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Author Topic: Do miners really think destroying Bitcoin will make them rich?  (Read 7460 times)
dinofelis
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March 26, 2017, 07:02:22 PM
 #101

I really mean it, you're a terrible communicator, like verbosity incarnate. Please work on using fewer words, and on presenting your thoughts clearly.

I don't have the time nor the energy for effort to be concise.  As I write on a keyboard about as fast as I talk, my stuff is always very verbose.
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March 26, 2017, 07:02:25 PM
 #102

I don't think they really care about bitcoin's future as in bitcoin in isolation. Delaying SegWit & LN means they're currently getting much bigger fees than before i.e. better profits which is what they're bothered about.

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March 26, 2017, 07:03:14 PM
 #103

I don't think they really care about bitcoin's future as in bitcoin in isolation. Delaying SegWit & LN means they're currently getting much bigger fees than before i.e. better profits which is what they're bothered about.

Amen.  Someone is getting it.
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March 26, 2017, 08:34:43 PM
 #104

I don't think they really care about bitcoin's future as in bitcoin in isolation. Delaying SegWit & LN means they're currently getting much bigger fees than before i.e. better profits which is what they're bothered about.

You cannot blame the miner, implementing segwit and LN will definitely cut their profit.  The first reaction of person should be protect their interest. Though as I read segwit and LN explanation, they are only for micro payment, I think it is somewhere around 0.042 BTC and larger than that can't be handled by LN by now but the possibility of increasing that is very possible, besides it is just a code.
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March 26, 2017, 10:51:10 PM
 #105

I don't think they really care about bitcoin's future as in bitcoin in isolation. Delaying SegWit & LN means they're currently getting much bigger fees than before i.e. better profits which is what they're bothered about.

That's partly true because they are monetizing Bitcoin transactions and it's gradually killing our beloved cryptocurrency.
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March 27, 2017, 04:13:36 AM
Last edit: March 27, 2017, 04:24:29 AM by deisik
 #106


As I said, you don't get it

1. The hashing stops the whole blockchain being recomputed and the ledger being entirely altered. Signatures have nothing to do with that. Zero.

The amount of hashing is the clumsy equivalent of a "digital signature" on the "original" so that it cannot get modified, exactly like the digital signature of a document proves that it is the original and not a modified version.  That was the analogy which visibly you didn't understand.

However, the difference between PoW and a digital signature is that with PoW, the "security" is not larger than the effort (waste) put in by the "signer" (the miners that made the first, original chain) and can hence be redone by anyone wishing to put more effort into it ; while with a genuine digital signature, the effort to make the signature is negligible, and the effort needed to fake it, is astronomical.   You essentially have that the cryptographic "proof of authenticity" with PoW is 1:1  (original/modified copy) while with a digital signature it is close to 0 : infinity.  That was my point

I think your point should be weighed critically

While I certainly agree with your inference that with PoW the effort to undo something essentially equals the effort it took to do it in the first place but it is still more consistent overall. What I mean to say is that there cannot be a loophole or vulnerability to somehow not to do the required amount of work to undo what's been already done. You would still need to spend the same amount of work, no matter what.

Absolutely not.  If you can reverse the hash function (that is, if you succeed in cracking it), you can redo a "proof of work" block chain in half an hour on a PC (well, maybe half a day Smiley ).  A cracked hash function is not a proof of work any more.  Note moreover that you do not need to *fully* crack the hash function, but to find simply a way for it to find precedents with leading zeros without having to "brute force" it.  The whole proof of work resides on the assumption that brute-forcing is the only way to find hashes with leading zeros

I'm not very familiar with cryptography in Bitcoin specifically (apart from basic concepts), thus bear with me. So even if you "crack" the hash function, you would still need to do absolutely the same amount of work now in direct order (as you yourself said in one of your previous posts), at least as far as I understand it. Besides, how are you go to crack hash function? Hash is basically a function that maps some data of arbitrary size (usually orders of magnitude greater in size than the resulting hash) to data of fixed size. Therefore, "cracking" a hash would pose to you conceptually the same task as finding a private key having only a public one, i.e. you will have to check myriads of possibilities that give you the same hash. And which one are you going to choose?

