Resolving Conflicts and Electing Leaders

1 Conflict Resolutions

Consider the R program:

{f <- function(L,k)}{{}{\ \ if(NROW(L)<=1) {}{\ \ \ \ k}{\ \ } else {}{\ \ \ \ b <- rbinom(NROW(L),1,0.5)}{\ \ \ \ L1 <- L[(1:NROW(L))*(1-b)]}{\ \ \ \ k <- f(L1,k <- k+1)}{\ \ \ \ L2 <- L[(1:NROW(L))*b]}{\ \ \ \ k <- f(L2,k <- k+1)}{\ \ }}{\ \ k}{}}

where the integer $k$ is initially $1$ , the list $L$ is initially ${1, 2, \dots, n}$ for simplicity’s sake, and $rbinom (\cdot, 1, 0.5)$ is the built-in R unbiased random Bernoulli ${0, 1}$ generator. Figure 1 exhibits a sample binary tree for $n = 5$ . The first round of coin tosses yields $b = (0, 1, 0, 1, 1)$ , where $0 = tail$ and $1 = head$ , and thus $L_{1} = (1, 3)$ , $L_{2} = (2, 4, 5)$ . On the left side, focusing on $L_{1}$ only, the next round of tosses yields $b^{'} = (1, 1)$ and thus $L_{1}^{'} = \emptyset$ , $L_{2}^{'} = {1, 3}$ . On the right side, focusing on $L_{2}$ only, the next round of tosses yields $b^{''} = (0, 0, 0)$ and thus $L_{1}^{''} = {2, 4, 5}$ , $L_{2}^{''} = \emptyset$ . Also, focusing on $L_{1}^{''}$ only, the next round of tosses yields $b^{'''} = (0, 0, 1)$ and thus $L_{1}^{'''} = {2, 4}$ , $L_{2}^{'''} = {5}$ . Termination occurs at the next level: any twig of the tree ceases to grow when its leaf contains $\leq 1$ items. The number of vertices (including empty vertices) is $13$ . We write $X_{5} = 13$ . The labeled ordering of vertices is determined by watching $k$ increase, from $1$ to $13$ , as the algorithm progresses. Smaller values of $k$ occur on the left side because recursive splitting starts with $L_{1}$ , $L_{2}^{'}$ , … and thereafter continues with $L_{2}$ , $L_{1}^{''}$ , $L_{1}^{'''}$ , ….

Figure 1: Strongly binary tree for conflict resolution.

The probability generating function for

X_{n}

, given

n

, is likewise recursive [3]:

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

f_{0} (z) = f_{1} (z) = z .

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

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

we have first moment

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

that is,

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

where $g_{k} = f_{k}^{'} (1)$ and $g_{0} = g_{1} = 1$ . Clearly $g_{2} = 5$ and $g_{3} = 23 / 3$ . Differentiating again:

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

we have second factorial moment

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

that is,

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

where $h_{k} = f_{k}^{''} (1)$ and $h_{0} = h_{1} = 0$ . Clearly $h_{2} = 28$ and $h_{3} = 548 / 9$

. Finally, we have variance

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

which is $8$ when $n = 2$ and $88 / 9$ when $n = 3$ .

Such recursions are helpful for small $n$ , but give no asymptotic information as $n \to \infty$ . It can be proved that [4, 5]

	$\frac{E (X_{n})}{n}$	$= \frac{2}{ln (2)} + δ_{1} (\frac{ln (n)}{ln (2)}) + o (1)$
		$\approx (2.8853900817...) + δ_{1} ({log}_{2} (n)),$

	$\frac{V (X_{n})}{n}$	$= \frac{1}{ln (2)} (1 + 8 \infty \sum k = 1 \frac{1}{(2^{k} + 1)^{2}}) + δ_{2} (\frac{ln (n)}{ln (2)}) + o (1)$
		$\approx (3.3834344923...) + δ_{2} ({log}_{2} (n))$

where $δ_{j} (x)$ is a periodic function of period $1$ with tiny amplitude and zero mean, for $j = 1$ , $2$ . Functions like these appear throughout the analysis of algorithms. We shall not specify these here nor later, but merely indicate their presence when required. An infinite series representation of the mean for arbitrary $n$ is [6, 7]

E (X_{n}) = 1 + 2 \infty \sum m = 0 2^{m} [1 - {(1 - 2^{- m})}^{n} - n 2^{- m} {(1 - 2^{- m})}^{n - 1}] .

