Welcome to The Riddler. Every week, I offer up problems related to the things we hold dear around here: math, logic and probability. Two puzzles are presented each week: the Riddler Express for those of you who want something bite-size and the Riddler Classic for those of you in the slow-puzzle movement. Submit a correct answer for either,^{1} and you may get a shoutout in the next column. Please wait until Monday to publicly share your answers! If you need a hint or have a favorite puzzle collecting dust in your attic, find me on Twitter.

## Riddler Express

By all accounts, Riddler Nation had a lot of fun hunting for the mysterious numbers a few weeks back. So here’s what we’re going to do: For the next four weeks, the Riddler Express will feature a similar puzzle that combines multiplication and logic. We’ll be calling these *CrossProducts*.

For your first weekly CrossProduct, there are five three-digit numbers — each belongs in a row of the table below, with one digit per cell. The products of the three digits of each number are shown in the rightmost column. Meanwhile, the products of the digits in the hundreds, tens and ones places, respectively, are shown in the bottom row.

135 | |||

45 | |||

64 | |||

280 | |||

70 | |||

3,000 | 3,969 | 640 |

Can you find all five three-digit numbers and complete the table?

The solution to this Riddler Express can be found in the following column.

## Riddler Classic

From Toby Berger comes a towering challenge:

Cassius the ape (a friend of Caesar’s) has gotten his hands on a Lucas’ Tower puzzle (also commonly referred to as the “Tower of Hanoi”). This particular puzzle consists of three poles and three disks, all of which start on the same pole. The three disks have different diameters — the biggest disk is at the bottom and the smallest disk is at the top. The goal is to move all three disks from one pole to any other pole, one at a time, but there’s a catch. At no point can a larger disk ever sit atop a smaller disk.

For *N* disks, the minimum number of moves is 2^{N}−1. (Spoiler alert! If you haven’t proven this before, give it a shot. It’s an excellent exercise in mathematical induction.)

But this week, the *minimum* number of moves is not in question. It turns out that Cassius couldn’t care less about solving the puzzle, but he is very good at following directions and understands a larger disk can never sit atop a smaller disk. With each move, he randomly chooses one among the set of valid moves.

On average, how many moves will it take for Cassius to solve this puzzle with three disks?

*Extra credit:* On average, how many moves will it take for Cassius to solve this puzzle in the general case of *N* disks?

The solution to this Riddler Classic can be found in the following column.

## Solution to last week’s Riddler Express

Congratulations to 👏 Edouard Duriez 👏 of Toulouse, France, winner of last week’s Riddler Express.

Last week, I had found four cubic blocks in a peculiar arrangement. Three of them were flat on the ground, with their corners touching and enclosing an equilateral triangle. Meanwhile, the fourth cube was above the other three, filling in the gap between them in a surprisingly snug manner. I even had photo evidence:

If each of the four cubes had side length 1, then how far above the ground was the bottommost corner of the cube on top?

One approach was to look at the portion of the cube on top that dipped below the surface of the other three cubes. This shape was a tetrahedron, whose base (on top) was an equilateral triangle with side length 1. (These sorts of connections between cubes and equilateral triangles are nothing new to The Riddler.) The remaining three faces of the tetrahedron were all congruent isosceles right triangles whose hypotenuse had length 1.

If only you could figure out the height of this tetrahedron, then you could subtract it from 1 to determine how high the bottommost corner was from the ground. The height — along with one of the legs of the isosceles right triangles and a line connecting the equilateral base’s centroid with one of its corners — formed yet another right triangle.

Putting the Pythagorean theorem into action, you found that the tetrahedron’s height was 1/√6. And that meant the bottommost corner was a height of **1−1/√6**, or about 0.592, off the ground.

Meanwhile, solver Thomas Stone picked out a different right triangle within the tetrahedron to calculate its height, ultimately arriving at the same answer:

Blocks are one thing. But if any architects are reading this, please make a building with this geometric arrangement of four cubes. Most of the math has already been done for you!

## Solution to last week’s Riddler Classic

Congratulations to 👏 Mark Bradwin 👏 of Seattle, Washington and 👏 Nic Tamburello 👏of Gaithersburg, Maryland, winners of last week’s Riddler Classic.

Last week, you were a contestant on the game show Lingo, where your objective was to determine a five-letter mystery word. You were told this word’s first letter, after which you had five attempts to guess the word. You were allowed to guess any five-letter word, even one that had a different first letter.

