Come viene generato il codice hash dei blocchi?

1. Prendi una rappresentazionein byte di unaparticolareintestazione diblocco: http://mainnet-node.tzscan.io/chains/main/blocks/head/header/raw . Ilformato dell'intestazione delblocco è descrittoin documenti .
2. Ottieni un hash digest BLAKE2b (32bit)
3. Anteponi duebyte "\ x01 \ x34" (sono responsabili della lettera "B")
4. Base58 codificalo con checksum

Non c'è casualità.

Ilpanettiere sceglie l'intestazione delblocco. Unpanettierepuòfacilmente costruiremoltiblocchi diversie iniettare soloilblocco con l'hashpiù desiderato,manipolando lagenerazione delnumero "casuale". Poiché unblocco Tezos richiede solo unafacileprova di lavoro,questo èpiùfacile chein Bitcoin o Ethereum,doveil consiglio "no,nonpuoigenerarenumeri casualiin questomodo" è stato datoper anni.

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.

Laparte 1 dellamia domanda ha ricevuto risposta da Michael. E la secondaparte? Beh,l'ho appena dato un'occhiata. Quello che hofatto è:

• estrae la serie completa di codici hash,dalblocco 1 all'ultimo (MOLTO LUNGO).
• estrarrei numeri contenutiin tali codici hash (ovvero rimuovere le lettere).
• analizzare la "casualità" di questa serie dinumeri contest appositamenteprogettati.

Itest suggeriscono che l'ipotesi di casualitànonpuòessere rifiutata al 5%. Èinteressantenotare che la distribuzione deinumeri èmolto distorta. Semmai,ci sipotrebbe aspettare che assomigliin qualchemodo a una distribuzione uniforme (notando cheil contrarionon è vero,cioè la distribuzione uniformenonimplica casualità). Ci sto ancora lavorando,ma unprimo sguardo suggerisce che la seriepotrebbeessere casuale.

Il codice Rper riprodurre l'analisiprecedente è di seguito:

 # 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)