This is a spin-off motivated by Lopsy's interesting variant of Gamow's lion and zebras puzzle. It arose from a line of enquiry that tried to extend the vertical run past 5000km (or characterise the distribution of zebras to limit the run to within 5000km). That line of inquiry was eventually discarded but the spin-off puzzle seemed interesting in its own right.
Imagine a one-dimensional universe inhabited by 101 spots. Every so often, they play their own friendly variant of the 'lion and zebras' game of precision tag.
One spot is selected as the lion and the other spots (zebras) scatter to wherever they wish.
Goal At the start of the game, the lion nominates a position for each zebra. The lion wins if the zebras occupy those positions simultaneously. The zebras win if they can force the game to continue indefinitely without the lion winning.
Target Zebra One zebra is selected as the target to start the game. At any time, the lion may change the target to any zebra that is between 100m and 200m away, inclusive. For example, it may target a zebra 100m or 150m away, but not one 50m, 99m, 201m or 1km away. There is exactly one target at any given time.
Moves The spots take turns to move up to 100m per turn. On the odd turns, only the lion may move. On the even turns, only the zebra that was the target at the start of the turn may move.
Zebra Move Constraint 1: Non-crossing The target must use its turn to ensure its own position never crosses to the other side of the lion during that turn and during the lion's next turn. For example, if the lion is 50m away and the target starts moving towards it and moves for 80m, their positions cross during that move. Other than this restriction, spots may pass through each other with no consequence. Note that it is permissible for the target to end up at the same position as the lion - this constraint just prohibits crossing to the other side of the lion.
Zebra Move Constraint 2: Attractor The zebras are friendly. Within the positions allowed by the non-crossing constraint, the target will move to the position that minimises its distance to the last zebra it passed.
Can the lion win? If so, how? If not, why not?
Does it make any difference if the targeting distance was slightly modified so that the lion cannot tag zebras exactly 100m or 200m away (that is, change it from a closed interval to an open interval)?
Here is an example of placing one zebra at its target point. The game starts with lion $L$ and zebras $X,Y,Z$, with initial target $X$ and initial positions $[L,X,Y,Z] = [-150,0,-100,120]$. All other zebras are far away. The nominated position for $X$ is 140.
Use the notation $L+x$ and $L-x$ to mean $L$ moves $x$ meters towards the positive or negative end respectively. Then we play this out as follows:
$L+100,X+100,L+100$ takes $[L,X]$ to $[50,100]$.
$X$ must move to somewhere between 150 $(=L+100)$ and 200 $(=X+100)$ and therefore passes $Z$. The nearest position to $Z$ in that range is 150, so $X$ only moves $X+50$ to 150. The spots are now at $[L,X,Y,Z] = [50,150,-100,120]$ and $Z$ is the last zebra seen by $X$.
$L-10$ allows $X$ to get closer to $Z$, which it does with $X-10$, bringing $X$ to its target position 140. $L$ is now at 40, and it changes the target to $Y$, which is now in range. This freezes the position of $X$.
In this example, $Y$ and $Z$ are at convenient positions. In the general case, each zebra starts at a random position.