Elegant vertex labelings with prime numbers

TL;DR

This paper explores a novel graph labeling method assigning odd primes to vertices, aiming to produce edge differences that cover the first even numbers, with a conjecture about paths and an algorithm for constructing such labelings.

Contribution

It introduces the concept of elegant prime labelings for graphs, especially trees and paths, and provides an algorithm to generate elegant paths for up to 3500 vertices.

Findings

Successfully generated elegant paths for all n up to 3500

Conjecture that all paths are elegant with prime labelings

Proposes an algorithm for constructing elegant labelings

Abstract

We consider graph labelings with an assignment of odd prime numbers to the vertices. Similarly to graceful graphs, a labeling is said to be elegant if the absolute differences between the labels of adjacent vertices describe exactly the first even numbers. The labels of an elegant tree with $n$ vertices are the first $n$ odd prime numbers and we want that the resulting edge labels are exactly the first even numbers up to $2n-2$. We conjecture that each path is elegant and we give the algorithm with which we got elegant paths of $n$ primes for all $n$ up to $n=3500$.

Equations

$$\begin{array}[]{rccl}\psi\;:&E&\longrightarrow&2\mathbb{N}^{*}\\ \\ &e=uv&\longrightarrow&\psi(e)=|\varphi(v)-\varphi(u)|\end{array}$$

$$m{d}^m\log d\underset{d\to \infty }{\sim }|\phi (v_0)-\phi (v_p)|\le \sum _{i=0}^{{}^{p-1}}|\phi (v_i)-\phi (v_{i+1})|\le 4m{d}^m\left(1+o(1)\right)$$

$$\left\{\left|p_{\sigma (i+1)}-p_{\sigma (i)}\right|;1\le i\le n-1\right\}=\left\{2,4,6,\dots ,2n-2\right\}.$$

$$F_g({\mathcal{Q}}^{\ast })=F_g(\mathcal{Q})\cup \left\{\delta \right\}\setminus \left\{|f({\mathcal{Q}}_2)-f({\mathcal{Q}}_1)|\right\}.$$

$$F_g({\mathcal{Q}}^{\ast })=F_g(\mathcal{Q})\cup \left\{\delta \right\}\setminus \left\{|\ell ({\mathcal{Q}}_2)-\ell ({\mathcal{Q}}_1)|\right\}.$$

$$F_g({\mathcal{Q}}^{\ast })=F_g(\mathcal{Q})\cup \left\{\delta \right\}\setminus \left\{|\ell ({\mathcal{Q}}_2)-f({\mathcal{Q}}_1)|\right\}.$$

$$F_g({\mathcal{Q}}^{\ast })=F_g(\mathcal{Q}).$$

$$\textbf{C}\,:\,\left\{\begin{array}[]{c}\left(|r-p|\in F_{g}({\cal Q})\text{ and }|r-p|\neq|q-r|\in F_{g}({\cal Q})\right)\\ \text{or}\\ \left(|r-p|\in F_{g}({\cal Q})\text{ and }|q-r|=\delta\right)\\ \text{or}\\ \left(|q-r|\in F_{g}({\cal Q})\text{ and }|r-p|=\delta\right).\end{array}\right.$$

$$\left\{\left|r-\ell ({\mathcal{Q}}_1)\right|,\left|f({\mathcal{Q}}_2)-r\right|\right\}\subset F_g(Q)\cup \left\{\left|q-\ell ({\mathcal{Q}}_1)\right|,\left|f({\mathcal{Q}}_2)-q\right|\right\}.$$

$$\left\{\left|r-f({\mathcal{Q}}_1)\right|,\left|f({\mathcal{Q}}_2)-r\right|\right\}\subset F_g(Q)\cup \left\{\left|q-\ell ({\mathcal{Q}}_1)\right|,\left|f({\mathcal{Q}}_2)-q\right|\right\}.$$

$${\mathcal{Q}}_{10}:=5\underset{2}{\text{––}}7\underset{4}{\text{––}}3\underset{16}{\text{––}}19\underset{12}{\text{––}}31\underset{20}{\text{––}}11\underset{18}{\text{––}}29\underset{8}{\text{––}}37\underset{14}{\text{––}}23\underset{6}{\text{––}}17,$$