Also, the constant $0.8458586230...$ (one-quarter of the variance as $n \to \infty$ ) appears in [8, 9, 10].

2 Leader Elections

Two parameters are examined in this section: height (length of the root-to-leaf path that gives the leader) and size (number of vertices in the associated weakly binary tree). We further discuss a variation of the election in which draws (involving exactly two competitors) constitute an additional way to stop the process.

2.1 Height

Consider the R program:

{f <- function(L,k)}{{}{\ \ if(NROW(L)<=1) {}{\ \ \ \ k}{\ \ } else {}{\ \ \ \ b <- rbinom(NROW(L),1,0.5)}{\ \ \ \ L1 <- L[(1:NROW(L))*(1-b)]}{\ \ \ \ k <- f(L1,k <- k+1)}{\ \ \ \ if(NROW(L1)==0) {}{\ \ \ \ \ \ L2 <- L[(1:NROW(L))*b]}{\ \ \ \ \ \ k <- f(L2,k)}{\ \ \ \ }}{\ \ }}{\ \ k}{}}

where the integer $k$ is initially $0$ (unlike before) and the list $L$ is initially ${1, 2, \dots, n}$ . Figure 2 exhibits a sample binary tree for $n = 5$ . The first round of coin tosses yields $b = (0, 0, 0, 1, 0)$ , where $0 = tail$ and $1 = head$ , and thus $L_{1} = (1, 2, 3, 5)$ , $L_{2} = (4)$ . Focusing on $L_{1}$ only, the next round of tosses yields $b^{'} = (1, 1, 1, 1)$ and thus $L_{1}^{'} = \emptyset$ , $L_{2}^{'} = {1, 2, 3, 5}$ . Note that we do not indicate nor count the empty vertex. Focusing on $L_{2}^{'}$ only, the next round of tosses yields $b^{''} = (1, 0, 0, 0)$ and thus $L_{1}^{''} = {2, 3, 5}$ , $L_{2}^{''} = {1}$ . Focusing on $L_{1}^{''}$ only, the next round of tosses yields $b^{'''} = (0, 0, 0)$ and thus $L_{1}^{'''} = {2, 3, 5}$ , $L_{2}^{'''} = \emptyset$ . Focusing on $L_{1}^{'''}$ only, the next round of tosses yields $b^{''''} = (0, 0, 1)$ and thus $L_{1}^{''''} = {2, 3}$ , $L_{2}^{''''} = {5}$ . Termination occurs at the next level: the leftmost twig of the tree contains exactly $1$ item. The height (number of steps separating the vertices labeled $1$ and $7$ ) is $6$ . We write $H_{5} = 6$ .

Figure 2: Weakly binary tree for leader election of height 6.

The probability generating function for $H_{n}$ , given $n$ , is [11]:

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

f_{0} (z) = f_{1} (z) = 1.

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

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

we have first moment

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

that is,

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

where $g_{k} = f_{k}^{'} (1)$ and $g_{0} = g_{1} = 0$ . Clearly $g_{2} = 2$ and $g_{3} = 7 / 3$ . Differentiating again:

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

we have second factorial moment

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

that is,

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

where $h_{k} = f_{k}^{''} (1)$ and $h_{0} = h_{1} = 0$ . Clearly $h_{2} = 4$ and $h_{3} = 50 / 9$ . Finally, we have variance $V (H_{n})$ which is $2$ when $n = 2$ and $22 / 9$ when $n = 3$ .

