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A Tale of Two Limits: An Extremal Pagerank ProblemJ. Farnan and F. Kenter

A Tale of Two Limits: An Extremal Pagerank Problemthanks: Submitted to the editors April 8, 2021. \fundingThis work was funded in part by the NSF-DMS-1720225

and Franklin H. J. Kenter    Joseph Farnan United States Naval Academy, Annapolis, MD, josephterencefarnan@gmail.com    Franklin H. J. Kenter United States Naval Academy, Annalpols, MD, . kenter@usna.edu
(September 1, 2025)

A Tale of Two Limits, An Extremal Pagerank Problem

and Franklin H. J. Kenter    Joseph Farnan United States Naval Academy, Annapolis, MD, josephterencefarnan@gmail.com    Franklin H. J. Kenter United States Naval Academy, Annalpols, MD, . kenter@usna.edu
(September 1, 2025)
Abstract

For a directed graph, the Pagerank algorithm emulates a random walker on the graph that occasionally “jumps” to a random vertex based on a jumping parameter α\alpha. Upon completion, the algorithm generates a stochastic vector whose entries correspond to the limiting probability that the walker will be at that vertex. This vector is a right eigenvector of a corresponding Markov trasition matrix. Undoubtedly, this vector can drastically change based upon the jumping parameter α\alpha.

In this article, we investigate the maximum possible discrepancy for different Pagerank vectors on the same unweighted directed (perhaps with loops) graph as measured by the 2-norm. We show that the limsup of this discrepancy can be as large as 6750\sqrt{\frac{67}{50}} using a very specific construction. (For contrast, the norm of the difference for any two stochastic vectors is at most 2\sqrt{2}.) Interestingly, on this construction this discrepancy occurs when α=1\alpha=1 and when α\alpha is very close to 1.

keywords:
directed graphs, Pagerank algorithm, spectral graph theory, extremal spectral graph theory
{AMS}

05C35, 05C50, 05C81

1 Introduction

1.1 Overview

One way to rank the relative importance of vertices on a directed graph (network) is to use the Pagerank algorithm. The Pagerank algorithm was developed by Page and Brin to address the conundrum between quantity versus quality in the context of webpages [2], as a webpage could be considered influential either because it is linked to by many sites or by a few, high-quality sites. For instance, many space-enthusiasts may link to a NASA, indicating its authority. In turn, NASA may link to a revered astrophysicist on its own, indicating its endorsement.

Pagerank and similar network algorithms have been developed for other contexts. In the of a sports league, the graph could have teams be the vertices and the directed edges represent a team losing against the other team. Previous work has utilized this idea of to rank NCAA College Football teams where it is often the case that some teams perform very well but with weaker opponents [3]. In social media, the graph would have users of vertices and the directed edges represent one user following or mentioning one another. In [6], the PageRank algorithm is used to provide an alternative measure of influence of Twitter users beyond simply counting the number of retweets. That is, who retweets our message is just as important as how many times it is retweeted.

Consider a directed graph, GG. For our purposes and throughout, GG may have loops and bidirectional edges (i.e., there is an arc from ii to jj and another arc from jj to ii). However, there will be no duplicate arcs (i.e., there is at most one arc from ii to jj). The Pagerank algorithm produces a vector of length nn (the number of vertices in GG), denoted π\pi, where each entry measures on the proportion of the time we would encounter each vertex by simply wandering around the graph randomly and occasionally jump around. Hence, the Pagerank vector is a stochastic vector.

To begin calculating the Pagerank vector, consider a directed graph GG and choose an α\alpha to be your jumping parameter, 0<α<10<\alpha<1. The parameter α\alpha will determine the proportion of the time we will follow an arc in our random process. Intuitively, α\alpha balances the impact of arcs from high-quality vertices and high quantities of arcs. Once α\alpha is chosen, randomly pick a vertex of GG uniformly at random to start, denoted vv. Then, with probability α\alpha, choose an out-neighbor of vv, uniformly at random, and walk to that vertex; and with probability 1α1-\alpha, jump to an another random vertex chosen uniformly at random. If vv has no out neighbors, then simply jump to a random vertex, chosen uniformly at random. Take the vertex we are on and declare that the new vv and repeat. The Pagerank vector is the limiting probability of finding ourselves at each vertex after an arbitrarily large number of steps. Intuitively, the more often we visit a vertex, the more important it is and the higher the (Page)rank it has!

For a given directed graph GG and α\alpha, we define the Pagerank vector with jumping parameter α\alpha

(πα)i=limtP(being at vertex i at step t)(\pi_{\alpha})_{i}=\lim_{t\to\infty}P(\text{being at vertex }i\text{ at step }t)

for each vertex ii.

The Pagerank can be interpreted using linear algebra in the context of Markov chains where the probability transition matrix for a directed graph GG, 𝐑\mathbf{R}, is given by

𝐑ij={αdegjout+1αn if degjout01notherwise\mathbf{R}_{ij}=\begin{cases}\frac{\alpha}{{deg}^{out}_{j}}+\frac{1-\alpha}{n}~\text{ if }~{deg}^{out}_{j}\neq 0\\ \frac{1}{n}~\text{otherwise}\end{cases}

Where degjoutdeg_{j}^{out} is the number of out neighbors of j. In which case, the Pagerank vector π\pi is a right eigenvector of 𝐑\mathbf{R} with eigenvalue 1

𝐑π=π\mathbf{R}\pi=\pi

and is the stationary distribution of the random walk.

The existence and uniqueness of π\pi guaranteed by the Perron-Frobenius Theorem for all 0α<10\leq\alpha<1 (see, for example, [4]). For α=1\alpha=1, πα\pi_{\alpha} is well-defined provided GG

  • is aperiodic (i.e., the greatest common divisor of the length of all cycles is 1),

  • is weakly connected and

  • there is a unique maximal induced subgraph that is strongly connected (perhaps with just a vertex with a single loop).

For simplicity, we will typically drop one of the subscripts of the Pagerank vector. We note that in this paper we will use subscripts to both the value of α\alpha and which vertex is being spoken of. For instance, we write πα=1\pi_{\alpha=1} to refer to the entire Pagerank vector when alpha=1alpha=1 but also we write πA\pi_{A} to denote to entry of the Pagerank vector corresponding to the vertex AA; in the later case, the context of α\alpha will be clear.

