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Comment by jedwards1211 | Hacker News Reader

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.

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Firstly, the number of prime numbers < n grows by roughly n/log(n)

So moving from 2^16 = 65536

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.

> 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

the ^2's are

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.

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

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.

> if 2^512 calculations costs you $8 then (2^512)^2 calculations costs you $64

“The earth is flat blah blah blah”

opposing side of an argument resorting to a strawman is their implicit admission they were wrong.

We're trying to help you understand, and you're stubbornly refusing to swallow your pride and think about what we're saying

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