It can be proved that [12, 13, 14]

	$E (H_{n})$	$= \frac{ln (n)}{ln (2)} + \frac{1}{2} + δ_{3} (\frac{ln (n)}{ln (2)}) + o (1)$
		$\approx {log}_{2} (n) + (0.5) + δ_{3} ({log}_{2} (n)),$

	$V (H_{n})$	$= \frac{1}{12} + \frac{π^{2}}{6 ln (2)^{2}} - \frac{γ^{2} + 2 γ_{1}}{ln (2)^{2}} + ε (\frac{ln (n)}{ln (2)}) + o (1)$
		$\approx (3.1166951643...) + ε ({log}_{2} (n))$

as $n \to \infty$ , where $γ_{1}$ is the first Stieltjes constant [15] and fluctuations provided by $ε (x)$ are symmetrical not about $0$ , but instead about [16]

1 \int 0 ε (x) d x = - \frac{1}{ln (2)^{2}} \infty \sum \begin{matrix} k = - \infty k \neq 0 \end{matrix} {∣ ∣ ∣ ∣ ζ (1 - \frac{2 π i k}{ln (2)}) Γ (1 - \frac{2 π i k}{ln (2)}) ∣ ∣ ∣ ∣}^{2} \approx - 1.856 \times 10^{- 10} .

Needless to say, the evaluation of $V (H_{n})$ is an impressive and difficult achievement.

2.2 Size

Consider the R program:

{f <- function(L,k)}{{}{\ \ if(NROW(L)<=1) {}{\ \ \ \ k}{\ \ } else {}{\ \ \ \ b <- rbinom(NROW(L),1,0.5)}{\ \ \ \ L1 <- L[(1:NROW(L))*(1-b)]}{\ \ \ \ L2 <- L[(1:NROW(L))*b]}{\ \ \ \ k <- f(L1,k <- k+(NROW(L1)!=0)+(NROW(L2)!=0))}{\ \ \ \ if(NROW(L1)==0) {}{\ \ \ \ \ \ k <- f(L2,k)}{\ \ \ \ }}{\ \ }}{\ \ k}{}}

where the integer $k$ is initially $1$ and the list $L$ is initially ${1, 2, \dots, n}$ . Figure 3 is almost identical to Figure 2 – it contains the same binary tree for $n = 5$ – but the labeling is different. The number of vertices (excluding empty vertices) is $11$ . We write $Y_{5} = 11$ . The labeled ordering of vertices is determined by watching $k$ increase, from $1$ to $11$ , as the algorithm progresses. We sweep across the tree horizontally (in rows) rather than hierarchically. In particular, the command

\begin{matrix} {k < - k+(NROW(L1)!=0)+(NROW(L2)!=0)} \end{matrix}

increases $k$ by $2$ if both vertices are non-empty and by only $1$ if otherwise.

Figure 3: Weakly binary tree for leader election of size 11.

The probability generating function for $Y_{n}$ , given $n$ , is [11]:

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

\begin{matrix} f_{0} (z) = 1, & f_{1} (z) = z . \end{matrix}

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

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

we have first moment

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

that is,

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

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

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

we have second factorial moment

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

that is,

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

where $h_{k} = f_{k}^{''} (1)$ and $h_{0} = h_{1} = 0$ . Clearly $h_{2} = 14$ and $h_{3} = 200 / 9$ . Finally, we have variance $V (Y_{n})$ which is $2$ when $n = 2$ and $133 / 36$ when $n = 3$ .

It can be proved that [11]

	$E (Y_{n})$	$= \frac{2 ln (n)}{ln (2)} + (2 - \frac{ln (π) - γ}{ln (2)}) + δ_{4} (\frac{ln (n)}{ln (2)}) + o (1)$
		$\approx 2 {log}_{2} (n) + (1.1812500478...) + δ_{4} ({log}_{2} (n))$

as $n \to \infty$ , but a comparable asymptotic expression for $V (Y_{n})$ remains unknown.