1.2 Problem Statement

Even on the same graph, choosing different values for α\alpha can result in substantially different Pagerank vectors. For example, when α=0\alpha=0, the walker always jumps, in which case πα\pi_{\alpha} is the uniform vector: (1/n,1/n,,1/n)(1/n,1/n,\ldots,1/n) where nn is the number of vertices in GG. On the other extreme, when α=1\alpha=1, the walker always follows the arcs (unless they are stuck), so naturally, vertices with more in-arcs will be visited more often. In this situation, it is possible that πα\pi_{\alpha} is concentrated at one vertex. This leads to the question: For the same graph, how different can these rankings be? Our main problem is to determine the maximum difference between Pagerank vector with one α\alpha and a Pagerank with a different α\alpha but on the same graph. Formally we ask:

Problem 1.1.

How large can

πα1πα22||\pi_{\alpha_{1}}-\pi_{\alpha_{2}}||_{2}

be over all directed graphs GG and α1\alpha_{1}, α2\alpha_{2} with 0α1,α210\leq\alpha_{1},\alpha_{2}\leq 1? Here, πα1\pi_{\alpha_{1}} is the Pagerank vector produced on the graph GG with jumping constant α1\alpha_{1}, and πα2\pi_{\alpha_{2}} is the ranking values returned by the Pagerank algorithm for the graph GG with jumping constant different α2\alpha_{2}.

We focus on the 2-norm as it is the traditional norm, though we briefly discuss the problem for other norms in Section 3 which we leave as an open problem for the reader.

As we will discover, this problem is difficult because it requires carefully constructed examples with precisely chosen parameters α1\alpha_{1} and α2\alpha_{2}. While we do not answer this problem explicitly, we provide a construction that provides an extremal value of 6750\sqrt{\frac{67}{50}}. Surprisingly, to achieve this, we use α1=1\alpha_{1}=1 and α2\alpha_{2} very close to 1. We believe 6750\sqrt{\frac{67}{50}} to be best possible.

1.3 Previous Similar Work

Based on the linear-algebraic interpretation of Pagerank mentioned previously, this problem is fundamentally one about the principal eigenvector of graph-theoretic matrices. Indeed, problems of a similar nature have been considered before. For an irreducible nonnegative matrix, the principal ratio is the ratio between the maximum and minimum entries of the principal eigenvector. For Markov transition matrices of simple directed graphs, Askoy, Chung, and Peng show that the principal ratio of a random walk can be superexponentially small [1]. For the adjacency matrix of undirected graphs, Tait proved that the “kite graph” (a complete graph with a long path) acheives the maximum principal ratio [9].

1.4 Important facts and properties of PageRank

To get our feet wet, here are some important properties of Pagerank vectors we use.

Fact 1.2.

For α=0\alpha=0, πα=(1/n,1/n,,1/n)\pi_{\alpha}=(1/n,1/n,\ldots,1/n)^{\prime}.

Where nn is the number of vertices in the graph. What this says is that, in practice, one should choose an α\alpha away from 0 so that the resulting values vertices are more differentiated. Indeed, the original α\alpha for the Pagerank algorithm was α=0.85\alpha=0.85 [2].

In the context of maximizing πα=1πα=x2\|\pi_{\alpha=1}-\pi_{\alpha=x}\|_{2}, this suggests that we would want to choose α\alpha’s away from 0. Our the extremal choices for α\alpha are closer to 1.

Also, in practice, people usually avoid choosing an α\alpha close to 1 because accurately computing πα1\pi_{\alpha\approx 1} numerically is challenging using methods such as the power method [7]. Our graphical plots fail to show the true behavior near α=1\alpha=1 because when kk is large, these numerical methods are not accurate enough! See for example figure 2

Fact 1.3.

π21||\pi||_{2}\leq 1.

Proof.

π2π1=1||\pi||_{2}\leq||\pi||_{1}=1.

Fact 1.4.

πα1||\pi_{\alpha}||_{\infty}\leq 1 or in other words, max(πi)1max(\pi_{i})\leq 1.

Fact 1.5.

πα1πα21||\pi_{\alpha_{1}}-\pi_{\alpha_{2}}||_{\infty}\leq 1.

Proof.

πα1πα2|max(|πα1|,|πα2|)|1||\pi_{\alpha_{1}}-\pi_{\alpha_{2}}||_{\infty}\leq|max(|\pi_{\alpha_{1}}|,|\pi_{\alpha_{2}}|)|_{\infty}\leq 1.

Fact 1.6.

π1π222||\pi_{1}-\pi_{2}||_{2}\leq\sqrt{2}

Proof.

π1π22π11+π212||\pi_{1}-\pi_{2}||_{2}\leq\sqrt{||\pi_{1}||_{1}+||\pi_{2}||_{1}}\leq\sqrt{2}. The first inequality holds because the π\pis are non-negative.

This last bound is important because it provides an upper bound for how high πα=1πα=x2\|\pi_{\alpha=1}-\pi_{\alpha=x}\|_{2} can be. While we are unable to achieve 21.4142\sqrt{2}\approx 1.4142, we are not too far from it: 67501.1576\sqrt{\frac{67}{50}}\approx 1.1576.

2 Main Result

Theorem 2.1.

For any δ>0\delta>0, there exists an unweighted directed graph and α1=1\alpha_{1}=1 and α2\alpha_{2} with

πα1πα26750δ.\|\pi_{\alpha_{1}}-\pi_{\alpha_{2}}\|\geq\sqrt{\frac{67}{50}}-\delta.

In particular, for the Pagerank vectors on the directed graph on Γ(k)\Gamma(k) (defined in the next subsection),

limkπα1=1πα2=11/k=6750.\lim_{k\to\infty}\|\pi_{\alpha_{1}=1}-\pi_{\alpha_{2}=1-1/k}\|=\sqrt{\frac{67}{50}}.

This answer is very interesting in several ways. First, the answer depends on the number of vertices in the graph in the sense that increasing the number of vertices in the graph enabled an slightly increased value of πα1πα22||\pi_{\alpha_{1}}-\pi_{\alpha_{2}}||_{2}. In this paper, we will show that limkπα1πα22\displaystyle\lim_{k\to\infty}||\pi_{\alpha_{1}}-\pi_{\alpha_{2}}||_{2} (where k+3k+3 is number of vertices in the graph for very large kk) goes to a certain size, but implicit in this is that increasing the number of vertices in the graph will allow a value closer to the limit. Also, the extremal difference occurs approximately when α=1\alpha=1 and 1>α11/k1>\alpha\geq 1-1/k. Figure 2 plots πα=1πα=x2\|\pi_{\alpha=1}-\pi_{\alpha=x}\|_{2} for a particular construction we give in the next section. The graph is beyond deceiving! When x=1x=1, the value for πα=1πα=x2=0\|\pi_{\alpha=1}-\pi_{\alpha=x}\|_{2}=0, but a large enough graph, the maximum value is achieved very very close to x=1x=1!

