=. L. , There are two cases: ? There is a unique smallest enclosing rectangle (SER), s. We call internal robot any robot that is not on a side of s. The following subcases are possible: (a) There is exactly one robot that is at a corner c of s. So, there is exactly one internal robot. It is the only robot allowed to move, its destination is the closest free node on a side of s having c as extremity

, There are exactly two robots at two corners c 1 , c 2 of s. The other two robots are internal. The internal robots that are the closest to an occupied corner of s move towards the closest free node on a side of s having c 1 or c 2 as an extremity (Alg. 2, Lines 2.11-2.13). So, a Twin configuration is reached in finite time and by

, The following subcases are possible: i. There is a unique robot r at distance d from a corner of s. r is the only robot allowed to move

, to the side of s where r i is and the side of s where r 1 or r 2 are. (n.b., d 3 = d 4 otherwise the configuration is regular.) Let min ? {3, 4} such that d min = min{d 3 , d 4 }. Then, r min moves toward c min (Alg. 2, Lines 2.19-2.25), and a Twin configuration is reached in finite time. By Lemma 5.4, we are done. If r 1 and r 2 are closest to a corner of s, but not the same, then r 1 and r 2 are the only robots allowed to move. Their destination is the adjacent free node towards the closest corner (see Alg. 2, Lines 2.26-2.27). In finite time, either we retrieve case 1(c)i, or the system reaches a configuration Twin (depending on the choices of the scheduler)

, Some of the robots have a safe neighboring node such that if they move to that node and they move alone, then the number of SERs decreases. Only these robots are allowed to move. If activated, they try to move to such a safe node (Alg. 2, Lines 2.01-2.04)

. Hence, sequential moves (which happen with a probability greater than or equal to p 3 (1 ? p) 9 ) to reach either a Twin configuration, or a configuration Isolated containing a unique SER. Hence, with positive probability the system reaches in finite time a Twin configuration and by, the worst case, the system needs three consecutive

=. L. , Let r 1 , r 2 , r 3 , and r 4 be the robots. Assume there is a unique smallest enclosing rectangle (SER), s. Observe that there are two robots, say r 1 and r 2 , that are on two parallel sides of s that are on -rings. Let d be the minimal distance between r 3 or r 4 and an -ring which is a side of s. If r 3 (resp. r 4 ) is at distance d from an -ring which is a side of s, it is allowed to move, According to the number of robots on L-rings, we have 5 cases: (a) Every L-ring contains at most one robot

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