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, 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.
This week, get ready to play a game of squares! You start with five shaded squares on an infinitely large grid, in the following formation:
This is generation 1. To go from one generation to the next, you consider every square’s eight neighbors (up, down, left, right and the four diagonal directions). If at least three of those squares are shaded, in the previous iteration, that square will be shaded in the next generation.
That said, here are the first four generations:
How many squares will be shaded in generation 10?
Extra credit: As N gets very, very large, approximately how many squares will be shaded in generation N (in terms of N)?
From Travis Henry comes some vehicular trouble:
You want to change the transmission fluid in your old van, which holds 12 quarts of fluid. At the moment, all 12 quarts are “old.” But changing all 12 quarts at once carries a risk of transmission failure.
Instead, you decide to replace the fluid a little bit at a time. Each month, you remove one quart of old fluid, add one quart of fresh fluid and then drive the van to thoroughly mix up the fluid. (I have no idea if this is mechanically sound, but I’ll take Travis’s word on this!) Unfortunately, after precisely one year of use, what was once fresh transmission fluid officially turns “old.”
You keep up this process for many, many years. One day, immediately after replacing a quart of fluid, you decide to check your transmission. What percent of the fluid is old?
Solution to last week’s Riddler Express
Congratulations to 👏 Mike Porter 👏 of Haddenham, United Kingdom, winner of last week’s Riddler Express.
Last week, you had a square piece of paper. You folded it in half along an axis parallel to two of its sides. You then folded it in half along another axis that was perpendicular to the original one, so that you again had a square that was one-quarter the size of the original square.
At this point, you made three cuts along three straight lines, all the way through the folded paper. As you unfolded the paper, what was the greatest number of separate pieces you could have?
After you made the cuts, your folded square was made up of some number of pieces. You even knew this number was at most seven, since that’s the greatest number of regions into which the plane can be partitioned by three lines. But what you had to figure out was how many times each of these seven regions could be duplicated by folding up the paper before cutting.
Consider the following example, in which the paper was folded into the top right quarter, after which you made the three cuts shown:
When you unfolded the paper, there would be one copy of the region labeled “1.” Meanwhile, the regions labeled “2” would each have two copies, either horizontal or vertical reflections of each other. Finally, the region labeled “4” would have four copies, one in each of the four quarters of the larger square.
The key to this puzzle was realizing that these copy counts (one, two and four) were determined by which sides of the smaller square — particularly the left and bottom sides — bounded which region. If a region was bounded by both the left and bottom sides, there was just one copy. If it was bounded by exactly one of the two sides, there were two copies. And if it was bounded by neither side, there were four copies.
The riddle was asking you to maximize the number of copies, which meant you wanted lots of regions with four copies and not so many regions with one copy. Now, at least one region had to contain the bottom-left corner of the smaller square, which implied it would result in one copy. But what about the six other regions? They could all have four copies, if you made the cuts like so:
That gave you a maximum of 25 copies.
For extra credit, instead of three cuts, you made N cuts. Now what was the greatest number of pieces you could have?
Mike, this week’s winner, extended the approach for three cuts, again having only one region bounded by the inner sides of the smaller square:
In general, N lines can partition the plane into at most N(N+1)/2+1 regions. One of those regions had to have one copy, while the remaining N(N+1)/2 had four. The maximum total was therefore 4N(N+1)/2+1, or 2N(N+1)+1 copies.
By the way, this quadratic relationship means you can generate loads of pieces without making too many cuts. For example if you have the endurance and precision required to make 100 cuts, you can generate up to 20,201 pieces!
Solution to last week’s Riddler Classic
Congratulations to 👏 Steve Curry 👏 of Albuquerque, New Mexico, winner of last week’s Riddler Classic.
Last week, readers of The Riddler Ellora Sarkar and Daniel Gomez got married! In keeping with the hexagonal design of their wedding backdrop, I presented the following puzzle:
The larger regular hexagon in the diagram below had a side length of 1.
What was the side length of the smaller regular hexagon?
Solver David Ding supposed that half the side length of the smaller hexagon was x. Next, David drew the following right triangle within the circle:
The hypotenuse of this triangle was the radius of the circle, which also happened to be the side length of the larger hexagon, 1. The triangle’s shorter leg, which was the left half of the topmost side of the smaller hexagon, was x/2. Meanwhile, the length of the remaining leg, which could be found with a little trigonometry, was √3/2 + √3x, or √3(x + 1/2). Next, David plugged these three values into the Pythagorean theorem, which gave the following quadratic equation (after some simplification): 13x2 + 12x − 1 = 0. This could be factored as (13x−1)(x+1) = 0. Rejecting the negative root meant that x, the side length of the smaller hexagon, was 1/13. It was kind of neat how after all this work, a nice little rational number popped out!
For extra credit, you were asked to find the side lengths of two more, even smaller, hexagons on top, as shown in the animation below:
By again applying the Pythagorean theorem, but this time to successively smaller hexagons, solver Laurent Lessard found a recurrence relation for one hexagon’s side length (xn+1) in terms of the adjacent, larger hexagon’s side length (xn):
xn+1 = (√(48+xn2) − √(48−12xn2))/13
In the puzzle, you were told that x1 was 1. And we just found that x2 was 1/13. Applying the above formula, x3 was approximately 4.27×10-4 and x4 was approximately 1.32×10-8.
You might have noticed that these values got really small, really fast. As Laurent showed, the number of hexagons and the side length of the smallest hexagon approached a double exponential relationship.
That’s the kind of detail that you need for a great wedding. Congratulations again to Ellora and Daniel!
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 [email protected].