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Why can't miners meet the difficulty by picking a low number for the block hash?

What is preventing me from simply constructing a hash that's lower than the current target?Why is the hash rate pumped near difficulty calculations?How is it that concurrent miners do not subvert each other's work?Difficulty for hash (reversed function of “does hash met difficulty/target”)Why didn't the difficulty adjust for the first 30,240 blocks?Why wouldn't miners change the protocol? (e.g. change difficulty)How is mining difficulty defined (formula)?Why can't a Bitcoin fork take over?When computing the hash, do you need anything after the number of difficulty bits?

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2

In terms of 'proof of work' my novice understanding is that miners are tying to come up with a 64-digit hexadecimal number, "hash," that is less than or equal to the target hash.

Can't miners always get lower than the target has by inputting "63 zeros followed by a 1?

Beginner here, I apologize for my simplistic understanding.

difficulty proof-of-work block-validity block-hash

share|improve this question

edited 8 hours ago

Murch♦

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jackjack

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• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Possible duplicate of What is preventing me from simply constructing a hash that's lower than the current target?
  
  – chytrik
  5 hours ago
  

  
  
  
  

add a comment | 

2

In terms of 'proof of work' my novice understanding is that miners are tying to come up with a 64-digit hexadecimal number, "hash," that is less than or equal to the target hash.

Can't miners always get lower than the target has by inputting "63 zeros followed by a 1?

Beginner here, I apologize for my simplistic understanding.

difficulty proof-of-work block-validity block-hash

share|improve this question

edited 8 hours ago

Murch♦

36.5k2727 gold badges120120 silver badges354354 bronze badges

asked 9 hours ago

jackjack

1111 bronze badge

New contributor

jack is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Possible duplicate of What is preventing me from simply constructing a hash that's lower than the current target?
  
  – chytrik
  5 hours ago
  

  
  
  
  

add a comment | 

In terms of 'proof of work' my novice understanding is that miners are tying to come up with a 64-digit hexadecimal number, "hash," that is less than or equal to the target hash.

Can't miners always get lower than the target has by inputting "63 zeros followed by a 1?

Beginner here, I apologize for my simplistic understanding.

difficulty proof-of-work block-validity block-hash

share|improve this question

edited 8 hours ago

Murch♦

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asked 9 hours ago

jackjack

1111 bronze badge

New contributor

jack is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|improve this question

edited 8 hours ago

Murch♦

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asked 9 hours ago

jackjack

1111 bronze badge

New contributor

jack is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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share|improve this question

edited 8 hours ago

Murch♦

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asked 9 hours ago

jackjack

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jack is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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edited 8 hours ago

Murch♦

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edited 8 hours ago

Murch♦

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jackjack

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asked 9 hours ago

jackjack

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2 Answers
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4

The hash is a result of computing two rounds of SHA256 on the block header. If you set the hash yourself, when other nodes try to verify the hash by performing the aforementioned computation, it will not (however, there is an infinitesimal chance) result in "63 zeros followed by a 1". I suggest reading about hash functions.

The only reliable way to create a valid hash is to continuously compute the hash on the actual block header, while modifying the contents of the block so that the block is still valid (e.g. nonce, transactions, block version, etc.), but gives you a lot of tries (guesses) until you get an output that is valid. It is basically brute forcing, so having fast, efficient mining hardware is an advantage.

See also: Block Hashing Algorithm

share|improve this answer

answered 9 hours ago

JBaczukJBaczuk

5,81211 gold badge55 silver badges2525 bronze badges

add a comment | 

1

The block hash is the digest of hashing the block header with SHA-256d. This digest must meet the difficulty requirement when interpreted as a number, and since it's unique is also used as an identifier for the block.

The hash cannot be picked arbitrarily, as that would require you to revert a hash function. This contradicts the pre-image resistance property of cryptographic hash functions which states:

Given a hash h, it is difficult to find a message m that for which h = hash(m).

In fact, the proof-of-work puzzle of trying to find a block which has a specific prefix is essentially requiring miners to solve a much easier sub-problem: finding a partial pre-image.

In conclusion, the only practical way for a miner to pick a block hash is by finding a block header whose digest meets the difficulty statement.

share|improve this answer

answered 8 hours ago

Murch♦Murch

36.5k2727 gold badges120120 silver badges354354 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Not a programmer here, so I'm trying to understand this on simple terms. Let's use numbers 0-9. Target hash is 6. One miner picks number 2, which is below the target hash, is this sufficient?
  
  – jack
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Where does this simple analogy break down?
  
  – jack
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  Yes, any number that is smaller than the target hash meets the difficulty. But they cannot just "pick" 2. They have to try many block templates until they find one which happens to hash to 2. :)
  
  – Murch♦
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @jack to be more explicit, the miner does not pick a hash. The miner picks a header, from which the hash is computed.
  
  – stewbasic
  48 mins ago
  

  
  
  
  

add a comment | 

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4

The hash is a result of computing two rounds of SHA256 on the block header. If you set the hash yourself, when other nodes try to verify the hash by performing the aforementioned computation, it will not (however, there is an infinitesimal chance) result in "63 zeros followed by a 1". I suggest reading about hash functions.

The only reliable way to create a valid hash is to continuously compute the hash on the actual block header, while modifying the contents of the block so that the block is still valid (e.g. nonce, transactions, block version, etc.), but gives you a lot of tries (guesses) until you get an output that is valid. It is basically brute forcing, so having fast, efficient mining hardware is an advantage.

See also: Block Hashing Algorithm

share|improve this answer

answered 9 hours ago

JBaczukJBaczuk

5,81211 gold badge55 silver badges2525 bronze badges

add a comment | 

4

The hash is a result of computing two rounds of SHA256 on the block header. If you set the hash yourself, when other nodes try to verify the hash by performing the aforementioned computation, it will not (however, there is an infinitesimal chance) result in "63 zeros followed by a 1". I suggest reading about hash functions.

