# \(k\)-Circle Formation and \(k\)-epf by Asynchronous Robots
## 4 Algorithm
### Moveto Destination
Until \(r_{i}\) reaches its destination point computed at time \(t\), it remains the candidate robot for \(f_{j}\)._

Proof of Lemma 6.: Let \(r_{i}\) compute its movement path \(P\) and destination point \(q(t)\) by the execution of \(MovetoDestination(C(t),f_{j},r_{i})\). Note that, \(q(t)\) is either a point on the circle \(C(f_{i},\rho)\) or on some saturated circle \(C(f_{u},\rho)\). Let \(t^{\prime}\) be an arbitrary point of time such that \(r_{i}(t^{\prime})\
eq q(t)\). At time \(t^{\prime}\), \(f_{j}\) remains an unsaturated fixed point. As a result, \(f_{j}\) remains a target fixed point at time \(t^{\prime}\). Lemma 4 guarantees that \(r_{i}\) has moved at least \(\delta\) amount closer to \(C(f_{j},\rho)\). Therefore, it remains the candidate robot for \(f_{j}\). Next, we consider the case when there are two candidate robots for a target fixed point. Since robots have an agreement on the directions and orientations of the \(y\)-axis,there can be at most two candidate robots at any point of time. Note that, in this case, the configuration would have a unique target fixed point, that lies on the \(y\)-axis. Lemma 7: _Let \(f_{j}\) be the target fixed point and \(r_{i}\) and \(r_{v}\) are the two selected candidate robots for \(f_{j}\) at time \(t\). Until at least one of them reaches its destination point computed at time \(t\), no other robot becomes a candidate robot. If one of the candidate robots have reached its destination point and the other one has not, then the other robot either continues its inherent motion towards its destination point (computed at time \(t\)) without any collision or gets selected as a candidate robot only when \(N_{j}(t)\) reduces by one._

Proof of Lemma 7.: Let \(t^{\prime}>t\) be an arbitrary point of time when at least one of the candidate robots has completed its LCM cycle. Without loss of generality, assume that \(r_{i}\) has completed its LCM cycle at \(t^{\prime}\). Let \(q(t)\) be the destination point and \(P\) be the movement path computed for \(r_{i}\) by \(MovetoDestination(C(t),f_{j},r_{i})\). Note that \(q(t)\) is a point either on the \(C(f_{j},\rho)\) or on some saturated \(C(f_{u},\rho)\). We have the following cases:

**Case 1. **: \(q(t)\) is a point on the circle \(C(f_{j},\rho)\). We have the following subcases:

**Subcase 1. **: \(r_{i}(t^{\prime})=q(t)\). Since \(r_{i}\) has reached its destination, the first part of the lemma follows. We have \(N_{j}(t^{\prime})=N_{j}(t)-1\). At \(t^{\prime}\), if \(r_{v}\) has also completed its LCM cycle and has not reached its destination point, then it becomes the next candidate robot for \(f_{j}\). If \(r_{v}\) is in motion, then being the only robot in motion within the annulus region between \(C(f_{j},\rho)\) and \(C(f_{j},d(f_{j},r_{v}(t^{\prime})))\), it continues its motion without any collision. Note that, in this case, no other robot will be selected for movement until \(r_{v}\) reaches its destination. **Subcase 2. **: \(r_{i}(t^{\prime})\
eq q(t)\). First consider that \(|d(f_{j},r_{i}(t^{\prime}))-\rho|>|d(f_{j},r_{v}(t^{\prime}))-\rho|\), i.e., robot \(r_{v}\) is closer to \(C(f_{j},\rho)\) than \(r_{i}\). At \(t^{\prime}\), either \(r_{v}\) has also completed its LCM cycle and has not reached its destination point or \(r_{v}\) is in motion. In both the cases, \(r_{v}\) remains a candidate robot for \(f_{j}\). The first part of the lemma follows for \(r_{v}\). Robot \(r_{i}\) will be selected as a candidate robot when \(r_{v}\) will reach \(C(f_{j},\rho)\). Next consider that \(|d(f_{j},r_{i}(t^{\prime}))-\rho|<|d(f_{j},r_{v}(t^{\prime}))-\rho|\), i.e., robot \(r_{i}\) is closer to \(C(f_{j},\rho)\) than \(r_{v}\). Robot \(r_{i}\) will be selected as a candidate robot. At \(t^{\prime}\), if \(r_{v}\) has also completed its LCM cycle, then it will become the candidate robot when \(r_{i}\) will reach \(C(f_{j},\rho)\).