Graphs: Analysis of Expose

om-graphs/graphs.tex

@@ -294,7 +294,7 @@ tree is defined as the sum of the ranks of all its nodes.
 The key claim about the complexity of the $\Splay$ operation is the access lemma:
 
 \theoremn{Access lemma}{
-Let~$t$ be the root of splay tree. Then $\Splay(v)$ costs $\O(r(t) - r(v))$ rotations.
+Let~$t$ be the root of splay tree. Then $\Splay(v)$ costs at most $3(r(t) - r(v)) + 1$ rotations.
 }
 
 By setting all weights to one, the size of the node is exactly the size of the
@@ -341,7 +341,9 @@ Apart from left and right son,
 we allow each node~$v$ in a splay tree to have multiple \em{middle sons}. Middle sons
 represent the thin edges leading into a vertex~$v$. More precisely, if there is a thin edge
 leading from a fat path~$A$ into the vertex~$v$, then the root of the splay tree representing~$A$
-is a middle son of~$v$. \TODO picture of a middle sons (left part: "true" structure of
+is a middle son of~$v$. To distinguish between middles sons and left and right son, we
+will call left and right son \em{proper sons}.
+\TODO picture of a middle sons (left part: "true" structure of
 fat paths, right part: representation with splay trees). 
 
 However, a node does not know about its middle sons. We do not store a pointer from a
@@ -377,8 +379,76 @@ of~$p$. Then we move to the next step, where vertex~$p$ takes the role of vertex
 In the end, when~$v$ and~$\Root(v)$, are within the same splay tree, we splay~$v$.
 
 \subsection{Analysis of \Expose}
+For the sake of analysis we split $\Expose(v)$ into three phases. Let $(r_1, p_1),\dots,
+(r_k, p_k)$ be the thin edges on the path from~$v$ to the root of the virtual tree, see
+Figure~\TODO. Let us
+also denote $p_0 = v$. In the first phase, we splay all of $p_0, \dots, p_k$ into the roots
+of their respective splay trees. After first phase, $p_k$ is the root of the virtual tree
+and $p_0, \dots, p_k$ form a path made of thin edges. In the second phase, we turn edges
+$(p_0, p_1), (p_1, p_2), \dots, (p_{k-1}, p_k)$ into proper splay tree edges and left sons
+of $p_1, \dots, p_k$ into middle sons, just as described in the previous section. Now that
+$p_0, \dots, p_k$  form a path inside one splay tree, we splay $p_0 = v$ into the root of
+that tree -- this is the third phase.
+
+\TODO picture of phases
+
+The key phase we need to deal with in our analysis the first phase. Third phase is a
+simple splay in a splay tree, which should have have $\O(\log n)$ amortized complexity,
+unless we break potential in previous phases, of course. Real cost of the second phase can
+be bounded by
+the real cost of the third phase, so the third phase can also pay for the second phase.
+However, we somehow have to make sure that swapping thin edges and splay edges does not
+break the potential stored in the respective trees. 
+
+The first phase, on the other hand, performs a series of splay operation and each of them
+possible costs up to $\Omega(\log n)$ amortized, if we use the basic bound on the
+complexity of splay. So, how are we going to achieve the complexity we promised?
+
+The trick is to set weights of nodes in splay trees in a way that the whole virtual tree
+basically acts as one large splay tree. Recall that in the analysis of the splay tree, we
+assign an arbitrary non-negative weight to each node of the tree. In our analysis, we set
+the weight of a node to be
+$$ w(v) = 1 + \sum_{u\in M(v)}  s(u),$$
+where $M(v)$ is the set of middle sons of~$v$. That is, size of~$v$, $s(v)$, is the number
+of nodes in the subtree below~$v$ \em{including all nodes reachable through middle sons}.
+
+We define a potential~$\Phi$ of the virtual tree~$T$ to be the sum of the potential of each
+splay tree. This also means that $\Phi = \sum_{v\in T} r(v)$, where $r(v) = \log(s(v))$ is
+the rank of~$v$. 
+
+Let us star with the third phase. We have only a single splay in a splay tree, so
+according to Access lemma we get amortized complexity $\O(r(p_k) - r(v) + 1)$, where the ranks
+are just before the third phase. Since size of any node is bound by the total number of
+vertices, we get amortized complexity of third phase to be $\O(\log n)$.
+
+\obs{Turning a proper son into a middle son and vice versa does not change the potential.}
+
+This observation ensures that second phase does not change the potential. Thus, third
+phase can completely pay for the second phase.
+
+Finally, the dreaded first phase. We perform $k$ splays in $k$ different splay trees.
+According to the Access lemma, the total amortized cost is at most
+$$\sum_{i=1}^{k+1} 3(r(r_i) - r(p_{i-1})) + 1,$$
+where $r_{k+1}$ is the root of the virtual tree before the first phase. Observe that this
+sum telescopes! Since $r_i$ is a middle son of $p_i$, we have $s(r_i) < s(p_i)$ and by
+monotonicity the same holds for the ranks. Using this we get that the amortized complexity
+is $\O(r(r_{k+1}) - r(v) + k) = \O(\log n + k)$. This is still not the desired logarithmic
+complexity as $k$ can be possibly quite large, perhaps $\Theta(n)$. But do not despair,
+the third phase can save us. Notice that $\O(k)$ can be upper-bounded by the real cost of
+the third phase. Thus, third phase will also pay for the $+k$ part of the first phase.
+
+There is one last subtle problem we have to deal with. The problem is that structural
+updates can change the potential of the virtual tree outside of the $\Expose$. 
+The simple case is $\Cut$. Since we remove a subtree from the virtual tree, we only
+decrease the potential in the original virtual tree and the potential of the new tree is
+paid by the potential of the removed subtree.
+
+$\Link(u,v)$ is slightly more complicated, since adding a subtree means increasing
+size of some nodes. However, notice that after $\Expose(u)$ the node~$u$ has no right
+son as it is the last vertex on the fat path and also the root of the virtual tree. Thus,
+by linking $u$ and $v$ we only increase the size of $u$ and we increase it by at most $n$,
+so only need to pay $\O(\log n)$ into the potential.
 
-\TODO
 
 \section{Application: Faster Dinic's algorithm}
 \TODO