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125

answers:

3

In explanations I've read about public key cryptography, it is said that some large number is come up with by multiplying together 2 extremely large primes. Since factoring the product of large primes is almost impossibly time-consuming, you have security.

This seems like a problem that could be trivially solved with rainbow tables. If you know the approximate size of primes used and know there are 2 of them, you could quickly construct a rainbow table. It'd be a mighty large table, but it could be done and the task could be parallelized across hardware.

Why are rainbow tables not an effective way to beat public key crypto based on multiplying large primes?

Disclaimer: obviously tens of thousands of crazy-smart security conscious people didn't just happen to miss for decades what I thought up in an afternoon. I assume I'm misunderstanding this because I was reading simplified layman explanations (eg: if more than 2 numbers are used) but I don't know enough yet to know where my knowledge gap is.

Edit: I know "rainbow table" relates to using pre-calculated hashes in a lookup table but the above sounds like a rainbow table attack so I'm using the term here.


Edit 2: As noted in the answers, there's no way to store just all of the primes, much less all of their products.

  • This site says there are about this many 512 bit primes: ((2^511) * 1) / (512 log(2)) = 4.35 × 10151
  • The mass of the sun is 2 × 1030 kg or 2 × 1033 g
  • That's 2.17 × 10124 primes per gram of the sun.
  • Qty. of 512 bit numbers that can fit in a kilobyte: 1 kb = 1024 bytes = 8192 bits / 512 = 16
  • That can fit in a terabyte: 16*1024*1024*1024 = 1.72 × 1010
  • Petabyte: 16*1024*1024*1024*1024 = 1.72 × 1013
  • Exabyte: 16*1024*1024*1024*1024*1024 = 1.72 × 1016

Even if 1 exabyte weighed 1 gram, we're nowhere close to reaching the 2.17 × 10124 needed to be able to fit all of these numbers into a hard drive with the mass of the sun

+1  A: 

I think the main problem is that rainbow tables pregenerated for certain algorithms use a rather "small" range (usually something in the range of 128 bits). This doesn't usually cover the whole range, but speeds the brute force process up. They usually consume some TB of space.

In prime factorization, primes are much larger (for secure RSA, 2048 bits are recommended). So the rainbow tables wouldn't be "mighty large", but impossible to store anywhere (using up like millions of TB of space).

Also, rainbow tables use hash chains too further speed up the process (Wikipedia has a good explanation) which can't be used for primes.

schnaader
The largest rainbow tables are of 10 characters alpha-numeric, which is around 57-bits of security, no where near 128-bits.
BlueRaja - Danny Pflughoeft
OK, so the difference is even more extreme. Thanks for clarification.
schnaader
+1  A: 

The primes used in RSA and Diffie-Hellman are typically on the order of 2512. In comparison, there are only about 2256 atoms in the known universe. That means 2512 is large enough to assign 2256 unique numbers to every atom in the universe.

There is simply no way to store/calculate that much data.


As an aside, I assume you mean "a large table of primes" - rainbow tables are specificly tailored for hashes, and have no real meaning here.

BlueRaja - Danny Pflughoeft
I don't follow how you got the final stat from the 2 preceding ones.
Dinah
@Dinah: 2^512 numbers / 2^256 atoms = 2^256 numbers/atom
BlueRaja - Danny Pflughoeft
Ah, of course. Thanks for indulging me.
Dinah
+9  A: 

From one of my favorite books ever, Applied Cryptography by Bruce Schneier

"If someone created a database of all primes, won't he be able to use that database to break public-key algorithms? Yes, but he can't do it. If you could store one gigabyte of information on a drive weighing one gram, then a list of just the 512-bit primes would weigh so much that it would exceed the Chandrasekhar limit and collapse into a black hole... so you couldn't retrieve the data anyway"

In other words, it's impossible or unfeasible, or both.

Tom
FWIW, I haven't weighed disk drives lately, but the 360GB drive I just stuck into my laptop is a whole lot lighter than a pound (c. 400g), so we're doing better than the gram per gigabyte. I wonder how many bit-lengths of prime numbers we'd need now to create a black hole, and if Bruce updated that in his latest book (gotta buy it sometime).
David Thornley
I think even at a TB per gram it would still far exceed the limit if my quick finger math is to be believed.
Tom
@Tom, you're correct. I did the math and appended the original question.
Dinah