# Mathematics of the impossible, Chapter 12, Data structures

And as promised, here’s a chapter on data structures… in a complexity theory course!

Data structures aim to maintain data in memory so as to be able to support various operations, such as answering queries about the data, and updating the data. The study of data structures is fundamental and extensive. We distinguish and study in turn two types of data structure problems: static and dynamic. In the former the input is given once and cannot modified by the queries. In the latter queries can modify the input; this includes classical problems such as supporting insert, search, and delete of keys.

### 12.1 Static data structures

Definition 12.1. A static data-structure problem is simply a function $f:\{0,1\}^n \to [q]^{m}$. A data structure for $f$ with space $s$, word size $w$ and time $t$ is a way to write $f$ as

\begin{aligned} f(x)=h(g(x)) \end{aligned}

where $g:\{0,1\} ^{n}\to [2^{w}]^{s}$, $h:[2^{w}]^{s}\to [q]^{m}$, and each output bit of $h$ depends on $\le t$ input words (we think of $s$ as divided into words of length $w$).

Here we have $n$ bits of input data about which we would like to answer $m$ queries. Often the queries and or the word size are boolean, i.e., $2^{w}=q=2$. Another typical setting is $q=2^{w}$. The data structure aims to accomplish this by storing the input into $s$ bits of memory. This map is arbitrary, with no bound on resources. But after that, each query can be answered very fast, by reading only $t$ words. In general, these words can be read adaptively. But for simplicity we focus on the case in which the locations are fixed by the data structure and the same for every input $x\in \{0,1\} ^{n}$.

Exercise 12.1. Consider the data structure problem $f:\{0,1\} ^{n}\to \{0,1\} ^{m}$ where $m=n^{2}$ and query $(i,j)\in \{1,2,\ldots ,n\}^{2}$ is the parity of the input bits from $i$ to $j$.

Give a data structure for this problem with $s=n$, $w=1$, and $t=2$.

Exercise 12.2. Show that any data-structure problem $f:\{0,1\} ^{n}\to \{0,1\} ^{m}$ has a data structure with $w=1$ and the following parameters:

(1) $s=m$ and $t=1$, and

(2) $s=n$ and $t=n$.

Exercise 12.3. Prove that for every $n$ and $m\le 2^{n/2}$ there exist functions $f:\{0,1\} ^{n}\to \{0,1\} ^{m}$ s.t. any data structure with space $s=m/2$ and $w=1$ requires time $t\ge n-c$.

By contrast, next we present the the best known impossibility result.

Definition 12.2. A function $f:[2]^{n}\to [q]^{m}$ is $d$-wise uniform if any $d$ output coordinates are uniform when the input to $f$ is uniform.

Theorem 12.1. [65] Let $f:[q]^{d}\to [q]^{q}$ be $d$-wise uniform. Let $q$ be a power of $2$ and $c\log q\le d\le q^{c}$. Then any data structure with $w=\log q$ using space $s$ (which recall is measured in words of $w$ bits) and time $t$ has:

\begin{aligned} t\geq c\frac {\log q}{\log (s/d)}. \end{aligned}

Interpreting the input as coefficients of a degree $d-1$ univariate polynomial over $\mathbb {F}_{q}$ and outputting its evaluations shows that such functions exists, and are in P. Below we give a surprising data structure that nearly matches the theorem.

To match previous parameters note that $n=d\log q=dw$, and $m=\log q$. Hence the bound is $t\ge c(\log m)/\log (sw/n)$. Note that $sw$ is the space of the data structure measured in bits. It follows that if $sw$ is linear in $n$ then $t\ge c\log m$. This result remains non-trivial for $s$ slightly super-linear. But remarkably, if $sw=n^{1+c}$ then nothing is known (for $m$ power in $n$ one only gets $t\ge c$).

ProofThe idea in the proof is to find a subset $S$ of less than $d$ memory cells that still allows us to answer $\ge d$ queries. This is impossible, since we can’t generate $d$ uniform outputs from less than $d$ memory cells.

Let $p:=1/q^{1/4t}$. Include each memory bit in $S$ with probability $p$, independently. By Theorem 2.7, $\mathbb {P}[|S|\ge cps]\le 2^{-cps}$.

