Recursive PGFs for BSTs and DSTs

1 Binary Search Trees

Consider the R program:

{f <- function(x,V,k)}{{}{\ \ if(NROW(V)==0) k <- 0}{\ \ else {}{\ \ \ \ u <- V[1]}{\ \ \ \ if(x==u) k <- 1}{\ \ \ \ else if(x<u) c(x,V,k <- 1+f(x,V <- V[V<u],k))}{\ \ \ \ else c(x,V,k <- 1+f(x,V <- V[V>u],k)) }{\ \ \ \ }}{\ \ k}{}}

where $V$ is a random permutation on ${1, 3, 5, \dots, 2 n - 1}$ and $k$ is initially $0$ . To model successful searches, let $x$

be a random odd integer satisfying

1 \leq x \leq 2 n - 1

. To model unsuccessful searches, let

x

be a random even integer satisfying

0 \leq x \leq 2 n

. This scenario is exactly as described in [4]. It is assumed, of course, that

x

and

V

are drawn independently with uniform sampling. We begin with even

x

, because this case is simpler, followed by odd

x

1.1 Unsuccessful Search

The probability generating function for $K_{n}$ , given $n$ , obeys a recursion [5]

\begin{matrix} f_{n} (z) = \frac{2 z + n - 1}{n + 1} f_{n - 1} (z), & n \geq 2; \end{matrix}

f_{1} (z) = z .

Note that $f_{n} (1) = 1$ always. Differentiating with respect to $z$ :

f_{n}^{'} (z) = \frac{2}{n + 1} f_{n - 1} (z) + \frac{2 z + n - 1}{n + 1} f_{n - 1}^{'} (z)

we have first moment

E (K_{n}) = f_{n}^{'} (1) = \frac{2}{n + 1} + f_{n - 1}^{'} (1)

that is,

g_{n} = \frac{2}{n + 1} + g_{n - 1}

where $g_{k} = f_{k}^{'} (1)$ and $g_{1} = 1$ . Clearly $g_{2} = 5 / 3$ and $g_{3} = 13 / 6$ . Differentiating again:

f_{n}^{''} (z) = \frac{4}{n + 1} f_{n - 1}^{'} (z) + \frac{2 z + n - 1}{n + 1} f_{n - 1}^{''} (z)

we have second factorial moment

E (K_{n} (K_{n} - 1)) = f_{n}^{''} (1) = \frac{4}{n + 1} f_{n - 1}^{'} (1) + f_{n - 1}^{''} (1),

that is,

h_{n} = \frac{4}{n + 1} g_{n - 1} + h_{n - 1}

where $h_{k} = f_{k}^{''} (1)$ and $h_{1} = 0$ . Clearly $h_{2} = 4 / 3$ and $h_{3} = 3$

. Finally, we have variance

V (K_{n}) = h_{n} - g_{n}^{2} + g_{n}

which is $2 / 9$ when $n = 2$ and $17 / 36$ when $n = 3$ . From (more typical) harmonic number-based exact expressions, it can be proved that [2, 6, 7]

E (K_{n}) = 2 ln (n) + 2 (γ - 1) + \frac{3}{n} + o (\frac{1}{n}),

V (K_{n}) = 2 ln (n) + 2 (γ - \frac{π^{2}}{3} + 1) + \frac{7}{n} + o (\frac{1}{n})

as $n \to \infty$ .

1.2 Successful Search

The probability generating function for $K_{n}$ , given $n$ , obeys a recursion

\begin{matrix} n^{2} f_{n} (z) = (n - 1) (2 z + n - 1) f_{n - 1} (z) + z, & n \geq 2; \end{matrix}

f_{1} (z) = z .

