Tanya Khovanova's Math Blog

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.

Puzzle. 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 Charlie suggests:

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.”

This is my favorite puzzle in the last issue of the Emissary, proposed by Peter Winkler.

Puzzle. Alice and Bob each have 100 dollars and a biased coin that flips heads with probability 51%. At a signal, each begins flipping his or her coin once per minute, and betting 1 dollar (at even odds) on each flip. Alice bets on heads; poor Bob, on tails. As it happens, however, both eventually go broke. Who is more likely to have gone broke first?
Follow-up question: As above, but this time Alice and Bob are flipping the same coin (biased 51% toward heads). Again, assume both eventually go broke. Who is more likely to have gone broke first?

Puzzle. You got two envelopes with two distinct real numbers. You chose one of them and open it. After you see the number you are allowed to swap envelopes. You win if at the end you pick the larger number. Find a strategy that gives you a probability more than 1/2 of winning.

Another cute puzzle found on Facebook.

Puzzle. A teacher wrote four positive numbers on the board and invited his students to calculate the product of any two. The students calculated only five of six products and these are the results: 2, 3, 4, 5, 6. What is the last product? What are the original four numbers?

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 answer was?

Puzzle. The length of the day today in Boston is the same as the length of the coming night tonight. What is the total length of both?

My friend Alice reminds me of me: she has two sons and she is never straight with her age. Or, maybe, she just isn’t very good with numbers.

Once I visited her family for dinner and asked her point blank, “How old are you?” Here is the rest of the conversation:

Alice: I am two times older than my younger son was 5 years ago.
Bob: My mom is 12 times older than my older brother.
Carl: My younger brother always multiplies every number he mentions by 24.
Bob: My older brother is 30 years older than me.
Carl: My mom is 8 times older than me.
Alice: My older son always multiplies every number he mentions by 2.

How old is everyone?

Last year, when I read an application file of Wayne Zhao to PRIMES, I got very excited because he liked puzzles. And I’ve always wanted to have a project about puzzles. After Wayne was accepted to PRIMES we started working together. Wayne chose to focus on a variation of Sudoku called Sudo-Kurve.

We chose a particular shape of Sudo-Kurve for this project, which ended up being very rewarding. It is called Cube Sudo-Kurve. The Cube Sudo-Kurve consists of three square blocks. The gray bent lines indicate how rows and columns continue. For example, the first row of the top left block becomes the last column of the middle block and continues to the first row of the bottom right block. As usual each row, column, and square region has to have 9 distinct digits.

Wayne and I wrote a paper Mathematics of a Sudo-Kurve, which has been published at Recreational Mathematics Magazine.

A Cube Sudo-Kurve needs at least 8 clues to have a unique solution. Here we have a puzzle with 8 clues that we designed for our paper.

I’ve been invited to help with the Puzzle Column at the MSRI newsletter Emissary. We prepared six puzzles for the Fall 2018 issue.

I love the puzzles there. Number 2 is a mafia puzzle that I suggested. Number 6 is a fun variation on the hat puzzle I wrote a lot about. Here is puzzle Number 3.

Puzzle. Let A = {1,2,3,4,5} and let P be the set of all nonempty subsets of A. A function f from P to A is a “selector” function if f(B) is in B, and f(B union C) is either equal to f(B) or f(C). How many selector functions are there?