来自截断MD5的ECDF图

Question

在这个链接中，它说截断的MD5是均匀分布的。我想用PySpark来验证它，并首先在Python中创建了1,000,000个UUID，如下所示。然后从MD5中截取了前三个字符。但是，我得到的图形与均匀分布的累积分布函数不相似。我尝试了UUID1和UUID4，结果也类似。如何正确地确认截断MD5的均匀分布？

import uuid
import numpy as np
import matplotlib.pyplot as plt
from statsmodels.distributions.empirical_distribution import ECDF
import pandas as pd
import pyspark.sql.functions as f
%matplotlib inline

### Generate 1,000,000 UUID1 

uuid1 = [str(uuid.uuid1()) for i in range(1000000)]  # make a UUID based on the host ID and current time
uuid1_df = pd.DataFrame({'uuid1':uuid1})
uuid1_spark_df =  spark.createDataFrame(uuid1_df)
uuid1_spark_df = uuid1_spark_df.withColumn('hash', f.md5(f.col('uuid1')))\
               .withColumn('truncated_hash3', f.substring(f.col('hash'), 1, 3))

count_by_truncated_hash3_uuid1 = uuid1_spark_df.groupBy('truncated_hash3').count()

uuid1_count_list = [row[1] for row in count_by_truncated_hash3_uuid1.collect()]
ecdf = ECDF(np.array(uuid1_count_list))
plt.figure(figsize = (14, 8))
plt.plot(ecdf.x,ecdf.y)
plt.show()

编辑： 我添加了直方图。如下图所示，它看起来更像正态分布。

  plt.figure(figsize = (14, 8))
  plt.hist(uuid1_count_list)
  plt.title('Histogram of counts in each truncated hash')
  plt.show()

- Fisseha Berhane

---

如果你只使用前三个字节，那么只有4096个值。为什么不跟踪每个值出现的次数，然后验证它们大致相同的数量呢？ - Mr. Llama

我在上述代码中添加了最大计数和最小计数，正如你所看到的，范围非常大。 - Fisseha Berhane

Answer 1

这里有一个快速而简单的方法来演示这个问题：

import hashlib
import matplotlib.pyplot as plt
import numpy as np
import random

def random_string(n):
    """Returns a uniformly distributed random string of length n."""
    return ''.join(chr(random.randint(0, 255)) for _ in range(n))

# Generate 100K random strings
data = [random_string(10) for _ in range(100000)]
# Compute MD5 hashes
md5s = [hashlib.md5(d.encode()).digest() for d in data]
# Truncate each MD5 to the first three characters and convert to int
truncated_md5s = [md5[0] * 0x10000 + md5[1] * 0x100 + md5[2] for md5 in md5s]

# (Rather crudely) compute and plot the ECDF    
hist = np.histogram(truncated_md5s, bins=1000)
plt.plot(hist[1], np.cumsum([0] + list(hist[0])))

- NPE

Answer 2

我之前分析的问题在于，我绘制的是截断哈希计数的直方图。正确的方法应该是将截断哈希从十六进制转换为十进制，并查看十进制的分布情况。

import uuid
import numpy as np
import matplotlib.pyplot as plt
from statsmodels.distributions.empirical_distribution import ECDF
import pandas as pd
import pyspark.sql.functions as f
from pyspark.sql.types import IntegerType
%matplotlib inline

### Generate 1,000,000 UUID1 

uuid1 = [str(uuid.uuid1()) for i in range(1000000)]  
uuid1_df = pd.DataFrame({'uuid1':uuid1})
uuid1_spark_df =  spark.createDataFrame(uuid1_df)
uuid1_spark_df = uuid1_spark_df.withColumn('hash', f.md5(f.col('uuid1')))\
           .withColumn('truncated_hash3', f.substring(f.col('hash'), 1, 3))\
           .withColumn('truncated_hash3_base10', f.conv('truncated_hash3', 16, 10).cast(IntegerType()))


truncated_hash3_base10_list = [row[0] for row in 
uuid1_spark_df.select('truncated_hash3_base10').collect()]
pd_df = uuid1_spark_df.select('truncated_hash3_base10').toPandas()
ecdf = ECDF(truncated_hash3_base10_list)
plt.figure(figsize = (8, 6))
plt.plot(ecdf.x,ecdf.y)
plt.show()

plt.figure(figsize = (8, 6))
plt.hist(truncated_hash3_base10_list)
plt.show()

- Fisseha Berhane