Attacking Hyenas

Question

$N$ Hyenas are standing on a plane region in a forest. At $t=-1$, they see dead meat nearby. Being selfish, at $t=0$, each Hyena attacks the Hyena which is closest to it. All pairwise distances between them are distinct.

Question:

At most how many Hyenas can remain unattacked?

Example:

$N=1$, answer is $1$ since there is no other Hyena to attack the first Hyena

$N=2$, answer is $0$ (both attack each other)

$N=3$, answer is $1$

$N=7$, answer is $5$

$N=10$, answer is $7$

Note:

1) The closest pair is always under attack for $N>1$.

2) For odd $N$, it is easy to see that there exists at least $1$ Hyena which is not under attack.

Reduced Version:

Solve for $N=100$

Answer 1

We can slightly rephrase this question as "For a given number of attacked hyenas, what is the maximum number of unattacked hyenas we can have?"
Particularly, since hyenas can be divided into clusters such that attacks only occur within a cluster, how many unattacked hyenas can there be in a cluster containing a given number of attacked hyenas?

We start with the observation that

since distances must be distinct, we can't have any configuration that requires exactly identical distances (e.g. a hyena having some attacking relationship with six hyenas). However, by careful management of arbitrarily small differences, we can get arbitrarily close to such a configuration.
For instance, the circles from a pair of attacked hyenas can admit eight more circles of exactly the same radius, but since we can't have exact configurations only seven hyenas will fit.
What we can do, is to take this partial configuration with six unattacked hyenas:
By shrinking our infinitesimals, we can make a circle that touches the blue attacked hyena and none of the others arbitrarily large - enough to fit up to four more unattacked hyenas, like so:
However, the configuration functions as a virtual fifth hyena, so if we wish to repeat this construction we can only have added three hyenas (the end result would resemble what you'd have if you replaced A and its three attackers with a sufficiently small copy of the entire group of eight). This construction allows us to have $3a+1$ unattacked hyenas in a cluster of $a$ attacked hyenas.

To justify this incremental approach, we only need that the distinctness of distances means that every configuration of attacked hyena can be built up this way, and having interference between attacked hyenas can only decrease the possible numbers of unattacked hyenas.

Of course, maximizing the proportion of unattacked hyenas requires us to maximize the contribution of the $+1$, which is achieved by maximizing the number of clusters. By doing so, we can get $7a$ unattacked hyenas for every $2a$ attacked ones (or $9a$ total), and $7a+3$ unattacked hyenas for every $2a+1$ attacked ones (or $9a+4$ total). The sequence of unattacked hyenas threfore goes $7k, 7k, 7k+1, 7k+2, 7k+3, 7k+3, 7k+4, 7k+5, 7k+6$ for $N=9k+r$ hyenas with $k$ positive.

Answer 2

The following presents the maximum number of unattacked hyenas alongside their corresponding ratio:

$\frac{7}{9}$

and here is the representation of this;

and here is a better representation

with the right distances.

If you would like to add more hyenas after this point, you need to create a new attacked hyena including the red and blue ones shown above; such as;

let's put another 1 (11 hyenas) and the result becomes

but if we would like to add one more,

we cannot use that new hyena anymore, because no place is left to put to attack that green one. we need to find a better way to do that with 12 hyenas.

such as

which results as

$\frac{9}{12}$

so if N goes to a big number; you can have separate

"9 of hyenas" as this is

and at the end you will end up with

Assuming you possess N Hyenas, the maximum number of Hyenas that can stay unattacked is $\frac{7N}{9}$

and

Assuming you have 18 hyenas, you could create two distinct groups of 9 hyenas each, completely separated from one another. This arrangement would leave you with 14 hyenas that are not under attack, and you can have the same ratio with 18 or any 9N one.