5 ms·

thats linear search in gnfs the search space is number fields, so you increase the number of the number fields by the square of what came before.

jedwards1211 2y ago
Right, so if it costs $8 to search the space for a 16-bit key, a proper equation for the cost of cracking an n-bit key is Cost(n) = search_space(n) * $8 / search_space(16) And search_space(x) = 2^x so Cost(n) = 2^n * $2^3 / 2^16 = $2^(n - 13) Cost(32) = $2^(32 - 13) = $524288 Cost(64) = $2^(64 - 13) = $2251799813685248 So it quickly becomes astronomically expensive. If you double the number of bits n you get Cost(2n) = $2^(2n - 13) You were assuming that Cost(32) = Cost(16)^2, in other words Cost(2n) = Cost(n)^2. But this equality doesn't hold: Cost(n)^2 = $2^(n - 13)^2 = $2^(2n - 26) This is significantly smaller than Cost(2n) = $2^(2n - 13) as stated above. An equation with different units on each side like "512 bit = $8" doesn't work mathematically and will lead to contradictory conclusions.
- DarkmSparks 2y ago
  Firstly, the number of prime numbers < n grows by roughly n/log(n) So moving from 2^16 = 65536 to 2^32 = 4294967296 Increases the size of the total potential search space from 5909 to 193635251, which is ~ 5909 x 32769 secondly, the reason it grows by only n^2, is you only need to search along the curve n = a x b - which is the "sieve" part. if 2^512 calculations costs you $8 then (2^512)^2 calculations costs you $64 Thirdly, your stupidly high costs are because you seem to think you need to check if 4817 x 5693 is a prime fractorisation of 26768069 when in fact you already know the prime factorisation by that point.
  - littlestymaar 2y ago
    > if 2^512 calculations costs you $8 then (2^512)^2 calculations costs you $64 “The earth is flat blah blah blah”
    - DarkmSparks 2y ago
      opposing side of an argument resorting to a strawman is their implicit admission they were wrong.
      - jedwards1211 2y ago
        We're trying to help you understand, and you're stubbornly refusing to swallow your pride and think about what we're saying
  - bmacho 2y ago
    > if 2^512 calculations costs you $8 then (2^512)^2 calculations costs you $64 Can you answer a) If 1 apple costs you 8$ then (1)^2 apple costs you: ??? b) If 10 apples cost you 8$ then (10)^2 apples costs you: ??? edit: between the price and the amount, usually there is a ~linear relationship. So if you can buy 2^512 something for 8$, then chances are that for 8 times the price you'll only get ~8 times more amount, and not 2^512 times more
    - DarkmSparks 2y ago
      the ^2's are https://en.m.wikipedia.org/wiki/Big_O_notation https://en.m.wikipedia.org/wiki/Big_O_notation The tldr is bigO gives you how the cost of apples changes with the number of apples in the worst case. its a rough simplification, the precise formula is close but not exactly that. the simplification is also actually (2n)^2 but in my defense I was going from memory of work from more than 2 decades ago (testing generated prime factors were good prime factors, overwhelmingly they were not). using your apples example if the bigO of eating apples is O(n^2), and it takes you 8 minutes to eat 2 apples, it will take you no more than 64 minutes to eat 4 apples.
  - jedwards1211 2y ago