In short, you can't crack a hash function

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March 27, 2017, 05:41:47 AM
 #107

Besides, how are you go to crack hash function? Hash is basically a function that maps some data of arbitrary size (usually orders of magnitude greater in size than the resulting hash) to data of fixed size. Therefore, "cracking" a hash would pose to you conceptually the same task as finding a private key having only a public one, i.e. you will have to check myriads of possibilities that give you the same hash. And which one are you going to choose?

In short, you can't crack a hash function

Of course you can.  What is a hash function essentially ?  It is a function f that maps a number in the set A = {0,...N-1} into the set B = {0,...M-1} where M << N ; f(n) = m.

The trick of a hash function is that the calculation of f(n) = m is rather simple, but that there's no way, given an m, to find an n other than by exhaustively trying all elements of A. 

Given that M << N, there are essentially on average Q = N / M elements of A that map onto a given value of m.
If you find a way to find easily the set of these Q elements, given m, without having to scan over all of A, you have cracked the hash function.

Usually, a hash function is already considered cracked when you can easily find ONE SINGLE element of the inverse image of m, but a fully cracked hash function gives you all of them.

If on top of that, I'm able to search quickly  in the set of solutions (for instance, if I can generate them in an ordered way), I don't have to run over each of the solutions, I have only logarithmic difficulty to search in this set of solutions the one that satisfies the conditions I need (for instance, the block header without nonce).

I can work backwards in the hashing tree towards those data points where I have liberty to change things, and search each time, logarithmically, whether a solution exists.

In fact, the solution set I would take would start from m = 0: infinite difficulty in bitcoin: the hash being zero !
Once I have the header solution set, I look (logarithmically) for those solutions that have the right previous block hash and a few other things that need to be fixed, and essentially look for those solutions giving a merkel tree hash and a nonce.

For a found merkel tree hash, I now apply the same technique where everything is fixed except the coinbase transaction: so I calculate backward a/the solutions for this coinbase transaction, imposing (searching) everything that is fixed in the coinbase transaction --> I find the/a solution with the/a right coinbase comment.

I've just produced a correct block with maximum difficulty !  This chain is unbeatable now, nobody can make a chain with more PoW !  My PoW is orders of magnitude bigger than what has ever been done on bitcoin.

Note that even if I fork off 100 blocks ago, the new chain where I orphaned 100 blocks has orders of magnitude more PoW, with a calculation that I can do on my PC, if I have:

1) a simple way to generate the ORDERED set of solutions n to f(n) = 0.
2) a quick way to search into that set without having to "enumerate" it ; that is: if I can generate "solution number p" in that set without having to calculate solution 1, solution 2, ....

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March 27, 2017, 05:54:20 AM
Last edit: March 27, 2017, 06:08:08 AM by deisik
 #108

Besides, how are you go to crack hash function? Hash is basically a function that maps some data of arbitrary size (usually orders of magnitude greater in size than the resulting hash) to data of fixed size. Therefore, "cracking" a hash would pose to you conceptually the same task as finding a private key having only a public one, i.e. you will have to check myriads of possibilities that give you the same hash. And which one are you going to choose?

In short, you can't crack a hash function

Of course you can.  What is a hash function essentially ?  It is a function f that maps a number in the set A = {0,...N-1} into the set B = {0,...M-1} where M << N ; f(n) = m.

The trick of a hash function is that the calculation of f(n) = m is rather simple, but that there's no way, given an m, to find an n other than by exhaustively trying all elements of A. 