$${\mathcal{Q}}_{10}:=5\underset{2}{\text{––}}7\underset{10}{\text{––}}17\underset{6}{\text{––}}23\underset{14}{\text{––}}37\underset{8}{\text{––}}29\underset{18}{\text{––}}11\underset{20}{\text{––}}31\underset{12}{\text{––}}19\underset{16}{\text{––}}3,$$

$${\mathcal{Q}}_{10}:=5\underset{2}{\text{––}}7\underset{10}{\text{––}}17\underset{4}{\text{––}}13\underset{6}{\text{––}}19\underset{12}{\text{––}}31\underset{20}{\text{––}}11\underset{18}{\text{––}}29\underset{8}{\text{––}}37\underset{14}{\text{––23}},$$

$${\mathcal{Q}}_{10}:=29\underset{8}{\text{––}}37\underset{14}{\text{––}}23\underset{18}{\text{––}}5\underset{2}{\text{––}}7\underset{10}{\text{––}}17\underset{4}{\text{––}}13\underset{6}{\text{––}}19\underset{12}{\text{––}}31\underset{20}{\text{––}}11.$$

$${\mathcal{Q}}_{10}:=17\underset{10}{\text{––}}7\underset{2}{\text{––5}}\underset{18}{\text{––}}23\underset{14}{\text{––}}37\underset{8}{\text{––}}29\underset{16}{\text{––}}13\underset{6}{\text{––}}19\underset{12}{\text{––}}31\underset{20}{\text{––}}11,$$

$${\mathcal{Q}}_{11}:=17\underset{10}{\text{––}}7\underset{4}{\text{––3}}\underset{2}{\text{––5}}\underset{18}{\text{––}}23\underset{14}{\text{––}}37\underset{8}{\text{––}}29\underset{16}{\text{––}}13\underset{6}{\text{––}}19\underset{12}{\text{––}}31\underset{20}{\text{––}}11.$$

Full text

1 Introduction

In this paper, we adapt the notion of graceful graphs by considering an assignment of odd prime numbers to the vertices. Let $G=(V,E)$ be a graph, we look for labelings of the vertices with distinct odd primes which induce edge labelings with all even integers from $2$ up to $2|E|$. For instance in the tree displayed in Fig. 1, the first twelve odd primes are assigned to the vertices and we get all the even positive integers up to $22$. We call elegant any graph for which there exists such a labeling. We refer to Galian [2] for a very detailed survey about graph labelings.

Let us denote the increasing sequence of all odd prime numbers by $p_{1},p_{2},\ldots,p_{r},\ldots$ and $\mathbb{P}_{n}=\{p_{1},p_{2},\ldots,p_{n}\}$. We now precise the definition of an elegant graph.

Definition 1

Let $G=(V,E)$ be an undirected graph without loop or multiple edge, with $n$ vertices and $r$ edges. We say that $G$ is elegant if there exists an injective map $\varphi:V\to\left\{p_{1},p_{2},\ldots,p_{r+1}\right\}$ such that the induced map

[TABLE]

is a one-to-one correspondence from $E$ to $\{2,4,6,\ldots,2r\}$.

The complete graphs up to $K_{4}$ are graceful and elegant: Their elegant labelings are respectively determined by $\varphi(V)=\{3,5\},\{5,7,11\}$ and $\{7,11,17,19\}$. As in the case of graceful graphs, it seems that no other elegant complete graphs exists. We display an elegant labeling of the Petersen graph in Fig. 2.

Of course, if the number of vertices is too weak relatively to the maximal degree of the graph, then the graph is probably not (or cannot) be elegant. For instance, the star graphs $S_{n}$ with $|V(S_{n})|=n$ are not elegant as soon as the center has more than $8$ adjacent vertices (in fact, we verified that the only elegant star graphs are $S_{2},$ $S_{3}$, $S_{5}$, $S_{6}$ and $S_{9}$). Nevertheless, in the case of trees, we could hope that for each integer $d\geq 2$, the answer to the following question ${\rm A}_{d}$ be positive.