We are still able to answer a query if all of its memory bits fall in $S$. The probability that this happens is $p^{t}=1/q^{1/4}$. We now claim that with probability $\ge 1/q^{c}$, we can still answer $\sqrt {q}$ queries. Indeed, let $B$ be the number of queries we cannot answer. We have $\mathbb {E}[|B|]\le q(1-q^{1/4})$. And so

\begin{aligned} \mathbb {P}[B\ge q(1-1/\sqrt {q})]\le \frac {1-q^{1/4}}{1-\sqrt {q}}\le 1-q^{c}. \end{aligned}

Thus, if the inequality $2^{-cps}\leq 1/q^{c}$ holds then there exists a set $S$ of $cps$ bits with which we can answer $\ge \sqrt {q}>d$ queries. Hence we reach a contradiction if

\begin{aligned} c\log q\le cps="" Because $d>c\log q$ by assumption, and increasing $s$ only make the problem easier, we reach a contradiction if $cps, and="" the="" result="" follows.="" QED

Next we show a conceptually simple data structure which nearly matches the lower bound. For simplicity we focus on data structures which use space $q^{\epsilon }$ – recall in this case the previous result does not give anything. We will show this is for good reasons, there are data structures where the time is constant. We will only show $c_{\epsilon }d$-wise independence, as opposed to $d$-wise, but the proof techniques next and above generalize to other settings of parameters.

Theorem 12.2. There is a map $f:[q]^{d}\to [q]^{q}$ which is $c_{\epsilon }d$-wise uniform and has a data structure with $w=\log q$ space $s=dq^{\epsilon }$ and time $c_{\epsilon }$, for any $\epsilon$ and $q$ which is a power of $2$.

To give a sense of the parameters, let for example $q=d^{10}$.

ProofWe fill the memory with $s$ evaluations of the input polynomial. Then we pick a random bipartite graph with $s$ nodes on the left and $q$ nodes on the right. Every node on the right side has degree $g$. We answer each query by summing the corresponding cells in $s$. Let $d':=d/g$. To show $d'$-wise uniformity it suffices to show that for any subset $R\subseteq [q]$ on the right-hand side of size $d'$, the sum of the corresponding memory cells is uniform in $\mathbb {F}_{q}$. For this in turn it suffices that $R$ has a unique neighbor. And for that, finally, it suffices that $R$ has a neighborhood of size greater than $\frac {g|R|}{2}$ (because if every element in the neighborhood of $R$ has two neighbors in $R$ then $R$ has a neighborhood of size less than $g|r| 2$).

Note here we are using that the neighborhood has size $\le gd'=d$, and so the memory is $d$-wise uniform.

We pick the graph at random and show that it has the latter property with non-zero probability.
[Standard calculations follow that wordpress has trouble displaying… wait for the book draft, I guess.]
QED

#### 12.1.1 Succinct data structures

Succinct data structures are those where the space is close to the minimum, $n$. Specifically, we let $s=n+r$ for some $r=o(n)$ called redundancy. Unsurprisingly, stronger bounds can be probed in this setting. But, surprisingly, again these stronger bounds were shown to be tight. Moreover, it was shown that improving the bounds would imply stronger circuit lower bounds.

To illustrate, consider the ECC problem $f:\{0,1\} ^{n}\to \{0,1\} ^{m}$ where $f$ is an error-correcting code (with constant relative distance) and $m$ is linear in $n$.

Theorem 12.3. [26] Any data-structure for the ECC problem with $w=1$ using space $n+r$ requires time $\ge cn/r$.

This is nearly matched by the following result.

Theorem 12.4. [81] There is an ECC problem s.t. for any $r$ it has a data structure with $w=1$, space $n+r$, and time $c(n/r)\log ^{3}n$.

Moreover, it was shown that proving a time lower bound of $(n/r)\log ^{c}n$ would imply new circuit lower bounds. The latter result refers to bounds on the number of wires in circuits with arbitrary gates. But the following connection with the standard circuit model is also known.

Theorem 12.5. [81] Let $f:\{0,1\} ^{n}\to \{0,1\} ^{am}$ be a function computable by bounded fan-in circuits with $bm$ wires and depth $b\log m$, for constants $a,b$. Then $f$ has a data structure with space $n+o(n)$ and time $n^{o(1)}$.

Hence, proving $n^{\epsilon }$ time lower bounds for succinct data structures would give functions that cannot be computed by linear-size log-depth circuits, cf. 9.3.

#### 12.1.2 Succincter: The trits problem

In this section we present a cute and fundamental data-structure problem with a shocking and counterintuitive solution. The trits problem is to compute $f:[3]^{n}\to (\{0,1\} ^{2})^{n}$ where on input $n$ “trits” (i.e., ternary elements) $(t_{1},t_{2},\ldots ,t_{n})\in [3]^{n}$ $f$ outputs their representations using two bits per trit.

Example 12.1. For $n=1$, we have $f(0)=00,f(1)=01,f(2)=10$.

