Procedure Rendezvous$()$

$//$ Initially, there are $2$ robots on the $x$ axis, each with a red

$\frac{\frac{?}{?}}{?}$ colored light

if $($your own light color is red$)$ then

if $($exactly one other robot with a red light is visible$)$ then

Assuming the distance to the other robot is $,$

move ${|}_{{?}_{?}}\frac{-1}{(2{)}_{{?}_{?}}}|$ units towards the other robot, and

change light color to blue, and terminate

end

Very often, the algorithm is expressed as a set of "condition$-$action" pairs. Typically the

condition is in terms of what a robot sees and the action results in the robot movement

and light setting. Thus evaluation of a condition involves the Look phase and part of the

Compute phase, and performance of the action involves the remainder of the computation

$($to determine the new position and light setting$)$ and the action of moving and the light

setting itself.

At each step, a robot that has not terminated, checks the conditions in order, picks the

first condition it satisfies and performs the action associated with that condition. If a robot

does not satisfy any of the conditions, then it performs no action. As an example the above

procedure can be written as the following condition action.

Condition $1$: Your own light color is red and you can see exactly one other robot with a

red light.

Action:$1$ Assuming the other robot is at a distance of $1,$ move ${|}_{{?}_{?}}\frac{-1}{(2{)}_{{?}_{?}}}|$ units towards the

other robot, change light color to blue and terminate.

Prove that the above algorithm avoids robot collision and results in the robots finally

being at a distance of at most $2$ units.