Tanya Khovanova's Math Blog

I have an old book which I value a great deal. The book is called 200 Problems in Linguistics and Mathematics and only 1,550 copies were printed in 1972. Luckily, a new extended edition just appeared on the web. Both editions are in Russian, so I decided to translate some of the problems into English. Here is a sample:

Problem 1. Here are phrases in Swahili with their English translations:

• atakupenda — He will love you.
• nitawapiga — I will beat them.
• atatupenda — He will love us.
• anakupiga — He beats you.
• nitampenda — I will love him.
• unawasumbua — You annoy them.

Translate the following into Swahili:

• You will love them.
• I annoy him.

Problem 2. You are given words in Swahili: mtu, mbuzi, jito, mgeni, jitu and kibuzi. Their translations in a different order are: giant, little goat, guest, goat, person and large river. Make the correspondence.

Problem 3. In Russian the middle name is the patronymic. Thus, the middle initial is the first letter of the father’s first name. And, as in many languages, the first initial is the first letter of the first name. Here are names of males in a family:

• A.N. Petrov
• B.M. Petrov
• G.K. Petrov
• K.M. Petrov
• K.T. Petrov
• M.M. Petrov
• M.N. Petrov
• N.M. Petrov
• N.K. Petrov
• N.T. Petrov
• T.M. Petrov

Draw the family tree of the Petrovs, given that every father has two sons, the patriarch of the family has four grandsons, and his sons have two grandsons each. Prove that the solution is unique.

Problem 4. In Latvian a noun can be one of two genders; furthermore, adjectives agree with nouns in gender, number and case. You are given phrases in either the nominative or the genitive case with their translations:

• silts ezers — warm lake
• melns lauva — black lion
• liela krāsns — big oven
• lielas jūras — big sea’s
• sarkana ezera — red lake’s
• melna kafija — black coffee
• sarkans putns — red bird
• liela kalna — big mountain’s
• sarkanas lapas — red leaf’s
• sarkana pils — red castle
• liels ezers — big lake
• melna putna — black bird’s
• liela lauvas — big lion’s
• silta jūra — warm sea
• melnas kafijas — black coffee’s
• liels kalns — big mountain

Indicate which words are nouns and which are adjectives. Divide Latvian nouns into two groups, so that each group contains words of the same gender.

Problem 5. The Portuguese language takes its roots from Latin. In this problem modern Portuguese words are written on the left and their roots (in Latin and other languages on the right). All the words on the left belong to one of three classes: ancient borrowing, early borrowing and late borrowing.

• chegar — plicare
• praino — plaine
• plátano — platanum
• chão — planum
• plebe — plebem
• cheio — plenum
• prancha — planche

For every Portuguese word, indicate which class it belongs to. (Note that in Portuguese “ch” is pronounced as “sh”.)

I love “Knights and Knaves” logic puzzles. These are puzzles where knights always tell the truth and knaves always lie. A beautiful variation of such a puzzle with Gnyttes and Mnaivvs was given at the 2009 MIT mystery hunt. In this puzzle people’s ability to tell the truth changes during the night depending on the sex partner. You will enjoy figuring out who is a Gnytte or a Mnaivv for each day, who is infected and who slept with whom on each night. Just remember that the ultimate answer to the puzzle is a word or a phrase. So there is one more step after you solve the entire logic part. You do not really need to do this last step, but you might as well. Here we go:

The Sexaholics of Truthteller Planet

Each inhabitant of Veritas 7, better known as the Truthteller Planet, manifests one of two mutations: Gnytte or Mnaivv. Gnyttes always tell the truth, and Mnaivvs always lie. Once born a Gnytte or Mnaivv, the inhabitant can never change…until now.

Veritas 7 is in the midst of an outbreak of a nasty virus dubbed Nallyums Complex II. If an infected inhabitant has sex with another inhabitant of that planet, each one can be converted from Gnytte to Mnaivv, or Mnaivv to Gnytte, as shown below:

• If the sex is heteroverific (1 Gnytte and 1 Mnaivv), both become Mnaivvs.
• If the sex is homoverific (2 Gnyttes or 2 Mnaivvs), both become Gnyttes.

