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Author Topic: [Quiz] Answer the Bitcoin question and earn merits! #1  (Read 709 times)
hosseinimr93
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March 19, 2024, 02:42:21 PM
 #21

hosseinimr93 After seeing your algorithm, I tried to do it in C++, but I don't understand why it's getting a minus. I tried with the help of that source, but the answer was a minus.
I am not familiar with C++, but I feel you are calculating the exp_term wrongly.



You are totally right BlackHatCoiner but those are 2 different things, one thing is the 51% attack, where the miners can modify confirmed blocks, and the second thing is the double spend where the user modifies the unconfirmed transaction.
I think you didn't understand BlackHatCoiner well.
Any mining pool can try to perform a 51% attack and you don't have to own more than 51% of the total hash power to do that. That's what BlackHatCoiner says. He didn't say anything about double-spending.

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franky1
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March 19, 2024, 07:11:40 PM
Last edit: March 19, 2024, 08:14:56 PM by franky1
 #22

sorry to have to correct you but
You are totally right BlackHatCoiner but those are 2 different things, one thing is the 51% attack, where the miners can modify confirmed blocks, and the second thing is the double spend where the user modifies the unconfirmed transaction.
I think you didn't understand BlackHatCoiner well.
Any mining pool can try to perform a 51% attack and you don't have to own more than 51% of the total hash power to do that. That's what BlackHatCoiner says. He didn't say anything about double-spending.

Question: The dominant mining pool, at the time of writing this, turns evil and uses its hash-power to reorg the chain. A customer of yours sends you a transaction. If you wait for 10 confirmations, what is the chance that the dominant mining pool successfully double-spends / reverses your transaction?

a chain re-org could include all original chain transactions but just change the pool reward address to the malicious pool.. thus win all the rewards for 10 blocks
but altering-removing transactions then allows transactions that were deemed as spent. to then be unspent, to then be able to be spent again later. so they are 2 different things and yes blackhat did mention double spends. not just chain re-orgs for personal gain of rewards

I DO NOT TRADE OR ACT AS ESCROW ON THIS FORUM EVER.
Please do your own research & respect what is written here as both opinion & information gleaned from experience. many people replying with insults but no on-topic content substance, automatically are 'facepalmed' and yawned at
BlackHatCoiner (OP)
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March 19, 2024, 08:01:34 PM
Last edit: March 19, 2024, 08:25:22 PM by BlackHatCoiner
Merited by vapourminer (1), hosseinimr93 (1)
 #23

You are totally right BlackHatCoiner but those are 2 different things, one thing is the 51% attack, where the miners can modify confirmed blocks, and the second thing is the double spend where the user modifies the unconfirmed transaction.
That is correct. The miner does not double-spend; it gives the user the option to double-spend. I think, though, that I have clarified it enough, because I mentioned "reversal" alternatively to "double-spend". Whether the miner double-spends or "simply" reverses, the result is that your coins are prone to double-spending.

hosseinimr93 After seeing your algorithm, I tried to do it in C++, but I don't understand why it's getting a minus. I tried with the help of that source, but the answer was a minus.
The reason is probably because you're subtracting from 0 (see the definition of result and which operations you're doing to it). I cannot see, though, if you're printing result or 1-result.

Here you go. I have commented the important parts.
Code:
#include <iostream>
#include <cmath>
using namespace std;

// constant number 'e'
const double Euler = std::exp(1.0);

// factorial of integer x
int fact(int x){

    int i, factorial = 1;
    for(i = 1; i <= x; i++)
        factorial *= i;

    return factorial;
}

int main(){
    // q = the proportion of hashrate the attacker possesses
    // z = the total confirmations
    double q = 0.4;
    int z = 6;

    double p = 1 - q;
    double lambda = z*q/p, sum = 0;

    int k;
    for(k = 0; k <= z; k++){
        // build the fraction first
        double fraction = pow(lambda, k) * pow(Euler, -lambda);
        fraction /= fact(k);

        // sum is sum + fraction * (...)
        sum += fraction * (1 - pow((q/p), z-k));

    }

    // print
    cout << 1 - sum << endl;
}

Compile with "g++ -o attack attack.cpp" and run with "./attack".

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BlackHatCoiner (OP)
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March 20, 2024, 11:15:02 AM
 #24

New quiz!

https://bitcointalk.org/index.php?topic=5489635.0

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