Note that $f_{n} (1) = 1$ always. Differentiating with respect to $z$ :

n^{2} f_{n}^{'} (z) = 2 (n - 1) f_{n - 1} (z) + (n - 1) (2 z + n - 1) f_{n - 1}^{'} (z) + 1

we have first moment

E (K_{n}) = f_{n}^{'} (1) = \frac{2 (n - 1) + (n - 1) (n + 1) f_{n - 1}^{'} (1) + 1}{n^{2}}

that is,

g_{n} = \frac{(2 n - 1) + (n^{2} - 1) g_{n - 1}}{n^{2}}

where $g_{k} = f_{k}^{'} (1)$ and $g_{1} = 1$ . Clearly $g_{2} = 3 / 2$ and $g_{3} = 17 / 9$ . Differentiating again:

n^{2} f_{n}^{''} (z) = 4 (n - 1) f_{n - 1}^{'} (z) + (n - 1) (2 z + n - 1) f_{n - 1}^{''} (z)

we have second factorial moment

E (K_{n} (K_{n} - 1)) = f_{n}^{''} (1) = \frac{4 (n - 1) f_{n - 1}^{'} (1) + (n - 1) (n + 1) f_{n - 1}^{''} (1)}{n^{2}},

that is,

h_{n} = \frac{4 (n - 1) g_{n - 1} + (n^{2} - 1) h_{n - 1}}{n^{2}}

where $h_{k} = f_{k}^{''} (1)$ and $h_{1} = 0$ . Clearly $h_{2} = 1$ and $h_{3} = 20 / 9$ . Finally, we have variance $V (K_{n})$ which is $1 / 4$ when $n = 2$ and $44 / 81$ when $n = 3$ .

It can be proved that [2, 5, 6, 8]

E (K_{n}) = 2 ln (n) + (2 γ - 3) + \frac{2 ln (n)}{n} + \frac{2 γ + 1}{n} + o (\frac{1}{n}),

	$V (K_{n})$	$= 2 ln (n) + 2 (γ - \frac{π^{2}}{3} + 2) - \frac{4 ln (n)^{2}}{n} + \frac{2 (5 - 4 γ) ln (n)}{n}$
		$+ (5 + 10 γ - 4 γ^{2} - \frac{2 π^{2}}{3}) \frac{1}{n} + o (\frac{1}{n})$

as $n \to \infty$ .

1.3 Total Path Length

The total (internal) path length $L_{n}$ is the sum of $K_{n} - 1$ taken over all odd integers $x$ from $1$ to $2 n - 1$ . It is not surprising that calculations are more involved here than before. The probability generating function for $L_{n}$ , given $n$ , obeys a recursion [6]

\begin{matrix} f_{n} (z) = \frac{z^{n - 1}}{n} n - 1 \sum k = 0 f_{k} (z) f_{n - 1 - k} (z), & n \geq 1; \end{matrix}

f_{0} (z) = 1.

Note that $f_{n} (1) = 1$ always. Differentiating with respect to $z$ :

f_{n}^{'} (z) = \frac{(n - 1) z^{n - 2}}{n} n - 1 \sum k = 0 f_{k} (z) f_{n - 1 - k} (z) + \frac{z^{n - 1}}{n} n - 1 \sum k = 0 [f_{k}^{'} (z) f_{n - 1 - k} (z) + f_{k} (z) f_{n - 1 - k}^{'} (z)]

we have first moment

	$E (L_{n})$	$= f_{n}^{'} (1) = \frac{n - 1}{n} \cdot n + \frac{1}{n} n - 1 \sum k = 0 [f_{k}^{'} (1) + f_{n - 1 - k}^{'} (1)]$
		$= n - 1 + \frac{2}{n} n - 1 \sum k = 0 f_{k}^{'} (1),$

that is,

g_{n} = n - 1 + \frac{2}{n} n - 1 \sum k = 0 g_{k}

where $g_{k} = f_{k}^{'} (1)$ and $g_{0} = 0$ . Clearly $g_{1} = 0$ , $g_{2} = 1$ , $g_{3} = 8 / 3$ and $g_{4} = 29 / 6$ . Differentiating again:

	$f_{n}^{''} (z)$	$= \frac{(n - 1) (n - 2) z^{n - 3}}{n} n - 1 \sum k = 0 f_{k} (z) f_{n - 1 - k} (z) + \frac{2 (n - 1) z^{n - 2}}{n} n - 1 \sum k = 0 [f_{k}^{'} (z) f_{n - 1 - k} (z) + f_{k} (z) f_{n - 1 - k}^{'} (z)]$
		$+ \frac{z^{n - 1}}{n} n - 1 \sum k = 0 [f_{k}^{''} (z) f_{n - 1 - k} (z) + 2 f_{k}^{'} (z) f_{n - 1 - k}^{'} (z) + f_{k} (z) f_{n - 1 - k}^{''} (z)]$

we have second factorial moment

	$E (L_{n} (L_{n} - 1))$	$= f_{n}^{''} (1)$
		$= (n - 1) (n - 2) + 2 (n - 1) [f_{n}^{'} (1) - n + 1] + \frac{2}{n} n - 1 \sum k = 0 f_{k}^{'} (1) f_{n - 1 - k}^{'} (1) + \frac{2}{n} n - 1 \sum k = 0 f_{k}^{''} (1),$

that is,

h_{n} = - (n - 1) n + 2 (n - 1) g_{n} + \frac{2}{n} n - 1 \sum k = 0 g_{k} g_{n - 1 - k} + \frac{2}{n} n - 1 \sum k = 0 h_{k}

where $h_{k} = f_{k}^{''} (1)$ and $h_{0} = 0$ . Clearly $h_{1} = 0$ , $h_{2} = 0$ , $h_{3} = 14 / 3$ and $h_{4} = 58 / 3$ . Finally, we have variance $V (L_{n})$ which is $2 / 9$ when $n = 3$ and $29 / 36$ when $n = 4$ .

It can be proved that [2, 5, 9]

E (L_{n}) = 2 n ln (n) + 2 (γ - 2) n + 2 ln (n) + (2 γ + 1) + o (1),

	$V (L_{n})$	$= (7 - \frac{2 π^{2}}{3}) n^{2} - 2 n ln (n) + (17 - 2 γ - \frac{4 π^{2}}{3}) n$
		$- 2 ln (n) + (5 - 2 γ - \frac{2 π^{2}}{3}) + o (1)$

as $n \to \infty$ .

1.4 Higher Moments

A third moment expression appears in [10] for successful search; analogous work for unsuccessful search remains undone. We focus on total (internal) path length $L_{n}$ for BSTs. The cumulants $κ_{2}$ , $κ_{3}$ , … , $κ_{8}$ of $L_{n}$ were exhaustively studied by Hennequin [11, 12]; these asymptotically satisfy

κ_{s} \sim [a_{s} + (- 1)^{s + 1} 2^{s} (s - 1)! ζ (s)] n^{s}

as $n \to \infty$ , where

{a_{s}}_{s}^{8} = {7, - 19, \frac{937}{9}, - \frac{85981}{108}, \frac{21096517}{2700}, - \frac{7527245453}{81000}, \frac{19281922400989}{14883750}} .

Hoffman & Kuba [13] obtained a complicated recurrence for an associated sequence of rationals [14, 15]:

{c_{s}}_{s}^{8} = {7, - 19, \frac{2260}{9}, - \frac{229621}{108}, \frac{74250517}{2700}, - \frac{30532750703}{81000}, \frac{90558126238639}{14883750}}

using what they called tiered binomial coefficients. While they utilized notation $(n, m)_{i}$ , we adopt $T (i, n, m)$ . It suffices to say that $T (0, n, m) = (\frac{n + m}{n})$ and a rich theory about $T (i, n, m)$ for $i \geq 1$ awaits discovery. We give Mathematica code for generating $c_{s}$ :

{f[i\_,x\_,y\_] := (1/(i+1-x-y)) (Binomial[i-x,i]/Binomial[i-x-y,i])}{T[i\_,n\_,m\_] := If[n+m > 0, Coefficient[Normal[\ \ }{\ Series[f[i,x,y], {x,0,n}, {y,0,m}]], x\char 94n y\char 94m], 1/(1+i)]}{c[s\_] := c[s] = ((s+1)/(s-1)) *}{\ Sum[Sum[Sum[If[k1+k2+k3 == s, Multinomial[k1,k2,k3] c[k1] c[k2] *}{\ \ Sum[Sum[Sum[If[n+m+p == k3,}{\ \ \ Sum[Multinomial[n,m,p] Binomial[m+k2,j] (-1)\char 94j (-2)\char 94(n+m) n! m! T[n+k1+j,n,m],}{\ \ \ {j,0,m+k2}], 0], }{\ \ {p,0,k3}], {m,0,k3}], {n,0,k3}], 0], }{\ {k3,0,s}], {k2,0,s-1}], {k1,0,s-1}] }{c[0] = 1;}{c[1] = 0;}

and code for generating $a_{s}$ , given $c_{1}$ , $c_{2}$ , … , $c_{s}$ :

