# \(k\)-Circle Formation and \(k\)-epf by Asynchronous Robots
## 4 Algorithm
### Moveto Destination
If \(r_{v}\) is in motion, then it continues its motion without any collision (As destination point and movement path computed by \(r_{i}\) and \(r_{v}\), respectively are separated by the \(y\)-axis and there are no other robots in the half-plane containing \(r_{v}\), which is in motion within the annulus region between \(C(f_{j},\rho)\) and \(C(f_{j},d(f_{j},r_{v}(t^{\prime})))\)). We have two possible cases. First, \(r_{v}\) will also reach \(C(f_{j},\rho)\). Second, if it stops before reaching \(C(f_{j},\rho)\), then it will become a candidate robot only when \(r_{i}\) will reach \(C(f_{j},\rho)\). **Case 2. **: \(q(t)\) is a point on some saturated circle \(C(f_{u},\rho)\). We have the following cases:

**Subcase 1. **: \(r_{i}(t^{\prime})=q(t)\). Since \(r_{i}\) has reached its destination, the first part of the lemma follows. At \(t^{\prime}\), since \(C(f_{u},\rho)\) contains \(k+1\) robots, the next candidate robot for \(f_{j}\) will be selected from \(C(f_{u},\rho)\). Note that, this robot position would have higher \(y\)-coordinate than \(q(t)\). If \(r_{v}\) has also completed its LCM cycle and has not reached its destination point, then it will become a candidate robot for \(f_{j}\) only when \(N_{j}(t^{\prime\prime})=N_{j}(t^{\prime})-1\) for some \(t^{\prime\prime}>t^{\prime}\). If \(r_{v}\) is in motion, then it continues its motion without any collision (It is the only robot, which is in motion within the annulus region between \(C(f_{j},\rho)\) and \(C(f_{j},d(f_{j},r_{v}(t^{\prime})))\) and below the point \(q(t)\)). **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 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 only when \(N_{j}(t)\) reduces by one. 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 and has not reached its destination point, then it will become a candidate robot only when \(N_{j}(t)\) reduces by one. If \(r_{v}\) is in motion, then it continues its motion without any collision (As destination point and path computed by \(r_{i}\) and \(r_{v}\), respectively, are separated by the \(y\)-axis and there are no other robots in motion within the annulus region between \(C(f_{j},\rho)\) and \(C(f_{j},d(f_{j},r_{v}(t^{\prime})))\) and below the point \(q(t)\)). We have two possible cases. First, \(r_{v}\) will also reach \(C(f_{u},\rho)\). Second, if it stops before reaching \(C(f_{u},\rho)\), then it will become a candidate robot only when \(N_{j}(t)\) reduces by one. Next, we consider the case when there are two target fixed points, one from each half-plane. Let \(f_{j}\) and \(f_{a}\) be the target fixed points at time \(t\). Let \(r_{i}\) and \(r_{b}\) be their respective candidate robots. We have \(V_{j}(t)=(n_{k}(t),N_{j}(t),d_{j}(t))\) and \(V_{a}(t)=(n_{k}(t),N_{a}(t),d_{a}(t))\). Lemma 8: _Let \(C(t)\) admit two target fixed points during an execution of \(AlgorithmOneAxis(C(t))\) and \(t^{\prime}>t\) be an arbitrary point of time when at least one candidate robot has completed its LCM cycle. For at least one target fixed point \(f_{i}\in\{f_{j},f_{a}\}\) and its candidate robot, \(d_{i}(t^{\prime})+\delta\leq d_{i}(t)\)._

Proof of Lemma 8.: Each target fixed point is unique in their respective half-planes. Execution of \(AlgorithmOneAxis(C(t))\) ensures that for each target fixed point, its candidate robot is selected from its respective half-planes. The circle formation process continues independently in both the half-planes.