Question ${\rm A}_{d}$: Let $d\geq 2$ be an integer. Is there an integer $N_{d}$ such that each tree of maximal degree $d$ and with more than $N_{d}$ vertices is elegant?

The answer to ${\rm A}_{d}$ is negative for $d$ large enough that is an easy consequence of the prime number theorem, see for instance [1, 3]: Let us consider a symmetric and elegant tree $T$ rooted at a vertex of degree $d$, with exactly $d^{k}$ vertices of degree $d+1$ at each level $k=1,\ldots,m-1$ and with $d^{m}$ leaves at level $m$. We have $n=|V(T)|=1+d+\cdots+d^{m}$ and then $p_{n}\underset{d\to\infty}{\sim}md^{m}\log d$. Moreover, the absolute difference between labels of two adjacent vertices is less than $2n-2\sim_{d\to\infty}2d^{m}$. We consider the path $v_{0}v_{1}\cdots v_{k}\cdots v_{p}$ on the tree $T$, from the vertex $v_{0}$ labeled by $\varphi(v_{0})=3$ and ended by the vertex $v_{p}$ labeled by $\varphi(v_{p})=p_{n}$. Since $p\leq 2m$, we get

[TABLE]

which is impossible when $d$ is large enough. Since $n\to\infty$ when $m\to\infty$, there exists a minimal integer $d_{1}$ such that $A_{d}$ is false for all $d\geq d_{1}$.

As soon as $d=3$, it is difficult to know if $A_{d}$ is true or false (and we are far from the value $d=e^{4}$ which appears in the previous proof). Let us illustrate this point with an example: Let ${\rm C}_{n}$ be the regular caterpillar with $n$ vertices of degree $3$ and $n+2$ leaves. On the one hand, up to $n=25$ a trivial stocchastic algorithm has given elegant labelings of $C_{n}$ only for $n=3,5,10,18,19,20,22$ but it is quite possible that $C_{n}$ be elegant for all $n$ large enough. Let us recall that Rosa [4] proved that all caterpillars are graceful. On the other hand, up to $n=25$, as soon as we add a supplementary leaf $w$ on any leaf $v\in V(C_{n})$ or if we supress anywhere one leaf of $V(C_{n})$, then surprisingly our program finds in a few seconds that the modified tree is elegant.

Fortunately, the case $d=2$ seems not to be resistant and we are confident that $N_{2}=2$:

Conjecture 1

For all $n\geq 2$, the path of length $n$ is elegant.

Let us recall that a path of $n$ vertices is elegant if there exists a permutation $\sigma\in S_{n}$ such that

[TABLE]

For instance, up to $n=10$, the following labelings are elegant:

• •

$3$ $\underset{2}{\text{\textendash\textendash}}$ $5$

• •

$5$ $\underset{2}{\text{\textendash\textendash}}$ $3$ $\underset{4}{\text{\textendash\textendash}}$ $7$

• •

$11$ $\underset{6}{\text{\textendash\textendash}}$ $5$ $\underset{2}{\text{\textendash\textendash}}$ 3 $\underset{4}{\text{\textendash\textendash}}$ $7$

• •

$13$ $\underset{6}{\text{\textendash\textendash}}$ $7$ $\underset{4}{\text{\textendash\textendash}}$ $11$ $\underset{8}{\text{\textendash\textendash}}$ $3$ $\underset{2}{\text{\textendash\textendash}}$ $5$

• •

$7$ $\underset{2}{\text{\textendash\textendash}}$ $5$ $\underset{6}{\text{\textendash\textendash}}$ $11$ $\underset{8}{\text{\textendash\textendash}}$ $3$ $\underset{10}{\text{\textendash\textendash}}$ $13$ $\underset{4}{\text{\textendash\textendash}}$ $17$