Note that the input ranges over $3^{n}$ elements, and so the minimum space of the data structure is $s=\lceil \log _{2}3^{n}\rceil =\lceil n\log _{2}3\rceil \approx n\cdot 1.584\ldots$ This will be our benchmark for space. One can encode the input to $f$ as before using bits without loss of generality, but the current choice simplifies the exposition.

##### Simple solutions:
• The simplest solution (cf. 12.2) to this problem is to use $2$ bits per $t_{i}$. With such an encoding we can retrieve each $t_{i}\in [3]$ by reading just $2$ bits (which is optimal). The space used is $s=2n$ and we have linear redundancy.
• Another solution (cf. again 12.2) to this problem is what is called arithmetic coding: we think of the concatenated elements as forming a ternary number between $0$ and $3^{n}-1$, and we write down its binary representation. To retrieve $t_{i}$ it seems we need to read all the input bits, but the space needed is optimal.
• For this and other problems, we can trade between these two extreme as follows. Group the $t_{i}$’s into blocks of $t$. Encode each block with arithmetic coding. The retrieval time will be $ct$ bits and the needed space will be $(n/t)\lceil t\log _{2}3\rceil \leq n\log _{2}3+n/t$ (assuming $t$ divides $n$). This is block-wise arithmetic coding. It provides a power trade-off between retrieval time and redundancy. (Using number-theoretic results on logarithmic forms, one can show [80] that this last inequality is tight up to changing $n/t$ into $n/t^{c}$.)

##### The shocking solution: An exponential (!) trade-off

We now present an exponential trade-off: retrieval time $ct$ bits and redundancy $n/2^{t}+c$. In particular, if we set $t=c\log n$, we get retrieval time $O(\log n)$ and redundancy $O(1)$. Moreover, the bits read are all consecutive, so with word size $w=\log n$ this can be implemented in constant time. To repeat, we can encode the trits with constant redundancy and retrieve each in constant time. This solution can also be made dynamic.

Theorem 12.6. [5321] The trits problem has a data structure with space $n\log _{2}3+n/2^{t}+c$ (i.e., redundancy $n/2^{t}+c$) and time $ct$, for any $t$ and with word size $w=1$. For word wise $w=\log n$ the time is constant.

Next we present the proof.

Definition 12.3 (Encoding and redundancy). An encoding of a set $A$ into a set $B$ is a one-to-one (a.k.a. injective) map $f:A\to B$. The redundancy of the encoding $f$ is $\log _{2}|B|-\log _{2}|A$|.

The following lemma gives the building-block encoding we will use.

Lemma 12.1. For all sets $X$ and $Y$, there is an integer $b$, a set $K$ and an encoding

\begin{aligned} f:\left (X\times Y\right )\rightarrow \left (\{0,1\} ^{b}\times K\right ) \end{aligned}

such that (1) $f$ has redundancy $\le c/\sqrt {|Y|}$, and (2) $x\in X$ can be recovered just by reading the $b$ bits in $f(x,y)$.

Note that (1) says that $b+\log |K|-\log |X|-\log |Y|\le c/\sqrt {|Y|}$. For (2) to hold we must have $b\ge \log |X|$. Combining this with the previous expression we obtain $\log |K|-\log |Y|\le c/\sqrt {|Y|}$. In particular we get that $|K|\le 2^{c}\cdot |Y|$ (in fact it will be the case that $|K|\le c\cdot \sqrt {|Y|}$, but the looser bound is sufficient).

The basic idea for proving the lemma is to break $Y$ into $C\times K$ and then encode $X\times C$ by using $b$ bits:

\begin{aligned} X\times Y\rightarrow X\times C\times K\rightarrow \{0,1\} ^{b}\times K. \end{aligned}

There is however a subtle point. If we insist on always having $|C|$ equal to, say, $\sqrt {|Y|}$ or some other quantity, then one can cook up sets that make us waste a lot (i.e., almost one bit) of space. The same of course happens in the more basic approach that just sets $Y=K$ and encodes all of $X$ with $b$ bits. The main idea will be to “reason backwards,” i.e., we will first pick $b$ and then try to stuff as much as possible inside $\{0,1\} ^{b}$. Still, our choice of $b$ will make $|C|$ about $\sqrt {|Y|}$.