Given that M << N, there are essentially on average Q = N / M elements of A that map onto a given value of m.
If you find a way to find easily the set of these Q elements, given m, without having to scan over all of A, you have cracked the hash function

It seems this is where you make an impossible assumption (I didn't read past that)

In other words, there cannot be even theoretically such a way that you could find all elements of the set A that map into the set B by using a shortcut unless you check all possible elements of A if they actually map into B (provided that A is an arbitrary set of random numbers, of course). Thereby, you still have to brute force elements of A, which roughly corresponds to finding a private key. Indeed, you may not need to find precisely all elements that fit into the hash, but it is essentially the same with asymmetric encryption as well. There may be more than just one private key that produces a certain public key but you don't need to find all of them. The first encounered will suffice

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March 27, 2017, 06:02:17 AM
 #109

I don't think they really care about bitcoin's future as in bitcoin in isolation. Delaying SegWit & LN means they're currently getting much bigger fees than before i.e. better profits which is what they're bothered about.

Bitcoin mining is a business. In a business, the proprietors are only concerned about making profit. They know that Bitcoin is not going to last forever. So their aim is to make as much profit as possible, before the coin becomes obsolete.

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March 27, 2017, 06:15:56 AM
 #110

I don't think they really care about bitcoin's future as in bitcoin in isolation. Delaying SegWit & LN means they're currently getting much bigger fees than before i.e. better profits which is what they're bothered about.

You cannot blame the miner, implementing segwit and LN will definitely cut their profit.  The first reaction of person should be protect their interest. Though as I read segwit and LN explanation, they are only for micro payment, I think it is somewhere around 0.042 BTC and larger than that can't be handled by LN by now but the possibility of increasing that is very possible, besides it is just a code.

You can blame them, and let me take an example : If the needs of the consumers change, the company has to adapt to the changes or they will not survive. All parties in this debate is clear on the fact that aggressive scaling must happen. The miners or let me rather say the ones supporting BU, has decided that it is a better solution to scale Bitcoin slowly, because it benefits them, when congestion push up the fees. 

SegWit & LN eliminates this problem, because it scales aggressively, so we will not have to deal with this problem again soon.

The problem is, these miners has had a taste of these higher fees and they want more. ^grrrrrr^   

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March 27, 2017, 07:52:58 AM
 #111

If I were a miner (too expensive for my taste) then I would configure my mining software to mine only 0-fee transactions unless it didn't fill a block in which case then I would include the lowest fee transactions until it filled a block up to the limit allowed.  I am not most people.

If the developers would develop a less expensive way to be a miner then I would appreciate it.  Perhaps they could let miners like me win ever so often (every tenth block?) if we produce a block full of 0-fee transactions?
i REALLY dont think people understand how mining works..

for one, its NOT THE MINERS that set the fee.. or choose to mine a certain fee, or really do anything..

everyone in this thread is saying MINERS this and MINERS that.. but, they should be saying MINING POOLS..

its the POOLS that set everything up.. and YES YOU CAN START YOUR OWN MINING POOL.. if you get enough miners on it, you can set it do mine zero fee tx's



If I am not mistaken miners they choose to join in a mining pool in order to mine more effectively. Therefore it's a matter of choice which is the pool they go with. Moreover, mining is an investment and as investment has not only profit but risk too. Yesterday, I was written an article where the author is a miner and had invested about $30K and was claiming that had to protect his investment. So I would like to ask; What is the difference between miner's investment and bitcoin's holder investment who bought 30 coins with $1000 per coin? Miners are securing bitcoin but bitcoin has value because people want it.
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March 27, 2017, 08:15:24 AM
 #112

I don't think they really care about bitcoin's future as in bitcoin in isolation. Delaying SegWit & LN means they're currently getting much bigger fees than before i.e. better profits which is what they're bothered about.

You cannot blame the miner, implementing segwit and LN will definitely cut their profit.  The first reaction of person should be protect their interest. Though as I read segwit and LN explanation, they are only for micro payment, I think it is somewhere around 0.042 BTC and larger than that can't be handled by LN by now but the possibility of increasing that is very possible, besides it is just a code.

