Archive for the ‘Puzzles’ Category.
Every year I write about latest MIT Mystery Hunt puzzles that might be appealing to mathematicians. Before diving into mathy puzzles, I would like to mention two special ones:
Unfortunately math wasn’t prominent this year:
- Food Court—This is a probability puzzle that is surprisingly uninspiring. There is no mystery: the puzzle page contains a list of probability problems of several famous types. But this puzzles can find great use in probability classes.
- Torsion Twirl—Mixture of dancing and equations. I love it.
- People Mover—Logical deduction at the first stage.
On the other hand, Nikoli-type puzzles were represented very well:
- The Ferris of Them All—Several different Nikoli puzzles on a wheel.
- Toddler Tilt—Not exactly a Nicoli puzzle, but some weird logic on a grid, some music too.
- The Dollhouse Tour—Not exactly a Nicoli puzzle, but some weird logic on a grid, some pictures too.
- The Nauseator—The first part of the puzzle is a huge nonogram.
- Domino Maze—A non-trivial Thinkfun puzzle.
- Backlot—Finding a path on a grid with a fractal structure.
- Whale—Variation on Rush Hour.
Some computer sciency puzzles:
A couple of puzzles with the mathy side hidden:
The academic year is over and my junior PRIMES STEP group finished their paper about a classification of magic SET squares. A magic SET square is a 3 by 3 square of SET cards such that each row, column, and diagonal is a set. See an example below. The paper is posted at the arXiv:2006.04764.
In addition to classifying the magic SET squares, my students invented the game of SET tic-tac-toe. It is played on nine cards that form a magic SET square. Two players take turns picking a card from the square. The first player who has a set wins.
One might think that this game is the same as tic-tac-toe, as a player wins as soon at they have cards from the same row, column, or diagonal. But if you build a magic SET square, you might notices that each magic SET square contains 12 sets. In addition to rows, columns, and diagonals, there are sets that form broken diagonals. The picture below shows all the sets in a magic SET square.
There are more ways to win in this game than in a regular tic-tac-toe game. My students proved that ties are impossible in this game. They also showed, that, if played correctly, the first player always wins.
I heard this puzzle many years ago, and do not remember the origins of it. The version below is from Peter Winkler’s paper Seven Puzzles You Think You Must Not Have Heard Correctly.
Jan and Maria have fallen in love (via the internet) and Jan wishes to
mail her a ring. Unfortunately, they live in the country of Kleptopia
where anything sent through the mail will be stolen unless it is
enclosed in a padlocked box. Jan and Maria each have plenty of padlocks,
but none to which the other has a key. How can Jan get the ring safely
into Maria’s hands?
I don’t know whether this puzzle appeared before the Diffie-Hellman key
exchange was invented, but I am sure that one of them inspired the
other. The official solution is that Jan sends Maria a box with the ring
in it and one of his padlocks on it. Upon receipt Maria affixes her own
padlock to the box and mails it back with both padlocks on it. When Jan
gets it, he removes his padlock and sends the box back, locked only
with Maria’s padlock. As Maria has her own key, she can now open it.
My students suggested many other solutions. I wonder if some of them can be translated to cryptography.
- Jan can send the ring in a padlock box that is made of cardboard. Maria can just cut the cardboard with a knife.
- Jan can use the magic of the Internet to send Maria schematics of the
key so she can either 3d print it or get a professional to forge it. If they are
afraid of the schematics getting stolen Jan can send the schematics after the
package has been delivered.
- Jan can use a digital padlock and send the code using the Internet.
- Jan can send it in a secret puzzle box that can be opened without a key.
- Maria can smash the padlock with a hammer.
Now that we’ve looked at the Padlock Puzzle, let’s talk about
cryptography. I have an imaginary student named Charlie who doesn’t know
the Diffie-Hellman key exchange. Charlie decided that he can adapt the
padlock puzzle to help Alice send a secret message to Bob. Here’s what
Suppose the message is M. Alice converts it to binary. Then she creates a
random binary key A and XORs it with M. She sends the result, M XOR A,
to Bob. Then Bob creates his own random key B and XORs it with what he
receives and sends the result, M XOR A XOR B, back to Alice. Alice XORs
the result with her key to get M XOR A XOR B XOR A = M XOR B and sends
it to Bob. Bob XORs it with his key to decipher the message.
Each sent message is equivalent to a random string. Intercepting it is
not useful to an evil eavesdropper. The scheme is perfect. Or is it?
Here is a logic puzzle.
Puzzle. You are visiting an island where all people know each
other. The islanders are of two types:
truth-tellers who always tell the truth and liars who always lie. You
meet three islanders—Alice, Bob, and Charlie—and ask each of
them, “Of the two other islanders here, how many are truth-tellers?”
Alice replies, “Zero.”
Bob replies, “One.” What will Charlie’s reply be?
The solution proceeds as follows. Suppose Alice is a truth-teller. Then
Bob and Charlie are liars. In this situation Bob’s statement is true,
which is a contradiction. Hence, Alice is a liar. It follows, that there
is at least one truth-teller between Bob and Charlie. Suppose Bob is a
liar. Then the statement that there is one truth-teller between Alice
and Charlie is wrong. It follows that Charlie is a liar. We have a
contradiction again. Thus, Alice is a liar and Bob is a truth-teller.
From Bob’s statement, we know that Charlie must be a truth-teller. That
means, Charlie says “One.”
But here is another solution suggested by my students that uses meta
considerations. A truth-teller has only one possibility for the answer,
while a liar can choose between any numbers that are not true. Even if
we assume that the answer is only one of three numbers—0, 1, or 2—then
the liar still has two options for the answer. If Charlie is a liar,
there can’t be a unique answer to this puzzle. Thus, the puzzle
question implies that Charlie is a truth-teller. It follows that Alice
must be lying and Bob must be telling the truth. And the answer is the
same: Charlie says, “One.”
I found this puzzle on Facebook:
Puzzle. Solve this:
1+4 = 5,
2+5 = 12,
3+6 = 21,
5+8 = ?
97% will fail this test.
Staring at this I decided on my answer. Then I looked at the comments:
they were divided between 34 and 45 and didn’t contain the answer that
initially came to my mind. The question to my readers is to explain the
answers in the comments and suggest other ones. Can you guess what my