为什么tuple（set（[（1，“ a”，“ b”，“ c”，“ z”，“ f”]））==元组（set（[（a，b，c） “ z”，“ f”，1]））85％的时间启用了哈希随机化？

Given Zero Piraeus’ answer to another question, we have that

x = tuple(set([1, "a", "b", "c", "z", "f"]))
y = tuple(set(["a", "b", "c", "z", "f", 1]))
print(x == y)

Prints True about 85% of the time with hash randomization enabled. Why 85%?

I’m going to assume any readers of this question to have read both:

• Zero Piraeus’ answer and

• My explanation of CPython’s dictionaries.

The first thing to note is that hash randomization is decided on interpreter start-up.

The hash of each letter will be the same for both sets, so the only thing that can matter is if there is a collision (where order will be affected).

By the deductions of that second link we know the backing array for these sets starts at length 8:

_ _ _ _ _ _ _ _

In the first case, we insert 1:

_ 1 _ _ _ _ _ _

and then insert the rest:

α 1 ? ? ? ? ? ?

Then it is rehashed to size 32:

    1 can't collide with α as α is an even hash
  ↓ so 1 is inserted at slot 1 first
? 1 ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ?

In the second case, we insert the rest:

? β ? ? ? ? ? ?

And then try to insert 1:

    Try to insert 1 here, but will
  ↓ be rehashed if β exists
? β ? ? ? ? ? ?

And then it will be rehashed:

    Try to insert 1 here, but will
    be rehashed if β exists and has
  ↓ not rehashed somewhere else
? β ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ?

So whether the iteration orders are different depends solely on whether β exists.

The chance of a β is the chance that any of the 5 letters will hash to 1 modulo 8 and hash to 1 modulo 32.

Since anything that hashes to 1 modulo 32 also hashes to 1 modulo 8, we want to find the chance that of the 32 slots, one of the five is in slot 1:

5 (number of letters) / 32 (number of slots)

5/32 is 0.15625, so there is a 15.625% chance¹ of the orders being different between the two set constructions.

Not very strangely at all, this is exactly what Zero Piraeus measured.

_{¹Technically even this isn’t obvious. We can pretend every one of the 5 hashes uniquely because of rehashing, but because of linear probing it’s actually more likely for “bunched” structures to occur… but because we’re only looking at whether a single slot is occupied, this doesn’t actually affect us.}