You can blame them, and let me take an example : If the needs of the consumers change, the company has to adapt to the changes or they will not survive. All parties in this debate is clear on the fact that aggressive scaling must happen. The miners or let me rather say the ones supporting BU, has decided that it is a better solution to scale Bitcoin slowly, because it benefits them, when congestion push up the fees. 

SegWit & LN eliminates this problem, because it scales aggressively, so we will not have to deal with this problem again soon.

The problem is, these miners has had a taste of these higher fees and they want more. ^grrrrrr^ 

This is not the only viable explanation

In fact, we would be perfectly fine if it was the only rationale behind the actions of the puppet masters of BU team and their cronies. But we can't completely write off the possibility that they are ultimately looking to destroy Bitcoin (the reasons for wanting that vary greatly). Obviously, they can't openly admit that since admitting that would instantly make all their efforts futile. So they should necessarily hide their intentions by disguising them through claims of making Bitcoin better, and we can only look at indirect evidence to find out their true intentions

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March 27, 2017, 08:36:46 AM
 #113

Besides, how are you go to crack hash function? Hash is basically a function that maps some data of arbitrary size (usually orders of magnitude greater in size than the resulting hash) to data of fixed size. Therefore, "cracking" a hash would pose to you conceptually the same task as finding a private key having only a public one, i.e. you will have to check myriads of possibilities that give you the same hash. And which one are you going to choose?

In short, you can't crack a hash function

Of course you can.  What is a hash function essentially ?  It is a function f that maps a number in the set A = {0,...N-1} into the set B = {0,...M-1} where M << N ; f(n) = m.

The trick of a hash function is that the calculation of f(n) = m is rather simple, but that there's no way, given an m, to find an n other than by exhaustively trying all elements of A.  

Given that M << N, there are essentially on average Q = N / M elements of A that map onto a given value of m.
If you find a way to find easily the set of these Q elements, given m, without having to scan over all of A, you have cracked the hash function

It seems this is where you make an impossible assumption (I didn't read past that)

In other words, there cannot be even theoretically such a way that you could find all elements of the set A that map into the set B by using a shortcut unless you check all possible elements of A if they actually map into B (provided that A is an arbitrary set of random numbers, of course).

Of course not.  You are making a circular argument here.  You take for granted the properties of a hash function, to prove that a hash function has these properties.

Let me give you a toy example.  Suppose that my hash function is:

f(X) = (K.X) modulo M + C

This has the "random" properties of a hash function, right ?

You can solve this equation into:

f(n) = m

becomes:

n = (m - C) / K modulo M + L . M

where the division is unique in the field modulo M (supposed to be a prime), or has a few solutions in the ring modulo M if M is not prime or no solution, easily determined by Euclid's extended algorithm.

We now have an ordered solution set for n, where L is the index in the solution set, eventually with a second index in the division solution set if M is not prime and there are multiple solutions to the inverse of K.

Of course this was a trivial hash function.   A cryptographic hash function is way more complicated and supposed NOT to lead to such a solution.  But the principle is possible, as I just showed.  I don't have to spend any work to find the L-th solution to my hash problem:

f(n) = m.

This is what I understand by "cracking the hash function completely": being able to have an algorithm that spits out solution number L of the ordered solution set.
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March 27, 2017, 09:07:59 AM
 #114

Besides, how are you go to crack hash function? Hash is basically a function that maps some data of arbitrary size (usually orders of magnitude greater in size than the resulting hash) to data of fixed size. Therefore, "cracking" a hash would pose to you conceptually the same task as finding a private key having only a public one, i.e. you will have to check myriads of possibilities that give you the same hash. And which one are you going to choose?

In short, you can't crack a hash function

Of course you can.  What is a hash function essentially ?  It is a function f that maps a number in the set A = {0,...N-1} into the set B = {0,...M-1} where M << N ; f(n) = m.