• •

$13$ $\underset{4}{\text{\textendash\textendash}}$ $17$ $\underset{10}{\text{\textendash\textendash}}$ $7$ $\underset{12}{\text{\textendash\textendash}}$ $19$ $\underset{8}{\text{\textendash\textendash}}$ $11$ $\underset{6}{\text{\textendash\textendash}}$ $5$ $\underset{2}{\text{\textendash\textendash}}$ $3$

• •

$13$ $\underset{10}{\text{\textendash\textendash}}$ $3$ $\underset{4}{\text{\textendash\textendash}}$ $7$ $\underset{2}{\text{\textendash\textendash}}$ $5$ $\underset{14}{\text{\textendash\textendash}}$ $19$ $\underset{8}{\text{\textendash\textendash}}$ $11$ $\underset{12}{\text{\textendash\textendash}}$ $23$ $\underset{6}{\text{\textendash\textendash}}$ $17$

• •

$11$ $\underset{2}{\text{\textendash\textendash}}$ $13$ $\underset{8}{\text{\textendash\textendash}}$ $5$ $\underset{14}{\text{\textendash\textendash}}$ $19$ $\underset{10}{\text{\textendash\textendash}}$ $29$ $\underset{12}{\text{\textendash\textendash}}$ $17$ $\underset{6}{\text{\textendash\textendash}}$ $23$ $\underset{16}{\text{\textendash\textendash}}$ $7$ $\underset{4}{\text{\textendash\textendash}}3$

• •

$19$ $\underset{12}{\text{\textendash\textendash}}$ $31$ $\underset{14}{\text{\textendash\textendash}}$ $17$ $\underset{6}{\text{\textendash\textendash}}$ $23$ $\underset{10}{\text{\textendash\textendash}}$ $13$ $\underset{16}{\text{\textendash\textendash}}$ $29$ $\underset{18}{\text{\textendash\textendash}}$ $11$ $\underset{8}{\text{\textendash\textendash}}$ $3$ $\underset{4}{\text{\textendash\textendash}}$ $7$ $\underset{2}{\text{\textendash\textendash}}$ $5$

2 Operations on admissible paths

In the sequel, a path of $l$ primes ${\cal Q={\cal Q}}_{l}=q_{1}q_{2}\ldots q_{l}$ represents the labeling of the path of length $l$ with the primes $q_{1},q_{2},\ldots,q_{l}$ in this order; we will identify the vertices of a path and their label $q_{i}$.

Definition 2

Let $n\geq 2$ be a given integer.

We say that a path ${\cal Q}_{l}=q_{1}q_{2}\ldots q_{l}$ is admissible if the primes $q_{i}\in\mathbb{P}_{n}$ are distinct and if the set $E_{l}$ of the $l-1$ gaps $|q_{i+1}-q_{i}|$ is a subset of cardinal $l-1$ of $\{2,4,\ldots,2n-2\}$. 2. 2.

With regard to a path ${\cal Q}_{l}=q_{1}q_{2}\ldots q_{l}$ with $l<n$, a prime $p$ in $\mathbb{P}_{n}$ is said to be free if it is not a vertex of ${\cal Q}_{l}$. A gap $2k\leq 2n-2$ is said to be free if it does not belong to $E_{l}$.

We also adopt the notations :

• •

If ${\cal Q}=q_{1}q_{2}\ldots q_{l}$ then $\overline{{\cal Q}}=q_{l}q_{l-1}\ldots q_{1}$.

• •

If ${\cal Q}_{1}=q_{1}q_{2}\ldots q_{l}$ and ${\cal Q}_{2}=q_{l+1}q_{l+2}\ldots q_{p}$ , then ${\cal Q}_{1}{\cal Q}_{2}=q_{1}\ldots q_{l}q_{l+1}\ldots q_{p}$ is the concatenation of ${\cal Q}_{1}$ and ${\cal Q}_{2}$. When we want to add a prime to a path ${\cal Q}={\cal Q}_{1}{\cal Q}_{2}$ respectively on the left end, between ${\cal Q}_{1}$ and ${\cal Q}_{2}$ or on the right end, we note these paths $p{\cal Q},$ ${\cal Q}_{1}p{\cal Q}_{2}$ or ${\cal Q}p$.

