Settlers in a circle

In this Riddler problem, the goal is to spread out settlements in a circle so that they are as far apart as possible:

Antisocial settlers are building houses on a prairie that’s a perfect circle with a radius of 1 mile. Each settler wants to live as far apart from his or her nearest neighbor as possible. To accomplish that, the settlers will overcome their antisocial behavior and work together so that the average distance between each settler and his or her nearest neighbor is as large as possible.

At first, there were slated to be seven settlers. Arranging that was easy enough: One will build his house in the center of the circle, while the other six will form a regular hexagon along its circumference. Every settler will be exactly 1 mile from his nearest neighbor, so the average distance is 1 mile.

However, at the last minute, one settler cancels his move to the prairie altogether (he’s really antisocial). That leaves six settlers. Does that mean the settlers can live further away from each other than they would have if there were seven settlers? Where will the six settlers ultimately build their houses, and what’s the maximum average distance between nearest neighbors?

Here is my solution:
[Show Solution]

Alice and Bob fall in love

In this interesting Riddler problem, we’re dealing with a possibly unbounded sequence of… children? Here it goes:

As you may know, having one child, let alone many, is a lot of work. But Alice and Bob realized children require less of their parents’ time as they grow older. They figured out that the work involved in having a child equals one divided by the age of the child in years. (Yes, that means the work is infinite for a child right after they are born. That may be true.)

Anyhow, since having a new child is a lot of work, Alice and Bob don’t want to have another child until the total work required by all their other children is 1 or less. Suppose they have their first child at time T=0. When T=1, their only child is turns 1, so the work involved is 1, and so they have their second child. After roughly another 1.61 years, their children are roughly 1.61 and 2.61, the work required has dropped back down to 1, and so they have their third child. And so on.

(Feel free to ignore twins, deaths, the real-world inability to decide exactly when you have a child, and so on.)

Five questions: Does it make sense for Alice and Bob to have an infinite number of children? Does the time between kids increase as they have more and more kids? What can we say about when they have their Nth child — can we predict it with a formula? Does the size of their brood over time show asymptotic behavior? If so, what are its bounds?

Here is an explanation of my derivation:
[Show Solution]

If you’re just interested in the answers to the questions, here they are:
[Show Solution]

Ostomachion coloring

The following problems appeared in The Riddler. And it features an interesting combination of an ancient game with the four-color theorem.

The famous four-color theorem states, essentially, that you can color in the regions of any map using at most four colors in such a way that no neighboring regions share a color. A computer-based proof of the theorem was offered in 1976.

Some 2,200 years earlier, the legendary Greek mathematician Archimedes described something called an Ostomachion (shown below). It’s a group of pieces, similar to tangrams, that divides a 12-by-12 square into 14 regions. The object is to rearrange the pieces into interesting shapes, such as a Tyrannosaurus rex. It’s often called the oldest known mathematical puzzle.

Your challenge today: Color in the regions of the Ostomachion square with four colors such that each color shades an equal area. (That is, each color needs to shade 36 square units.) The coloring must also satisfy the constraint that no adjacent regions are the same color.

Extra credit: How many solutions to this challenge are there?


Here are the details of how I solved the problem:
[Show Solution]

To jump straight to the solution (and pretty pictures!) see below.
[Show Solution]

Timing a stoplight just right

This Riddler is about how to perfectly time a stoplight, something we’ve all had to deal with!

You are driving your car on a perfectly flat, straight road. You are the only one on the road and you can see anything ahead of you perfectly. At time t=0, you are at Point A, cruising along at a speed of 100 kilometers per hour, which is the speed limit for the whole road. You want to reach Point C, exactly 4 kilometers ahead, in the shortest time possible. But, at Point B, 2 kilometers ahead of you, there is a traffic light.

At time t=0, the light is green, but you don’t know how long it has been green. You do know that at the beginning of each second, there is a 1 percent chance that the light will turn yellow. Once it turns yellow, it remains yellow for 5 seconds and then turns red for 20 seconds. Your car can accelerate or decelerate at a maximum rate of 2 meters per second-squared. You must always drive at or below the speed limit. You can pass through the intersection when the traffic light is yellow, but not when it is red.

What is the best strategy to reach your destination as soon as possible?

Here is my solution:
[Show Solution]

Non-intersecting chessboard paths

This Riddler classic puzzle is about finding non-intersecting paths on a chessboard:

First, how long is the longest path a knight can travel on a standard 8-by-8 board without letting the path intersect itself?

Second, there are unorthodox chess pieces that don’t exist in the standard game, which are known as fairy chess pieces. What are the longest nonintersecting paths that can be taken by the camel (which moves like a knight, except 3 squares by 1 square), the zebra (3 by 2), and the giraffe (4 by 1)?

This is a very challenging problem, and there doesn’t appear to be any way to solve it via some clever observation or simplification. Of course, we can try to come up with ever longer tours by hand, but we’ll never know for sure that we have found the longest one.

Much like the recent Pokemon Go problem, we must resort to computational means to obtain a solution. In this case, however, the problem is “small enough” that we can find exact solutions!

Here are some optimal tours:
[Show Solution]

If you’re interested in the details of how I found my solutions, read on:
[Show Solution]