This occurs if either party or both parties have the disease. The disease itself is not transmitted via sex, which is some relief.

Several members of a Veritas 7 village have just contracted Nallyums Complex II. Below are statements from the 15 village residents, taken over the 5-day period since the outbreak. Interviews were taken each morning, and sex occured only at night. Each night, residents paired off to form seven separate copulating couples, with one individual left out. No individual was left out for more than one night.

Can you identify the infected individuals and track their pattern of sexual activity?

A note on wording: If someone refers to sex with someone who was a Gnytte or Mnaivv, they are referring to the individual’s truth-telling status just before sex. If a clue says that two individuals had sex, it means they had sex with each other. “Mutation” refers to the individual’s current status as Gnytte or Mnaivv.

Interviews Day 1

• Artoo: Etrusco is not infected.
• Bendox: Cravulon and Flav are not the same mutation today.
• Cravulon: Either Artoo or Flav is a Gnytte today.
• Dent: Jax-7 and I are both Mnaivvs today.
• Etrusco: Greasemaster is a Mnaivv today.
• Flav: There are at least five Mnaivvs today.
• Greasemaster: Murgatroid is a Gnytte today.
• Holyoid: Etrusco is either an infected Mnaivv or an uninfected Gnytte today.
• Irono: Etrusco is a Mnaivv today.
• Jax-7: Among Artoo, Greasemaster, and Nebulose, exactly one is a Gnytte today.
• Killbot: Bendox is a Gnytte today.
• Lexx: Holyoid and Irono are not the same mutation today.
• Murgatroid: There are at least eight Gnyttes today.
• Nebulose: Murgatroid is a Mnaivv today.
• Oliver: Lexx is a Gnytte today.

Surveillance Night 1

Security cameras revealed that Jax-7 did not have sex with anyone last night, and that Nebulose and Murgatroid had sex.

Interviews Day 2

• Artoo: Last night, I did not have sex with an infected individual.
• Bendox: Last night, Oliver had sex with someone who was a Mnaivv.
• Cravulon: There are at least six Gnyttes today.
• Dent: If there are only five Gnyttes today, then Cravulon and Oliver had sex last night.
• Etrusco: Last night, Bendox had sex with someone who was a Mnaivv.
• Flav: Neither Killbot nor her partner last night is infected.
• Greasemaster: Last night, either Cravulon and Bendox had sex, or Oliver and Etrusco had sex, but not both.
• Holyoid: Last night, I had sex with someone who was a Gnytte.
• Irono: Last night, I did not have sex with Flav.
• Jax-7: Last night, Artoo had sex with Dent.
• Killbot: Last night, I had sex with someone who was a Mnaivv.
• Lexx: Cravulon is infected.
• Murgatroid: Nebulose is not infected.
• Nebulose: Last night, either Cravulon or Flav had sex with Etrusco.
• Oliver: Last night, Irono had sex with someone who was a Mnaivv.

Surveillance Night 2

Security cameras revealed that Holyoid did not have sex with anyone last night, and that Irono and Oliver had sex.

Interviews Day 3

• Artoo: Last night, I had sex with the individual who had sex with Murgatroid on Night 1.
• Bendox: Last night, I had sex with an uninfected individual.
• Cravulon: Last night, Jax-7 had sex with the individual who had sex with Lexx on Night 1.
• Dent: Last night, I had sex with someone who was a Mnaivv.
• Flav: Last night, I had sex with an uninfected individual.
• Holyoid: Neither Oliver nor Irono is infected.
• Irono: Last night, the individual who had sex with Oliver on Night 1 had sex with someone who was a Mnaivv.
• Jax-7: There are more than seven Gnyttes today.
• Killbot: Bendox and Murgatroid are the same mutation today.
• Lexx: Last night, the individual who had sex with Flav on Night 1 had sex with an infected individual.
• Nebulose: The individual who had sex with Etrusco on Night 1 is a Mnaivv today.
• Oliver: Last night, Lexx had sex with the individual who had sex with Bendox on Night 1.

