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Old 2015-05-22, 06:58   #1
tha
 
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Default Two very suspect results

As per thread title:

exponent 67151753

exponent 67077133
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Old 2015-05-22, 11:45   #2
manfred4
 
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That is obvious a fault result, that we had discussed here already some times. That result happens, when some iteration gets the result -1 (stored in hex as 0xFFF...FF) and from there every next iteration will be -1 as well (because (-1)^2 - 2 = -1)
Same thing happens with the number 2 (or 0x000...002) which is (2)^2-2 = 2.
Results like that can be, even if they have an error code of 0 be marked as suspect, because I think those two results are not possible without an error in the calculation.
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Old 2015-05-22, 12:06   #3
retina
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Quote:
Originally Posted by manfred4 View Post
... the result -1 (stored in hex as 0xFFF...FF) ...
Nope. In gets stored as 1111...11110 in binary. In hex it would be either of the two values: 0x1FFF...FFE or 0x7FFF...FFE
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Old 2015-05-22, 15:36   #4
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Quote:
Originally Posted by retina View Post
Nope. In gets stored as 1111...11110 in binary.
Gets stored as 1111...1111 in binary.

EDIT: Try (unsigned) -1. If a standard integer, it'll give 4294967295 = 1111...1111 in binary.

Last fiddled with by legendarymudkip on 2015-05-22 at 15:38
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Old 2015-05-22, 16:23   #5
Madpoo
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Quote:
Originally Posted by tha View Post
They're from the same user, so I'm betting they have a very funky system.

They also submitted this one recently (a double-check):
http://www.mersenne.org/report_expon...7767619&full=1

At least it verified okay.
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Old 2015-05-22, 16:29   #6
LaurV
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Quote:
Originally Posted by legendarymudkip View Post
Gets stored as 1111...1111 in binary.

EDIT: Try (unsigned) -1. If a standard integer, it'll give 4294967295 = 1111...1111 in binary.
Nope, it gets stored as 11111...1110, as retina said. You forget we do modulo 2^p-1, and NOT modulo 2^n, as you see in your pocket calculator. Also, it can not start with F or 3 in hex, because it always has an ODD number of bits, as p is always prime, therefore odd. So it can only start with 7 or 1 in hex, followed by many F's and one last E. For clarity, 11111...111 in binary is zero mod 2^p-1, and not -1 mod 2^p-1.
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Old 2015-05-22, 16:46   #7
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Quote:
Originally Posted by LaurV View Post
Nope, it gets stored as 11111...1110, as retina said. You forget we do modulo 2^p-1, and NOT modulo 2^n, as you see in your pocket calculator. Also, it can not start with F or 3 in hex, because it always has an ODD number of bits, as p is always prime, therefore odd. So it can only start with 7 or 1 in hex, followed by many F's and one last E. For clarity, 11111...111 in binary is zero mod 2^p-1, and not -1 mod 2^p-1.
Yes, you are right (as is he), I was thinking about -1 in general rather than what a -1 meant here. Thanks for the correction!
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