The trick of a hash function is that the calculation of f(n) = m is rather simple, but that there's no way, given an m, to find an n other than by exhaustively trying all elements of A. 

Given that M << N, there are essentially on average Q = N / M elements of A that map onto a given value of m.
If you find a way to find easily the set of these Q elements, given m, without having to scan over all of A, you have cracked the hash function

It seems this is where you make an impossible assumption (I didn't read past that)

In other words, there cannot be even theoretically such a way that you could find all elements of the set A that map into the set B by using a shortcut unless you check all possible elements of A if they actually map into B (provided that A is an arbitrary set of random numbers, of course).

Of course not.  You are making a circular argument here.  You take for granted the properties of a hash function, to prove that a hash function has these properties

I won't pour mathematical formulas here since no one will try to get into them anyway

I will just give a simple (relatively) task. Say, you have some vulnerable information that you want to protect. For example, it may be access keys to your exchange account which allow to manage your funds and which you want to protect. You set a password, get its hash and then encode the keys using this hash by applying exclusive disjunction on the keys (logical XOR). XOR operation is required so that you could do the process in the reverse order (i.e. decrypt the keys using your password) but it is inconsequential as such. Now you have stolen the encrypted keys, you know how they get encrypted, so how are going to decrypt them? Hash function is known, let it be sha:512. I guess you would have to check possible combinations sequentially until you find the keys that fit

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March 27, 2017, 09:31:23 AM
 #115

Besides, how are you go to crack hash function? Hash is basically a function that maps some data of arbitrary size (usually orders of magnitude greater in size than the resulting hash) to data of fixed size. Therefore, "cracking" a hash would pose to you conceptually the same task as finding a private key having only a public one, i.e. you will have to check myriads of possibilities that give you the same hash. And which one are you going to choose?

In short, you can't crack a hash function

Of course you can.  What is a hash function essentially ?  It is a function f that maps a number in the set A = {0,...N-1} into the set B = {0,...M-1} where M << N ; f(n) = m.

The trick of a hash function is that the calculation of f(n) = m is rather simple, but that there's no way, given an m, to find an n other than by exhaustively trying all elements of A.  

Given that M << N, there are essentially on average Q = N / M elements of A that map onto a given value of m.
If you find a way to find easily the set of these Q elements, given m, without having to scan over all of A, you have cracked the hash function

It seems this is where you make an impossible assumption (I didn't read past that)

In other words, there cannot be even theoretically such a way that you could find all elements of the set A that map into the set B by using a shortcut unless you check all possible elements of A if they actually map into B (provided that A is an arbitrary set of random numbers, of course).

Of course not.  You are making a circular argument here.  You take for granted the properties of a hash function, to prove that a hash function has these properties

I won't pour mathematical formulas here since no one will try to get into them anyway

I will just give a simple (relatively) task. Say, you have some vulnerable information that you want to protect. For example, it may be access keys to your exchange account which allow to manage your funds and which you want to protect. You set a password, get its hash and then encode the keys using this hash by applying exclusive disjunction on the keys (logical XOR). XOR operation is required so that you could do the process in the reverse order (i.e. decrypt the keys using your password) but it is inconsequential as such. Now you have stolen the encrypted keys, you know how they get encrypted, so how are going to decrypt them? Hash function is known, let it be sha:512. I guess you would have to check possible combinations sequentially until you find the keys that fit

You have just re-invented a one-time pad encryption which cannot be cracked of course.  The hash function is not even necessary.  You only needed a random number (the one-time pad), and you used the entropy of your password, and the hash function, to use this to generate a random number serving as one time pad.  Even worse, I can NEVER find your keys, even with infinite amount of work, because I will never know when I found them.

You haven't used the irreversibility of the hash function here.  You only used it to transform some entropy (that of your password) into another random number.