{Sum[(-1)\char 94(j-1) (j-1)! BellY[s, j, Table[c[i], {i,1,s-j+1}]], {j,1,s}]}

This final line employs a well-known expression for cumulants in terms of partial (or incomplete) Bell polynomials of central moments.

2 Digital Search Trees

Consider the R program:

{f <- function(x,M,p,k)}{{\ \ }{\ \ q <- NCOL(M)}{\ \ if(NROW(M)==0) k <- 0}{\ \ else {}{\ \ \ \ if(all(x==matrix(M[1,],ncol=q))) k <- 1}{\ \ \ \ else {}{\ \ \ \ \ \ M <- matrix(M[-1,],ncol=q)}{\ \ \ \ \ \ M <- matrix(M[M[,p]==x[p],],ncol=q) }{\ \ \ \ \ \ k <- 1+f(x,M,p <- p+1,k)}{\ \ \ \ \ \ }}{\ \ \ \ }}{\ \ k}{}}

where $M$ is a random binary $n \times ℓ$ matrix with $n$ distinct rows, $p$ is initially $1$ and $k$ is initially $0$ . It is usually assumed [2, 16] that $ℓ = \infty$ , from which the row-distinctness requirement follows almost surely (imagining the rows as binary expansions of $n$ independent Uniform $[0, 1]$ numbers). If instead $ℓ = n$ , as exploratively specified in [17], then the matrix $M$ would need to be generated carefully to avoid duplicate keys. To model successful searches, let $x$ be a random row of $M$ . To model unsuccessful searches, let $x$ be a random binary $ℓ$

-vector that is not a row of

M

2.1 Unsuccessful Search

The probability generating function for $K_{n}$ , given $n$ , is

⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ \begin{matrix} \frac{1}{2} z + \frac{1}{2} z^{2} & if n = 2, \frac{1}{4} z + \frac{5}{8} z^{2} + \frac{1}{8} z^{3} & if n = 3, \frac{1}{8} z + \frac{19}{32} z^{2} + \frac{17}{64} z^{3} + \frac{1}{64} z^{4} & if n = 4, \frac{1}{16} z + \frac{65}{128} z^{2} + \frac{195}{512} z^{3} + \frac{49}{1024} z^{4} + \frac{1}{1024} z^{5} & if n = 5 \end{matrix}

for $ℓ = \infty$ and

⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ \begin{matrix} \frac{2}{3} z + \frac{1}{3} z^{2} & if n = 2, \frac{2}{7} z + \frac{2}{3} z^{2} + \frac{1}{21} z^{3} & if n = 3, \frac{8}{65} z + \frac{302}{455} z^{2} + \frac{22}{105} z^{3} + \frac{1}{273} z^{4} & if n = 4, \frac{52}{899} z + \frac{7384}{13485} z^{2} + \frac{34502}{94395} z^{3} + \frac{26}{} 899 z^{4} + \frac{1}{6293} z^{5} & if n = 5 \end{matrix}

for $ℓ = n$ . A closed-form expression exists [2] for $f_{n} (z)$ when $ℓ = \infty$ , but a corresponding simple recursive formula does not evidently materialize. Section 3 contains verification of these polynomial expressions.