We wonder about the cross-covariance of $H_{n}$ and $Y_{n}$ , whose calculation would require a bivariate generating function.

2.3 Draws

Let us alter the rules so that a draw between two persons is allowed (if precisely two persons are left, then they both are declared leaders). The third line of the two preceding R programs is simply replaced by

\begin{matrix} {if(NROW(L) < =2) {} \end{matrix}

and the initial conditions of the two preceding ${g_{n}, h_{n}}$ recurrences are also changed. For height, we now have $g_{0} = g_{1} = g_{2} = 0$ (implying $g_{3} = 4 / 3)$ and $h_{0} = h_{1} = h_{2} = 0$ (implying $h_{3} = 8 / 9$ ); thus

	$E ({~ H}_{n})$	$= \frac{ln (n)}{ln (2)} + (\frac{1}{2} - \frac{π^{2}}{12 ln (2)}) + δ_{5} (\frac{ln (n)}{ln (2)}) + o (1)$
		$\approx {log}_{2} (n) - (0.6865691104...) + δ_{5} ({log}_{2} (n)) .$

For size, we now have $g_{0} = 0$ , $g_{1} = g_{2} = 1$ (implying $g_{3} = 10 / 3)$ and $h_{0} = h_{1} = h_{2} = 0$ (implying $h_{3} = 74 / 9$ ); thus

	$E ({~ Y}_{n})$	$= \frac{2 ln (n)}{ln (2)} + ⎛ ⎜ ⎝ 2 - \frac{ln (π) - γ + \frac{π^{2}}{8}}{ln (2)} ⎞ ⎟ ⎠ + δ_{6} (\frac{ln (n)}{ln (2)}) + o (1)$
		$\approx 2 {log}_{2} (n) - (0.5986036178...) + δ_{6} ({log}_{2} (n)) .$

It would seem that the fraction $π^{2} / 16$ given within the formula for $E ({~ Y}_{n})$ in [11] is incorrect and should be $π^{2} / 8$ instead.

3 Coin Tosses

Before moving on to more complicated examples, it may be worthwhile to contemplate the most elementary recursive program imaginable:

{f <- function(L,k)}{{}{\ \ if(NROW(L)==0) {}{\ \ \ \ k}{\ \ } else {}{\ \ \ \ b <- rbinom(NROW(L),1,0.5)}{\ \ \ \ L1 <- L[(1:NROW(L))*(1-b)]}{\ \ \ \ k <- f(L1,k <- k+1)}{\ \ }}{\ \ k}{}}

mentioned at the end of [17], where $k$ is initially $0$ . Note that the third line here is

\begin{matrix} {if(NROW(L) < =0) {} \end{matrix}

unlike what we have seen before. People iteratively toss all coins which show tails until only heads are visible. Omitting details, it follows that the procedural height has generating function

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

f_{0} (z) = 1

therefore

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

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

where $g_{0} = 0$ (implying $g_{1} = 2$ , $g_{2} = 8 / 3$ , $g_{3} = 22 / 7)$ and $h_{0} = 0$ (implying $h_{1} = 4$ , $h_{2} = 64 / 9$ , $h_{3} = 1420 / 147$ ). From this [18]

	$E ({^H}_{n})$	$= \frac{ln (n)}{ln (2)} + (\frac{1}{2} + \frac{γ}{ln (2)}) + δ_{7} (\frac{ln (n)}{ln (2)}) + o (1)$
		$\approx {log}_{2} (n) + (1.3327461772...) + δ_{7} ({log}_{2} (n)),$

	$V ({^H}_{n})$	$= (\frac{1}{12} + \frac{π^{2}}{6 ln (2)^{2}}) + δ_{8} (\frac{ln (n)}{ln (2)}) + o (1)$
		$\approx (3.5070480758...) + δ_{8} ({log}_{2} (n))$

as $n \to \infty$ .