• •

If ${\cal Q}=q_{1}q_{2}\ldots q_{l}$ then $f({\cal Q})=q_{1}$, $\ell({\cal Q})=q_{l}$ and ${\rm length}({\cal Q})=l$.

• •

If ${\cal Q}=q_{1}q_{2}\ldots q_{l}$ is admissible, $F_{p}({\cal Q})$ is the set of free primes for ${\cal Q}$ and $F_{g}({\cal Q})$ is the set of free gaps for ${\cal Q}$.

In the algorithm described in Section 3, we randomly apply transformations on admissible paths of length $l<n$ primes in order to find other admissible paths. Our aim is either to improve the length $l$ by adding a prime to the path, or to substitute a prime of the path for a free prime, or simply to shuffle the path. Proposition 1 gives two elementary tools $A1$-$A2$ and $A3$-$A4$ for shuffling an admissible path:

Proposition 1

Let $n\geq 3$ and ${\cal Q}={\cal Q}_{1}{\cal Q}_{2}$ be an admissible path of $l<n$ primes. We denote by $\delta=|f({\cal Q}_{2})-\ell({\cal Q}_{1})|$ the gap between ${\cal Q}_{1}$ and ${\cal Q}_{2}$.

1. A1.

If $|f({\cal Q}_{2})-f({\cal Q}_{1})|\in F_{g}({\cal Q})$, then the path ${\cal Q}^{*}=\overline{{\cal Q}}_{1}{\cal Q}_{2}$ is admissible and

[TABLE] 2. A2.

If $|\ell({\cal Q}_{2})-\ell({\cal Q}_{1})|\in F_{g}({\cal Q})$, then the path ${\cal Q}^{*}={\cal Q}_{1}\overline{{\cal Q}}_{2}$ is admissible and

[TABLE] 3. A3.

If $|\ell({\cal Q}_{2})-f({\cal Q}_{1})|\in F_{g}({\cal Q})$, then the path ${\cal Q}^{*}=Q_{2}Q_{1}$ is admissible and

[TABLE] 4. A4.

If $|\ell({\cal Q}_{2})-f({\cal Q}_{1})|=\delta$, then the path ${\cal Q}^{*}=Q_{2}Q_{1}$ is admissible and

[TABLE]

Remark 1** (Insertion of a free gap)**

In Algorithm 1 decribded in Section 3 and when we get an admissible transformation from ${\cal Q}$ to ${\cal Q}^{*}$ with a property $A_{i}$, we try to insert a free prime in ${\cal Q}^{*}$: If $r$ is a new free prime for ${\cal Q}^{*}$, we can try to insert it directly with Proposition 2. If a gap $\delta$ has become free, we can try to insert in ${\cal Q}^{*}$ by Proposition 2, any prime $r=s\pm\delta\in F_{p}({\cal Q})$ if ${\cal Q}^{*}={\cal Q}_{1}{\cal Q}_{2}$ with $s=\ell({\cal Q}_{1})$ or $s=f({\cal Q}_{2})$. Obviously, we also test in the algorithm if one of the paths $r{\cal Q}$ or ${\cal Q}r$ is admissible.

For instance, let us consider $n=7$ and the admissible path ${\cal Q}_{6}=7\textendash 19\textendash 17\textendash 11\textendash 3\textendash 13$. The last free gap is $4=|11-7|$ and we can apply the transformation given by A1 with ${\cal Q}_{1}=7\textendash 19\textendash 17$ and ${\cal Q}_{2}=11\textendash 3\textendash 13$, we get ${\cal Q}^{*}=\overline{{\cal Q}}_{1}{\cal Q}_{2}=17\textendash 19\textendash 7\textendash 11\textendash 3\textendash 13$ and the last free gap is now $\delta=6$. The last free prime is $r=5=11-\delta$ and we can insert $r$ by the third point of Proposition 2 with ${\cal Q}_{1}=17\textendash 19\textendash 17\textendash 11$ and ${\cal Q}_{2}=3\textendash 13$: We get an admissible and elegant path of seven primes with the admissible transformation ${\cal Q}_{6}={\cal Q}_{1}{\cal Q}_{2}\to{\cal Q}_{7}={\cal Q}_{1}r\overline{{\cal Q}_{2}}=17\textendash 19\textendash 7\textendash 11\textendash 5\textendash 13\textendash 3$.**

