Approximate sorting of data streams with limited storage

Abstract

We consider the problem of approximate sorting of a data stream (in one pass) with limited internal storage where the goal is not to rearrange data but to output a permutation that reflects the ordering of the elements of the data stream as closely as possible. Our main objective is to study the relationship between the quality of the sorting and the amount of available storage. To measure quality, we use permutation distortion metrics, namely the Kendall tau, Chebyshev, and weighted Kendall metrics, as well as mutual information, between the output permutation and the true ordering of data elements. We provide bounds on the performance of algorithms with limited storage and present a simple algorithm that asymptotically requires a constant factor as much storage as an optimal algorithm in terms of mutual information and average Kendall tau distortion. We also study the case in which only information about the most recent elements of the stream is available. This setting has applications to learning user preference rankings in services such as Netflix, where items are presented to the user one at a time.

Appendix

1.1 Proof of (10)

Lemma 6

We have

$$\begin{aligned} \sum _{k=1}^{n-m+1}\left( {\begin{array}{c}n-k-1\\ m-2\end{array}}\right) k\ln k\le \left( {\begin{array}{c}n\\ m\end{array}}\right) \left( H_{n}-H_{m}+1-\frac{m}{n}\right) . \end{aligned}$$

Proof

To prove the upper bound on $\sum _{k=2}^{n-m+1}\left( {\begin{array}{c}n-k-1\\ m-2\end{array}}\right) k\lg k$, we use Abel’s identity Apostol (1976, Theorem 4.2), which states that for an arithmetic function a, a real number x, and a function f with a continuous derivative on $\left[ 1,x\right]$, we have

$$\begin{aligned} \sum _{1\le k\le x}a\left( k\right) f\left( k\right) =A\left( x\right) f\left( x\right) -\int _{1}^{x}A\left( y\right) f'\left( y\right) dy, \end{aligned}$$

where $A\left( y\right) =\sum _{1\le k\le y}a\left( k\right)$ for $y\in {\mathbb {R}}$. To use Abel’s identity, we let $a\left( k\right) =k\left( {\begin{array}{c}n-k-1\\ m-2\end{array}}\right)$ and $f\left( k\right) =\ln k$. Hence,

$$\begin{aligned} A\left( y\right)&=\sum _{1\le k\le y}k\left( {\begin{array}{c}n-k-1\\ m-2\end{array}}\right) \\&=\left( {\begin{array}{c}n\\ m\end{array}}\right) -\frac{(\left\lfloor y\right\rfloor (m-1)+n)}{m}\left( {\begin{array}{c}n-\left\lfloor y\right\rfloor -1\\ m-1\end{array}}\right) , \end{aligned}$$

for $y\ge 1$ and $A\left( y\right) =0$ for $y<1$. By Abel’s identity, we have

$$\begin{aligned} \sum _{k=1}^{n-m+1}\left( {\begin{array}{c}n-k-1\\ m-2\end{array}}\right) k\ln k=A(n-m+1)f(n-m+1)-\int _{y=1}^{n-m+1}A(y)f'(y)dy. \end{aligned}$$

The first term on the right side equals $\left( {\begin{array}{c}n\\ m\end{array}}\right) \ln \left( n-m+1\right)$ and for the second term, we find

$$\begin{aligned}&\int _{y=1}^{n-m+1}A(y)f'(y)dy\\&\quad =\int _{y=1}^{n-m+1}\left( \left( {\begin{array}{c}n\\ m\end{array}}\right) -\frac{(\left\lfloor y\right\rfloor (m-1)+n)}{m}\left( {\begin{array}{c}n-\left\lfloor y\right\rfloor -1\\ m-1\end{array}}\right) \right) \frac{1}{y}dy\\&\quad =\left( {\begin{array}{c}n\\ m\end{array}}\right) \ln \left( n-m+1\right) -\int _{y=1}^{n-m+1} \frac{(\left\lfloor y\right\rfloor (m-1)+n)}{m}\left( {\begin{array}{c}n-\left\lfloor y\right\rfloor -1\\ m-1\end{array}}\right) \frac{1}{y}dy. \end{aligned}$$

Hence,

$$\begin{aligned} \sum _{k=1}^{n-m+1}\left( {\begin{array}{c}n-k-1\\ m-2\end{array}}\right) k\ln k=\int _{y=1}^{n-m+1}\frac{(\left\lfloor y\right\rfloor (m-1)+n)}{m}\left( {\begin{array}{c}n-\left\lfloor y\right\rfloor -1\\ m-1\end{array}}\right) \frac{1}{y}dy. \end{aligned}$$

