Update README.md

Author: aneubeck

crates/consistent-hashing/README.md

@@ -81,18 +81,37 @@ In the remainder of this section we prove that the `consistent_choose_k` algorit
 Let's define `M(k,n) = consistent_choose_max(_, k, n)` and `S(k, n) := consistent_choose_k(_, k, n)` as short-cuts for some arbitrary fixed `key`.
 We assume that `consistent_hash(key, k, n)` computes `k` independent consistent hash functions.
 
+### Property 1
+
 Since `M(k, n) < n` and `S(k, n) = {M(k, n)} ∪ S(k - 1, M(k, n))` for `k > 1`, `S(k, n)` constructs a strictly monotonically decreasing sequence. The sequence outputs exactly `k` elements which therefore must all be distinct which proves property 1 for `k <= n`.
 
 Properties 2, 3, and 4 can be proven via induction as follows.
 
+### Property 4
+
 `k = 1`: We expect that `consistent_hash` returns a single uniformly distributed node index which is consistent in `n`, i.e. changes the hash value with probability `1/(n+1)`, when `n` increments by one. In our implementation, we use an `O(1)` implementation of the jump-hash algorithm. For `k=1`, `consistent_choose_k(key, 1, n)` becomes a single function call to `consistent_choose_max(key, 1, n)` which in turn calls `consistent_hash(key, 0, n)`. I.e. `consistent_choose_k` inherits the all the desired properties from `consistent_hash` for `k=1` and all `n>=1`.
 
 `k → k+1`: `M(k+1, n+1) = M(k+1, n)` iff `M(k, n+1) < n` and `consistent_hash(_, k, n+1-k) < n - k`. The probability for this is `(n+1-k)/(n+1)` for the former by induction and `(n-k)/(n+1-k)` by the assumption that `consistent_hash` is a proper consistent hash function. Since both these probabilities are assumed to be independent, the probability that our initial value changes is `1 - (n+1-k)/(n+1) * (n-k)/(n+1-k) = 1 - (n-k)/(n+1) = (k+1)/(n+1)` proving property 4.
 
+### Property 3
+
 Property 3 is trivially satisfied if `S(k+1, n+1) = S(k+1, n)`. So, we focus on the case where `S(k+1, n+1) != S(k+1, n)`, which implies that `n ∈ S(k+1, n+1)` as largest element.
 We know that `S(k+1, n) = {m} ∪ S(k, m)` for some `m` by definition and `S(k, n) = S(k, u) ∖ {v} ∪ {w}` by induction for some `u`, `v`, and `w`. Thus far we have `S(k+1, n+1) = {n} ∪ S(k, n) = {n} ∪ S(k, u) ∖ {v} ∪ {w}`.
 
 If `u = m`, then `S(k+1, n) = {m} ∪ S(k, m) ∖ {v} ∪ {w}` and `S(k+1, n+1) = {n} ∪ S(k, n) = {n} ∪ S(k, m) ∖ {v} ∪ {w}` and the two differ exaclty in the elemetns `m` and `n` proving property 3.
 
 If `u ≠ m`, then `consistent_hash(_, k, n) = m`, since that's the only way how the largest values in `S(k+1, n)` and `S(k, n)` can differ. In this case, `m ∉ S(k+1, n+1)`, since `n` (and not `m`) is the largest element of `S(k+1, n+1)`. Furthermore, `S(k, n) = S(k, m)`, since `consistent_hash(_, i, n) < m` for all `i < k` (otherwise there is a contradiction).
-Putting it together leads to `S(k+1, n+1) = {n} ∪ S(k, m)` and `S(k+1, n) = {m} ∪ S(k, m)` which differ exactly in the elements `n` and `m` which concludes the proof.
+Putting it together leads to `S(k+1, n+1) = {n} ∪ S(k, m)` and `S(k+1, n) = {m} ∪ S(k, m)` which differ exactly in the elements `n` and `m` which concludes the proof.
+
+### Property 2
+
+The final part is to prove property 2. This time we have an inducation over `k` and `n`.
+As before, induction start for `k=1` and for all `n>0` is inherited from the `consistency_hash` implementation. The case `n=k` is also trivially covered, since the only valid set are the numbers `{0, ..., k-1}` which the algorithm correctly outputs. So, we only need to care about the induction step where `k>1` and `n>k`.
+
+We need to prove that `P(i ∈ S(k+1, n+1)) = (k+1)/(n+1)` for all `0 <= i <= n`. Property 3 already proves the case `i = n`. Furthermore we know that `P(n ∈ S(k+1, n+1)) = (k+1)/(n+1)` and vice versa  `P(n ∉ S(k+1, n+1)) = 1 - (k+1)/(n+1)`. Let's consider those two cases separately.
+
+`n ∈ S(k+1, n+1)`: By the definition of `S`, we know that `S(k+1, n+1) = {n} ∪ S(k, n)`. `P(i ∈ S(k+1, n+1)) = P(i ∈ S(k, n)) P(n ∈ S(k+1, n+1)) = k/n * (k+1)/(n+1)` for all `0 <= i < n`.
+
+`n ∉ S(k+1, n+1)`: Once more by definition, `S(k+1, n+1) = S(k+1, n)` in this case. `P(i ∈ S(k+1, n+1)) = P(i ∈ S(k+1, n)) P(n ∉ S(k+1, n+1)) = (k+1)/n * (1 - (k+1)/(n+1))` for all `0 <= i < n`.
+
+Summing both cases together leads to `P(i ∈ S(k+1, n+1)) = k/n * (k+1)/(n+1) + (k+1)/n * (1 - (k+1)/(n+1)) = k/n * (k+1)/(n+1) + k/n * (1 - (k+1)/(n+1)) + 1/n * (1 - (k+1)/(n+1)) = k/n * (k+1)/(n+1) + k/n - k/n * (k+1)/(n+1) + 1/n - 1/n * (k+1)/(n+1) = k/n + 1/n - 1/n * (k+1)/(n+1) = (k+1)/n - (k+1)/(n+1)/n = (k+1)/n * (1 - 1/(n+1)) = (k+1)/(n+1)` for all `0 <= i < n` which concludes the proof.