Proposition 2** (Insertion of a free prime)**

Let $n\geq 3$ and ${\cal Q=Q}_{1}{\cal Q}_{2}$ be an admissible path of $l<n$ primes and let $r\in F_{p}({\cal Q})$. We denote by C the condition

[TABLE]

If C is true for $\delta=|f({\cal Q}_{2})-l({\cal Q}_{1})|$ with

$p=\ell({\cal Q}_{1})$* and $q=f({\cal Q}_{2})$, then the path ${\cal Q}^{*}=Q_{1}rQ_{2}$ is admissible;* 2. 2.

$p=f({\cal Q}_{1})$* and $q=f({\cal Q}_{2})$, then the path ${\cal Q}^{*}=\overline{{\cal Q}_{1}}r{\cal Q}_{2}$ is admissible;* 3. 3.

$p=\ell({\cal Q}_{1})$* and $q=\ell({\cal Q}_{2})$, then the path ${\cal Q}^{*}={\cal Q}_{1}r\overline{{\cal Q}_{2}}$ is admissible.*

In order to substitute a prime $p$ of a path ${\cal Q}$ for a free prime, we can consider $36$ transformations $A_{5},A_{6},\ldots,A_{40}:{\cal Q}\in X\rightarrow{\cal Q}^{*}\in Y$ where $X=\left\{q{\cal Q}_{1}{\cal Q}_{2},{\cal Q}_{1}q{\cal Q}_{2},\right.$ $\left.{\cal Q}_{1}{\cal Q}_{2}q\right\}$ and $Y=$ $\left\{r{\cal Q}_{1}{\cal Q}_{2},r{\cal Q}_{1}{\cal\overline{Q}}_{2},r{\cal\overline{Q}}_{1}{\cal Q}_{2},r{\cal\overline{Q}}_{1}\overline{{\cal Q}}_{2},{\cal Q}_{1}r{\cal Q}_{2},{\cal Q}_{1}r{\cal\overline{Q}}_{2},{\cal\overline{Q}}_{1}r{\cal Q}_{2},{\cal\overline{Q}}_{1}r\overline{{\cal Q}}_{2},\right.$

${\cal Q}_{1}{\cal Q}_{2}r,{\cal Q}_{1}{\cal\overline{Q}}_{2}r,\left.{{\cal\overline{Q}}_{1}{\cal Q}_{2}r,\cal\overline{Q}}_{1}\overline{{\cal Q}}_{2}r\right\}.$ The following proposition details only the transformations ${\cal Q}={\cal Q}_{1}q{\cal Q}_{2}$ $\to{\cal Q}^{*}={\cal Q}_{1}r{\cal Q}_{2}$ and ${\cal Q}={\cal Q}_{1}q{\cal Q}_{2}\to{\cal Q}^{*}={\cal\overline{Q}}_{1}r{\cal Q}_{2}$, but it is trivial to find for which conditions the other transformations $A_{i}$ give admissible paths.

Proposition 3

Let $n\geq 3$ and ${\cal Q}={\cal Q}_{1}q{\cal Q}_{2}$ be an admissible path of $l<n$ primes. Let $r$ be a free prime.

The path ${\cal Q}^{*}={\cal Q}_{1}r{\cal Q}_{2}$ is admissible if $\left|r-\ell({\cal Q}_{1})\right|\neq\left|f({\cal Q}_{2})-r\right|$ and

[TABLE] 2. 2.