We proceed as follows:

$$\begin{aligned}&\sum _{k=1}^{n-m+1} \left( {\begin{array}{c}n-k-1\\ m-2\end{array}}\right) k\ln k\\&\quad \le \int _{y=1}^{n-m+1}\frac{(\left\lfloor y\right\rfloor (m-1)+n)}{m}\left( {\begin{array}{c}n-\left\lfloor y\right\rfloor -1\\ m-1\end{array}}\right) \frac{1}{\left\lfloor y\right\rfloor }dy\\&\quad =\sum _{k=1}^{n-m}\frac{k(m-1)+n}{m}\left( {\begin{array}{c}n-k-1\\ m-1\end{array}}\right) \frac{1}{k}\\&\quad =\frac{m-1}{m}\sum _{k=1}^{n-m}\left( {\begin{array}{c}n-k-1\\ m-1\end{array}}\right) +\frac{n}{m}\sum _{k=1}^{n-m}\left( {\begin{array}{c}n-k-1\\ m-1\end{array}}\right) \frac{1}{k}\\&\quad =\frac{m-1}{m}\left( {\begin{array}{c}n-1\\ m\end{array}}\right) +\frac{n}{m}\left( {\begin{array}{c}n-1\\ m-1\end{array}}\right) \left( H_{n-1}-H_{m-1}\right) \\&\quad =\frac{m-1}{m}\left( {\begin{array}{c}n-1\\ m\end{array}}\right) +\left( {\begin{array}{c}n\\ m\end{array}}\right) \left( H_{n}-H_{m}\right) +\frac{1}{m}\left( {\begin{array}{c}n-1\\ m-1\end{array}}\right) \left( \frac{n-m}{m}\right) \\&\quad =\left( {\begin{array}{c}n-1\\ m\end{array}}\right) +\left( {\begin{array}{c}n\\ m\end{array}}\right) \left( H_{n}-H_{m}\right) = \left( {\begin{array}{c}n\\ m\end{array}}\right) \left( H_{n}-H_{m}+1-\frac{m}{n}\right) , \end{aligned}$$

where we have used the fact that for nonnegative integers $\ell ,j$, we have

$$\begin{aligned} \sum _{i=1}^{\ell -j}\left( {\begin{array}{c}\ell -i\\ j\end{array}}\right) \frac{1}{i}= \left( {\begin{array}{c}\ell \\ j\end{array}}\right) \left( H_{\ell }-H_{j}\right) , \end{aligned}$$

proved below, to obtain the third equality.

To prove

$$\begin{aligned} \sum _{i=1}^{\ell -j}\left( {\begin{array}{c}\ell -i\\ j\end{array}}\right) \frac{1}{i}= \left( {\begin{array}{c}\ell \\ j\end{array}}\right) \left( H_{\ell }-H_{j}\right) \end{aligned}$$

let us write it as

$$\begin{aligned} \sum _{i=j+1}^{\ell }\frac{\left( \ell -i+j\right) !}{\left( \ell -i\right) !} \frac{1}{i-j}=\frac{\ell !}{\left( \ell -j\right) !}\sum _{i=j+1}^{\ell }\frac{1}{i}\cdot \end{aligned}$$

(19)

The proof is by induction. The equality (19) holds for $j=0$ as both sides reduce to $\sum _{i=1}^{\ell }\frac{1}{i}$. As induction hypothesis, suppose (19) holds for a certain value of j. We show that it also holds for $j+1$. We have

$$\begin{aligned}&\sum _{i=j+2}^{\ell } \frac{\left( \ell -i+j+1\right) !}{\left( \ell -i\right) !}\frac{1}{i-j-1}\\&\quad =\sum _{i=j+1}^{\ell -1}\frac{\left( \ell -i+j\right) !}{\left( \ell -i-1 \right) !}\frac{1}{i-j}\\&\quad =\sum _{i=j+1}^{\ell }\frac{\left( \ell -i\right) \left( \ell -i+j \right) !}{\left( \ell -i\right) !}\frac{1}{i-j}\\&\quad =\sum _{i=j+1}^{\ell }\frac{\left( \ell -j\right) \left( \ell -i+j \right) !}{\left( \ell -i\right) !}\frac{1}{i-j}-\sum _{i=j+1}^{\ell } \frac{\left( i-j\right) \left( \ell -i+j\right) !}{\left( \ell -i\right) !}\frac{1}{i-j}\\&\quad =\left( \ell -j\right) \sum _{i=j+1}^{\ell }\frac{\left( \ell -i+j \right) !}{\left( \ell -i\right) !}\frac{1}{i-j}-\sum _{i=j+1}^{\ell } \frac{\left( \ell -i+j\right) !}{\left( \ell -i\right) !}\\&\quad \mathop {=}\limits ^{\mathsf {(a)}}\left( \ell -j\right) \frac{\ell !}{\left( \ell -j\right) !} \sum _{i=j+1}^{\ell }\frac{1}{i}-j!\sum _{i=j+1}^{\ell } \left( {\begin{array}{c}\ell -i+j\\ j\end{array}}\right) \\&\quad =\frac{\ell !}{\left( \ell -j-1\right) !}\sum _{i=j+1}^{\ell } \frac{1}{i}-j!\sum _{i=j}^{\ell -1}\left( {\begin{array}{c}i\\ j\end{array}}\right) \\&\quad =\frac{\ell !}{\left( \ell -j-1\right) !}\sum _{i=j+1}^{\ell } \frac{1}{i}-\frac{\ell !}{(j+1)\left( \ell -j-1\right) !}\\&\quad =\frac{\ell !}{\left( \ell -j-1\right) !}\sum _{i=j+2}^{\ell }\frac{1}{i}, \end{aligned}$$