Surveillance Night 3

Security cameras revealed that Dent and Jax-7 had sex.

Interviews Day 4

• Artoo: There are at most seven Gnyttes today.
• Cravulon: Last night, I had sex with someone who has never been a Gnytte.
• Dent: The two sex partners of an individual who was a Mnaivv on the first three days had sex last night.
• Etrusco: Last night, Oliver did not have sex.
• Flav: Last night, I had sex with an infected individual.
• Greasemaster: There are exactly eight infected individuals.
• Killbot: None of the individuals who have been left out of the sexual activity is infected.
• Murgatroid: Last night, I had sex with someone who was a Gnytte on Day 1.
• Nebulose: Last night, I had sex with someone who has had sex with Artoo.
• Oliver: Last night, an individual who had sex with Holyoid during one of the first two nights had sex with an individual who had sex with Jax-7 during one of the first two nights.

Surveillance Night 4

Security cameras have been vandalized.

Interviews Day 5

• Artoo: Last night, I had sex with a Mnaivv, but not the individual who, on Night 3, had sex with the individual who, on Night 2, had sex with the individual who, on Night 1, had sex with Dent.
• Bendox: Last night, Dent had sex with the individual who, on Night 1, had sex with the individual who, on Night 3, had sex with the individual who, on Night 2, had sex with Artoo.
• Cravulon: Last night, Holyoid had sex with the individual who, on Night 3, had sex with the individual who, on Night 2, had sex with the individual who, on Night 1, had sex with Murgatroid.
• Dent: Jax-7 is a Gnytte today.
• Flav: Last night, Cravulon did not have sex with the individual who, on Night 1, had sex with the individual who, on Night 2, had sex with the individual who, on Night 3, had sex with Cravulon.
• Jax-7: Last night, Greasemaster did not have sex.
• Nebulose: Last night, Killbot either had sex with an uninfected individual or did not have sex.

I do not remember where I dug this logic puzzle out from:

Folks living in Trueton always tell the truth. Those who live in Lieberg, always lie. People living in Alterborough alternate strictly between truth and lie. One night 911 got a call: “Fire, help!” The operator couldn’t identify the phone number, so he asked, “Where are you calling from?” The reply was Lieberg.
Assuming no one had overnight guests from another town, where should the firemen go?

After you have solved this problem, you will see that the sentence “Fire, help!” is true. I wonder, if this statement were a lie, how would we interpret it? It could be that it is just a joke and there is no fire and therefore no need for the police. Or it could be that help is needed — perhaps for a robbery, but not for a fire.

However, when you solve this puzzle, you’ll find out that the “Fire, help!” statement is true, so you do not need to wonder what it would mean if the statement were a lie. But I wondered, and as a result I invented several new puzzles.

Here is the first one:

Let’s say that people will call the police only if something is happening and there are only two things that could be happening: fire or robbery. Suppose that when people calling the police need to lie, they replace fire with robbery, and vice versa. Suppose also that when asked about location, people will not say, &quotI am not from Lieberg,” as they could have, but will always reply with one word, which is a name of one of three towns. So, there are two possibilities for the first statement — Fire, Help! or Robbery, Help! — and three possible answers to the question about location. — Trueton, Lieberg and Alterborough. My puzzle in this case is: What answer to the location question will give the biggest headache to the police?

We can branch the original puzzle out in a different way. Here is my second puzzle:

We can assume that only fire, not robbery, could happen in this remote place and when people call the police they either say “Fire, help!” or “We do not have a fire, thank you for your time.” Let us again assume that people will call the police only if something is happening. In this case, what combinations of first and second sentences of the callers will never happen?