We believe the value 6750\sqrt{\frac{67}{50}} is the best possible:

Conjecture 2.2.

For any unweighted directed graph (perhaps with loops and perhaps bidirected arcs) and two jumping constants α1\alpha_{1} and α2\alpha_{2}, we have,

πα1πα2<6750.\|\pi_{\alpha_{1}}-\pi_{\alpha_{2}}\|<\sqrt{\frac{67}{50}}.

We will give our reasoning and intuition for this conjecture in Section 3

2.1 Our Construction, Γ(k)\Gamma(k)

Our construction is a long “ladder” with k+3k+3 vertices: C1,C2,B1,,Bk,C_{1},C_{2},B_{1},\ldots,B_{k}, and AA. It is illustrated in graph Figure 1 and is constructed as follows. Vertex C1C_{1} has an arc to itself and to C2C_{2}. C2C_{2} has an arc back to C1C_{1}, a loop to itself and a path to B1B_{1}. B1B_{1} has an arc back to C1C_{1} and C2C_{2}, and an arc to B2B_{2}. Likewise, for all i<ki<k, BiB_{i}, has arcs back to both C1C_{1} and C2C_{2} and an arc to Bi+1B_{i+1}. Then, BkB_{k} which has an arc back to C1C_{1} and C2C_{2} and to a vertex AA. However, vertex AA only has a loop to itself.

This construction will end up concentrating the random walkers either in AA or in C1C_{1} and C2C_{2}, depending on the value of the jumping constant α\alpha.

When α=1\alpha=1, the random walker will always follow the arcs. The walker will climb up and fall down “the ladder”, but inevitably, they will eventually make it to AA and become stuck. Therefore, when α=1\alpha=1 we have πA=1\pi_{A}=1.

On the other hand, for other, carefully chosen, values of α\alpha close to 1, the random walker will frequently “fall down” to C1C_{1} and C2C_{2}. In which case, it is next to impossible for the walker to get to AA. And even if the walker gets to AA, since α<1\alpha<1, the walker is able to avoid becoming stuck. Within the proof of Theorem 2.1, we show that πA0\pi_{A}\to 0 as kk\to\infty for α=11/k\alpha=1-1/k.

AABkB_{k}\cdotsB1B_{1}C2C_{2}C1C_{1}
Figure 1: Construction of the family of graphs used in the proof of Theorem 2.1

2.2 Proof of Theorem 2.1

Our proof is outlined as follows. We will start by explicitly computing πα1=1\pi_{\alpha_{1}=1} for Γ(k)\Gamma(k) for any k>0k>0. We then compute πα2=11/k\pi_{\alpha_{2}=1-1/k} and show that πC1\pi_{C_{1}} and πC225\pi_{C_{2}}\to\frac{2}{5} as kk\to\infty. We then analyze πBi\pi_{B_{i}} and show that iπBi2\sum_{i}\pi_{B_{i}}^{2} can be approximated by a geometric series in terms of πC2\pi_{C_{2}}. As a result, we will have that as kk\to\infty, πα1=1πα2=11/k2267/50\|\pi_{\alpha_{1}=1}-\pi_{\alpha_{2}=1-1/k}\|_{2}^{2}\to 67/50.  

2.2.1 Computing πα=1\pi_{\alpha=1}

Lemma 2.3.

When α=1\alpha=1, πA=1\pi_{A}=1, and πv=0\pi_{v}=0 for any vertex vAv\neq A.

Proof 2.4.

The loop at AA ensures that when α=1\alpha=1, a “random-walker” must follow their way to AA and remain there. Therefore, πA\pi_{A} has a value of 1 at vertex AA and 0 at all other vertices.

Now, the more difficult task, of carefully computing πα2=11/k\pi_{\alpha_{2}=1-1/k} remains.

2.2.2 Computing πA\pi_{A} when α=11/k\alpha=1-1/k

We will show that as kk\to\infty,

πC125,πC225, and πA0.\pi_{C_{1}}\to\frac{2}{5},\quad\pi_{C_{2}}\to\frac{2}{5},\text{ and }\quad\pi_{A}\to 0.

Our main technique for computing πv\pi_{v} is to consider change with each time step for sufficiently large tt. We let Δπv=u(𝐑v,uπu)πv\Delta\pi_{v}=\sum_{u}\left(\mathbf{R}_{v,u}\pi_{u}\right)-\pi{v}. Effectively, Δπv\Delta\pi_{v} is “rate in” - “rate out”.

Lemma 2.5.

For any vertex vv, the stationary distribution obeys Δπv=0\Delta\pi_{v}=0.

\square  
We can apply the previous lemma to compute πA\pi_{A}.

For a random walker to reach AA there are 4 disjoint possibilities:

  • kk steps ago, the random walker was at C2C_{2} and did not jump within the intervening kk steps from C2C_{2} to A.

  • 1 step ago, the walker was at AA and then they took the loop back to AA.

  • kik-i steps ago, the walker jumped to BiB_{i} and the made it up the “ladder” to AA without jumping again, or

  • the walker could have directly jumped to AA.

Therefore, we have

ΔπA=((11k)kπC2(1/3)k+(11k)πA+i=1k1k(1/3)ki(11k)ki3+k)πA.\Delta\pi_{A}=\left(\left(1-\frac{1}{k}\right)^{k}\pi_{C_{2}}(1/3)^{k}+\left(1-\frac{1}{k}\right)\pi_{A}+\sum_{i=1}^{k}\frac{\frac{1}{k}(1/3)^{k-i}\left(1-\frac{1}{k}\right)^{k-i}}{3+k}\right)-\pi_{A}.

The first term accounts for the first possibility. The second term accounts for the second possibility. The sum accounts for the last two possibilities. Finally, πA-\pi_{A} falls from the definition of ΔπA\Delta\pi_{A} and contains all of the walkers leaving AA even if they come back.

This simplifies to

ΔπA=((11k)kπC2(1/3)k+1kπA+i=1k1k(1/3)ki(11k)ki3+k).\Delta\pi_{A}=\left(\left(1-\frac{1}{k}\right)^{k}\pi_{C_{2}}(1/3)^{k}+-\frac{1}{k}\pi_{A}+\sum_{i=1}^{k}\frac{\frac{1}{k}(1/3)^{k-i}(1-\frac{1}{k})^{k-i}}{3+k}\right).

Claim:

πA0 as k\pi_{A}\to 0\text{ as }k\to\infty
Proof 2.6.