where for $\mathsf {(a)}$ we have used the induction hypothesis. $\square$

1.2 Proof of (14)

In this subsection, we prove (14) as follows

$$\begin{aligned}&P \left( D_{u,v}\left( X,Y_{1}''\right) \right) \\&\quad \mathop {=}\limits ^{\mathsf {(a)}}\frac{1}{t}\sum _{a=0}^{ \left\lfloor (3t-2)/4\right\rfloor }\frac{\left( {\begin{array}{c}a+2\\ 2\end{array}}\right) +\left( {\begin{array}{c}t-a+1\\ 2\end{array}}\right) }{2\left( {\begin{array}{c}t+2\\ 2\end{array}}\right) }+ \frac{1}{t}\sum _{a=\left\lfloor (3t-2)/4\right\rfloor +1}^{t-1}\frac{\left( {\begin{array}{c}a+2\\ 3\end{array}}\right) +\left( {\begin{array}{c}t+2\\ 3\end{array}}\right) -\left( {\begin{array}{c}t-a+2\\ 3\end{array}}\right) }{2a\left( {\begin{array}{c}t+2\\ 2\end{array}}\right) }\\&\quad =\frac{1}{t^{3}+O\left( t^{2}\right) }\biggl [\sum _{a=0}^{\left\lfloor (3t-2)/4\right\rfloor }\left( \left( {\begin{array}{c}a+2\\ 2\end{array}}\right) +\left( {\begin{array}{c}t-a+1\\ 2\end{array}}\right) \right) \\&\qquad +\, \sum _{a=\left\lfloor (3t-2)/4\right\rfloor +1}^{t-1}\frac{1}{a}\left( \left( {\begin{array}{c}a+2\\ 3\end{array}}\right) +\left( {\begin{array}{c}t+2\\ 3\end{array}}\right) -\left( {\begin{array}{c}t-a+2\\ 3\end{array}}\right) \right) \biggl ]\\&\quad =\frac{1}{t^{3}+O\left( t^{2}\right) }\biggl [\frac{\left( 3t/4 \right) ^{3}}{6}+\frac{t^{3}-\left( t/4\right) ^{3}}{6}+\frac{t^{3}-\left( 3t/4\right) ^{3}}{18}+O\left( t^{2}\right) \\&\qquad +\, \sum _{a=\left\lfloor (3t-2)/4\right\rfloor +1}^{t-1} \frac{t^{3}-\left( t-a\right) ^{3}+O\left( t^{2}\right) }{6a}\biggl ]\\&\quad =\frac{1}{1+O\left( t^{-1}\right) }\left( \frac{307}{1152}+\frac{1}{6} \sum _{\left\lfloor (3t-2)/4\right\rfloor +1}^{t-1}\left( \frac{3}{t} -\frac{3a}{t^{2}}+\frac{a^{2}}{t^{3}}+O\left( \frac{1}{at}\right) \right) \right) \\&\quad =\frac{1}{1+O\left( t^{-1}\right) }\left( \frac{307}{1152} +\frac{1}{6}\left( \frac{3}{4}-\frac{21}{32}+\frac{37}{192} +O\left( \frac{1}{t}\right) \right) \right) \\&\quad =\frac{181}{576}+o(1), \end{aligned}$$

where $\mathsf {(a)}$ follows from (13) and where we have used $\sum _{a=0}^{k+O\left( 1\right) }\left( {\begin{array}{c}a+O\left( 1\right) \\ 2\end{array}}\right) =\frac{k^{3}}{6}+O\left( k^{2}\right)$ and $\frac{1}{a}\left( {\begin{array}{c}a+2\\ 3\end{array}}\right) =\frac{1}{3}\left( {\begin{array}{c}a+2\\ 2\end{array}}\right)$.