My third puzzle is a more complicated version of the second puzzle:

Let’s assume that fires happen with the same probability in every town and that the cost of sending a team of firefighters is identical for every town. Furthermore, the ferocity of all the fires is minuscule and the cost of sending a team is the same, whether or not it turns out that there is a fire. If the police think that the call could have come from several places, they send several teams and the cost multiplies accordingly. The police are obsessed with making charts and the most important number they analyze is the cost they incur, depending on the content of the first sentence of each call.
Given that there are frequent fires, what could the ratio of the cost to police for the calls “Fire, help!” be compared to those that begin with “We do not have a fire, thank you for your time”?

Computational Linguistics Olympiads started in Moscow in 1962. Finally in 2007 the US caught up and now we have the NACLO — North American Computational Linguistics Olympiad.

The problems from past Soviet Olympiads are hard to find, so here I present a translation from Russian of a sample problem from the Moscow Linguistics Olympiad website:

You are given sentences in Niuean language with their translations into English:

1. To lele e manu. — The bird will fly.
2. Kua fano e tama. — The boy is walking.
3. Kua koukou a koe. — You are swimming.
4. Kua fano a ia. — He is walking.
5. Ne kitia he tama a Sione. — The boy saw John.
6. Kua kitia e koe a Pule. — You are seeing Pule.
7. To kitia e Sione a ia. — John will see him.
8. Ne liti e ia e kulï. — He left the dog.
9. Kua kai he kulï e manu. — The dog is eating the bird.

Translate the following sentences into Niuean:

1. John swam.
2. You will eat the dog.
3. Pule is leaving you.
4. The bird will see the boy.
5. The dog is flying.

Here is a baby puzzle. On Monday the baby said A, on Tuesday AU, on Wednesday AUUA, on Thursday AUUAUAAU. What will she say on Saturday?

You can see that this very gifted baby increases her talking capacity twice each day. The first half of what she says repeats her speech of the day before; and the second half is like the first half, but switches every A and U. If the baby continues indefinitely, her text converges to an infinite sequence that mathematicians call the Thue-Morse sequence (A010060). Of course, mathematicians use zeroes and ones instead of A and U, so the sequence looks like 0110100110010110100….

This sequence has many interesting properties. For example, if you replace every zero by 01 and every one by 10 in the Thue-Morse sequence, you will get the Thue-Morse sequence back. You can see that this is so if you code A in the baby’s speech by 01 and U by 10. Thus the Thue-Morse sequence is a fixed point under this substitution. Moreover, the only two fixed points under this substitution are the Thue-Morse sequence and its negation (A010059).

The Thue-Morse sequence possesses many other cool properties. For example, the sequence doesn’t contain substrings 000 and 111. Actually any sequence built from the doubles 01 and 10 can’t contain the triples 000 and 111 because we switch the digit after every odd-indexed term of such a sequence. A more general and less trivial statement is also true for the Thue-Morse sequence: it doesn’t contain any cubes. That is, it doesn’t contain XXX, where X is any binary string.

I stumbled upon this sequence when I was playing with evil and odious numbers invented by John H. Conway. A number is evil if the number of ones in its binary expansion is even, and odious if it’s odd. We can define a function, called the odiousness of a number, in the following way: odiousness(n) is one, if n is odious and 0 otherwise. We can apply the odiousness function to a sequence of non-negative integers term-wise. Now I can describe the Thue-Morse sequence as the odiousness of the sequence of non-negative numbers. Indeed, the odiousness of the number 2ⁿ + k is opposite of the odiousness of k, if k is less than 2ⁿ. That means if we already know the odiousness of the numbers below 2ⁿ, the next 2ⁿ terms of the odiousness sequence is the bitwise negation of the first 2ⁿ terms. So odiousness is built the same way as the Thue-Morse sequence, and you can easily check that the initial terms are the same too.

Let me consider as an example the sequence which is the odiousness of triangular numbers A153638: 0, 1, 0, 0, 0, 0, 1, 1, 0, 0, 1, 0…. What can we say about this sequence? We can say that the number of zeroes is infinite, as all the terms with indices of the form 2^n-1(2ⁿ+1) are zeroes. Also, the number of ones is infinite because all the terms with indices of the form 2^2n-1(2^2n-1-1) are ones.

