Score:1

Crypto

is it efficient if blockchain uses double Hash algorithms $H_1$ and $H_2$?

Don Freecs

10/9/22, 3:10 PM

I wonder is it efficient to use less Target condition and double hash algorithms with different target (or the same target with different Hash algorithms) and one nonce in a Block.

Example

Target 1 for Hash1 H1 is 3zeroes, 000F543D... Target 2 for Hash2 H2 is 4zeroes, 0000FSDF...?

1 + 8

hash

cryptocurrency

Maarten Bodewes

10/9/22, 3:34 PM

I got to understand that the value of the hash as (unsigned) integer needs to be below a certain value. That should be enough to specify how efficient mining is. With leading zeros you can only use powers of two, but that problem disappears if you compare the entire value.

Don Freecs

10/9/22, 3:45 PM

@MaartenBodewes indeed, but don't forget changing the nonce means the changing of the hash result, but what about the hardness? getting for $n$ hash algorithm a specific amount of leading zeroes each, by just one nonce? instead of wanting a specific amount of zeroes, does dividing in two/or plus hashes helps to effects the hardness of the problem?

kodlu

10/9/22, 3:50 PM

and you are comparing the hardness to a single hash with 3+4=7 zeroes? Is that the question?

Don Freecs

10/9/22, 3:55 PM

@kodlu No, My question about what if we modified the block to contain more hash algorithm instead of one, how it affects the systems, and precisely the hardness. 3+4 = 7 it is just one of the examples to give a new perspective ... Sorry for my problem of using the language it is not my native. Thanks

kodlu

10/9/22, 3:56 PM

ok, but you want to use the same nonce, right?

Don Freecs

10/9/22, 3:59 PM

Yes. may be we can generalize this question to $n$ hashes algorithm for a positive integer $n$ (Vector) or from your question what if we use extra nonce ($m$)

Ievgeni

10/11/22, 12:27 PM

What do you mean by efficient? From which point of view?

Don Freecs

10/11/22, 1:24 PM

@levgeni what is the impact if we add more than one hash algorithm and decrease the target leading zeroes...

Score:1

Crypto

Ievgeni

10/11/22, 2:58 PM

I'm supposing that both functions are enough secure (i.e the output seems random, and there is no more efficient attack to find preimage than brute-forcing).

The idea for proof of work is based on the following assumption. Find an $x$ such that $H(y|x) =O^\lambda w$, for a fixed $y$ takes a time $\approx2^\lambda$.

Then if you suppose that $H_1$, and $H_2$ are "independent" (finding a solution for one hash function doesn't help you to find a solution for the other one), then solving the two puzzle will take a time $\approx2^{\lambda_1} + 2^{\lambda_2}$. Notice that's much smaller than $2^{\lambda_1 + \lambda_2}$.

Thus : solving two independent puzzles with parameters $\lambda_1$ and $\lambda_2$ is much easier than solving one puzzle with parameter $\lambda_1 + \lambda_2$.

Does it answer to your question?

0 + 11

Don Freecs

10/11/22, 3:18 PM

Thanks for the explanation , what if we choose $\lambda_1$ and $\lambda_2$ such that $2 ^{\lambda_1 } + 2 ^{\lambda_2} \approx 2^{\lambda_1 + \lambda_2}$ ?

Ievgeni

10/11/22, 3:21 PM

It's possible only if one of the $\lambda$'s is much smaller than the other one. Then the puzzle is quite equivalent to "invert" only one of the function.

Don Freecs

10/11/22, 8:16 PM

it us clearly now that saving the amount of leading zero makes the problem harder right???

Ievgeni

10/11/22, 8:22 PM

I'm not sure to understand what do you mean?

Don Freecs

10/11/22, 8:24 PM

this time we ask for the same amount of zeroes in each Hash function even using only one hash function or twice or more...

Ievgeni

10/11/22, 8:32 PM

Then the amount of time will be $k2^\lambda$, with $\lambda$ the number of zero, and $k$ the number of hash functions considered.

Don Freecs

10/11/22, 8:34 PM

thanks sir, one last question does this useful???

Ievgeni

10/12/22, 6:26 AM

I don't think so, because, it's noticeable that the witness would have size $\approx k\lambda$, then it's better to use only one function with parameter $\log(k) + \lambda$, then the puzzle would have approximately the same hardness with a quite smaller witness (of size $\approx \log(k) + \lambda$).

Ievgeni

10/14/22, 10:43 AM

@Nour-eddineRAHMANI I thought about your proposition, and it could have a great interest, if we choose to compute the SAME witness for both hash function, then the size of the witness is still $\lambda$, and the time of computation becomes $2^{k\lambda}$, then it becomes much more efficient (in term of size of witness).

Don Freecs

10/16/22, 1:35 PM

Sorry for the late response, I am busy with master thesis, I'll be happy to study more about blockchain, and your discussion was helpful to me, my respects.

Ievgeni

10/16/22, 3:44 PM

No worries (:- ) )

Elon Musk

I sit in a Tesla and translated this thread with Ai:

EN: is it efficient if blockchain uses double Hash algorithms $H_1$ and $H_2$?

TH: จะมีประสิทธิภาพหรือไม่หากบล็อกเชนใช้อัลกอริธึมแฮชสองเท่า $H_1$ และ $H_2$

RO: este eficient dacă blockchain folosește algoritmi Hash dublu $H_1$ și $H_2$?

RU: эффективно ли, если блокчейн использует алгоритмы двойного хеширования $H_1$ и $H_2$?

VI: có hiệu quả không nếu blockchain sử dụng thuật toán Hash kép $H_1$ và $H_2$?

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