For the stationary distribution,

ΔπA=(11k)kπC2(1/3)kπA/k+i=1k1k(1/3)ki(11k)k3+k.\Delta\pi_{A}=(1-\frac{1}{k})^{k}\pi_{C_{2}}(1/3)^{k}-\pi_{A}/k+\sum_{i=1}^{k}\frac{\frac{1}{k}(1/3)^{k-i}(1-\frac{1}{k})^{k}}{3+k}.

Observe that the series

i=1k(1/k(1/3)ki(11/k)k3+k).\sum_{i=1}^{k}\left(\frac{1/k~(1/3)^{k-i}(1-1/k)^{k}}{3+k}\right).

is a finite geometric series that’s bounded above by its infinite sum which is

3/k2(3+k).\frac{3/k}{2(3+k)}.

And solving for πA\pi_{A} we have

πA<(11/k)kπC2(1/3)k1/k+32(3+k).\pi_{A}<\frac{(1-1/k)^{k}\pi_{C_{2}}(1/3)^{k}}{1/k}+\frac{3}{2(3+k)}.

Since 0πC210\leq\pi_{C_{2}}\leq 1 and 1/k>01/k>0, we can simplify the above equation

(2.1) πA<(11/k)k(1/3)k1/k+32(3+k)<(1/3)k1/k+32(3+k)=k3k+32(k+1).\pi_{A}<\frac{(1-1/k)^{k}(1/3)^{k}}{1/k}+\frac{3}{2(3+k)}<\frac{(1/3)^{k}}{1/k}+\frac{3}{2(3+k)}=\frac{k}{3^{k}}+\frac{3}{2(k+1)}.

Therefore, as kk\to\infty, πA0\pi_{A}\to 0

2.2.3 Relating πC1\pi_{C_{1}}, πC2\pi_{C_{2}} and iπBi\sum_{i}\pi_{B_{i}} when α=11/k\alpha=1-1/k

Claim:

πC1=πC2\pi_{C_{1}}=\pi_{C_{2}}

and

πC1=1/k3+k+(11/k)i=1kπBi3+πC1/3+(11/k)πC1/2\pi_{C_{1}}=\frac{1/k}{3+k}+(1-1/k)\cdot\frac{\sum_{i=1}^{k}\pi_{B_{i}}}{3}+\pi_{C_{1}}/3+(1-1/k)\cdot\pi_{C_{1}}/2

and hence

πC2=1/k3+k+(11/k)i=1kπBi3+πC2/3+(11/k)πC2/2\pi_{C_{2}}=\frac{1/k}{3+k}+(1-1/k)\cdot\frac{\sum_{i=1}^{k}\pi_{B_{i}}}{3}+\pi_{C_{2}}/3+(1-1/k)\cdot\pi_{C_{2}}/2
Proof 2.7.

Now, consider ΔπC1\Delta\pi_{C_{1}} and ΔπC2.\Delta\pi_{C_{2}}. For C1C_{1}, the vertex at the bottom of the chain of vertices, we have

ΔπC1=1/k3+k+(11/k)i=1kπBi3+πC2/3+(11/k)πC1/2πC1\Delta\pi_{C_{1}}=\frac{1/k}{3+k}+(1-1/k)\cdot\frac{\sum_{i=1}^{k}\pi_{B_{i}}}{3}+\pi_{C_{2}}/3+(1-1/k)\cdot\pi_{C_{1}}/2-\pi_{C_{1}}

This follows similarly to the case for AA and πA\pi_{A} previously. The first term is represents mass jumping from some vertex to C1C_{1}; the second term is the probability mass that is at BiB_{i} and returns to C1C_{1}; the third term is the mass from C2C_{2} that goes back to C1C_{1}; the fourth term is the probability that C1C_{1} gets from its self loop; and the fifth term is everything leaving C1C_{1}.

For ΔπC1=0\Delta\pi_{C_{1}}=0, we have

πC1=1/k3+k+(11/k)i=1kπBi3+πC2/3+(11/k)πC1/2\pi_{C_{1}}=\frac{1/k}{3+k}+(1-1/k)\cdot\frac{\sum_{i=1}^{k}\pi_{B_{i}}}{3}+\pi_{C_{2}}/3+(1-1/k)\cdot\pi_{C_{1}}/2

Similarly for πC2\pi_{C_{2}}, we have

ΔπC2=1/k3+k+(11/k)i=1kπBi3+(11/k)πC2/3+(11/k)πC1/2πC2\Delta\pi_{C_{2}}=\frac{1/k}{3+k}+(1-1/k)\cdot\frac{\sum_{i=1}^{k}\pi_{B_{i}}}{3}+(1-1/k)\cdot\pi_{C_{2}}/3+(1-1/k)\cdot\pi_{C_{1}}/2-\pi_{C_{2}}

and

(2.2) πC2=1/k3+k+(11/k)i=1kπBi3+(11/k)πC2/3+(11/k)πC1/2\pi_{C_{2}}=\frac{1/k}{3+k}+(1-1/k)\cdot\frac{\sum_{i=1}^{k}\pi_{B_{i}}}{3}+(1-1/k)\cdot\pi_{C_{2}}/3+(1-1/k)\cdot\pi_{C_{1}}/2

The symmetry for the equations for πC1\pi_{C_{1}} and πC2\pi_{C_{2}} give that

πC1=πC2\pi_{C_{1}}=\pi_{C_{2}}

Now, let us relate πC2\pi_{C_{2}} to the πBi\pi_{B_{i}}.

We can substitute πC2\pi_{C_{2}} in for πC1\pi_{C_{1}} in the equation for πC2\pi_{C_{2}}

πC2=1/k3+k+(11/k)i=1kπBi3+(11/k)πC2/3+(11/k)(πC2)/2\pi_{C_{2}}=\frac{1/k}{3+k}+(1-1/k)\cdot\frac{\sum_{i=1}^{k}\pi_{B_{i}}}{3}+(1-1/k)\cdot\pi_{C_{2}}/3+(1-1/k)\cdot(\pi_{C_{2}})/2
(2.3) πC2=1/k3+k+(11/k)i=1kπBi3+(11/k)πC256\pi_{C_{2}}=\frac{1/k}{3+k}+(1-1/k)\cdot\frac{\sum_{i=1}^{k}\pi_{B_{i}}}{3}+\frac{(1-1/k)\cdot\pi_{C_{2}}\cdot 5}{6}

Finally, let us each consider the πBi\pi_{B_{i}}.