Obviously, we can define the evilness of a number or of a sequence with non-negative terms. Namely, the evilness of a number is 1 if the number is evil, and 0 if it is not. The evilness is the bitwise negation of the odiousness. The evilness of the sequence of non-negative integers is the negation of the Thue-Morse sequence. The odiousness sequence of any sequence of zeroes and ones is the sequence itself, and the evilness sequence is its negation.

I would like to define an inverse odiousness operation on binary sequences. Many different sequences can have the same odiousness sequence. In such a case mathematicians usually define the inverse operation as a minimal non-negative sequence whose odiousness is the given sequence. Obviously, the minimal inverse of a binary sequence is the sequence itself, and thus not very interesting. I suggest that we define the inverse as a minimal increasing sequence. In this case the odiousness inverse of the Thue-Morse sequence is the sequence of non-negative numbers.

For example, let me describe the inverse odiousness of the sequence of all ones. Naturally, all the numbers in the sequence must be odious, and by minimality property this is the sequence of odious numbers A000069: 1, 2, 4, 7, 8, 11, 13, 14, 16, 19…. Analogously, the odiousness inverse of the sequence of all zeroes is the sequence of evil numbers A001969: 0, 3, 5, 6, 9, 10, 12, 15, 17, 18, 20….

Let us find the odiousness inverse of the alternating sequence A000035: 0, 1, 0, 1, 0, 1…. This is the lexicographically smallest sequence of numbers changing putridity. By the way, “putridity” is the term suggested by John Conway to encompass odiousness and evilness the same way as parity encompasses oddness and evenness.

The odiousness inverse of the alternating sequence is the sequence A003159: 0, 1, 3, 4, 5, 7, 9, 11, 12, 13…. By my definition we can describe this sequence as indices of terms of the Thue-Morse sequence that are different from the previous term. This sequence can be described in many other ways. For example, the official definition in the OEIS is that this sequence consists of numbers whose binary expansion ends with an even number of zeroes. It is fun to prove that this is the case. It is also fun to show that this sequence can be built by adding numbers to it that are not doubles of previous terms.

Let us look at the first differences of the previous sequence. This is the sequence A026465: 1, 2, 1, 1, 2, 2, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2… — the length of n-th run of identical symbols in the Thue-Morse sequence. As we know that the Thue-Morse sequence doesn’t contain three ones or three zeroes in a row, we can state that the terms of this sequence will continue to be ones or twos.

You can define putridity sequences for any non-negative sequence. Which of them are interesting? I do not know, but I know which of them are not very interesting. For example, the putridity of pronic (oblong) numbers sequence is the same as the putridity of the triangular numbers sequence. This is because pronic numbers are twice triangular numbers and putridity is independent of factors of two. Another uninformative putridity sequence is the odiousness of the powers of two. This sequence consists only of ones.

I bet that my readers can find putridity sequences that are interesting.

I ran an experiment. I copied multiple choices from the 2007 AMC 8 into a file and asked my son Sergei to try to guess the answers, looking only at the choices. I allowed him to keep several choices. The score I assigned depended on the number of leftover choices. If the leftover choices didn’t contain the right answer, the score for the problem was zero. Otherwise, it was scaled according to the number of choices he left. For example, if he had four choices left and the right answer was among them he got 1/4 of a point. Here are the choices:

1. 9, 10, 11, 12, 13.
2. 2/5, 1/2, 5/4, 5/3, 5/2.
3. 2, 5, 7, 10, 12.
4. 12, 15, 18, 30, 36.
5. 24, 25, 26, 27, 28.
6. 7, 17, 34, 41, 80.
7. 25, 26, 29, 33, 36.
8. 3, 4.5, 6, 9, 18.
9. 1, 2, 3, 4, cannot be determined.
10. 13, 20, 24, 28, 30.
11. Choose picture: I, II, III, IV, cannot be determined.
12. 1:1, 6:5, 3:2, 2:1, 3:1.
13. 503, 1006, 1504, 1507, 1510.
14. 5, 8, 13, 14, 18.
15. a+c < b, ab < c, a+b < c, ac < b, b/c = a.
16. Choose picture: 1, 2, 3, 4, 5.
17. 25, 35, 40, 45, 50.
18. 2, 5, 6, 8, 10.
19. 2, 64, 79, 96, 131.
20. 48, 50, 53, 54, 60.
21. 2/7, 3/8, 1/2, 4/7, 5/8.
22. 2, 4.5, 5, 6.2, 7.
23. 4, 6, 8, 10, 12.
24. 1/4, 1/3, 1/2, 2/3, 3/4.
25. 17/36, 35/72, 1/2, 37/72, 19/36.

