June 4, 2023

Brighton Journal

Complete News World

Can you save your crew?

Welcome to The Riddler. Every week I present problems related to the things we hold dear here: math, logic, and probability. Two puzzles are offered each week: Riddler Express for those of you who want something bite-sized and Riddler Classic for those of you who are into the slow puzzle movement. Submit a correct answer to either one, and you might get a shout out to the next column. Please wait until Monday to share your answers publicly! If you need a hint or have a favorite puzzle gathering dust in the attic, Find me on Twitter Or send me an email.

Riddler Express

For Easter, you and your family decide to decorate 10 beautiful eggs. She pulls a carton of fresh eggs from the refrigerator and removes 10 eggs. There are two eggs left in the carton, and you put them back in the fridge.

The next day, she opens the carton again to find that the positions of the eggs have somehow shifted – or so she thinks. Maybe the Easter Bunny has been snooping around your fridge?

The 12 holes in the carton are arranged in a symmetrical six-by-two row at a 180-degree rotation, and the eggs are indistinguishable from each other. How many distinct ways to put two eggs in this carton? (Note: Placing two eggs in the two leftmost slots should be considered the same as placing them in the two rightmost slots, since you can switch between these arrangements by rotating the carton 180 degrees.)

Extra credit: Instead of two eggs left, suppose you have other indistinguishable numbers of eggs between zero and 12. How many distinct ways are there to put these eggs in the carton?

Submit your answer

Riddler Classic

From Nis Jørgensen comes a picaresque mystery of a captain and crew:

You are the captain of a crew of three (not including you): Geordie, Sydney, and Alandra. Your ship has been captured by a previously unknown enemy, who decides to return your ship if you can win a simple game.

Each of the three crew members will be issued a number between zero and one, chosen at random and standardized within that range. As the captain, your goal is to guess who has the largest number.

The catch is that you can only ask a yes or no question to each crew member. Depending on the answer to the question you ask the first crew member, you can update the question you ask the second. Similarly, based on the answers to the first two questions, you can update the third question you ask. But in the end, you still have to guess which crew member has the most.

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What’s your optimal strategy, and what are your chances of getting your ship back?

Submit your answer

Another solution to Riddler Express

Congratulations to Sweet Tea Dorminy of Greenville, South Carolina, last week’s Riddler Express winner.

Express was introduced last week by high school student Max Misterka, winner of the 2023 Regeneron Science Talent Search. Max and I were playing a game where we secretly chose a number. Let’s call Max’s number M And my number z. After we both revealed our numbers, it was Max’s score Mzwhile my result zM. Whoever gets the highest score wins.

When we played recently, Max and I picked different integers. Surprisingly, we tied – there was no winner! What numbers did we choose?

Because Max and I tied, that means integers M And z Equality satisfies Mz = zM. by taking M-th and z– Roots to both sides, that gave you M1 /M = z1 /z. At this point, it was worth taking a closer look at the job F(x) = x1 /x. After all, after M1 /M = z1 /z Means the same thing as having F(M) = F(z).

This function is incremented for small values ​​of xto reach the maximum value when x It was about 2.718 (ie H). After this maximum, the function decreased forever, asymmetrically approaching 1. Because the function was increasing and then decreasing, with no further change of direction in between, it means either M or z must be less than Hwhile the other number should have been greater than H. Let’s say so M The number was the smallest.

At this point, there weren’t many options: M It must be either 1 or 2. If M was 1, then you need 1z = z1Which means z It was also equal to 1. Since he said the riddle M And z were distinct, this was not a viable solution. if M was instead 2, then you need 2z = z2. Certainly, this equation has two solutions: z = 2 (which again did not yield distinct numbers) and z = 4. Thus, the only two integers I could choose were Max and 2 and 4such as 24 = 42.

To get extra credit, you had to analyze another round of the game where Max and I picked positive numbers that weren’t necessarily whole numbers. I told Max my number without knowing his, at which point he told me the match was again a draw. I replied: “Ah, that means we must have chosen the same number!” What number did we choose?

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Mathematically, this means that F(M) = F(z) implied it M and z were equal. for any value M between 1 and HThere was a return z more than H Such that F(M) = F(z). so for F(M) = F(z) to suggest M = z, Given that they were at least 1, both M And z It had to be H. Alternatively, as noted by Fernando Mendez’s solution, both can be any positive number less than or equal to 1.

Facing a high school student who was at the top of his class in math and science, all I can say is that I’m glad I tied (instead of lost) both times we played this game.

A solution to the latest Riddler Classic

Congratulations to Jason Weinrib of Phoenix, AZ, who won last week’s Riddler Classic.

Last week, I was introduced to the game-like game Sudoku Star Battle. In the five-star version of the game, you’re trying to fill a 21×21 grid with stars according to certain rules:

  • Each row must contain exactly five stars.
  • Each column must contain exactly five stars.
  • Each selected area in bold must contain exactly five stars.
  • There cannot be two adjacent stars horizontally, vertically or diagonally.

For example, here was a solved gamepad:

In this example, the stars appeared to be evenly distributed throughout the board, although there were some gaps. In particular, this forum had 20 empty 2×2 squares, highlighted below:

Some of these regions overlapped binary in two – however, they are still considered distinct.

In a solved board of Star Battle, what is the minimum and maximum number of 2×2 empty squares?

At first glance, this seemed like a rather complex combinatorial puzzle, or perhaps something that would require a lot of simulation. But as it turns out, you can figure this out with some relatively straightforward algebra!

Solver N. Scott Cardell started by acquiring a plot of land. Each of the 21 rows contained five stars, meaning there were 105 stars in total. Meanwhile, there are 202, or 400, a total of two-by-two squares in the grid. Since stars cannot be adjacent, this means that any 2×2 square contains at most one star.

Now a star on one of the four corners has occurred in exactly one of these 2×2 squares, while a star on one edge has happened on two such squares and a star on the inside of the grid has happened on four such squares. If there are c corner stars, H Edge stars and I Inner stars, the number of squares were 2 by 2 that had a star on it c +2H +4I. Since there are 400 2×2 squares in total, the number of squares without The star was 400 – (c +2H +4I).

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Since the total number of stars was 105, that is c + H + I = 105 or I = 105 – Hc. Furthermore, since each edge (like any other row or column) has five stars, with the corner stars counting for two edges, the H +2c = 20 or H = 20 – 2c.

At this point you can algebraically eliminate the variables from the expression for the number of empty squares 2 x 2, 400 – (c +2H +4I). Delivery 105 – Hc to I gave you 3c +2H – 20. Finally, connect 20-2c to H gave you 20 – c.

After all that work, this was a surprisingly simple result. To find the number of empty 2 x 2 squares, all you have to do is count the number of stars that were in the four corners and subtract that from 20. Sure enough, this was consistent with the solved Star Battle game. In the original puzzle: There were no stars in the corner, and there were 20 empty 2×2 squares.

So what is the answer? The minimum number of empty squares was two by two 16, when the four corners contained stars. was the max 20, when the four corners were devoid of stars. (In my opinion, this puzzle turned out to be simpler than it seemed at first – in contrast to Star Battle itself, which is much more difficult than it seems).

Do you want more puzzles?

Well, aren’t you lucky? There is an entire book full of the best mysteries from this column and some never-before-seen scrapes. It’s called “The Riddler” and it is In stores now!

Do you want to submit a puzzle?

Email Zach Wissner-Gross at [email protected].