ΔπBi=1/k3+k+(11/k)πBi13πBi\Delta\pi_{B_{i}}=\frac{1/k}{3+k}+(1-1/k)\cdot\frac{\pi_{B_{i-1}}}{3}-\pi_{B_{i}}

And thus for the stationary distribution,

πBi=1/k3+k+(11/k)πBi13\pi_{B_{i}}=\frac{1/k}{3+k}+(1-1/k)\cdot\frac{\pi_{B_{i-1}}}{3}

And also for B1B_{1},

πB1=1/k3+k+(11/k)πC23\pi_{B_{1}}=\frac{1/k}{3+k}+(1-1/k)\cdot\frac{\pi_{C_{2}}}{3}
πBi>(11/k)πBi13\pi_{B_{i}}>(1-1/k)\frac{\pi_{B_{i-1}}}{3}

and

πB1>(11/k)πC23.\pi_{B_{1}}>(1-1/k)\frac{\pi_{C_{2}}}{3}.

So by induction,

(2.4) πBi>(11/k)iπC23i\pi_{B_{i}}>(1-1/k)^{i}\frac{\pi_{C_{2}}}{3^{i}}

and

i=1kπBi>i=1k(11/k)iπC23i=(πC2πC2(11/k3)k)(111/k3)\sum_{i=1}^{k}\pi_{B_{i}}>\sum_{i=1}^{k}(1-1/k)^{i}\frac{\pi_{C_{2}}}{3^{i}}=\frac{(\pi_{C_{2}}-\pi_{C_{2}}(\frac{1-1/k}{3})^{k})}{(1-\frac{1-1/k}{3})}

This equation comes from viewing the previous equations as a finite geometric series.

2.2.4 Showing πC1,πC225\pi_{C_{1}},\pi_{C_{2}}\to\frac{2}{5}

We will now evaluate the limit with our choice α=11k\alpha=1-\frac{1}{k}. We already know that as kk\to\infty, πA0\pi_{A}\to 0; now we show that πC12/5\pi_{C_{1}}\to 2/5, πC22/5\pi_{C_{2}}\to 2/5.

Claim:

πC1=πC225\pi_{C_{1}}=\pi_{C_{2}}\to\frac{2}{5}
Proof 2.8.

We will show that limkπC225\lim_{k\to\infty}\pi_{C_{2}}\leq\frac{2}{5} and that limkπC225\lim_{k\to\infty}\pi_{C_{2}}\geq\frac{2}{5}

Let’s start with

πC1+πC2+πA+i=1kπBi=1.\pi_{C_{1}}+\pi_{C_{2}}+\pi_{A}+\sum_{i=1}^{k}\pi_{B_{i}}=1.

and so

limk[πC1+πC2+πA+i=1kπBi]=1.\lim_{k\to\infty}\left[\pi_{C_{1}}+\pi_{C_{2}}+\pi_{A}+\sum_{i=1}^{k}\pi_{B_{i}}\right]=1.

By (2.2) we can substitute

πC1\displaystyle\pi_{C_{1}} =\displaystyle= πC2\displaystyle\pi_{C_{2}}

into the above:

limk[2πC2+πA+i=1kπBi]=1.\lim_{k\to\infty}\left[2\pi_{C_{2}}+\pi_{A}+\sum_{i=1}^{k}\pi_{B_{i}}\right]=1.

and by distributing the limits:

2[limkπC2]+[limkπA]+[limki=1kπBi]=1.2\left[\lim_{k\to\infty}\pi_{C_{2}}\right]+\left[\lim_{k\to\infty}\pi_{A}\right]+\left[\lim_{k\to\infty}\sum_{i=1}^{k}\pi_{B_{i}}\right]=1.

Since πA0\pi_{A}\to 0, We can find the total ranking as

i=1kπBi+πC1+πC21.\sum_{i=1}^{k}\pi_{B_{i}}+\pi_{C_{1}}+\pi_{C_{2}}\to 1.

To find an upper bound for πC2\pi_{C_{2}} we can sum the Pagerank values in C2C_{2}, C1C_{1} and the value of the πBi\pi_{B_{i}}s that comes from C2C_{2}. This gives us:

1>πC1+((πC2πC2(11/k3)k)(111/k3)1>\pi_{C_{1}}+\frac{((\pi_{C_{2}}-\pi_{C_{2}}\cdot(\frac{1-1/k}{3})^{k})}{(1-\frac{1-1/k}{3})}

By taking the limit of both sides

limk1limk[πC1+(πC2πC2(11/k3)k(111/k3)]\ \lim_{k\to\infty}1\geq\lim_{k\to\infty}\left[\pi_{C_{1}}+\frac{(\pi_{C_{2}}-\pi_{C_{2}}\cdot(\frac{1-1/k}{3})^{k}}{(1-\frac{1-1/k}{3})}\right]
1limk[πC1]+limk[(πC2πC2(11/k3)k(111/k3)]1\geq\lim_{k\to\infty}\left[\pi_{C_{1}}\right]+\lim_{k\to\infty}\left[\frac{(\pi_{C_{2}}-\pi_{C_{2}}\cdot(\frac{1-1/k}{3})^{k}}{(1-\frac{1-1/k}{3})}\right]

Splitting the fraction and substituting πC2\pi_{C_{2}} in for πC1\pi_{C_{1}}

1>limk(πC2)+[limkπC2(111/k3)]limk[πC2(11/k3)k(111/k3)]1>\lim_{k\to\infty}(\pi_{C_{2}})+\left[\lim_{k\to\infty}\frac{\pi_{C_{2}}}{(1-\frac{1-1/k}{3})}\right]-\lim_{k\to\infty}\left[\frac{\pi_{C_{2}}\cdot(\frac{1-1/k}{3})^{k}}{(1-\frac{1-1/k}{3})}\right]

Taking the limits

1limkπC2+limkπC22301\geq\lim_{k\to\infty}\pi_{C_{2}}+\lim_{k\to\infty}\frac{\pi_{C_{2}}}{\frac{2}{3}}-0

And so by solving for limkπC2\lim_{k\to\infty}\pi_{C_{2}},

2/5limkπC22/5\geq\lim_{k\to\infty}\pi_{C_{2}}

We now find a lower bound for πC2\pi_{C_{2}}.