ProofPick any two sets $X$ and $Y$, where $|Y|>1$ without loss of generality. Define $b:=\left \lceil \log _{2}\left (|X|\cdot \sqrt {|Y|}\right )\right \rceil$, and let $B:=\{0,1\} ^{b}$. To simplify notation, define $d:=2^{b}/|X|$. Note $c\sqrt {|Y|}\le d\le c\sqrt {|Y|}.$

How much can we stuff into $B$? For a set $C$ of size $|C|=\left \lfloor |B|/|X|\right \rfloor$, we can encode elements from $X\times C$ in $B$. The redundancy of such an encoding can be bounded as follows:

To calculate the total redundancy, we still need to examine the encoding from $Y$ to $C\times K$. Choose $K$ of size $|K|=\left \lceil |Y|/|C|\right \rceil$, so that this encoding is possible. With a calculation similar to the previous one, we see that the redundancy is:

The total redundancy is then $c/\sqrt {|Y|}$, which gives (1).

For (2), it is clear from the construction that any $x\in X$ can be recovered from the element of $B$ only. QED

Proof of Theorem 12.6 Break the ternary elements into blocks of size $t$: $(t'_{1},t'_{2},\ldots ,t'_{n/t})\in T_{1}\times T_{2}\times \ldots \times T_{n/t}$, where $T_{i}=[3]^{t}$ for all $i$. The encoding, illustrated in Figure 1, is constructed as follows, where we use $f_{L}$ to refer to the encoding guaranteed by Lemma Lemma 12.1.

Compute $f_{L}(t'_{1},t'_{2})=(b_{1},k_{1})\in B_{1}\times K_{1}$.

For $i=2,\ldots ,n/t-1$ compute $f_{L}(k_{i-1},t'_{i+1}):=(b_{i},k_{i})\in B_{i}\times K_{i}$.

Encode $k_{n/t-1}$ in binary as $b_{n/t}$ using arithmetic coding.

The final encoding is $(b_{1},b_{2},\ldots ,b_{n/t})$. We now compute the redundancy and retrieval time.

Redundancy: From (1) in Lemma 12.1, the first $n/t-1$ encodings have redundancy $c3^{-t/2}\le 1/2^{ct}$. For the last (arithmetic) encoding, the redundancy is at most $1$. So the total redundancy is at most $\displaystyle \left (\frac {n}{t}-1\right )\cdot \frac {1}{2^{ct}}+1=\frac {n}{2^{ct}}+c$. One can visualize this as a “hybrid argument” transforming a product of blocks of ternary elements into a product of blocks of binary elements, one block at the time.

Retrieval Time: Say that we want to recover some $t_{j}$ which is in block $t'_{i}$. To recover block $t'_{i}$, Lemma 12.1 guarantees that we only need to look at $b_{i-1}$ and $b_{i}$. This is because $k_{i-1}$ can be recovered by reading only $b_{i}$, and $t'_{i}$ can be recovered by reading $k_{i-1}$ and $b_{i-1}$. Thus to complete the proof it suffices to show that each $b_{i}$ has length $ct$.

This is not completely obvious because one might have thought that the $K_{i}$ become larger and larger, and so we apply the lemma to larger and larger inputs and the $B_{i}$ get large too. However, recall that each $|K_{i}|\le c|T_{i}|=c3^{t}$ from the comment after the statement of Lemma 12.1. Hence, every time we apply the lemma on an input of size at most $s\le 3^{ct}$. Since the lemma wastes little entropy (by (1) in Lemma 12.1), none of its outputs can be much larger than its input, and so $|B_{i}|=2^{ct}$.

QED

### 12.2 Dynamic data structures

We now study dynamic data structures. As we mentioned, here the input is not fixed but can be modified by the queries.

Definition 12.4. Fix an error-correcting code $\text {ECC}:\{0,1\} ^{n}\to \{0,1\} ^{m}$ where $m\le cn$ and $\Delta (\text {ECC}(x),\text {ECC}(y))\ge c$ for any $x\ne y$ in $\{0,1\} ^{n}$. Here $\Delta (u,v)$ is the relative distance, the fraction of bit positions where $u$ and $v$ differ.

The $\text {ECC}$ problem asks to support operations, starting with the all-zero message:

$M(i,b)$ for $i\in \{1,2,\ldots ,n\}$ and $b\in \{0,1\}$ which sets bit $i$ of the message to $b$, and

$C(i)$ for $i\in \{1,2,\ldots ,m\}$ which returns bit $i$ of the codeword corresponding to the current message.

The time of a dynamic data structure is the maximum number of read/write operations in memory cells required to support an operation.

Theorem 12.7. The $\text {ECC}$ problem requires time $t\ge c\log _{w}n\ge (c\log n)/\log \log n$ for cell size $w:=\log n$.

One might wonder if stronger bounds can be shown for this problem. But in fact there exist codes for which the bounds are nearly tight.