    Okay I hadn't read about how many calculations it takes to cover the search space, so my equations aren't right about that. However, you will do yourself a big favor if you take the time to understand why this is wrong: > if 2^512 calculations costs you $8 then (2^512)^2 calculations costs you $64 The cost per calculation is some constant C: Cost(n calcs) = n calcs * C Therefore, Cost(n^2 calcs) = n^2 calcs * C In this example, C = $8 / 2^512 = $2^-509 So Cost(2^512^2) = Cost(2^1024) = 2^1024 * $2^-509 = $2^425
    - DarkmSparks 2y ago
      I wouldnt argue with a claim of $2425 tbh. The $8 will vary, and the actual cost function completely depends on the implimentation, its definitely possible to do worse, very likely possible to do better - there was rumors a few years ago that some Riemann surface based math can do it in O(1), but I know nothing about Riemann surfaces so can't judge their veracity.
littlestymaar 2y ago
Do you even read?
- bmacho 2y ago
  > if 2^512 calculations costs you $8 then (2^512)^2 calculations costs you $64 This and the stuff they've been writing about suggest that we are observing a mind that used to know things, but suffered some serious damages/degrade over time. That's always so sad.
- DarkmSparks 2y ago
  Did you? you linked https://en.wikipedia.org/wiki/General_number_field_sieve https://en.wikipedia.org/wiki/General_number_field_sieve already That gives the worst case complexity right at the top. > 2^16 [1] 65536 > n<-2^32 > exp(((64/9)^(1/3)+1)(log(n)^(1/3))(log(log(n))^(2/3))) [1] 38178499 > 38178499/65536 [1] 582 2^16 -> 2^32 ~ x 2^9 > n<-2^64 > exp(((64/9)^(1/3)+1)(log(n)^(1/3))(log(log(n))^(2/3))) [1] 84794674511 > 84794674511/38178499 [1] 2221 2^32-> 2^64 ~ x 2^11 > n<-2^128 > exp(((64/9)^(1/3)+1)(log(n)^(1/3))(log(log(n))^(2/3))) [1] 2.507616e+15 > 2.507616e+15/84794674511 [1] 29572 2^64 -> 2^128 ~ x 2^15 So in GNFS 2^32 -> 2^64 complexity doesn't increase 4294967296 times, worst case it increases 2221 times _In practice_ the cost grows closer to (nbits)^2 mostly because as an "embarrassingly parallel" algorithm you get significant benefits from cached results (quickly discard entire number fields that were previously calculated). if O(x) = 8 then O(x)^2 = 8^2 I did miss a x2 earlier because e.g. 128 bits is 128^2 (16384 rather than 29572) harder than 64 bits, not 64^2 So its (2xO(x))^2 = (2 x 8)^2 = $256
  - jedwards1211 2y ago
    function ops(n) { return Math.exp (((64/9) * (1/3) + 1) * (Math.log(n) * (1/3)) * (Math.log(Math.log(n)) * (2/3))) } > ops(2*16) 121106.42245436447 > ops(2*32) 38178499.24944067 > ops(2*32) / ops(2*16) 315.24751929508244 So if ops(2*16) costs $8, then ops(2*32) costs $8 * ops(2*32) / ops(2*16) = $2521.98. Far more than $8^2. The cost reaches the millions for 64 bits, and ~$165 trillion for 128 bits: > 8 * ops(2*64) / ops(2*16) 5601332.962726709 (far more than $8^2^2) > 8 * ops(2*128) / ops(2*16) 165647073370.16437 (far more than $8^2^2^2) Note that this is increasing faster than the number of bits squared: > ops(2*32) / 32 * 2 37283.690673281904 > ops(2*64) / 64 * 2 20701824.831895076 > ops(2*128) / 128 * 2 153052707259.34015 As the wiki page says, it's super-polynomial in the number of bits. If you still disagree with all of this, can you explain what's wrong with this method of calculating the worst-case cost of factoring the number n? Cost(n) = ops(n) * Cost(2^16) / ops(2^16) Or what you don't understand about this way of calculating it?

