## Thanos and Particles Riddles

Here are two riddles, the solutions of which I will discuss next week. Interested readers are welcome to email me solutions.

### Thanos Problem

Thanos, the all-powerful supervillain, can snap his fingers and destroy half of all the beings in the universe.

Suppose now there are  Thanoses, each snapping his fingers in a sequence, one after the other. When a Thanos snaps his finger, each being (Thanos or human) dies independently with probability .

Question: Out of  people on Earth, how many can we expect to still be alive at the end? More formally, let  be the probability that a human being survives. Initial values: .

1. Show that , that is,  is bounded between positive constants.

2. Does  converge? If so, to what limit?

3. Does  converge? If so, to what limit?

Note: this problem was posted (with fewer details) by Oliver Roederon on the 538 blog, but without an analytical solution. The plot is due to Laurent Lessard (github).

Particles Problem

Suppose there are  particles in the unit square. Initially one particle is awake and all others are sleeping. Each awake particle moves in the unit square at speed  in a direction you prescribe and wakes up any sleeping particle it encounters. The particles that are awake move simultaneously and particles can change direction at any point in time.

Question: Show that you can wake up all the particles by time Note: I learned this problem from Maria Gringlaz.

## Lake Covering Problem

A circular lake of radius  meters must be covered by rectangular boards of length  meters.
Rectangular boards of all widths are available, but the cost of a covering is the total width of the boards used. Clearly the lake can be covered by using parallel boards of total width  meters.
Is there a more efficient covering possible? Here are examples of two competing coverings.

## Solution to Particles Riddle

In a previous post I discussed the Particles riddle, a solution of which is as follows. Cover the unit square by four closed subsquares of side length 1/2, denoted  as in the figure. Suppose that the initially awake particle is located at , and there are sleeping particles in  (the case when some of these are unoccupied is easy to handle.) The particle at  wakes up a particle at some point  by time . During the next  time units, one of these two particles can travel to wake a particle in some  in , and at the same time, the other  can travel to wake a particle at some  in . In another  time unit, one of the particles from  can travel to  and one of the particles at  can travel to . Thus by time , there will be an awake particle in each of the originally occupied subsquares . The argument can now be iterated in each of these subsquares. Repeated subdivision will yield a geometric series  as an upper bound for the time needed to reach all  particles. The constant 6.2  can certainly be improved; we will not try to optimize this constant. See also a similar solution posted by Sang-il Oum here.

The argument above readily generalizes to higher dimensional cubes (where the constant will depend on dimension) and to other self-similar sets. Compactness and convexity are not enough, however: Let  be an orthonormal sequence in Hilbert space. Then the closed convex hull of  is compact, but the time needed to reach the first  vertices  is of order .

A metric space  is called a  if the distance between any two points equals the length of the shortest curve between them (In this case one also says that  is an instrinsic metric on ). In particular, a convex set in  a  Banach space is a length space; for the particle problem, it is natural to restrict attention to length spaces. The assumption that the particles travel at most at unit speed means that their paths are Lipschitz maps from  to  with Lipschitz constant .

A large class of length spaces where any finite collection of points can be reached in bounded time are  length spaces.  A metric space  is called    if there is some doubling constant   such that every ball  in  of any radius , can be covered by at most  balls of radius . See https://en.wikipedia.org/wiki/Doubling_space.

The following statement is easily proved by induction on :  Given  particles, one of them awake, in a doubling length space with doubling constant  and diameter , all particles can be woken up by time . (As before, awake particles travel at unit speed).