Let’s start with (2.2):

πC2=1/k3+k+(11/k)i=1kπBi3+(11/k)πC256\pi_{C_{2}}=\frac{1/k}{3+k}+(1-1/k)\cdot\frac{\sum_{i=1}^{k}\pi_{B_{i}}}{3}+\frac{(1-1/k)\cdot\pi_{C_{2}}\cdot 5}{6}

By removing the first term,

πC2>(11/k)i=1kπBi3+5(11/k)πC26)\pi_{C_{2}}>(1-1/k)\frac{\sum_{i=1}^{k}\pi_{B_{i}}}{3}+\frac{5(1-1/k)\pi_{C_{2}}}{6})

Combining the πC2\pi_{C_{2}} terms,

πC2πC256+5πC26k>(11/k)i=1kπBi3\pi_{C_{2}}-\frac{\pi_{C_{2}}\cdot 5}{6}+\frac{5\pi_{C_{2}}}{6k}>(1-1/k)\cdot\frac{\sum_{i=1}^{k}\pi_{B_{i}}}{3}

Simplifying

πC26+5πC26k>(11/k)i=1kπBi3\frac{\pi_{C_{2}}}{6}+\frac{5\pi_{C_{2}}}{6k}>(1-1/k)\cdot\frac{\sum_{i=1}^{k}\pi_{B_{i}}}{3}

Therefore,

(2.5) i=1kπBi<(πC26+5πC26k)3(11/k)\sum_{i=1}^{k}\pi_{B_{i}}<\left(\frac{\pi_{C_{2}}}{6}+\frac{5\pi_{C_{2}}}{6k}\right)\cdot\frac{3}{(1-1/k)}

By applying this,

πC1+πC2+πA+i=1kπBi=1\pi_{C_{1}}+\pi_{C_{2}}+\pi_{A}+\sum_{i=1}^{k}\pi_{B_{i}}=1
1\displaystyle 1 <\displaystyle< πC2+πC2+k3k+12k\displaystyle\pi_{C_{2}}+\pi_{C_{2}}+\frac{k}{3^{k}}+\frac{1}{2k}
+(πC26+5πC26k)3(11/k)\displaystyle+\left(\frac{\pi_{C_{2}}}{6}+\frac{5\pi_{C_{2}}}{6k}\right)\frac{3}{(1-1/k)}

Which will help us to find πC2\pi_{C_{2}} since

1\displaystyle 1 <\displaystyle< πC2+πC2+πC22(11/k)+k3k+12k+5πC22k2\displaystyle\pi_{C_{2}}+\pi_{C_{2}}+\frac{\pi_{C_{2}}}{2(1-1/k)}+\frac{k}{3^{k}}+\frac{1}{2k}+\frac{5\pi_{C_{2}}}{2k-2}

Or solving for πC2\pi_{C_{2}},

πC2>25πC22(kk11)k3k12k5πC22k2\pi_{C_{2}}>\frac{2}{5}-\frac{\pi_{C_{2}}}{2}(\frac{k}{k-1}-1)-\frac{k}{3^{k}}-\frac{1}{2k}-\frac{5\pi_{C_{2}}}{2k-2}

Let d=πC22(kk11)+k3k+12k+5πC22k2d=\frac{\pi_{C_{2}}}{2}(\frac{k}{k-1}-1)+\frac{k}{3^{k}}+\frac{1}{2k}+\frac{5\pi_{C_{2}}}{2k-2}. We do this because as k,d0k\to\infty,d\to 0 and because πC2>25d\pi_{C_{2}}>\frac{2}{5}-d which implies:

limkπC1=limkπC2=25\lim_{k\to\infty}\pi_{C_{1}}=\lim_{k\to\infty}\pi_{C_{2}}=\frac{2}{5}

2.2.5 Computing πα=11/k2\|\pi_{\alpha=1-1/k}\|^{2}

To compute πα=11/k2\|\pi_{\alpha=1-1/k}\|^{2}, we are not necessarily concerned with what i=1kπBi\sum_{i=1}^{k}\pi_{B_{i}} is equal to, rather we are more concerned about what i=1kπBi2\sum_{i=1}^{k}\pi_{B_{i}}^{2} is equal to.

Using the fact that πC1=πC2\pi_{C_{1}}=\pi_{C_{2}} and πC22/5\pi_{C_{2}}\to 2/5 and πC2+i=1kπBi\pi_{C_{2}}+\sum_{i=1}^{k}\pi_{B_{i}} goes to a geometric series we now have the following,

Lemma 2.9.
limkπα=11/k221750\lim_{k\to\infty}\|\pi_{\alpha=1-1/k}\|_{2}^{2}\to\frac{17}{50}

Proof 2.10.

As before, we will provide an lower bound and then an upper bound for limkπα=11/k22.\lim_{k\to\infty}\|\pi_{\alpha=1-1/k}\|_{2}^{2}.

Let us begin with the lower bound.

πα=11/k22\displaystyle\|\pi_{\alpha=1-1/k}\|_{2}^{2} =\displaystyle= πA2+πC12+πC22+ikπBi2\displaystyle\pi_{A}^{2}+\pi_{C_{1}}^{2}+\pi_{C_{2}}^{2}+\sum_{i}^{k}\pi_{B_{i}}^{2}
>\displaystyle> πA2+πC12+πC22+i=1k(11/k)iπC23i\displaystyle\pi_{A}^{2}+\pi_{C_{1}}^{2}+\pi_{C_{2}}^{2}+\sum_{i=1}^{k}(1-1/k)^{i}\frac{\pi_{C_{2}}}{3^{i}}
=\displaystyle= πA2+πC12+πC22πC22(11/k3)2k1(11/k3)2\displaystyle\pi_{A}^{2}+\pi_{C_{1}}^{2}+\frac{\pi_{C_{2}}^{2}-\pi_{C_{2}}^{2}(\frac{1-1/k}{3})^{2k}}{1-\left(\frac{1-1/k}{3}\right)^{2}}

Where the first inequality follows from (2.4). The second equality applies the finite geometric series.

So by applying the limit on both sides, we have

limkπα=11/k22\displaystyle\lim_{k\to\infty}\|\pi_{\alpha=1-1/k}\|_{2}^{2} \displaystyle\geq limk(πA2+πC12+πC22πC22(11/k3)2k1(11/k3)2)\displaystyle\lim_{k\to\infty}\left(\pi_{A}^{2}+\pi_{C_{1}}^{2}+\frac{\pi_{C_{2}}^{2}-\pi_{C_{2}}^{2}(\frac{1-1/k}{3})^{2k}}{1-\left(\frac{1-1/k}{3}\right)^{2}}\right)
=\displaystyle= limkπA2+limkπC12+limkπC22πC22(11/k3)2k1(11/k3)2\displaystyle\lim_{k\to\infty}\pi_{A}^{2}+\lim_{k\to\infty}\pi_{C_{1}}^{2}+\lim_{k\to\infty}\frac{\pi_{C_{2}}^{2}-\pi_{C_{2}}^{2}(\frac{1-1/k}{3})^{2k}}{1-\left(\frac{1-1/k}{3}\right)^{2}}
=\displaystyle= 0+425+49258\displaystyle 0+\frac{4}{25}+\frac{4\cdot 9}{25\cdot 8}

To find an upper bound, we only need an upper bound for the πBi\pi_{B_{i}}. We can observe that

πBi=(1/3)iπC2(11k)i+n=1i1(1/3)in(11k)in1k)3+k.\pi_{B_{i}}=(1/3)^{i}\pi_{C_{2}}\left(1-\frac{1}{k}\right)^{i}+\sum_{n=1}^{i-1}\frac{(1/3)^{i-n}(1-\frac{1}{k})^{i-n}\cdot\frac{1}{k})}{3+k}.

