Asynchronous Gathering in Rings with 4 Robots

Abstract

In this paper we consider the gathering of oblivious mobile robots in a n-node ring. In this context, the single class of configurations left open in the most recent study [2] is $\mathcal {SP}4$ (a special class of configurations with only four robots).

We present an algorithm to solve some of the most intricate configurations in $\mathcal {SP}4$, those that can lead to a change of the axis of symmetry. Our approach lays the methodological bases for closing the remaining open cases for $\mathcal {SP}4$ solvability.

A Appendix: Case Studies

In this section we investigate the gathering for some specific values of ring size n. For each value we identify the configurations for which our algorithm solves the problem and the configurations that are still open. We consider here only rings of size $n=4x+3$. For rings of size $n=4x+1$, the set $\mathcal {SP}4_g$ is empty and therefore we do not solve any $\mathcal {SP}4$ configurations in that case.

1.1 A.1 Ring of Size 7

There are 4 towerless configurations ($|\mathcal {I}| = 4$). There is no obviously-non-gatherable configuration ($|\mathcal {NG}|=0$). The set of admissible configurations contains 1 asymmetrical and 1 symmetrical ($|\mathcal {A}|=1+1=2$). There are therefore 2 configurations of $\mathcal {SP}4$:

• $C_1$ defined by $(a_1,b_1,a_1,c_1) = (0,3,0,0)$. $C_1\not \in \mathcal {SP}4_g$ since $c_1\ne b_1-1$.

• $C_2$ defined by $(a_2,b_2,a_2,c_2) = (1,1,1,0)$. $C_2\in \mathcal {SP}4_g$.

There is an algorithm solving the gathering problem if the initial configuration belongs to $\mathcal {A}\cup \{C_2\}$. The problem is open if the initial configuration is $C_1$.

1.2 A.2 Ring of Size 11

$|\mathcal {I}| = 20$, $|\mathcal {NG}|=0$, $|\mathcal {A}|=10+4=14$, and $|\mathcal {SP}4|=6$ partitioned as follows:

• $C_1$ defined by $(a_1,b_1,a_1,c_1) = (0,7,0,0)$. $C_1\not \in \mathcal {SP}4_g$ since $c_1\ne b_1-1$.

• $C_2$ defined by $(a_2,b_2,a_2,c_2) = (1,5,1,0)$. $C_2\not \in \mathcal {SP}4_g$ since $c_2\ne b_2-1$.

• $C_3$ defined by $(a_3,b_3,a_3,c_3) = (0,5,0,2)$. $C_3\not \in \mathcal {SP}4_g$ since $c_3\ne b_3-1$.

• $C_4$ defined by $(a_4,b_4,a_4,c_4) = (2,3,2,0)$. $C_4\not \in \mathcal {SP}4_g$ since $c_4\ne b_4-1$.

• $C_5$ defined by $(a_5,b_5,a_5,c_5) = (3,1,3,0)$. $C_5\in \mathcal {SP}4_g$.

• $C_6$ defined by $(a_6,b_6,a_6,c_6) = (1,3,1,2)$. $C_6\not \in \mathcal {SP}4_g$ since $a_6= b_6-2$.

There is an algorithm solving the gathering problem if the initial configuration belongs to $\mathcal {A}\cup \{C_5\}$. The problem is open if the initial configuration is $C_1$, $C_2$, $C_3$, $C_4$, or $C_6$.

1.3 A.3 Ring of Size $n=4x+3$ for $x\in \{1,2,3,\ldots \}$

$|\mathcal {I}| = \frac{(n^2-1)(n-3)}{48}$, $|\mathcal {NG}|=0$, $|\mathcal {A}|=\frac{(n-1)(n-3)(n-5)}{48}+\frac{(n-3)^2}{16}=\frac{(n-4)(n-3)(n+5)}{48}$, and $|\mathcal {SP}4|=\frac{(n-3)(n+1)}{16}$ are partitioned as follows:

• $C_{b,c}$ defined by $(\frac{n-4-b-c}{2},b,\frac{n-4-b-c}{2},c)$ for $b\in \{1,3,5,7,\ldots ,n-4\}$ and $c\in \{0,2,4,\ldots ,b-1\}$. All $C_{b,c}$ for $c\ne b-1$ do not belong to $\mathcal {SP}4_g$.

• Depending on the parity of x, there exists or not a configuration $C_{b,c}$ such that $c=b-1$ and $\frac{n-4-b-c}{2}=a=b-2$:

  • If x is odd; there is no such configuration. There are x configurations in $\mathcal {SP}4_g$: $C_b=C_{b,b-1}$ defined by $(\frac{n-3}{2}-b,b,\frac{n-3}{2}-b,b-1)$ for $b\in \{1,3,\ldots ,2x-1\}$.

  • If x is even; there is a unique such configuration $C_{b,c}$ for $b=x+1$. There are $x-1$ configurations in $\mathcal {SP}4_g$:

    $C_b=C_{b,b-1}$ defined by $(\frac{n-3}{2}-b,b,\frac{n-3}{2}-b,b-1)$ for $b\in \{1,3,5,7,\ldots ,2x-1\}{\setminus }\{x+1\}$.

Note that two of our algorithms for $C_b$ and $C_{b'}$ are incompatible if $b+b'=\frac{n+1}{2}=2x+2$. In conclusion there is an algorithm solving the gathering problem if the initial configuration belongs to the set:

• $\mathcal {A}\cup \{C_1\}\cup \{C_3 \text { or } C_{2x-1}\}\cup \{C_5 \text { or } C_{2x-3}\}\cup \{C_7 \text { or } C_{2x-5}\}\cup \ldots \cup \{C_x \text { or } C_{x+2}\}$, if x is odd.

• $\mathcal {A}\cup \{C_1\}\cup \{C_3 \text { or } C_{2x-1}\}\cup \{C_5 \text { or } C_{2x-3}\}\cup \{C_7 \text { or } C_{2x-5}\}\cup \ldots \cup \{C_{x-1} \text { or } C_{x+3}\}$, if x is even.

The problem is open if the initial configurations belong to $\{C_{b,c}\}$ for $c\ne b-1$, or if two incompatible configurations are included in the set of initial configurations (such as the set $\{C_3,C_{2x-1}\}$).