The only reliable way to create a valid hash is to continuously compute the hash on the actual block header, while modifying the contents of the block so that the block is still valid (e.g. nonce, transactions, block version, etc.), but gives you a lot of tries (guesses) until you get an output that is valid. It is basically brute forcing, so having fast, efficient mining hardware is an advantage.

See also: Block Hashing Algorithm

share|improve this answer

answered 9 hours ago

JBaczukJBaczuk

5,81211 gold badge55 silver badges2525 bronze badges

add a comment | 

The hash is a result of computing two rounds of SHA256 on the block header. If you set the hash yourself, when other nodes try to verify the hash by performing the aforementioned computation, it will not (however, there is an infinitesimal chance) result in "63 zeros followed by a 1". I suggest reading about hash functions.

The only reliable way to create a valid hash is to continuously compute the hash on the actual block header, while modifying the contents of the block so that the block is still valid (e.g. nonce, transactions, block version, etc.), but gives you a lot of tries (guesses) until you get an output that is valid. It is basically brute forcing, so having fast, efficient mining hardware is an advantage.

See also: Block Hashing Algorithm

share|improve this answer

answered 9 hours ago

JBaczukJBaczuk

5,81211 gold badge55 silver badges2525 bronze badges

share|improve this answer

answered 9 hours ago

JBaczukJBaczuk

5,81211 gold badge55 silver badges2525 bronze badges

answered 9 hours ago

JBaczukJBaczuk

5,81211 gold badge55 silver badges2525 bronze badges

answered 9 hours ago

JBaczukJBaczuk

5,81211 gold badge55 silver badges2525 bronze badges

1

The block hash is the digest of hashing the block header with SHA-256d. This digest must meet the difficulty requirement when interpreted as a number, and since it's unique is also used as an identifier for the block.

The hash cannot be picked arbitrarily, as that would require you to revert a hash function. This contradicts the pre-image resistance property of cryptographic hash functions which states:

Given a hash h, it is difficult to find a message m that for which h = hash(m).

In fact, the proof-of-work puzzle of trying to find a block which has a specific prefix is essentially requiring miners to solve a much easier sub-problem: finding a partial pre-image.

In conclusion, the only practical way for a miner to pick a block hash is by finding a block header whose digest meets the difficulty statement.

share|improve this answer

answered 8 hours ago

Murch♦Murch

36.5k2727 gold badges120120 silver badges354354 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Not a programmer here, so I'm trying to understand this on simple terms. Let's use numbers 0-9. Target hash is 6. One miner picks number 2, which is below the target hash, is this sufficient?
  
  – jack
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Where does this simple analogy break down?
  
  – jack
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  Yes, any number that is smaller than the target hash meets the difficulty. But they cannot just "pick" 2. They have to try many block templates until they find one which happens to hash to 2. :)
  
  – Murch♦
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @jack to be more explicit, the miner does not pick a hash. The miner picks a header, from which the hash is computed.
  
  – stewbasic
  48 mins ago
  

  
  
  
  

add a comment | 

1

The block hash is the digest of hashing the block header with SHA-256d. This digest must meet the difficulty requirement when interpreted as a number, and since it's unique is also used as an identifier for the block.

The hash cannot be picked arbitrarily, as that would require you to revert a hash function. This contradicts the pre-image resistance property of cryptographic hash functions which states:

Given a hash h, it is difficult to find a message m that for which h = hash(m).

In fact, the proof-of-work puzzle of trying to find a block which has a specific prefix is essentially requiring miners to solve a much easier sub-problem: finding a partial pre-image.

In conclusion, the only practical way for a miner to pick a block hash is by finding a block header whose digest meets the difficulty statement.

share|improve this answer

answered 8 hours ago

Murch♦Murch

36.5k2727 gold badges120120 silver badges354354 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Not a programmer here, so I'm trying to understand this on simple terms. Let's use numbers 0-9. Target hash is 6. One miner picks number 2, which is below the target hash, is this sufficient?
  
  – jack
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Where does this simple analogy break down?
  
  – jack
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  Yes, any number that is smaller than the target hash meets the difficulty. But they cannot just "pick" 2. They have to try many block templates until they find one which happens to hash to 2. :)
  
  – Murch♦
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @jack to be more explicit, the miner does not pick a hash. The miner picks a header, from which the hash is computed.
  
  – stewbasic
  48 mins ago
  

  
  
  
  

add a comment | 

The block hash is the digest of hashing the block header with SHA-256d. This digest must meet the difficulty requirement when interpreted as a number, and since it's unique is also used as an identifier for the block.

The hash cannot be picked arbitrarily, as that would require you to revert a hash function. This contradicts the pre-image resistance property of cryptographic hash functions which states:

Given a hash h, it is difficult to find a message m that for which h = hash(m).

In fact, the proof-of-work puzzle of trying to find a block which has a specific prefix is essentially requiring miners to solve a much easier sub-problem: finding a partial pre-image.

In conclusion, the only practical way for a miner to pick a block hash is by finding a block header whose digest meets the difficulty statement.

share|improve this answer

answered 8 hours ago

Murch♦Murch

36.5k2727 gold badges120120 silver badges354354 bronze badges

share|improve this answer

answered 8 hours ago

Murch♦Murch

36.5k2727 gold badges120120 silver badges354354 bronze badges

answered 8 hours ago

Murch♦Murch

36.5k2727 gold badges120120 silver badges354354 bronze badges

answered 8 hours ago

Murch♦Murch

36.5k2727 gold badges120120 silver badges354354 bronze badges