Braille puzzle

This week’s Fiddler is a counting problem about the Braille system.

Braille characters are formed by raised dots arranged in a braille cell, a three-by-two array. With six locations for dots, each of which is raised or unraised, there are 26, or 64, potential braille characters.

Each of the 26 letters of the basic Latin alphabet (shown below) has its own distinct arrangement of dots in the braille cell. What’s more, while some arrangements of raised dots are rotations or reflections of each other (e.g., E and I, R and W, etc.), no two letters are translations of each other. For example, only one letter (A) consists of a single dot – if there were a second such letter, A and this letter would be translations of each other.

Of the 64 total potential braille characters, how many are in the largest set such that no two characters consist of raised dot patterns that are translations of each other?

Extra Credit In addition to six-dot braille, there’s also an eight-dot version. But what if there were even more dots? Let’s quadruple the challenge by doubling the size of the cell in each dimension. Consider a six-by-four array, where a raised dot could appear at each location in the array. Of the 224 total potential characters, how many are in the largest set such that no two characters are translations of each other?

My solution:
[Show Solution]

Making something out of nothing

This week’s Fiddler is a problem about composing functions. Here it goes:

Consider $f(n) = 2n+1$ and $g(n) = 4n$. It’s possible to produce different whole numbers by applying combinations of $f$ and $g$ to $0$. How many whole numbers between $1$ and $1024$ (including $1$ and $1024$) can you produce by applying some combination of $f$’s and $g$’s to the number $0$?

Extra Credit: Now consider the functions $g(n) = 4n$ and $h(n) = 1−2n$. How many integers between $-1024$ and $1024$ (including $-1024$ and $1024$) can you produce by applying some combination of $g$’s and $h$’s to the number $0$?

My solution:
[Show Solution]


This week’s Riddler Classic is about Lotería, also known as Mexican bingo!

A thousand people are playing Lotería, also known as Mexican bingo. The game consists of a deck of 54 cards, each with a unique picture. Each player has a board with 16 of the 54 pictures, arranged in a 4-by-4 grid. The boards are randomly generated, such that each board has 16 distinct pictures that are equally likely to be any of the 54.

During the game, one card from the deck is drawn at a time, and anyone whose board includes that card’s picture marks it on their board. A player wins by marking four pictures that form one of four patterns, as exemplified below: any entire row, any entire column, the four corners of the grid and any 2-by-2 square.

Four four-by-four grids are shown. In the first grid, the third row has four blue markers. In the second grid, the second column has four blue markers. In the third grid, the four corner squares are marked. And in the fourth grid, the two middle squares in the third and fourth columns are marked, forming a smaller two-by-two square.

After the fourth card has been drawn, there are no winners. What is the probability that there will be exactly one winner when the fifth card is drawn?

My solution:
[Show Solution]

Shared birthdays

This week’s Riddler Classic is a challenging counting problem about shared birthdays.

Suppose people walk into a room, one at a time. Their birthdays happen to be randomly distributed throughout the 365 days of the year (and no one was born on a leap day). The moment two people in the room have the same birthday, no more people enter the room and everyone inside celebrates by eating cake, regardless of whether that common birthday happens to be today.

On average, what is the expected number of people in the room when they eat cake?

Extra credit: Suppose everyone eats cake the moment three people in the room have the same birthday. On average, what is this expected number of people?

My solution:
[Show Solution]

Knocking down the last bowling pin

This week’s Riddler classic is a tough one! Here is a paraphrased version of the problem.

Imagine $n^2$ bowling pins arranged in a rhombus. The image to the right illustrates the case $n=3$. We knock down the topmost pin. When any pin gets knocked down, each of the (up to two) pins directly behind it has a probability $p$ of being knocked over (independently of each other). We are interested in the probability that the bottommost pin gets knocked down in the limit of large $n$.

The original problem specifically asked about $p=0.5$ and $p=0.7$.

My solution:
[Show Solution]

Tower of goats

This week’s Riddler classic is a counting problem. Can the goats fit in the tower?

A tower has 10 floors, each of which can accommodate a single goat. Ten goats approach the tower, and each goat has its own (random) preference of floor. Multiple goats can prefer the same floor. One by one, each goat walks up the tower to its preferred room. If the floor is empty, the goat will make itself at home. But if the floor is already occupied by another goat, then it will keep going up until it finds the next empty floor, which it will occupy. But if it does not find any empty floors, the goat will be stuck on the roof of the tower. What is the probability that all 10 goats will have their own floor, meaning no goat is left stranded on the roof of the tower?

My solution:
[Show Solution]

Squid game

This week’s Riddler Classic is Squid Game-themed!

There are 16 competitors who must cross a bridge made up of 18 pairs of separated glass squares. Here is what the bridge looks like from above:

To cross the bridge, each competitor jumps from one pair of squares to the next. However, they must choose one of the two squares in a pair to land on. Within each pair, one square is made of tempered glass, while the other is made of normal glass. If you jump onto tempered glass, all is well, and you can continue on to the next pair of squares. But if you jump onto normal glass, it will break, and you will be eliminated from the competition.

The competitors have no knowledge of which square within each pair is made of tempered glass. The only way to figure it out is to take a leap of faith and jump onto a square. Once a pair is revealed — either when someone lands on a tempered square or a normal square — all remaining competitors take notice and will choose the tempered glass when they arrive at that pair.

On average, how many of the 16 competitors will make it across the bridge?

Here is my solution.
[Show Solution]

And here is a much better solution!
[Show Solution]

Flawless war

his week’s Riddler Classic has to do with the card game “War”. Here is the problem, paraphrased:

War is a two-player game in which a standard deck of cards is first shuffled and then divided into two piles with 26 cards each; one pile for each player. In every turn of the game, both players flip over and reveal the top card of their deck. The player whose card has a higher rank wins the turn and places both cards on the bottom of their pile. Assuming a deck is randomly shuffled before every game, how many games of War would you expect to play until you had a game that lasted just 26 turns (with no ties; a flawless victory)?

Here is my solution:
[Show Solution]

Cutting a ruler into pieces

This week’s Riddler Classic is a paradoxical question about cutting a ruler into smaller pieces.

Recently, there was an issue with the production of foot-long rulers. It seems that each ruler was accidentally sliced at three random points along the ruler, resulting in four pieces. Looking on the bright side, that means there are now four times as many rulers — they just happen to have different lengths. On average, how long are the pieces that contain the 6-inch mark?

With four cuts, each piece will be on average 3 inches long, but that can’t be the answer, can it?

Here is my solution:
[Show Solution]

Connect the dots

This week’s Riddler Classic is a problem about connecting dots to create as many non-intersecting polygons as possible. Here is the problem:

Polly Gawn loves to play “connect the dots.” Today, she’s playing a particularly challenging version of the game, which has six unlabeled dots on the page. She would like to connect them so that they form the vertices of a hexagon. To her surprise, she finds that there are many different hexagons she can draw, each with the same six vertices.

What is the greatest possible number of unique hexagons Polly can draw using six points?

(Hint: With four points, that answer is three. That is, Polly can draw up to three quadrilaterals, as long as one of the points lies inside the triangle formed by the other three. Otherwise, Polly would only be able to draw one quadrilateral.)

Extra Credit: What is the greatest possible number of unique heptagons Polly can draw using seven points?

Here is my solution:
[Show Solution]