4 Finding the Maximum

Consider the R program:

{f <- function(L,k,r)}{{}{\ \ if(NROW(L)==0) {}{\ \ \ \ c(k,r)}{\ \ } else if(NROW(L)==1) {}{\ \ \ \ c(k,r <- max(L,r))}{\ \ } else {}{\ \ \ \ b <- rbinom(NROW(L),1,0.5)}{\ \ \ \ L0 <- L[(1:NROW(L))*b]}{\ \ \ \ L1 <- L[(1:NROW(L))*(1-b)]}{\ \ \ \ V <- f(L1,k <- k+1,r)}{\ \ \ \ k <- V[1]; r <- V[2]}{\ \ \ \ L2 <- L0[L0>r]}{\ \ \ \ V <- f(L2,k <- k+1,r)}{\ \ \ \ k <- V[1]; r <- V[2]}{\ \ }}{\ \ c(k,r)}{}}

where the integer $k$ is initially $0$ and the list $L$ is initially ${1, 2, \dots, n}$ . We find a comparison of a conflict resolution tree in Figure 4 and its corresponding maximum-finding tree in Figure 5 to be helpful. Note that the vertex count for the latter is only $14$ , which is less than the vertex count of $16$ determined for the former. The discrepancy arises because, at the instant the $5^{th}$ vertex $3$ becomes the leader and $r$ consequently increases from $0$ to $3$ , both the $6^{th}$ vertex $1$ becomes empty and the $8^{th}$ & $9^{th}$ vertices $2456$ are reduced to $456$ . It follows that the $10^{th}$ vertex $24$ is reduced to $4$ , which cannot have any descendants, eliminating two twigs. Also, $r$ increases from $3$ to $4$ and, subsequently, from $4$ to $6$ ; while the penultimate vertex becomes empty, this does not further alter the count. We write $X_{6} = 14$ .

Figure 4: Conflict resolution tree (to be compared with next figure).

Figure 5: Maximum-finding tree (to be compared with previous figure).

The probability generating function for $X_{n}$ , given $n$ , is [19, 20, 21, 22]:

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

f_{0} (z) = f_{1} (z) = z .

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

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

we make use of the identity

n \sum k = 1 k \sum j = 1 a_{k j} = n \sum j = 1 n \sum k = j a_{k j} .

From this follows the first moment:

	$E (X_{n})$	$= f_{n}^{'} (1) = 2^{1 - n} + 2^{- n} f_{n}^{'} (1) + 2^{- n} (- 1 + 2^{n}) + 2^{- n} n \sum k = 1 2^{- 1 + k} f_{n - k}^{'} (1) + 2^{- n} n \sum j = 1 (\frac{n}{j}) f_{j}^{'} (1)$
		$= 1 - 2^{- n} + 2^{1 - n} + 2^{- n} f_{n}^{'} (1) + n \sum k = 1 2^{- 1 - n + k} f_{n - k}^{'} (1) + 2^{- n} n - 1 \sum j = 0 (\frac{n}{j}) f_{j}^{'} (1) - 2^{- n} + 2^{- n} f_{n}^{'} (1)$
		$= 1 + 2^{1 - n} f_{n}^{'} (1) + n - 1 \sum i = 0 2^{- i - 1} f_{i}^{'} (1) + 2^{- n} n - 1 \sum j = 0 (\frac{}{n}) j f_{j}^{'} (1)$

setting $i = n - k$ , hence $i + 1 = 1 + n - k$ and $0 \leq i \leq n - 1$ when $1 \leq k \leq n$ . That is,

g_{n}

= 1 + 2^{1 - n} g

Resolving Conflicts and Electing Leaders

Sorting Real Numbers in O(n√( n)) Time and Linear Space

Recursive PGFs for BSTs and DSTs

Broadcasts on Paths and Cycles

A complete system of deduction for Sigma formulas

Communication-Efficient String Sorting

Data Lakes, Clouds and Commons: A Review of Platforms for Analyzing and Sharing Genomic Data

Countably Infinite Multilevel Source Polarization for Non-Stationary Erasure Distributions

1 Conflict Resolutions