The first term above contains all of the probability mass that has gone through C2C_{2} since it last jumped and the second term contains the probability mass that has not (i.e., it jumped onto vertex BjB_{j} for j<ij<i).

By overestimating the second term of the equation above with an infinite geometric series, it is bounded by 3/k2(3+k)\frac{3/k}{2(3+k)}. Ergo

πBi<(1/3)iπC2(11k)i+3/k2(3+k).\pi_{B_{i}}<(1/3)^{i}\pi_{C_{2}}\left(1-\frac{1}{k}\right)^{i}+\frac{3/k}{2(3+k)}.

However, we are concerned about πBi2\pi_{B_{i}}^{2} so by expanding we have

πBi2<(1/3)2iπC22(11/k)2i+(1/3)iπC2(11/k)i3/k2(3+k)+(3/k2(3+k))2.\pi_{B_{i}}^{2}<(1/3)^{2i}\pi_{C_{2}}^{2}(1-1/k)^{2i}+(1/3)^{i}\pi_{C_{2}}(1-1/k)^{i}\frac{3/k}{2(3+k)}+\left(\frac{3/k}{2(3+k)}\right)^{2}.

Since (1/3)iπC2(11/k)i<1(1/3)^{i}\pi_{C_{2}}(1-1/k)^{i}<1 we know that

πBi2<(1/3)2iπC22(11/k)2i+3/k2(3+k)+(3/k2(3+k))2.\pi_{B_{i}}^{2}<(1/3)^{2i}\pi_{C_{2}}^{2}(1-1/k)^{2i}+\frac{3/k}{2(3+k)}+(\frac{3/k}{2(3+k)})^{2}.

So

πα=11/k22\displaystyle\|\pi_{\alpha=1-1/k}\|_{2}^{2}
=\displaystyle= πC12+(πC22+i=1kπBi2)+πA2.\displaystyle\pi_{C_{1}}^{2}+\left(\pi_{C_{2}}^{2}+\sum_{i=1}^{k}\pi_{B_{i}}^{2}\right)+\pi_{A}^{2}.
<\displaystyle< πC12+(πC22+i=1k[(1/3)2iπC22(11/k)2i+3/k2(3+k)+(3/k2(3+k))2])+πA2.\displaystyle\pi_{C_{1}}^{2}+\left(\pi_{C_{2}}^{2}+\sum_{i=1}^{k}\left[(1/3)^{2i}\pi_{C_{2}}^{2}(1-1/k)^{2i}+\frac{3/k}{2(3+k)}+(\frac{3/k}{2(3+k)})^{2}\right]\right)+\pi_{A}^{2}.
=\displaystyle= πC12+(πC22+i=1k(1/3)2iπC22(11/k)2i)+k3/k2(3+k)+k(3/k2(3+k))2+πA2.\displaystyle\pi_{C_{1}}^{2}+\left(\pi_{C_{2}}^{2}+\sum_{i=1}^{k}(1/3)^{2i}\pi_{C_{2}}^{2}(1-1/k)^{2i}\right)+k\frac{3/k}{2(3+k)}+k(\frac{3/k}{2(3+k)})^{2}+\pi_{A}^{2}.
<\displaystyle< πC12+πC221(1/9)+k(3/k2(3+k)+(3/k2(3+k))2)+πA2.\displaystyle\pi_{C_{1}}^{2}+\frac{\pi_{C_{2}}^{2}}{1-(1/9)}+k\left(\frac{3/k}{2(3+k)}+\left(\frac{3/k}{2(3+k)}\right)^{2}\right)+\pi_{A}^{2}.

where the first inequality follows using the previous estimate of πBi\pi_{B_{i}}, the second inequality follows from overestimating the geometric sum as an infinite series.

So by applying the limit, we have

limkπα=11/k22limk[425+49258+32(3+k)+9/k4(3+k)2+πA2]\lim_{k\to\infty}\|\pi_{\alpha=1-1/k}\|_{2}^{2}\leq\lim_{k\to\infty}\left[\frac{4}{25}+\frac{4\cdot 9}{25\cdot 8}+\frac{3}{2(3+k)}+\frac{9/k}{4(3+k)^{2}}+\pi_{A}^{2}\right]

which goes to

425+49258=1750\frac{4}{25}+\frac{4\cdot 9}{25\cdot 8}=\frac{17}{50}

as kk\to\infty.

2.2.6 Computing πα=1πα=11/k2\|\pi_{\alpha=1}-\pi_{\alpha=1-1/k}\|_{2}

Since when α=1\alpha=1 vertex πA=1\pi_{A}=1, and α=11/k\alpha=1-1/k, πA0\pi_{A}\to 0, the sum

πα=1πα=11/k2=i=1(πα=1,vπα=11/k,v)2\|\pi_{\alpha=1}-\pi_{\alpha=1-1/k}\|_{2}=\sum_{i=1}(\pi_{\alpha=1,v}-\pi_{\alpha=1-1/k,v})^{2}

has either πα=1,v0\pi_{\alpha=1,v}\to 0 or πα=11/k,v0\pi_{\alpha=1-1/k,v}\to 0, so

limkπα=11/kπα=12=limkπα=11/k2+limkπα=12.\lim_{k\to\infty}\|\pi_{\alpha=1-1/k}-\pi_{\alpha=1}\|^{2}=\lim_{k\to\infty}\|\pi_{\alpha=1-1/k}\|^{2}+\lim_{k\to\infty}\|\pi_{\alpha=1}\|^{2}.

We have

limkπα=11/kπα=1πα=11/k2+πα=121750+1=6750\lim_{k\to\infty}\|\pi_{\alpha=1-1/k}-\pi_{\alpha=1}\|\to\sqrt{\|\pi_{\alpha=1-1/k}\|^{2}+\|\pi_{\alpha=1}\|^{2}}\to\sqrt{\frac{17}{50}+1}=\sqrt{\frac{67}{50}}

\triangle

Indeed the deceiving graph from Figure 2 shows πα=11/kπα=167501.1575\|\pi_{\alpha=1-1/k}-\pi_{\alpha=1}\|\to\sqrt{\frac{67}{50}}\approx 1.1575 as α1\alpha\to 1.

