# Can You Pack The Ellipses?

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 or send me an email.

## Riddler Express

From Paul R. Pudaite comes speed-skating stumper:

In long-track speed skating’s mass start competition, points awarded at three intermediate sprints (laps 4, 8 and 12) and one final sprint (lap 16). The first three skaters to finish each intermediate sprint earn 3 points, 2 points and 1 point, respectively. For the final sprint, the first six skaters to cross the line respectively earn 60, 40, 20, 10, 6 and 3 points.

In the end, competitors are ranked by how many points they accumulated during the race. When skaters with the same point total, whoever crossed the finish line first is ranked higher.

A particular skating competition consists of two semifinal heats, with eight skaters from each heat advancing to the final.

What is the highest possible point total for a skater who fails to advance to the final?

## Riddler Classic

For whatever reason, there is once again intrigue on social media concerning optimized packing. In particular, I have seen several tweets and even a web comic over the last month about packing unit squares into as small a larger square as possible. In particular, the case of 17 squares is known for its lack of symmetry and general aesthetic appeal:

There have been similar packing puzzles on The Riddler before, but with this renewed interest, I figured it was time for another one. But circles and squares are so cliche.

This week, try packing three *ellipses* with a major axis of length 2 and a minor axis of length 1 into a larger ellipse with the same aspect ratio. What is the smallest such larger ellipse you can find? Specifically, how long is its major axis?

*Extra credit: *Instead of three smaller ellipses, what about other numbers of ellipses?

## Solution to the last Riddler Express

Congratulations to ÑÐ±âÑÐ±Ã·âÐ±âÐ±â¤ÐÐ±â Jim Woest ÑÐ±âÑÐ±Ã·âÐ±âÐ±â¤ÐÐ±â of Huntington Beach, California, winner of last week’s Riddler Express.

Last week, you had four squares to place on a large, flat table. You could place the squares so that their edges aligned, but their interiors could not overlap.

Your goal was to position the squares so that you could trace as many rectangles as possible using the edges of the squares. For example, if you had two squares instead of four, you could have placed the squares side by side, as shown below:

With this arrangement, it was possible to trace three rectangles: the square on the left, the square on the right and the larger rectangle around both squares.

How would you have arranged four squares to get as many rectangles as possible? And what was this number of rectangles?

Most solvers started by examining a few arrangements of squares. For example, you could have arranged the four squares in a two-by-two grid:

Here, there were four one-by-one squares, two rows, two columns and one larger two-by-two square. In total, there were nine distinct rectangles. But it was possible to do better.

Next, consider the four-by-one arrangement:

Again, there were four one-by-one squares, as well as three two-by-one rectangles, two three-by-one rectangles and one four-by-one rectangle. In total, there were 10 distinct rectangles. But it was possible to do better.

The two aforementioned arrangements had four contiguous squares. However, this wasn’t required by the puzzle. In fact, the mind-blowing optimal arrangement had four squares that only touched each other at the corners. Here was that arrangement, with the four squares colored for clarity:

While this may have looked like five squares, there was no square in the middle — that was just empty space. Nevertheless, the edges of the surrounding four rectangles could still be used to form rectangles. This time, there were *five* one-by-one squares, two horizontal two-by-one rectangles, two vertical two-by-one rectangles, one horizontal three-by-one rectangle and one vertical three-by-one rectangle. In total, there were **11** distinct rectangles.

Readers were evenly split between answers of 10 and 11. Solver Cass from Denver, Colorado, correctly answered 11 but still wasn’t convinced this was the solution, writing, “I think this is the answer, but I also think that, in the end, I have yet lost. Perhaps Riddler is defeated, perhaps he is not. But even if he is, I am not Batman. I am Joker now.”

Cass — this week, you *are* Batman!

## Solution to the last Riddler Classic

Congratulations to ÑÐ±âÑÐ±Ã·âÐ±âÐ±â¤ÐÐ±â Ryan Lafitte ÑÐ±âÑÐ±Ã·âÐ±âÐ±â¤ÐÐ±â of Tucker, Georgia, winner of last week’s Riddler Classic.

Last week, you tried your hand at a version of solitaire played in southern Italy with a deck of 40 Neapolitan cards, with four suits numbered from 1 to 10. The deck was shuffled and then cards were turned over one at a time. Flipping over the first card you said “one,” the second card “two” and the third card “three.” You repeated this, saying “one” for the fourth card, “two” for the fifth card and “three” for the sixth card. You continued your way through the deck, until you at last said “one” for the 40th card.

If at any point the number you said matched the value of the card you flipped over, you lost.

What was your probability of winning the game?

This was a veritable combinatorial challenge. There were essentially four different card types to consider:

- Four 1s, which could not occur in 14 places in the deck
- Four 2s, which could not occur in 13 places in the deck
- Four 3s, which could not occur in 13 places in the deck
- 28 other cards, which could occur anywhere in the deck

Several readers came up with a decent approximation for the correct solution. While flipping through the deck, you will at some point encounter the 12 cards numbered 1, 2 or 3. For each of these, you have a roughly two-in-three chance of saying the number *not* indicated on the card. That meant your probability of winning was approximately (2/3)^{12}, or 0.77 percent.

Austin Shapiro of Oakland, California put this approximation to the test by running 10 million solitaire simulations. It turned out your chances were slightly better than predicted: roughly **0.83 percent**.

In my opinion, the exact calculation wasn’t particularly enlightening. It required you to consider all 40!/(4!·4!·4!·28!) — or 193,584,473,082,000 — cases, and figure out which among them had all the 1s, 2s and 3s in the right spots.

Solver Starvind worked through the detailed combinatorics here and even extended the puzzle so that victory allowed for a certain number of matched numbers greater than zero. According to Starvind, you should consider a variant of this solitaire that allows for up to three matches, which increases your chances of winning to almost 40 percent.

## Want more puzzles?

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.