Welcome to The Riddler. Every week, I offer up problems related to the things we hold dear around here: math, logic and probability. There are two types: Riddler Express for those of you who want something bite-sized, and Riddler Classic for those of you in the slow-puzzle movement. Submit a correct answer for either,^{1} and you may get a shoutout (chosen at random) in next week’s column. If you need a hint, you can try asking me nicely on Twitter.

## Riddler Express

From Po-Shen Loh, a math professor at Carnegie Mellon and the founder of education site Expii, a Star Wars engineering puzzle:

With “Rogue One” out, perhaps you’ve been inspired to build your own Death Star.^{2} Let’s say the fictional Death Star, a spherical galactic superweapon that can destroy planets with its laser, has a diameter of about 150 kilometers, with a two-meter-wide exhaust port at the bottom. Its design is incredibly energy efficient, because nearly all of the energy generated by the reactor is converted into the superlaser’s light, and only a tiny fraction is released as exhaust heat.

After some research, you learn that it is possible to build a Death Star replica 50 meters in diameter, with a one-meter-wide exhaust port. But you don’t want to stop at 50 meters: you want to take the blueprint and scale it as large as possible (e.g., to hundreds of kilometers). There is a significant hitch, though: Power output scales with the volume, while surface area scales only with the radius squared. If you tried to scale the 50-meter replica up, using the same rate of exhaust surface area per unit power, at what radius would the entire surface of the sphere be required for exhaust? (This would become a theoretical upper limit on the radius, and therefore on power output.)

## Riddler Classic

Also from Po, a Star Wars blaster battle puzzle:

In Star Wars battles, sometimes a severely outnumbered force emerges victorious through superior skill. You panic when you see a group of nine stormtroopers coming at you from very far away in the distance. Fortunately, they are notoriously inaccurate with their blaster fire, with only a 0.1 percent chance of hitting you with each of their shots. You and each stormtrooper fire blasters at the same rate, but you are K times as likely to hit your target with each shot. Furthermore, the stormtroopers walk in a tight formation, and can therefore create a larger target area. Specifically, if there are N stormtroopers left, your chance of hitting one of them is \((K \cdot \sqrt{N})/1000\). Each shot has an independent probability of hitting and immediately taking out its target. For approximately what value of K is this a fair match between you and the stormtroopers (where you have 50 percent chance of blasting all of them)?

## Solution to last week’s Riddler Express

Congratulations to 👏 Kevin Roust 👏 of Chicago, winner of last week’s Express puzzle!

An earthquake knocked six labeled rocks off a shelf and onto the floor of a geology museum. A janitor, wanting to clean the floor, puts the rocks back behind the six labels, but in random order. The probability that all six rocks are placed behind their correct labels is 1/6!, or 1/720. (There are 720 total possible arrangements.) What are the probabilities that exactly five, four, three, two, one and zero rocks are correctly placed?

Five rocks: **zero**. If five rocks are in their correct places, then the sixth rock must be as well, so there are no cases in which exactly five rocks are correctly placed.

Four rocks: **15/720**. If four rocks are in their correct places, then that means two rocks have been flipped. There are “6 choose 2” or 15 ways to flip two rocks. The mathematical formulas for the number of combinations are well known: Specifically, there are \(\frac{N!}{K!(N-K)!}\) different ways to choose K objects from a larger set of N objects.

Three rocks: **40/720**. If three rocks are in their correct places, then that means the other three rocks have been mixed up. There are “6 choose 3” or 20 ways to select the three mixed up rocks. But they could be mixed up in different ways. Say these three rocks were meant to be in order: ABC. There are two ways they could all be in the wrong places: BCA and CAB. So 20 times two is 40.

Two rocks: **135/720**. First, choose the four misplaced rocks: there are “6 choose 4” or 15 ways to do so. Then count the ways to misplace four rocks. Suppose they belong in the order ABCD. They could all be misplaced nine different ways: BADC, BCDA, BDAC, CADB, CDAB, CDBA, DABC, DCBA and DCAB. Nine times 15 is 135.

One rock: **264/720**. There are six ways to choose the rock that’s in the correct place. (Equivalently, there are six ways to choose the five rocks that aren’t in the correct place.) For the five misplaced rocks, there are 44 ways to misplace them all. Six times 44 is 264.

Zero rocks: **265/720**. Because we know there are 720 total possible arrangements, we can simply subtract those we’ve already counted: 720-1-0-15-40-135-264 = 265.

## Solution to last week’s Riddler Classic

Congratulations to 👏 Gabriel Burns 👏 of Des Moines, Iowa, winner of last week’s Classic puzzle!

A series of N wires snakes from the top of a tall tower down to the bottom, hopelessly tangled behind a wall as they go. At the top of the tower, the wire ends are helpfully labeled from 1 to N. At the bottom, however, the wire ends are unlabeled. Armed only with the ability to tie any pairs of wires together at the bottom, and a circuit detector to test at the top of the tower whether two wires are connected, how many trips up the long and winding stairs do you need to correctly identify all the unlabeled wires at the bottom?

If N=1, you need zero trips — you already know what the single wire coming out the bottom is. If N=2, your task is impossible — you will never be able to surely identify which of the wires coming out at the bottom of the tower is which.

But if N>2, you need only **two trips** to the top of the tower to correctly identify all the wires on the bottom.

Here’s how to do it: Label the wires on the bottom with your own, arbitrary labels, say B1, B2, B3, up through BN. At the bottom of the tower, tie B1 to B2, B3 to B4, B5 to B6, and so on. If N is odd, simply leave the last wire (BN) unconnected. If N is even, leave the last two wires (BN-1 and BN) unconnected. Then go upstairs and test all the connections, keeping a log of which pairs of wires at the top are connected below.

Go back downstairs and untie all your connections. This time, tie BN to B1, B2 to B3, B4 to B5, and so on, shifting by one the ties you’d made before. (Again, leave one wire — BN-1 — unconnected if N is odd, and two — BN-1 and BN-2 — if N is even.) Go upstairs one more time and test all the connections, again logging which circuits were complete.

You’re essentially done! If you check your log carefully, you will set off a chain reaction of deductions. First, you know which wire upstairs is the one you labeled BN downstairs, because it was not connected on your first trip but was connected on your second trip. Given that, you know which wire B1 is because it was connected to BN on your second trip. Then you know which B2 is because it was connected to B1 on your first trip. B3 was connected to B2 on your first trip, B4 to B3 on your second trip, and so one down the list of wires, alternating between what you learned on your first and second trips to the top of the tower.

You can find further excellent explanations from Dennis Roos and Laurent Lessard.

## Want to submit a riddle?

Email me at oliver.roeder@fivethirtyeight.com.