1 - p)^n. Thus, the probability that it occurs at least once is 1 - (1 - p)^n. According to http://www.mersenne.org/status.htm, GIMPS has double-checked a total of 315,179. So for your first question, p = 1/(16^16) and n = 315,179. According to Mathematica, the probability of having at least one such error due to chance is thus roughly: 1.7085887826090294136983459361845392 * 10^-14 That's less than one in 58.5 trillion, nothing to get too worried about (if you trust the assumption I pointed out at the beginning). Now, to answer your second question, we solve for n in the equation: 1 - (1 - p)^n = 0.01 where p = 1/(16^16) (1 - p)^n = 0.99 take the log of both sides: n*log(1 - p) = log 0.99 n = (log 0.99)/(log (1 - p)) Mathematica gives: 1.853959733443683384907575121974264716245215072 * 10^17 Assuming GIMPS double-checked every possible Mersenne number with a prime exponent in order from (2^2) - 1 upwards, in order for the probability of an error to reach 1%, GIMPS would need to search every such exponent up to about: M(7.87 * 10^18) In my opinion, a far greater threat is that a user will deliberately submit incorrect residues to get GIMPS credit. To do this, he could choose any residue he wanted, submit it once manually as a first time LL test, then resubmit it manually as a double-check. This could be difficult to catch, especially if he created many user accounts, and never submitted a first time test and its double check using the same account.   2004-12-27, 02:19 #7 dave_0273   Oct 2003 Australia, Brisbane 2·5·47 Posts Isn't there the little "check sum" or whatever it is at the end of each result line? I am not exactly sure what it is used for but I always assumed it was there so that people couldn't submit "made up" results.   2004-12-27, 03:55   #8
ColdFury

Aug 2002

26·5 Posts Quote:
 Isn't there the little "check sum" or whatever it is at the end of each result line?
The checksum can be forged.   2004-12-27, 09:29   #9
dave_dm

May 2004

1208 Posts Quote:
 Originally Posted by jinydu First, I would like to point out that it seems Uncwilly and dave_dm have made an unspoken assumption: that an erroneous residue is equally likely to take on any hexadecimal value. This assumption is necessary to come up with figures at all (unless a more realistic model is available), but it should be kept in mind that there may be some types of CPU errors may not fit this assumption.
Yup :) The figure I worked out used the same argument as jinydu but instead I worked out the probability of an undetected hardware error, having misparsed the question.

So with p = 0.01, the probability of two residues matching but some hardware error occuring is

q = 2 * p * (1 - p) * 2^(-64) + p^2 * 2^(-64)

i.e. {one hardware error} + {two hardware errors}

Then log(0.99) / log(1 - q) gives the 9.3 x 10^18.

Dave   2004-12-27, 14:19 #10 Uncwilly 6809 > 6502   """"""""""""""""""" Aug 2003 101×103 Posts 23×5×239 Posts Ok, so 10^18 is "Ain't gonna happen", while 10^40 is mathametically impossible. :)   2004-12-27, 14:22   #11
jinydu

Dec 2003
Hopefully Near M48

2·3·293 Posts Quote:
 Originally Posted by jinydu In my opinion, a far greater threat is that a user will deliberately submit incorrect residues to get GIMPS credit. To do this, he could choose any residue he wanted, submit it once manually as a first time LL test, then resubmit it manually as a double-check. This could be difficult to catch, especially if he created many user accounts, and never submitted a first time test and its double check using the same account.
Isn't someone going to come to defend the current system? Seriously, there must be some safeguard against this, right?   Thread Tools Show Printable Version Email this Page Similar Threads Thread Thread Starter Forum Replies Last Post gd_barnes Conjectures 'R Us 16 2014-08-07 02:11 ATH PrimeNet 11 2010-06-03 06:38 masser Sierpinski/Riesel Base 5 44 2006-09-24 17:19 Unregistered PrimeNet 9 2006-03-26 05:48 TheJudger Data 4 2005-04-04 08:54

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