2.2 Successful Search

The probability generating function for $K_{n}$ , given $n$ , is

⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ \begin{matrix} \frac{1}{2} z + \frac{1}{2} z^{2} & if n = 2, \frac{1}{3} z + \frac{1}{2} z^{2} + \frac{1}{6} z^{3} & if n = 3, \frac{1}{4} z + \frac{7}{16} z^{2} + \frac{9}{32} z^{3} + \frac{1}{32} z^{4} & % if n = 4, \frac{1}{5} z + \frac{3}{8} z^{2} + \frac{11}{32} z^{3} + \frac{5}{64} z^{4} + \frac{1}{320} z^{5} & if n = 5 \end{matrix}

for $ℓ = \infty$ and

⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ \begin{matrix} \frac{1}{2} z + \frac{1}{2} z^{2} & if n = 2, \frac{1}{3} z + \frac{11}{21} z^{2} + \frac{1}{7} z^{3} & if n = 3, \frac{1}{4} z + \frac{9}{20} z^{2} + \frac{39}{140} z^{3} + \frac{3}{140} z^{4} & if n = 4, \frac{1}{5} z + \frac{1707}{4495} z^{2} + \frac{23561}{67425} z^{3} + \frac{4657}{67425} z^{4} + \frac{39}{22475} z^{5} & if n = 5 \end{matrix}

for $ℓ = n$ . A closed-form expression exists [1, 2] for $f_{n} (z)$ when $ℓ = \infty$ , but a corresponding simple recursive formula again does not materialize. Means and variances for $ℓ = \infty$ and those for $ℓ = n$ unsurprisingly become closer as $n$ increases.

2.3 Total Path Length

The total (internal) path length $L_{n}$ is the sum of $K_{n} - 1$ taken over all rows $m$ of $M$ . It is not surprising that calculations are more involved here than before. Assume that $ℓ = \infty$ . The probability generating function for $L_{n}$ , given $n$ , obeys a recursion [18]

\begin{matrix} f_{n} (z) = z^{n - 1} 2^{1 - n} n - 1 \sum k = 0 (\frac{n - 1}{k}) f_{k} (z) f_{n - 1 - k} (z), & n \geq 1; \end{matrix}

f_{0} (z) = 1.

Note that $f_{n} (1) = 1$ always. Differentiating with respect to $z$ :

	$f_{n}^{'} (z)$	$= (n - 1) z^{n - 2} 2^{1 - n} n - 1 \sum k = 0 (\frac{}{n - 1}) k f_{k} (z) f_{n - 1 - k} (z)$
		$+ z^{n - 1} 2^{1 - n} n - 1 \sum k = 0 (\frac{n - 1}{}) k [f_{k}^{'} (z) f_{n - 1 - k} (z) + f_{k} (z) f_{n - 1 - k}^{'} (z)]$

we have first moment

E (L_{n}) = f_{n}^{'} (1) = n - 1 + 2^{2 - n} n - 1 \sum k = 0 (\frac{n - 1}{k}) f_{k}^{'} (1)

that is,

g_{n} = n - 1 + 2^{2 - n} n - 1 \sum k = 0 (\frac{n - 1}{k}) g_{k}

where $g_{k} = f_{k}^{'} (1)$ and $g_{0} = 0$ . Clearly $g_{1} = 0$ , $g_{2} = 1$ , $g_{3} = 5 / 2$ and $g_{4} = 35 / 8$ . Differentiating again:

	$f_{n}^{''} (z)$	$= (n - 1) (n - 2) z^{n - 3} 2^{1 - n} n - 1 \sum k = 0 (\frac{n - 1}{k}) f_{k} (z) f_{n - 1 - k} (z)$
		$+ (n - 1) z^{n - 2} 2^{2 - n} n - 1 \sum k = 0 (\frac{}{n - 1}) k [f_{k}^{'} (z) f_{n - 1 - k} (z) + f_{k} (z) f_{n - 1 - k}^{'} (z)]$
		$+ z^{n - 1} 2^{1 - n} n - 1 \sum k = 0 (\frac{n - 1}{}) k [f_{k}^{''} (z) f_{n - 1 - k} (z) + 2 f_{k}^{'} (z) f_{n - 1 - k}^{'} (z) + f_{k} (z) f_{n - 1 - k}^{''} (z)]$

we have second factorial moment

	$E (L_{n} (L_{n} - 1))$	$= f_{n}^{''} (1)$
		$= (n - 1) (n - 2) + 2 (n - 1) [f_{n}^{'} (1) - n + 1]$
		$+ 2^{2 - n} n - 1 \sum k = 0 (\frac{n - 1}{k}) f_{k}^{'} (1) f_{n - 1 - k}^{'} (1) + 2^{2 - n} n - 1 \sum k = 0 (\frac{n - 1}{k}) f_{k}^{''} (1),$

