Update README.md

Author: aneubeck

crates/consistent-hashing/README.md

@@ -128,16 +128,16 @@ 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 = 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 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.
+`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 the former is `(n+1-k)/(n+1)` by induction and `(n-k)/(n+1-k)` for the latter 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 exactly in the elemetns `m` and `n` proving property 3.
+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 exactly in the elements `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.