块的哈希码如何生成？

1. 获取特定块头的字节表示形式: http://mainnet-node.tzscan.io/chains/main/blocks/head/header/raw .块标题的格式在the 中进行了描述文档.
2. 获取其BLAKE2b(32位)哈希摘要
3. 添加两个字节" \ x01 \ x34"(它们负责" B"字母)
4. Base58使用校验和对其进行编码

没有随机性.

面包师选择块标题.面包师可以轻松地构造许多不同的块,并仅注入具有所需哈希值的块,从而操纵"随机"数的生成.由于Tezos区块仅需要简单的工作证明,因此比比特币或以太坊中的建议要简单得多,在比特币或以太坊中,建议"不,您不能以这种方式生成随机数"已经提出了多年.

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.

我的问题的第1部分已由Michael回答.那第二部分呢?好吧,我只是看了一眼.我所做的是:

• 提取完整的哈希码系列,从块1到最新的(非常长).
• 提取此类哈希码中包含的数字(即删除字母).
• 通过专门设计的测试来分析这一系列数字的"随机性".

测试表明随机假设不能在5％时被拒绝.有趣的是,数字的分布非常偏斜.如果有的话,可能会希望它有点像均匀分布(请注意相反的说法不正确,即均匀分布并不意味着随机性).我仍然对此感到不安,但乍一看似乎表明该系列可能是随机的.

再现以上分析的R代码如下:

 # 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)