Como o código de hash de blocos é gerado?

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1. Tome uma representação debytes de um determinado cabeçalho debloco: http://mainnet-node.tzscan.io/chains/Main/blocos/cabeça/cabeçalho/cru . Oformato do cabeçalho dobloco é descritoem o docs .
2. obter um hashblake2b (32bit) digere
3. Prepare doisbytes '\ x01 \ x34' (eles são responsáveis ​​pela letra "B"
4. base58 o codifique com soma de verificação

Não há aleatoriedade.

O Bakerescolhe o cabeçalho dobloco. Umpadeiropodefacilmente construirmuitosblocos diferentese injetar apenas obloco com o hashmais desejado,manipulando suageração denúmeros "aleatórios". Porque umbloco de Tezos requer apenas umaprovafácil detrabalho,isso émaisfácil do queem Bitcoin ou Ethereum,onde o conselho "Não,vocênãopodegerarnúmeros aleatórios dessamaneira"foi dado há anos.

1. Take a byte representation of a particular block header: http://mainnet-node.tzscan.io/chains/main/blocks/head/header/raw. The format of the block header is described in the docs.
2. Get a BLAKE2b (32bit) hash digest of it
3. Prepend two bytes '\x01\x34' (they are responsible for "B" letter)
4. Base58 encode it with checksum

There is no randomness.

The baker chooses the block header. A baker can easily construct many different blocks and inject only the block with the most desired hash, manipulating your 'random' number generation. Because a Tezos block only requires an easy proof of work, this is easier than in Bitcoin or Ethereum, where the advice "no, you can't generate random numbers that way" has been given for years.

Parte 1 daminhaperguntafoi respondidapor Michael. E a segundaparte? Bem,eutive uma olhadanisso. O queeufiz é:

• Extraia a série completa de códigos de hash,dobloco 1para omais recente (muito longo).
• Extraia osnúmeros contidosem tais códigos de hash (i.e. Remover letras).
• analisar a "aleatoriedade" desta série denúmeros comtestespropositalmenteprojetados.

Ostestes sugerem que a hipótese da aleatoriedadenãopode ser rejeitadaem 5%. Curiosamente,a distribuição denúmeros émuito distorcida. Se alguma coisa,pode-seesperar que seja umpoucoparecido com uma distribuição uniforme (percebendo que o contrárionão é verdade,isto é,a distribuição uniformenãoimplica aleatoriedade). Eu aindaestoubatendonisso,mas umaprimeira olhada sugere que a sériepode ser aleatória.

O código Rpara reproduzir a análise acimaestá abaixo:

 # Initialise stuff

remove(list = ls())    
options(timeout = 1000000) # in case request times-out

library(jsonlite)
library(ggplot2)
library(randtests)

# Get maximum number of blocks

last_block <- fromJSON("https://api6.tzscan.io//v3/head")
N <- last_block$level
blocks <- seq(1,N,by = 1)
hashs <- vector(mode="character", length = N)

# Download all hash codes (timer and print included, for analysis) VERY LONG

start <- proc.time()
for (i in 1:N) {
  url <- paste0("https://api6.tzscan.io//v3/block_hash_level/",i)
  hashs[i] <- fromJSON(url)
  print(i)
}
finish <- proc.time() - start

# Remove letters

n_hashs <- as.numeric(gsub("\\D+","", hashs, perl = TRUE))
df <- data.frame(value=n_hashs)
names(df)[names(df) == "df.value....." ] <- "value"

# Plot (definitively not a uniform distribution, for any level of zooming in)

ggplot(df, aes(x=value)) + geom_histogram() 

# Randomness tests (indicating the number sequence is not random)

bartels.rank.test(df$value)
cox.stuart.test(df$value)
difference.sign.test(df$value)
rank.test(df$value)
runs.test(df$value)
turning.point.test(df$value)
 

Part 1 of my question was answered by Michael. What about the second part? Well, I just had a look at it. What I did is:

• extract the full series of hash codes, from block 1 to the latest (VERY LONG).
• extract the numbers contained in such hash codes (i.e. remove letters).
• analyse the "randomness" of this series of numbers with purposely designed tests.

The tests suggest that the randomness hypothesis cannot be rejected at the 5%. Interestingly, the distribution of numbers is very skewed. If anything, one might expect it to be somewhat resembling a uniform distribution (noticing that the contrary is not true, i.e. uniform distribution does not imply randomness). I'm still pounding on this, but a first look suggest that the series might be random.

The R code to reproduce the above analysis is below:

# Initialise stuff

remove(list = ls())    
options(timeout = 1000000) # in case request times-out

library(jsonlite)
library(ggplot2)
library(randtests)

# Get maximum number of blocks

last_block <- fromJSON("https://api6.tzscan.io//v3/head")
N <- last_block$level
blocks <- seq(1,N,by = 1)
hashs <- vector(mode="character", length = N)

# Download all hash codes (timer and print included, for analysis) VERY LONG

start <- proc.time()
for (i in 1:N) {
  url <- paste0("https://api6.tzscan.io//v3/block_hash_level/",i)
  hashs[i] <- fromJSON(url)
  print(i)
}
finish <- proc.time() - start

# Remove letters

n_hashs <- as.numeric(gsub("\\D+","", hashs, perl = TRUE))
df <- data.frame(value=n_hashs)
names(df)[names(df) == "df.value....." ] <- "value"

# Plot (definitively not a uniform distribution, for any level of zooming in)

ggplot(df, aes(x=value)) + geom_histogram() 

# Randomness tests (indicating the number sequence is not random)

bartels.rank.test(df$value)
cox.stuart.test(df$value)
difference.sign.test(df$value)
rank.test(df$value)
runs.test(df$value)
turning.point.test(df$value)