that is,

h_{n} = - (n - 1) n + 2 (n - 1) g_{n} + 2^{2 - n} n - 1 \sum k = 0 (\frac{}{n - 1}) k g_{k} g_{n - 1 - k} + 2^{2 - n} n - 1 \sum k = 0 (\frac{n - 1}{k}) h_{k}

where $h_{k} = f_{k}^{''} (1)$ and $h_{0} = 0$ . Clearly $h_{1} = 0$ , $h_{2} = 0$ , $h_{3} = 4$ and $h_{4} = 61 / 4$ . Finally, we have variance $V (L_{n})$ which is $1 / 4$ when $n = 3$ and $31 / 64$ when $n = 4$ .

Define constants

\begin{matrix} α = \infty \sum j = 1 \frac{1}{2^{j} - 1}, & β = \infty \sum j = 1 \frac{1}{{(2^{j} - 1)}^{2}}, & Q = \infty \prod j = 1 (1 - \frac{1}{2^{j}}) . \end{matrix}

Let $Q_{l}$ denote the $l^{th}$ partial product of $Q$ and

φ (x) = ⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩ \begin{matrix} \frac{x - ln (x) - 1}{(x - 1)^{2}} & if% x \neq 1, \frac{1}{2} & if x = 1. \end{matrix}

It can be proved that [18, 19]

	$E (L_{n})$	$= \frac{n ln (n)}{ln (2)} + n (\frac{γ - 1}{ln (2)} + \frac{1}{2} - α + δ_{1} (n)) + \frac{ln (n)}{ln (2)}$
		$+ (\frac{2 γ - 1}{2 ln (2)} + \frac{5}{2} - α) + δ_{2} (n) + O (\frac{ln (n)}{n}),$

V (L_{n}) = n (C + δ_{3} (n)) + O (\frac{ln (n)^{2}}{n})

as $n \to \infty$ , where

C = \frac{Q}{ln (2)} \sum j, k, l \geq 0 \frac{(- 1)^{j}}{Q_{j} Q_{k} Q_{l}} 2^{- j (j + 1) / 2 - k - l} φ (2^{- j - k} + 2^{- j - l}) = 0.266003645 4....

This expression for $C$ is, needless to say, a stunning result.

Assuming instead that $ℓ = n$ , all we currently possess are PGFs for small $n$ :

⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ \begin{matrix} z & if n = 2, \frac{4}{7} z^{2} + \frac{3}{7} z^{3} & if n = 3, \frac{4}{5} z^{4} + \frac{4}{35} z^{5} + \frac{3}{35} z^{6} & if n = 4, \frac{8984}{13485} z^{6} + \frac{3136}{13485} z^{7} + \frac{364}{4495} z^{8} + \frac{52}{4495} z^{9} + \frac{39}{4495} z^{10} & if n = 5. \end{matrix}

A deeper understanding of finite-key DSTs would be welcome.

2.4 Some Combinatorics

We focus on unsuccessful searches, for both infinite keys ( $ℓ = \infty$ ) and finite keys ( $ℓ = n$ ). Let us examine the coefficients of $z^{n}$ and $z$ for simplicity. The digital search trees appearing in Figure 1 for $n = 2$ proceed from matrices

(\begin{matrix} M x \end{matrix}) = ⎛ ⎜ ⎜ ⎝ \begin{matrix} a & b 1 & c 1 & d \end{matrix} ⎞ ⎟ ⎟ ⎠, ⎛ ⎜ ⎜ ⎝ \begin{matrix} a & b 0 & c 0 & d \end{matrix} ⎞ ⎟ ⎟ ⎠, ⎛ ⎜ ⎜ ⎝ \begin{matrix} a & b 0 & c 1 & d \end{matrix} ⎞ ⎟ ⎟ ⎠, ⎛ ⎜ ⎜ ⎝ \begin{matrix} a & b 1 & c 0 & d \end{matrix} ⎞ ⎟ ⎟ ⎠

respectively. When $ℓ = \infty$ , the indicated keys are merely abbreviations (two leading bits in an infinite sequence); hence the keys are automatically distinct; thus