jedwards1211 2y ago

Right, so if it costs $8 to search the space for a 16-bit key, a proper equation for the cost of cracking an n-bit key is Cost(n) = search_space(n) * $8 / search_space(16) And search_space(x) = 2^x so Cost(n) = 2^n * $2^3 / 2^16 = $2^(n - 13) Cost(32) = $2^(32 - 13) = $524288 Cost(64) = $2^(64 - 13) = $2251799813685248 So it quickly becomes astronomically expensive. If you double the number of bits n you get Cost(2n) = $2^(2n - 13) You were assuming that Cost(32) = Cost(16)^2, in other words Cost(2n) = Cost(n)^2. But this equality doesn't hold: Cost(n)^2 = $2^(n - 13)^2 = $2^(2n - 26) This is significantly smaller than Cost(2n) = $2^(2n - 13) as stated above. An equation with different units on each side like "512 bit = $8" doesn't work mathematically and will lead to contradictory conclusions.

DarkmSparks 2y ago
Firstly, the number of prime numbers < n grows by roughly n/log(n) So moving from 2^16 = 65536 to 2^32 = 4294967296 Increases the size of the total potential search space from 5909 to 193635251, which is ~ 5909 x 32769 secondly, the reason it grows by only n^2, is you only need to search along the curve n = a x b - which is the "sieve" part. if 2^512 calculations costs you $8 then (2^512)^2 calculations costs you $64 Thirdly, your stupidly high costs are because you seem to think you need to check if 4817 x 5693 is a prime fractorisation of 26768069 when in fact you already know the prime factorisation by that point.
- littlestymaar 2y ago
  > if 2^512 calculations costs you $8 then (2^512)^2 calculations costs you $64 “The earth is flat blah blah blah”
  - DarkmSparks 2y ago
    opposing side of an argument resorting to a strawman is their implicit admission they were wrong.
    - jedwards1211 2y ago
      We're trying to help you understand, and you're stubbornly refusing to swallow your pride and think about what we're saying
- bmacho 2y ago
  > if 2^512 calculations costs you $8 then (2^512)^2 calculations costs you $64 Can you answer a) If 1 apple costs you 8$ then (1)^2 apple costs you: ??? b) If 10 apples cost you 8$ then (10)^2 apples costs you: ??? edit: between the price and the amount, usually there is a ~linear relationship. So if you can buy 2^512 something for 8$, then chances are that for 8 times the price you'll only get ~8 times more amount, and not 2^512 times more
  - DarkmSparks 2y ago
    the ^2's are https://en.m.wikipedia.org/wiki/Big_O_notation https://en.m.wikipedia.org/wiki/Big_O_notation The tldr is bigO gives you how the cost of apples changes with the number of apples in the worst case. its a rough simplification, the precise formula is close but not exactly that. the simplification is also actually (2n)^2 but in my defense I was going from memory of work from more than 2 decades ago (testing generated prime factors were good prime factors, overwhelmingly they were not). using your apples example if the bigO of eating apples is O(n^2), and it takes you 8 minutes to eat 2 apples, it will take you no more than 64 minutes to eat 4 apples.
- jedwards1211 2y ago
  Okay I hadn't read about how many calculations it takes to cover the search space, so my equations aren't right about that. However, you will do yourself a big favor if you take the time to understand why this is wrong: > if 2^512 calculations costs you $8 then (2^512)^2 calculations costs you $64 The cost per calculation is some constant C: Cost(n calcs) = n calcs * C Therefore, Cost(n^2 calcs) = n^2 calcs * C In this example, C = $8 / 2^512 = $2^-509 So Cost(2^512^2) = Cost(2^1024) = 2^1024 * $2^-509 = $2^425
  - DarkmSparks 2y ago
    I wouldnt argue with a claim of $2425 tbh. The $8 will vary, and the actual cost function completely depends on the implimentation, its definitely possible to do worse, very likely possible to do better - there was rumors a few years ago that some Riemann surface based math can do it in O(1), but I know nothing about Riemann surfaces so can't judge their veracity.

littlestymaar 2y ago

Do you even read?

bmacho 2y ago
> if 2^512 calculations costs you $8 then (2^512)^2 calculations costs you $64 This and the stuff they've been writing about suggest that we are observing a mind that used to know things, but suffered some serious damages/degrade over time. That's always so sad.
DarkmSparks 2y ago
Did you? you linked https://en.wikipedia.org/wiki/General_number_field_sieve https://en.wikipedia.org/wiki/General_number_field_sieve already That gives the worst case complexity right at the top. > 2^16 [1] 65536 > n<-2^32 > exp(((64/9)^(1/3)+1)(log(n)^(1/3))(log(log(n))^(2/3))) [1] 38178499 > 38178499/65536 [1] 582 2^16 -> 2^32 ~ x 2^9 > n<-2^64 > exp(((64/9)^(1/3)+1)(log(n)^(1/3))(log(log(n))^(2/3))) [1] 84794674511 > 84794674511/38178499 [1] 2221 2^32-> 2^64 ~ x 2^11 > n<-2^128 > exp(((64/9)^(1/3)+1)(log(n)^(1/3))(log(log(n))^(2/3))) [1] 2.507616e+15 > 2.507616e+15/84794674511 [1] 29572 2^64 -> 2^128 ~ x 2^15 So in GNFS 2^32 -> 2^64 complexity doesn't increase 4294967296 times, worst case it increases 2221 times _In practice_ the cost grows closer to (nbits)^2 mostly because as an "embarrassingly parallel" algorithm you get significant benefits from cached results (quickly discard entire number fields that were previously calculated). if O(x) = 8 then O(x)^2 = 8^2 I did miss a x2 earlier because e.g. 128 bits is 128^2 (16384 rather than 29572) harder than 64 bits, not 64^2 So its (2xO(x))^2 = (2 x 8)^2 = $256
- jedwards1211 2y ago
  function ops(n) { return Math.exp (((64/9) * (1/3) + 1) * (Math.log(n) * (1/3)) * (Math.log(Math.log(n)) * (2/3))) } > ops(2*16) 121106.42245436447 > ops(2*32) 38178499.24944067 > ops(2*32) / ops(2*16) 315.24751929508244 So if ops(2*16) costs $8, then ops(2*32) costs $8 * ops(2*32) / ops(2*16) = $2521.98. Far more than $8^2. The cost reaches the millions for 64 bits, and ~$165 trillion for 128 bits: > 8 * ops(2*64) / ops(2*16) 5601332.962726709 (far more than $8^2^2) > 8 * ops(2*128) / ops(2*16) 165647073370.16437 (far more than $8^2^2^2) Note that this is increasing faster than the number of bits squared: > ops(2*32) / 32 * 2 37283.690673281904 > ops(2*64) / 64 * 2 20701824.831895076 > ops(2*128) / 128 * 2 153052707259.34015 As the wiki page says, it's super-polynomial in the number of bits. If you still disagree with all of this, can you explain what's wrong with this method of calculating the worst-case cost of factoring the number n? Cost(n) = ops(n) * Cost(2^16) / ops(2^16) Or what you don't understand about this way of calculating it?