It is a pity that you didn't look at the simple math I posted, because it explained exactly what I wanted to illustrate.  Note that you could use my toy hash function in your example, and your system would be just as secure.  Because the irreversibility of a hash function is not needed in your case.
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March 27, 2017, 09:41:23 AM
 #116

Besides, how are you go to crack hash function? Hash is basically a function that maps some data of arbitrary size (usually orders of magnitude greater in size than the resulting hash) to data of fixed size. Therefore, "cracking" a hash would pose to you conceptually the same task as finding a private key having only a public one, i.e. you will have to check myriads of possibilities that give you the same hash. And which one are you going to choose?

In short, you can't crack a hash function

Of course you can.  What is a hash function essentially ?  It is a function f that maps a number in the set A = {0,...N-1} into the set B = {0,...M-1} where M << N ; f(n) = m.

The trick of a hash function is that the calculation of f(n) = m is rather simple, but that there's no way, given an m, to find an n other than by exhaustively trying all elements of A. 

Given that M << N, there are essentially on average Q = N / M elements of A that map onto a given value of m.
If you find a way to find easily the set of these Q elements, given m, without having to scan over all of A, you have cracked the hash function

It seems this is where you make an impossible assumption (I didn't read past that)

In other words, there cannot be even theoretically such a way that you could find all elements of the set A that map into the set B by using a shortcut unless you check all possible elements of A if they actually map into B (provided that A is an arbitrary set of random numbers, of course).

Of course not.  You are making a circular argument here.  You take for granted the properties of a hash function, to prove that a hash function has these properties

I won't pour mathematical formulas here since no one will try to get into them anyway

I will just give a simple (relatively) task. Say, you have some vulnerable information that you want to protect. For example, it may be access keys to your exchange account which allow to manage your funds and which you want to protect. You set a password, get its hash and then encode the keys using this hash by applying exclusive disjunction on the keys (logical XOR). XOR operation is required so that you could do the process in the reverse order (i.e. decrypt the keys using your password) but it is inconsequential as such. Now you have stolen the encrypted keys, you know how they get encrypted, so how are going to decrypt them? Hash function is known, let it be sha:512. I guess you would have to check possible combinations sequentially until you find the keys that fit

You have just re-invented a one-time pad encryption which cannot be cracked of course.  The hash function is not even necessary.  You only needed a random number (the one-time pad), and you used the entropy of your password, and the hash function, to use this to generate a random number serving as one time pad.  Even worse, I can NEVER find your keys, even with infinite amount of work, because I will never know when I found them

It is assumed that you can check them

In reality, you could use them to try to access the funds and if you are allowed, then you have found the correct key pair. So you could assume that you can check the keys instantly (this won't work in real life but we are conceptualizing the process). But that doesn't in the least defy the fact that you would still have to check each key one by one. Regarding using a random number, I don't really see how you would be able to decode the encrypted access keys later if this random number is not saved somewhere which makes the whole idea of using a random number worthless (as I get it)

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March 27, 2017, 11:31:06 AM
 #117

I don't think that miners, woke up one day and try to hurt btc and people who using it. Someone force them to do that, if there is someone. And I believe that they want to play with btc, make more bitcoins. That won't be good. Do you know why there is economic crisis?  Bcs governments can make as much money as they want. If they want billions, will have them. Fiat money has no value anymore bcs of that. But there are only 21M btc, and people know that they won't be more, that's why btc has a value now, same as gold. There is a dirty game behind of all this, and if we con't do something now, it won't end up well.
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March 27, 2017, 11:32:05 AM
 #118

But that doesn't in the least defy the fact that you would still have to check each key one by one. Regarding using a random number, I don't really see how you would be able to decode the encrypted access keys later if this random number is not saved somewhere which makes the whole idea of using a random number worthless (as I get it)

My point is that in your example, your key is encrypted with a "one time pad", which is, by definition, a random number that YOU know, and that the enemy (me) doesn't.  However, instead of "remembering" the one time pad directly, you prefer remembering a (random) password, and you have to TRANSFORM that random number (your password) into another random number of different format.  To do this entropy transformation, you use a hash function, but you don't NEED a hash function for that ; any injective function from the password set into the "one time pad" set is good enough.  You don't need the irreversibility of that function (which is the property of a hash function: practical irreversibility).  As I showed you, my function f(n) = m = (K n) mod M + C, which is a "randomly looking function" that is perfectly reversible, could do just as well for your case.  The fact that f(n) would be "cracked" and made reversible, doesn't alter the security of your key protection, which is simply protected by the entropy of your password, that simply had to be mathematically transformed in the right form factor to be XORed with your key.