Refer to caption
Figure 2: A plot of x=αx=\alpha versus y=πα=1πα=xy=||\pi_{\alpha={1}}-\pi_{\alpha=x}|| using the construction from Theorem 2.1, Γ(k)\Gamma(k), for sufficiently large kk. The resolution of the plot is unable to capture that |πα=1πα=x||0|\pi_{\alpha={1}}-\pi_{\alpha=x}||\to 0 as x1x\to 1.

3 The Open Problem

Originally, in this study, our construction had only one “CC” vertex. This would lead to a limiting value of 43=1+(12)2+(122)2+1.15470\sqrt{\frac{4}{3}}=\sqrt{1+\left(\frac{1}{2}\right)^{2}+\left(\frac{1}{2^{2}}\right)^{2}+\ldots}\approx 1.15470 which is just a tad less than 67501.15758\sqrt{\frac{67}{50}}\approx 1.15758.

This would seem to suggest that adding more “CC” vertices (which we call the set C) would be beneficial as in Figure 3. This would result in a larger total probability mass among the CiC_{i}. However, splitting the probability mass among more vertices would result in a smaller 2-norm when squaring. For instance, if πα1\pi_{\alpha_{1}} was concentrated at one vertex whereas πα2\pi_{\alpha_{2}} was concentrated on more than two other vertices equally, the resulting difference would be at most 1+3132=43\sqrt{1+3\frac{1}{3^{2}}}=\sqrt{\frac{4}{3}}, which is worse than our result.

AABkB_{k}\cdotsB1B_{1}CiC_{i}\cdotsC1C_{1}
Figure 3: A more general construction with more CC vertices. The optimal value of πα1πα2\|\pi_{\alpha_{1}}-\pi_{\alpha_{2}}\| occurs for two CC vertices.

Consider a similar construction to the path we had earlier but with a varying number of vertices in CC. This general construction is the same as the previous version except we can add a fixed amount of vertices that are copies (in that they have the same in-arcs and out-arcs) of C1C_{1} except they do not have an arc to B1B_{1} and they have an arc to and from every other vertex in CC.

We will give an informal argument why having two vertices in CC is the best for this more general construction. As in the proofs above, for α\alpha close to (but not equal to) 11, the probability mass always sums to 11 and approximately forms a geometric series, so we have

1(m1)πCi+πCi11m+11\approx(m-1)\pi_{C_{i}}+\frac{\pi_{C_{i}}}{1-\frac{1}{m+1}}

where πCi\pi_{C_{i}} is the Pagerank each vertex in any such vertex in CC and mm is the number of such vertices. By solving for πCi\pi_{C_{i}} we have

πCim1+m2.\pi_{C_{i}}\approx\frac{m}{1+m^{2}}.

Based on this result, we can make a formula for π22\|\pi\|_{2}^{2}

π22\displaystyle\|\pi\|_{2}^{2} \displaystyle\approx mπCi2+i=1kπBi2\displaystyle m\pi_{C_{i}}^{2}+\sum_{i=1}^{k}\pi_{B_{i}}^{2}
\displaystyle\approx mπCi2+i=1k1(m+1)2iπCi2\displaystyle m\pi_{C_{i}}^{2}+\sum_{i=1}^{k}\frac{1}{(m+1)^{2i}}\pi_{C_{i}}^{2}
\displaystyle\approx πCi2[m1+111(1+m)2]\displaystyle\pi_{C_{i}}^{2}\left[m-1+\frac{1}{1-\frac{1}{(1+m)^{2}}}\right]
=\displaystyle= m2(m2+1)2[m1+111(1+m)2]\displaystyle\frac{m^{2}}{\left(m^{2}+1\right)^{2}}\left[m-1+\frac{1}{1-\frac{1}{(1+m)^{2}}}\right]
=\displaystyle= m4+2m3+m(m+2)(m2+1)2\displaystyle\frac{m^{4}+2m^{3}+m}{(m+2)\left(m^{2}+1\right)^{2}}

As a result, we have that

πα1πα21+m4+2m3+m(m+2)(m2+1)2\|\pi_{\alpha_{1}}-\pi_{\alpha_{2}}\|\approx\sqrt{1+\frac{m^{4}+2m^{3}+m}{(m+2)\left(m^{2}+1\right)^{2}}}

This function reaches a maximum of approximately 1.360390\sqrt{1.360390} at approximately m=1.445036m=1.445036, but since this problem only accepts discrete answers the optimum is either at m=1m=1 or m=2m=2. For m=1m=1 we have 43\sqrt{\frac{4}{3}}, and for m=2m=2, we have 6750\sqrt{\frac{67}{50}} slightly higher as shown in Figure 4.

Refer to caption
Figure 4: Plot of f(x)=1+x4+2x3+x(x+2)(x2+1)2f(x)=\sqrt{1+\frac{x^{4}+2x^{3}+x}{(x+2)\left(x^{2}+1\right)^{2}}} with the values for x=1,2,x=1,2,\ldots indicated. For positive integral values f(x)f(x) is maximized at x=2x=2 indicating that 2 “C” vertices provides for an optimal construction.

We can also ask about the supremum values of πα1πα2p\|\pi_{\alpha_{1}}-\pi_{\alpha_{2}}\|_{p} for different p1p\geq 1 Our construction achieves the supremum values for both the 11- and the \infty-norms, The maximum that can be possibly achieved for the \infty norm for the difference between two Pagerank vectors is 1, and these constructions achieve that because when α=1\alpha=1, πA=1\pi_{A}=1 and when α=11/k\alpha=1-1/k, πA0\pi_{A}\to 0. Similarly, the maximum that can be possibly achieved for the 11-norm for the difference between two Pagerank vectors is 2, and our constructions achieve that because when α=1\alpha=1, πA=1\pi_{A}=1 and when α=11/k\alpha=1-1/k, πA0\pi_{A}\to 0 and so the probability mass is distributed among the other vertices and so πα1πα211\|\pi_{\alpha_{1}}-\pi_{\alpha_{2}}\|_{1}\to 1. However, it is remains unclear if our construction is optimal for 1<p<1<p<\infty.

We would hope the argument above is convincing that 6750\sqrt{\frac{67}{50}} is perhaps the best possible for the 2-norm, but this is far from a proof! In fact, it highlights why optimization is difficult for discrete combinatorial objects. There is much work on optimizing a graph based on eigenvalues and/or eigenvectors [1, 8, 9], and even among them, there are several cases where the optimal graph follows a canonical construction in small cases but becomes different in larger cases [5].

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