But also, maybe if you get some free time, go solve this
Partial results
(old result) N lions can always move radially outwards and ensure 100-N zebras are inside their convex hull or circle
if N lions try to trap 1 zebra, then zebra obviously wins for N=2,3,4. not yet sure what happens for N=6 or above (N=5 is annoying to analyse)
assume a double-zebra is a zebra that moves at half speed (imagine it is two zebras stuck to each other). a lion can always catch a double-zebra. (zebras only have an advantage because the zebras can always separate from each other)
assume a double-lion is a lion that moves at half speed (imagine it is two lions stuck to each other). a zebra encircled by N double-lions at some large radius D, can always escape the double-lions. (for lions to encircle a zebra and trap it successfully, it is necessary (but not sufficient) for the lions to be willing to separate from each other.) if D is small I'm unsure what happens
Holy fuck, wait did I solve it?
Assume an origin, split the plane into 100 pie slices centred at origin, all with equal angle. Assume one zebra in each pie slice at a large distance D from origin.
lemma 1 - it is not possible for 100 lions in a pie slice to drive the zebra of that slice towards the centre repeatedly
lemma 2 - it is not possible for 100 lions in a pie slice to drive the zebra of that slice towards the edge of the pie slice
if I can prove 1 and 2, I can prove zebras win
I don't know, 1 and 2 both just feel intuitively true to me, but I don't have proof. If you send one lion to chase the zebra, you'll basically end up in a stalemate situation. If you send two lions to chase the zebra, the midpoint of the lions will move at half speed compared to the zebra. If a point is chasing the zebra at half its speed, not only can the zebra outrun it, but also it can complete an entire circle around this point if it wanted to without being caught, it is just a question of how large a circle it needs and whether this fits inside the pie slice.
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