TITLE: Understanding Proof of Hopcroft & Karps Matching Algorithm
QUESTION [5 upvotes]: Hopcroft & Karps algorithm to compute a maximum matching takes $\mathcal O(mn^{1/2})$ time, which is composed by $\mathcal O(n^{1/2})$ iterations and each iteration taking $\mathcal O(m)$.
In my lecture script it says $\mathcal O(n^{1/2})$ iterations is bases on the fact that there are "relativly" short $M$-augmenting paths of maximal length of $2\cdot \lfloor |M|/ (s - |M|) \rfloor +  1$.
I am having trouble understanding this upper bound for a path's length.
Remarks:

$G = (V, E)$ with $V = U \cup W$ bipartite graph
$s \le n = |V|$ cardinality of a maximum matching $S$
$M$ any matching in $G$

Moreover, I know that

After every iteration there are no shortest path of length $l$ anymore.
shortest $M$-augmenting paths extend by the additive factor of 2, since they always have odd length.
Matching $|M \triangle P| =  |M| + 1$ where $\triangle$ is the symmetric difference of a matching $ M$ and a $M$-augmenting Path $P$

REPLY [2 votes]: Note that at a given stage taking the symmetric difference between the partial matching and an optimal matching gives disjoint augmenting paths and alternating cycles.
At iteration $l$ all augmenting paths are of length bigger or equal to $l$.
At this stage if there are $s-|M|$  disjoint augmenting paths and say the $i$th path uses $L_i$ matched edges, then $L_i \geq (l-1)/2$ matched edges, and thus 
$(s-|M|) (l-1)/2 \leq \sum_i L_i \leq |M|$
and solving for $l$ tells you how large $l$ could possibly be.