It is clear that if you keep all choices, your score will be 5, which is the expected score for AMC if you are randomly guessing the answers. Sergei’s total score was 7.77, which is noticeably better than the expected 5.

There were two questions where Sergei felt that he knew the answer exactly. First was question number two with choices: 2/5, 1/2, 5/4, 5/3, 5/2. All but one of the choices has a 5 in it, so 1/2 must be wrong. Numbers 2/5 and 5/2 are inverses of each other, so if organizers expect you to make a mistake by inverting the right answer, then one of these choices must be the right answer. But 5/4 and 5/3 are better suited as a miscalculation of 5/2, than of 2/5. His choice was 5/2, and it was correct. The second question for which he was sure of the answer was question 19, with his answer 79. I still do not have a clue why.

Sergei’s result wasn’t too much better than just guessing. That means that AMC 8 organizers do a reasonably good job of hiding the real answer. You can try it at home and see if you can improve on Sergei’s result. I will publish the right answers as a comment to this essay in a week or so.

Browsing Braingle I stumbled upon a standard probability puzzle which is very often misunderstood:

Suppose I flip two coins without letting you see the outcome, and I tell you that at least one of the coins came up heads. What is the probability that the other coin is also heads?

The standard “wrong” answer is 1/2. Supposedly, the right answer is 1/3. Here is the explanation for that “right” answer:

For two coins there are four equally probable outcomes: HH, HT, TH and TT. Obviously, TT is excluded in this case, and of the remaining three possibilities only one has two heads.

Here is the problem with this problem. Suppose I flip two coins without letting you see the outcome. If I get one head and one tail, what will I tell you? I can tell you that at least one of the coins came up heads. Or, I can tell you that at least one of the coins came up tails. The fact that I can tell you different things changes the a posteriori probabilities.

You need to base your calculation not only on your knowledge that there are only three possibilities for the outcome: HH, HT and TH, but also on the conditional probabilities of these outcomes, given what I told you. I claim that the initial problem is undefined and the answer depends on what I decide to say in each different case.

Let us consider the first of two strategies I might use:

I flip two coins. If I get two heads, I tell you that I have at least one head. If I get two tails I tell you that I have at least one tail. If I get one head and one tail, then I will tell you one of the above with equal probability.

Given that I told you that I have at least one head, what is the probability that I have two heads? I leave it to my readers to calculate it.

Suppose I follow the other strategy:

I flip two coins. If I get two tails, I say, “Oops. It didn’t work.” Otherwise, I say that I have at least one head.

Given that I told you that I have at least one head, what is the probability that I have two heads? If you calculate answers for both strategies correctly, you will have two different answers. That means the problem is not well-defined in the first place.

When I came to the US, I heard about Mensa — the high IQ society. My IQ had never been tested, so I was curious. I was told that there was a special IQ test for non-English speakers and that my fresh immigrant status and lack of English knowledge was not a problem. I signed up.

There were two tests. One test had many rows of small pictures, and I had to choose the odd one out in each row. That was awful. The test was English-free, but it wasn’t culture-free. I couldn’t identify some of the pictures at all. We didn’t have such things in Russia. I remember staring at a row of tools that could as easily have been from a kitchen utensil drawer as from a garage tool box. I didn’t have a clue what they were.

But the biggest problem was that the idea of crossing the odd object out seems very strange to me in general. What is the odd object out in this list?

Cow, hen, pig, sheep.