The path ${\cal Q}^{*}={\cal\overline{Q}}_{1}r{\cal Q}_{2}$ is admissible if $\left|r-f({\cal Q}_{1})\right|\neq\left|f({\cal Q}_{2})-r\right|$ and

[TABLE]

3 An algorithm for finding elegant paths

We have found elegant labelings for all paths up to $n=3500$ vertices with the algorithm we describe below. In Algorithm 1, we construct incrementally a sequence of paths: from a path ${\cal Q}_{l}$ of $l$ primes, we try either to add a prime or to modify the path ${\cal Q}_{l}$ without improving its length. The integer $n$ being fixed, our aim is to obtain $l=n$.

The first part 1-4 of Algorithm 1 is a trivial greedy algorithm, we simply try to add free primes on the ends of ${\cal Q}_{l}$. In the While statement of the step 5, we intensively and randomly use the transformations $A_{i}$ given in Section 2. When $m$ reaches $N$ without giving an elgant path of length $n$, we quit and we start a new run of Algorithm 1. With $N=40n$, we get with Algorithm 1, all the elegant paths of $n$ primes from $n=2$ up to $n=200$ in less than 5 seconds with one core of a 3.6 GHz processor.

If $n\geq 200$, we accelerate the calculations with Algorithm 2 in which we have chosen $N=20n$ et $c_{0}=20$: when a run reaches $l\geq n-\delta$, we do not give up the path if $l<n$ but we suppress the right end $q_{l}$ of the path ${\cal Q}_{l}$. We take for instance $\delta=1$ for $n=200$ and $\delta=5$ from $n=2000$ up to $n=3500$. This can be seen as a perturbation on a non-optimal and rigid configuration. When $n\in[2000,2100]$, the average run-time to find one elegant path with Algorithm 2 is 4 minutes; when $n\in[3400,3500]$, this average run-time becomes $40$ minutes. It is surprising to find a solution so easily among $3500!>10^{10000}$ permutations.

Algorithm 1

Randomly choose a prime $q_{1}$ in $\mathbb{P}_{n}$ and set $l:=1$, ${\cal Q}_{l}:=q_{1}$;

$l_{1}:=0;$

While $l_{1}\neq l$ and $l<n$ do

•

$l_{1}:=l_{1}+1$;

•

If we find $p\in F_{p}({\cal Q}_{l})$ s.t. ${\cal Q}_{l}p$ (or resp. $p{\cal Q}_{l}$) is admissible then we set ${\cal Q}_{l+1}:={\cal Q}_{l}p$ (or resp. ${\cal Q}_{l+1}:=p{\cal Q}_{l}$) and $l:=l+1$;

If $l=n$ then return the elegant path ${\cal Q}_{n}$ and quit.

While $l<n$ and $m<N$ do

•

Randomly choose $i\in\{1,2,3\}$.

Case $i=1$ : Look for an admissible transformation ${\cal Q}^{*}=\overline{{\cal Q}_{1}}{\cal Q}_{2}$ or ${\cal Q}_{1}\overline{{\cal Q}_{2}}$ of ${\cal Q}_{l}$ with A1 or A2. If we succeed in changing ${\cal Q}_{l}$, we set ${\cal Q}_{l}:={\cal Q}^{*}$. Then, we try to insert a free prime in this new path ${\cal Q}_{l}$ with Remark 1 and Prop. 2; if we succeed in this, we set $l:=l+1$ and ${\cal Q}_{l}:=\cal Q^{*}$, the path $\cal Q^{*}$ having been given by Prop. 6.

Case $i=2$ : If $|\ell({\cal Q}_{l})-f({\cal Q}_{l})|\in F_{g}({\cal Q}_{l})$ then randomly choose $u\in\{1,\ldots,l-1\}$ and set ${\cal Q}_{1}:=q_{1}\cdots q_{u}$, ${\cal Q}_{2}:=q_{u+1}\cdots q_{l}$ and ${\cal Q}_{l}:={\cal Q}_{2}{\cal Q}_{1}$. Then, if we succeed in inserting a free prime in this new ${\cal Q}_{l}$ with Remark 1 and Prop. 2, we set $l:=l+1$ and ${\cal Q}_{l}:=\cal Q^{*}$, the path $\cal Q^{*}$ having been given by Prop. 2.