After each of your guesses, you were told which letters of your guess were also in the mystery word and whether any of the letters were in the correct position. In the example below, T was in the correct position (remember, the first letter was provided to you), while A and C were in the mystery word but not in the correct positions.

For this example, here was how you might have figured out the mystery word (TACOS) using all five guesses:

The mystery word and guesses could have also contained multiple instances of a letter. For example, the mystery word MISOS contains one O, so a guess with more than one O (like MOSSO) would only have had the first O marked as correct (but in this case, in the wrong position).

As a contestant, your plan was to make a mockery of the game show by adopting a bold strategy: No matter what, before you were even told what the first letter of the mystery word was, you had decided what your first four guesses would be. Then, with your fifth guess, you would use the results of your first four guesses (and your encyclopedic knowledge of five-letter words!) to determine all remaining possibilities for the mystery word. If multiple mystery words were still possible, you would pick one of these at random. You had to assume that the mystery word was selected randomly from this word list, which was also the list your guesses had to be chosen from.

Which four five-letter words would you have chosen to maximize your chances of victory?

Coming up with any old five-letter words wasn’t particularly challenging. But to pick four *good* words that also worked well together, you pretty much had to tackle last week’s extra credit, which asked you to determine your chances of victory given your choice of four words.

This turned out to be a coding exercise. The most challenging part was getting all those edge cases right, particularly when there were repeated letters in one of your guesses or in the mystery word itself. A neat result that came out of all this was that if there were *N* five-letter words in the dictionary (here, *N* happened to be 8,636) and there were *K* distinct states that the Lingo board took on in terms of correct letters and the given first letter, your chances of guessing the correct word were exactly *K*/*N*.

In any case, the puzzle’s creator, Vince Vatter, and myself independently wrote code to score sequences of five-letter words and verified that our code always returned the same results. (Translation: Vince helped me debug my code.) So if you think you were scored in error, it’s far more likely that you were in error than both Vince and myself. (Translation: Vince didn’t screw up, you did.)

Prior to submitting the puzzle, Vince himself found a sequence of words (WIDTH, BARES, CLOMP, GUNKY) that won a whopping 95.79 percent of the time. But this week’s winners did even better! Without further ado, here were all the submitters who did better than 95 percent:

##### Could you maximize your chances at Riddler Lingo?

Top performers at Riddler Lingo by their win percentage, plus their winning five-letter words

Rank | Name | Words | Win % |
---|---|---|---|

1 | Mark Bradwin | BINTS, CLOAK, GYRED, WHUMP | 95.96 |

1 | Nic Tamburello | BINTS, GYRED, WHUMP, CLOAK | 95.96 |

3 | Jenny Mitchell | BLINK, CHOMP, GUDES, WARTY | 95.90 |

4 | James Anderson | BARES, GUNKY, WIDTH, CLOMP | 95.79 |

5 | Q P Liu | CARES, BUMPY, KLONG, WIDTH | 95.74 |

5 | Nikhil Mahajan | CARES, KLONG, BUMPY, WIDTH | 95.74 |

7 | Michael Engen | CLIPT, GYBED, KHOUM, WARNS | 95.48 |

8 | David Devore | CHOMP, FURAN, GYBED, KILTS | 95.40 |

9 | Dallas Trinkle | BLOTS, CAGER, DINKY, WHUMP | 95.39 |

10 | James Bach | FUMED, BLOCS, PINKY, GARTH | 95.08 |

As part of his strategy, Mark only considered word combinations with 20 unique letters — a trend you can see up and down the leaderboard. Nic, meanwhile, automated his search by using simulated annealing.

Finally, since writing a scoring algorithm was the majority of the work required for this puzzle, I wanted to give a special shoutout to all submitters who correctly scored their own submission in the extra credit. In addition to most of those on the leaderboard, this list included: David Ding, Mike Onigman, Laurynas Navidauskas, Bryce Wargin, Peter Ji and Paulina Leperi.

Even without knowing the first letter in advance, Riddler Nation has truly put Lingo’s contestants to shame. Well done, all!

## Want more riddles?

Well, aren’t you lucky? There’s a whole book full of the best puzzles from this column and some never-before-seen head-scratchers. It’s called “The Riddler,” and it’s in stores now!

## Want to submit a riddle?

Email Zach Wissner-Gross at riddlercolumn@gmail.com.