Comment le code de hachage de blocs est généré?

1. Prenez une représentation d'octets d'unbloc deblocparticulier: http://mainnet-node.tzscan.io/chains/Main/Blocs/Head/Header/RAW . Leformat de l'en-tête deblocest décrit dans Le Docs .
2. Obtenez un digest de hachage de Blake2b (32bits)
3. Prepend Deux octets '\ x01 \ x34' (ils sont responsables de la lettre "B")
4. base58 le codez avec de la somme de contrôle

Iln'y apas de hasard.

Leboulanger choisit l'en-tête debloc. Unboulangerpeutfacilement construire denombreuxblocs différentset injecter uniquement lebloc avec le hachage leplus souhaité,manipuler votregénération denombres «aléatoire». Étant donné qu'unbloc Tezosnécessite unepreuve detravailfacile,c'estplusfacile que dans Bitcoin ouenthérum,où le conseil "Non,vousne pouvezpasgénérer denombres aléatoires de cettefaçon" a été donnée depuis des années.

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.

Lapartie 1 dema question a été réponduepar Michael. Qu'enest-il de la deuxièmepartie? Ehbien,je viens de regarder ça. Ce quej'aifaitest:

• Extraire la série complète de codes de hachage,dubloc 1 au dernier (très long).
• extraire lesnombres contenus dans detels codes de hachage (c'est-à-dire supprimer des lettres).
• analyser le "aléatoire" de cette série denombres avec destests conçus à dessein.

Lestests suggèrent que l'hypothèse de hasardne peut être rejetée à 5%. Faitintéressant,la distribution de chiffresesttrès asymétrique. Si quelque chose,onpourrait s'attendre à ce qu'il ressemble unpeu à une distribution uniforme (remarquant que le contrairen'estpas vrai,c'est-à-dire une distribution uniformen'impliquepas aléatoire). Jefrappetoujours sur cela,mais unpremier regard suggère que la sériepourrait être aléatoire.

Le code Rpour reproduire l'analyse ci-dessusest ci-dessous:

 # 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)