The standard answer is supposed to be hen, as it is the only bird. But that is not the only possible correct answer. For example, pig is the only one whose meat is not kosher. And, look, sheep has five letters while the rest have three.

Thus creative people get fewer points. That means, IQ tests actually measure how standard and narrow your mind is.

The second test asked me to continue patterns. Each page had a three-by-three square of geometric objects. The bottom right corner square, however, was empty. I had to decide how to continue the pattern already established by the other eight squares by choosing from a set of objects they provided.

This test is similar to continuing a sequence. How would you continue the sequence 1,2,3,4,5,6,7,8,9? The online database of integer sequences has 1479 different sequences containing this pattern. The next number might be:

• 10, if this is the sequence of natural numbers;
• 1, if this is the sequence of the digital sums of natural numbers;
• 11, if this the sequence of palindromes;
• 0, if this is the sequence of digital products of natural numbers;
• 13, if this is the sequence of numbers such that 2 to their powers doesn’t contain 0;
• 153, if this is the sequence of numbers that are sums of fixed powers of their digits;
• 22, if this is the sequence of numbers for which the sum of digits equals the product of digits; or
• any number you want.

Usually when you are asked to continue a pattern the assumption is that you are supposed to choose the simplest way. But sometimes it is difficult to decide what the testers think the simplest way is. Can you replace the question mark with a number in the following sequence: 31, ?, 31, 30, 31, 30, 31, … You might say that the answer is 30 as the numbers alternate; or, you might say that the answer is 28 as these are the days of the month.

Towards the end of my IQ test, the patterns were becoming more and more complicated. I could have supplied several ways to continue the pattern, but my problem was that I wasn’t sure which one was considered the simplest.

When I received my results, I barely made it to Mensa. I am glad that I am a member of the society of people who value their brains. But it bugs me that I might not have been creative enough to fail their test.

Browsing the Internet, I stumbled upon a coin puzzle which I slightly shrank to emphasize my point:

Carl flipped two coins and was asked if at least one of the two coins landed “heads up”. He replied, “Yes. In fact the first coin I flipped landed heads up.” What is the chance that Carl’s coins both landed heads up?

The standard answer is 1/2, because there are only two possibilities for the coin flips: HH and HT. But how do we know that these possibilities are equally probable?

The answer depends on what we expect Carl to say when he flips two heads. My personal assumption is that Carl is a perfectionist and always volunteers extra information. If Carl gets two heads, I would expect him to say, “Yes. In fact both coins I flipped landed heads up.” In this case the answer to the puzzle is 0.

Another strange but reasonable assumption is that upon flipping two heads, there is an equal probability that Carl would say either, “Yes. In fact the first coin I flipped landed heads up;” or, “Yes. In fact the second coin I flipped landed heads up.” In this case, the answer to the puzzle is 1/3.

I could describe an assumption for Carl’s answering strategy that leads to the puzzle’s answer of 1/2, but it looks too artificial to me.

This puzzle is not well-defined, but unfortunately there are many versions of it floating around the Internet with incorrect solutions.

Here is a standard logic puzzle:

A criminal is sentenced to death. He is allowed to make one last statement. If the statement is true, the criminal will be sent to the electric chair. It the statement is false, he will be hanged. Can you suggest a good piece of advice for this man?

I can offer many pieces of advice to this man. The simplest thing is to keep silent. Or he can communicate without making statements, like asking, “Can I have some crème brûlée, please?”

One can argue that the puzzle implies that it’s a favor to allow the prisoner to make a last statement, but without it he will die anyway. In this case the standard piece of advice to this man would be to create a paradoxical situation by saying, “I will be hanged.”

Another, less standard, idea is to state something that is very difficult to check. For example, to give the exact number of planets in our galaxy, or posit that P = NP. My son, Sergei, suggested saying that “Schrödinger’s cat is dead.”

But the most popular idea among my AMSA students is to say, “I am sorry.” I’m not 100% sure that they mean it as a statement that is impossible to check. Maybe they think that these words can do magic and save lives. Or maybe it could be the best thing for a criminal to say before dying.