P {K_{2} = 2} = \frac{2 \cdot 2^{4}}{2^{6}} = \frac{1}{2},

P {K_{2} = 1} = \frac{2 \cdot 2^{4}}{2^{6}} = \frac{1}{2}

where $2^{(n + 1) n}$ is the count of $(n + 1) \times n$ binary matrices. When $ℓ = n$ , however, key-distinctness must be manually enforced. We obtain the condition

(a, b) \neq (1, c), (a, b) \neq (1, d) & (1, c) \neq (1, d)

which is equivalent to $a = 0 & d = 1 - c$ and gives $4$ possibilities; also the condition

(a, b) \neq (0, c), (a, b) \neq (1, d) & (0, c) \neq (1, d)

which is equivalent to $(1 - a) c + a d = 1 - b$ and gives $8$ possibilities; therefore

P {{~ K}_{2} = 2} = \frac{2 \cdot 4}{4! / 1!} = \frac{1}{3},

P {{~ K}_{2} = 1} = \frac{2 \cdot 8}{4! / 1!} = \frac{2}{3}

where $(2^{n})! / (2^{n} - n - 1)!$ is the count of permutations of $2^{n}$ objects, taken $n + 1$ at a time.

Figure 1: Two linear cases and two triangular cases for $n = 2$ .

For $n = 3$ and $ℓ = \infty$ , using Figures 2 and 3, we have

P {K_{3} = 3} = \frac{4 \cdot 2^{7}}{2^{12}} = \frac{1}{8},

P {K_{3} = 1} = \frac{2 \cdot 2^{9}}{2^{12}} = \frac{1}{4}

but when $ℓ = n$ instead, we have

P {{~ K}_{3} = 3} = \frac{4 \cdot 20}{8! / 4!} = \frac{1}{21},

P {{~ K}_{3} = 1} = \frac{2 \cdot 240}{8! / 4!} = \frac{2}{7} .

Figure 2: Four linear cases for $n = 3$ ; note that two are reflections of the others.

Figure 3: Two triangular cases for $n = 3$ ; note that one is a reflection of the other.

For $n = 4$ and $ℓ = \infty$ , using Figures 4, 5 and 6, we have

P {K_{4} = 4} = \frac{8 \cdot 2^{11}}{2^{20}} = \frac{1}{64},

P {K_{4} = 1} = \frac{4 \cdot 2^{14} + 4 \cdot 2^{14}}{2^{20}} = \frac{1}{8}

but when $ℓ = n$ instead, we have

P {{~ K}_{4} = 4} = \frac{8 \cdot 240}{16! / 11!} = \frac{1}{273},

P {{~ K}_{4} = 1} = \frac{4 \cdot 6912 + 4 \cdot 9216}{16! / 11!} = \frac{8}{65} .

The emergence of bi-triangular cases at $n = 4$ complicates our study for $n \geq 5$ . A similar argument for coefficients of $z^{2}$ , … , $z^{n - 1}$ , as well as for successful searches, is possible.

Figure 4: Eight linear cases for $n = 4$ (these four cases plus their reflections).

Figure 5: Four triangular cases for $n = 4$ (these two cases plus their reflections).

Figure 6: Four bi-triangular cases for $n = 4$ (these two cases plus their reflections).

Third and fourth moment expressions appear in [20] for unsuccessful search on infinite keys. The covariance between two random distinct successful search costs within the same tree is apparently $\sim D / n$ as $n \to \infty$ , where [18]

D = C - \frac{1}{12} - \frac{π^{2}}{6 ln (2)^{2}} + α + β = - 0.4970105417....

Verifying this interesting result via simulation remains open. What can be said about the cost covariance for two distinct unsuccessful searches? What can be said about the cost covariance given a successful search and an unsuccessful search?

3 Acknowledgements

I am grateful to Markus Kuba and Sumit Kumar Jha for helpful discussions, and to David Penman for providing [12] (which at one time was available at http://algo.inria.fr/).

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Steven Finch

MIT Sloan School of Management

Cambridge, MA, USA

steven_finch@harvard.edu

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