Theorem 12.8. [81]There exists codes for which the ECC problem can be solved in time $c\log ^{2}n$ with cell size $w=1$.

The technique in the proof of Theorem 12.7 is from [24] and can be applied to many other natural problems, leading to tight results in several cases, see Exercise ??. It is not far from the state-of-the art in this area, which is $\log ^{1+c}n$ [42].

Proof of Theorem 12.7 Pick $x\in \{0,1\} ^{n}$ uniformly and $i\in \{1,2,\ldots ,m\}$ uniformly, and consider the sequence of operations

\begin{aligned} M(1,x_{1}),M(2,x_{2}),\ldots ,M(n,x_{n}),C(i). \end{aligned}

That is, we set the message to a uniform $x$ one bit at a time, and then ask for a uniformly selected bit of the codeword $\text {ECC}(x)$, which we also denote by $C_{x}=C_{x}(1),C_{x}(2),\ldots ,C_{x}(n)$.

We divide the $n$ operations $M(i,x_{i})$ into consecutive blocks, called epochs. Epoch $e$ consists of $n/w^{3e}$ operations. Hence we can have at least $E:=c\log _{w}n$ epochs, and we can assume that we have exactly this many epochs (by discarding some bits of $n$ if necessary).

The geometrically decaying size of epochs is chosen so that the number of message bits set during an epoch $e$ is much more than all the cells written by the data structure in future epochs.

A key idea of the proof is to see what happens when the cells written during a certain epoch are ignored, or reverted to their contents right before the epoch. Specifically, after the execution of the $M$ operations, we can associate to each memory cell the last epoch during which this cell was written. Let $D^{e}(x)$ denote the memory cells of the data structure after the first $n$ operations $M$, but with the change that the cells that were written last during epoch $e$ are replaced with their contents right before epoch $e$. Define $C_{x}^{e}(i)$ to be the result of the data structure algorithm for $C(i)$ on $D^{e}(x)$, and $C_{x}^{e}=C_{x}^{e}(1),C_{x}^{e}(2),\ldots ,C_{x}^{e}(n)$.

Let $t(x,i,e)$ equal $1$ if $C(i)$, executed after the first $n$ operations $M$, reads a cell that was last written in epoch $e$, and $0$ otherwise. We have

\begin{aligned} t\ge \max _{x,i}\sum _{e}t(x,i,e)\ge \mathbb {E}_{x,i}\sum _{e}t(x,i,e)=\sum _{e}\mathbb {E}_{x,i}t(x,i,e)\ge \sum _{e}\mathbb {E}_{x}\Delta (C_{x},C_{x}^{e}),~~~~(12.1) \end{aligned}

where the last inequality holds because $C_{x}^{e}(i)\ne C_{x}(i)$ implies $t(x,i,e)\ge 1$.

We now claim that if $t\le w$ then $\mathbb {E}_{x}\Delta (C_{x},C_{x}^{e})\ge c$ for every $e$. This concludes the proof.

In the remainder we justify the claim. Fix arbitrarily the bits of $x$ set before Epoch $e$. For a uniform setting of the remaining bits of $x$, note that the message ranges over at least

\begin{aligned} 2^{n/w^{3e}} \end{aligned}

codewords. On the other hand, we claim that $C_{x}^{e}$ ranges over much fewer strings. Indeed, the total number of cells written in all epochs after $e$ is at most

\begin{aligned} t\sum _{i\ge e+1}n/w^{3i}\le ctn/w^{3(e+1)}. \end{aligned}

We can describe all these cells by writing down their indices and contents using $B:=ctn/w^{3e+2}$ bits. Note that this information can depend on the operations performed during Epoch $e$, but the point is that it takes few possible values overall. Since the cells last changed during Epoch $e$ are reverted to their contents before Epoch $e$, this information suffices to describe $D^{e}(x)$, and hence $C_{x}^{e}$. Therefore, $C_{x}^{e}$ ranges over $\le 2^{B}$ strings.

For each string in the range of $C_{x}^{e}$ at most two codewords can have relative distance $\le c$, for else you’d have two codewords at distance $\le 2c$, violating the distance of the code.

Hence except with probability $2\cdot 2^{B}/2^{n/w^{3e}}$ over $x$, we have $\Delta (C_{x},C_{x}^{e})\ge c$. If $t_{M}\le w$ then the first probability is $\le 0.1$, and so $\mathbb {E}_{x}\Delta (C_{x},C_{x}^{e})\ge c$, proving the claim. QED

Exercise 12.4. Explain how to conclude the proof given the claim.

### 12.3 Notes

The exposition of the trits problem is from [76].

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