¿Cómo se genera el código hash de los bloques?

1. Tome una representaciónen bytes de unencabezado debloqueen particular: http://mainnet-node.tzscan.io/chains/main/blocks/head/header/raw . Elformato delencabezado delbloque se describeen el documentos .
2. Obtenga un resumen de hash BLAKE2b (32bits)
3. Anteponer dosbytes '\ x01 \ x34' (son responsables de la letra "B")
4. Codifíqueloen Base58 con suma de comprobación

No hay aleatoriedad.

Elpanaderoeligeelencabezado delbloque. Unpanaderopuede construirfácilmentemuchosbloques diferentese inyectar soloelbloque conel hashmás deseado,manipulando sugeneración denúmeros 'aleatorios'. Debido a que unbloque de Tezos solo requiere unaprueba detrabajofácil,estoesmásfácil queen Bitcoin o Ethereum,dondeel consejo "no,no sepuedengenerarnúmeros aleatorios deesamanera" se ha dado durante años.

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.

Michael respondió a laparte 1 demi pregunta. ¿Y la segundaparte? Bueno,acabo deecharle un vistazo. Lo que hicefue:

• extraer la serie completa de códigos hash,desdeelbloque 1 hastael último (MUY LARGO).
• extraer losnúmeros contenidosen dichos códigos hash (es decir,eliminar letras).
• analice la "aleatoriedad" deesta serie denúmeros conpruebas diseñadasexpresamente.

Laspruebas sugieren que la hipótesis de la aleatoriedadno sepuede rechazar al 5%. Curiosamente,la distribución denúmerosestámuy sesgada. Entodo caso,unopodríaesperar que separezca unpoco a una distribución uniforme (notando que lo contrarionoes cierto,es decir,la distribución uniformenoimplica aleatoriedad). Todavíaestoyinsistiendoen esto,pero unprimer vistazo sugiere que la seriepodría ser aleatoria.

El código Rpara reproducirel análisis anterior seencuentra a continuación:

 # 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)