Whether I can calculate back from the one time pad to the password, or not, doesn't matter in your case.  So the fact that reversing the hash function doesn't destroy your system, is non sequitur.
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March 27, 2017, 12:05:26 PM
 #119

I don't think that miners, woke up one day and try to hurt btc and people who using it. Someone force them to do that, if there is someone. And I believe that they want to play with btc, make more bitcoins. That won't be good. Do you know why there is economic crisis?  Bcs governments can make as much money as they want. If they want billions, will have them. Fiat money has no value anymore bcs of that. But there are only 21M btc, and people know that they won't be more, that's why btc has a value now, same as gold. There is a dirty game behind of all this, and if we con't do something now, it won't end up well.

There is always some force behind anyone doing anything

But just knowing this doesn't tell us anything in particular. If they want to simply make more bitcoins (in this subtle way bypassing the fixed limit of 21M bitcoins to be mined), why are there rumors that they are preparing a 51% attack against Bitcoin? Somehow this doesn't add up well to your theory. Given how many bugs the BU code contains, they don't seem to be particularly interested to actually bring BU to life. That says that they might in fact be looking to destroy Bitcoin, not just mine more bitcoins in search of profits

My point is that in your example, your key is encrypted with a "one time pad", which is, by definition, a random number that YOU know, and that the enemy (me) doesn't.  However, instead of "remembering" the one time pad directly, you prefer remembering a (random) password, and you have to TRANSFORM that random number (your password) into another random number of different format.  To do this entropy transformation, you use a hash function, but you don't NEED a hash function for that ; any injective function from the password set into the "one time pad" set is good enough.  You don't need the irreversibility of that function (which is the property of a hash function: practical irreversibility).  As I showed you, my function f(n) = m = (K n) mod M + C, which is a "randomly looking function" that is perfectly reversible, could do just as well for your case.  The fact that f(n) would be "cracked" and made reversible, doesn't alter the security of your key protection, which is simply protected by the entropy of your password, that simply had to be mathematically transformed in the right form factor to be XORed with your key

Okay, I see you point. As I said, I'm not very familiar with Bitcoin cryptography to continue this discussion in depth

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March 27, 2017, 12:43:34 PM
 #120

I don't think that miners, woke up one day and try to hurt btc and people who using it. Someone force them to do that, if there is someone. And I believe that they want to play with btc, make more bitcoins. That won't be good. Do you know why there is economic crisis?  Bcs governments can make as much money as they want. If they want billions, will have them. Fiat money has no value anymore bcs of that. But there are only 21M btc, and people know that they won't be more, that's why btc has a value now, same as gold. There is a dirty game behind of all this, and if we con't do something now, it won't end up well.

+1

That's what all the "sudden" seismic political movements around the world these days are probably all about. There's no proper proof, of course, but it fits the long term pattern of history: when the old monetary system begins to lose control and break down, tyrants begin the biggest wars they can summon up.

Are we heading for big changes to the monetary system, an economic depression, and huge wars? It's not impossible, but let's not wish for it by over-interpreting the current situation. At the same time, diligent behaviour is necessary, just in case.

Above all, cryptocurrency could really help cut off the money that today's politicians need to wage major wars. We need to cut them off from cryptocurrency, and fortunately, Satoshi designed Bitcoin that way from the outset. Maybe, (dare I say it) this time will be different

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