046e25546c13ec88a610d456a38ce4680281f803
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Martin Hafskjold Thoresen
5 years ago
b066471

Write Lower Bounds. Should probably revisit this one, as I don't understand what's happening towards the end.

1 files changed,105insertions(+),8deletions(-) M book.tex

M book.tex => book.tex +105~~-8~~

@@ 977,7 977,8 @@ We end up with the same query time $O(\log^{d-1} n + k)$, since the procedure is\chapter{Connectivity in Dynamic Graphs} \chapter{Connectivity in Dynamic Graphs}% \label{ch:connectivity} \topics{Dynamic connectivity on trees, Euler tour trees}. Before starting, we point out that this chapter is subject to fewer proofs, and more stated results.@@ 1160,15 1161,111 @@ The general problem of maintaining a minimum spanning forest can be solved dynam\chapter{Lower Bounds} \topics{Dynamic Partial Sums, Dynamic Connectivity.} In this chapter we will look at lower bounds for basic computational problems. We will use the \emph{Cell Probe} model, in which we get computation ``for free'' while we pay for memoroy accesses. \section{Dynamic Partial Sum}% \label{sec:partial-sum} The problem is to maintain an array $A$ of length $n$ containing numbers, while supporting two operations: \code{Update}{i, x}, which sets $A[i] = x$, and \code{Query}{i} where we query the prefix sum $A[1] + \cdots + A[i]$. We will show that there is a lower bound of $\Omega(\log n)$ on either \algo{Update} or \algo{Query}. In the proof we will use the universe ${[n]}$ and $+$ as operator, but any group of polynomial size with an arbitrary operator works. The idea of the proof is to construct an access sequence where we update with random numbers, which will break any data structure. Imagine we list out the queries done through time, and build a mental binary tree over them. We want the sequence to be \emph{maximally interleaved}, which means that any access in a left subtree affects a query in a right subtree for the same node. The way we choose this sequence is by iterating $i$ from $0$ through $n$, and reverse its bits to get the index. We assume $n$ is a power of two. For instance, if $n = 8$, we get the sequence \[\texttt{0b000},\ \texttt{0b001},\ \texttt{0b010},\ \texttt{0b011},\ \texttt{0b100},\dots \implies 0, 4, 2, 6, 1,\dots\] The operations we do on each cell is as follows: we first \algo{Query} the number there, and then we set it to a totally random number. The argument we will make is that the information transfer guarantees our lower bound. We claim that for a given node in our mental tree, $\Omega(T)$ memory values are written in the left subtree and then read in the right subtree, where $T$ is the size of the subtrees. This implies that $\Omega(n \log n)$ work is begin done, simply by moving $\log n$ bits of information (one number) for each of the $n$ operations. We prove the claim: each $x_i$ has $\Theta(\log n)$ bits of entropy, since they are totally random numbers. Let $R$ be the memory cells read in the right subtree, and $W$ the memory cells written in the left subtree. We see that if we are given the state of the data structure before performing the operations in the left subtree and the operationg we are going to perform in the right subtree, we can recover the operations done in the left subtree, without knowing what they are. That is, by having the state before and the queries on the right we can recover the $x_i$ we are writing in the left. To see this, we consider again the cast where $n = 8$, so the left tree is the sequence $0, 4, 2, 6$ and the right tree is $1,5,3,7$. $Q(1) = x_0 \text{\ (since $x_1 = 0$)}$, so we know $x_0$ which was just written. $Q(5) = x_0 + x_2 + x_4 + x_1$; we have to wait for this one. $Q(3) = x_0 + x_2 + x_1$: we know $x_1$ and $x_0$, so now we know $x_2$, which gets us $x_4$ from $Q(5)$. $Q(7)$ gives is $x_6$. Now we can explicitly encode $|R \cap W| \log n$ bits, and use this to extract $\Omega(T \log n)$ bits of entropy: the $x_i$s. Since we have extracted $\Omega(T \log n)$ bits of entropy from $|R \cap W| \log n$ bits, we have $|R \cap W| = \Theta(T)$, which proves the claim. \section{Dynamic Connectivity} We have already looked at this problem in Chapter~\ref{ch:connectivity}, and we claimed that there is a lower bound of $\Omega(\log n)$ on dynamic connectivity. We will now prove it. The graph we will consider is laid out on a grid, such that adjacent columns have a perfect matching; the graph can be seen as a composition of permutations. We have $n$ vertices, and the grid is a perfect square, so there are $\sqrt{n}$ permutations permuting $\sqrt{n}$ elements. Each permutation requires $\sqrt{n} \log n$ bits. The operations we consider is updaing the $i$th perimutation, and querying that a prefix of the permutations is equivalent to a given permutation. These are \emph{block operations}, in the sense that they do not act on single edges or vertices, but groups of them; more specifically they correspond to $O(\sqrt{n})$ regular operations. This can only make our problem easier, since we now can apply amortization schemes, or somehow expolit that we have information about nearby operations. The reason we do not simply query for the total permutation is that this is hard to do using connectivity queries: we would have to search for it, using a lot of queries. We claim that $\sqrt{n}$ updates and $\sqrt{n}$ verify sums require $\Omega(\sqrt{n} \sqrt{n} \log n)$ cell probes, which implies a lower bound of $\Omega(\log n)$ per operation, since we do $\sqrt{n}$ block operations, which all corresponds to $\sqrt{n}$ graph operations. \subsection{The Proof} \todo{wtf is this} Similar to in Section~\ref{sec:partial-sum} we will consider the interleaving access pattern. We will look at how much information has to be carried over from the left to the right subtree for a given node. We claim that that every node in a right subtree have to do $\Omega(l\sqrt{n})$ expected cell probes reading cells that were written in the left subtree, where $l$ is the number of leaves in the subtree. In addition, we sum this lower bound over every node in the tree. Note that there is no double counting of reads, since we only count a read in the \LCA{} for the read and the latest write. Since every leaf is in $\log n$ subtrees, and the size of the tree is $O(\sqrt{n})$ we have $\sum l = O(\log n \sqrt{n})$, so we end up with $\Omega(\sqrt{n} \sqrt{n} \log n)$, which is the claim. Now we have to prove the claim from the previous paragraph. We can do this in a similar manner as in Section~\ref{sec:partial-sum}. The left subtree for a given node contains $l/2$ updates with $l/2$ independent, totally random permutations. An encoding of these updates must use $\Omega(l\sqrt{n} \log n)$ bits ($l$ updates, which are $\sqrt{n}$ numbers of size $n$). We show that if the claim is false, there is a smaller encoding, which contradicts information theory. We will encode verified sums in the right subtree with which we can recover the permutations in the left subtree. This is done by encoding the \emph{query} permutations on the right. However, what we assume to know is a little different this time. We still assume we know everything before the left subtree, but we can not assume to know what we are passing into the queries on the right, because this is exactly what we are trying to figure out! In fact, these two things convey the same information, as both is reconstrucable from the other. However, this time we know that the queries, which asks if a composition prefix of the permutations is the same as the given permutation, always answer ``Yes''. \todo{how does this work??} Short version: We end up also encoding a separator of the sets $R\setminus W$ and $W\setminus R$, where $R$ and $W$ are the cells read and writte to in the right and left subtree respectively. Then we can simulate all possible input permutations and use the separator to quit early if we see that the cell accessed is not right. The encoding size ends up begin $\Omega(l \sqrt{n} \log n)$, which implies that $|R| + |W| = \Omega(l \sqrt{n} \log n)$, which was the claim we needed to prove. Dynamic partial sums, dynamic connectivity. \chapter{Integer Structures}