Else there exists $u\in\{1,\ldots,l-1\}$ s.t. $|q_{u+1}-q_{u}|=|l({\cal Q}_{l})-f({\cal Q}_{l})|$ and we set ${\cal Q}_{l}:={\cal Q}_{2}{\cal Q}_{1}$ with ${\cal Q}_{1}=q_{1}\cdots q_{u}$ and ${\cal Q}_{2}=q_{u+1}\cdots q_{l}$.

Case $i=3$ : Look for a transformation among $A_{5},\ldots,A_{40}$ which gives an admissible path ${\cal Q}^{*}$. If we succeed in modifying ${\cal Q}_{l}$, we set ${\cal Q}_{l}:={\cal Q}^{*}$. If we succeed in inserting a free prime in this new ${\cal Q}_{l}$ with Remark 1 and Prop. 2, we set $l:=l+1$ and ${\cal Q}_{l}:=\cal Q^{*}$.

•

$m:=m+1$.

Return the path ${\cal Q}_{l}$ which is elegant if $l=n$.

Algorithm 2

$l:=0$;

While $l<n$ do

Do Algorithm 1 which gives an admissible path ${\cal Q}_{l}$ of length $l$;

If $l=n$ then return the elegant path ${\cal Q}_{n}$;

If $n-\delta\leq l<n$ then

c:=0;

While $l<n$ and $c<c_{0}$ do

(a)

Suppress the last prime $q_{l}$ of ${\cal Q}_{l}$ and set $l:=l-1$;

(b)

Do Step 5 of Algorithm 1 and set c:=c+1;

(c)

If $l=n$ then return the elegant path ${\cal Q}_{n}$.

Let us detail an example of a possible run of Algorithm 1 in the case $n=11$.

• •

Steps $1-4$ of Algorithm 1: We randomly choose $q_{1}=5$, $q_{2}=7,\;q_{3}=3,\ldots$, and we get

[TABLE]

the last free prime is $13$ and the free gap is $10$.

• •

Step $5$ of Algorithm 1: We randomly choose values for $i$ in $\{1,2,3\}$ :

• –

$i=1$: The gap $|17-7|=10$ is free and we apply the transformation ${\cal Q}={\cal Q}_{1}{\cal Q}_{2}\rightarrow{\cal Q}^{*}={\cal Q}_{1}\overline{{\cal Q}_{2}}$ with ${\cal Q}_{1}=5-7$ and ${\cal Q}_{2}=3-19-31-11-29-37-23-17$. We get

[TABLE]

the last free gap is now $4$.

• –

$i=3$ : We apply the transformation ${\cal Q}={\cal Q}_{1}{\cal Q}_{2}q\rightarrow{\cal Q}^{*}={\cal Q}_{1}r\overline{{\cal Q}_{2}}$ with ${\cal Q}_{1}=5-7-17$, ${\cal Q}_{2}=23-37-29-11-31-19$, $q=3$ and $r=13$. We find

[TABLE]

the free prime is now $3$ and the free gap is $16$.

• –

$i=2$ : Since the gap between the two ends is $|23-5|=18\notin F_{g}({\cal Q})$, we apply ${\cal Q}={\cal Q}_{1}{\cal Q}_{2}\rightarrow{\cal Q}^{*}={\cal Q}_{2}{\cal Q}_{1}$ with ${\cal Q}_{1}=5-7-17-13-19-31-11$ and ${\cal Q}_{2}=29-37-23$. We get

[TABLE]

• –

$i=1$ : The gap $|29-13|=16$ is free and we apply the transformation ${\cal Q}={\cal Q}_{1}{\cal Q}_{2}\rightarrow{\cal Q}^{*}=\overline{{\cal Q}_{1}}{\cal Q}_{2}$ with ${\cal Q}_{1}=29-37-23-5-7$ and ${\cal Q}_{2}=17-13-19-31-11$. We get

[TABLE]

and the last free gap is $4$. We can place the prime $3$ between $7$ and $5$ and we get an elegant path of eleven primes:

[TABLE]

Acknowledgements

Thanks to Shalom Eliahou for his encouragment and more.