Tanya Khovanova's Math Blog

We’ve all been hearing about parallel computing, and now it has turned up in a coin-weighing puzzle invented by Konstantin Knop.

“We have N indistinguishable coins. One of them is fake and it is not known whether it is heavier or lighter, but all genuine coins weigh the same. There are two balance scales that can be used in parallel. Each weighing lasts a minute. What is the largest number of coins N for which it is possible to find the fake coin in five minutes?”

This puzzle reminds me of another coin-weighing problem, where in a similar situation you need to find a fake coin by using one scale with four pans. The answer in this variation would be 5⁵ = 3125. We can divide coins in five groups with the same number of coins and put four groups on the scale. If one of the groups is different (heavier or lighter), then this group contains the fake coin. Otherwise, the leftover group contains the fake coin. This way each weighing reduces the pile with the fake coin by a factor of five. One scale with four pans gives you more information than two scales with two pans used in parallel. We can conclude that Knop’s puzzle should require at least the same number of weighings as the four-pan puzzle for the same number of coins. So we can expect the answer to Knop’s puzzle will not be bigger than 3125. But what will it be?

Once when I was working at Telcordia, I received a phone call from my doctor’s office. Here is how it went:

— Are you Tanya Khovanova?
— Yes.
— You should come here immediately and redo your blood test ASAP.
— What’s going on?
— Your blood count shows that you are dead.
— If I’m dead, then what’s the hurry?
Given that I wasn’t dead, the conclusion was that there had been a mistake in the test. If there had been a mistake, the probability that something was wrong after the test was the same as it was before the test. There was no hurry.

I started my Yellow Road plan on February 9 when I was 245.2 pounds.

I decided that my first target weight would be my actual weight on February 9: 245.2. Every day this target weight goes down by 0.1 pounds. I weigh myself every morning and compare my actual weight to my target weight. My actions depend on the difference.

My Yellow Zone is plus or minus one pound of my target weight. My Green Zone means I am doing even better: my weight is less than my target weight minus one pound. My Red Zone means that I am not doing so great: my weight is more than my target weight plus a pound.

If I am within the Yellow corridor, I continue building my healthy habits as I have been doing. If I am in the Green Zone, I can afford to digress from healthy habits and indulge myself a bit. If I am in the Red Zone, I have to reduce my evening meals to apples only, which I do not particularly like. The Red Zone has different shades: if I am one pound over my target weight I have to start my apple restrictions after 8:00 pm. If I am two pounds over, then after 6:00 pm, and so on.

Today I am ten pounds lighter. In the process, I have made these discoveries:

Writing down my weight daily is very important. When I look at yesterday’s number and today’s number, I start thinking about what caused the increase or decrease. Now I have more clarity about which foods are better for me.

I have a better picture of how much I should eat. One day I held a party and I didn’t eat much. In fact, I had only one small desert. I didn’t feel full and went to bed feeling proud of myself. The next morning I weighed myself and was surprised to find I had gained three pounds. The amount of food I should be eating is much smaller than I expected. I think it may be three times less than what I was used to eating. My plan might not be aggressive enough. Currently when I am in my lightest Red Zone, I have to eat just apples after 8:00 pm. I discovered that I can still gain weight with this regime.

A half-empty stomach is not such a bad feeling. I was so afraid of starting a plan where I might feel hungry. Now I discovered that there are several hours between my first signal that I should eat and real hunger. My first sense that I should eat something might not be actual hunger at all. I do experience a light feeling in my stomach, but now I am starting to learn to enjoy it.

The system works for me. That’s the bottom line. For the first time in my life, I found a way to lose weight. All my friends ask me about this system. I explain that this Yellow Road plan is not a panacea. My plan is based on many other things that I did before. If it continues working, I promise to discuss it further: to analyze what exactly works and why.

My next step is to adjust my plan in light of my discoveries. From now on, I’ll eat only apples after 8:00 pm—not as an exception, but as a rule.

Here is a problem from the 2012 Moscow Olympiad:

There were n people at a meeting. It appears that any two people at the meeting shared exactly two common acquaintances.

• Prove that all the people have exactly the same number of common acquaintances at this meeting.
• Show that n can be greater than 4.

My question is: Why 4? I can answer that myself. If in a group of four people any two people share exactly two common acquaintances, then all four people know each other. So in this Olympiad problem, the author wanted students to invent a more intricate example.

Let’s take this up a notch and work on a more difficult problem.

There were n people at a meeting. It appears that any k people at the meeting shared exactly k common acquaintances.

• Prove that all the people have exactly the same number of common acquaintances at this meeting.
• Is it possible that n can be greater than 2k?

I stumbled upon an article, Winners Live Longer, that says:

“When 524 nominees for the Nobel Prize were examined and compared to the actual winners from 1901 to 1950, the winners lived longer by 1.4 years. Why? It seems just having won and knowing you are on top gives you a boost of 1.8% to your life expectancy.”

This goes on top of the pile of Bad Conclusions From Statistics. With any kind of awards where people can be nominated several times, winners on average would live longer. The reason is that nominees who die early lose their chance to be nominated again and to win.

I wonder what would happen if we were to compare Fields medal nominees and winners. There is a cut off age of 40 for receiving a Fields medal. If we compare the life span of Fields medal winners and nominees who survived past 40, we might get a better picture of how winning affects life expectancy.

Living a long life increases your chances of getting a Nobel Prize, but doesn’t help you get a Fields medal.

I wish I could write four papers in three weeks. The title just means that I submitted four papers to the arXiv in the last three weeks—somehow, after the stress of doing my taxes ended, four of my papers converged to their final state very fast. Here are the papers with their abstracts:

• On k-visibility graphs (with Matthew Babbitt and Jesse Geneson). We examine several types of visibility graphs in which sightlines can pass through k objects. For k ≥ 1 we improve the upper bound on the maximum thickness of bar k-visibility graphs from 2k(9k−1) to 6k, and prove that the maximum thickness must be at least k+1. We also show that the maximum thickness of semi-bar k-visibility graphs is between the ceiling of 2(k+1)/3 and 2k. Moreover we bound the maximum thickness of rectangle k-visibility graphs. We also bound the maximum number of edges and the chromatic number of arc and circle k-visibility graphs. Furthermore we give a method for finding the number of edges in semi-bar k-visibility graphs based on skyscraper puzzles.
• Skyscraper Numbers (with Joel Brewster Lewis). We introduce numbers depending on three parameters which we call skyscraper numbers. We discuss properties of these numbers and their relationship with Stirling numbers of the first kind, and we also introduce a skyscraper sequence.
• Connected Components of Underlying Graphs of Halving Lines (with Dai Yang). In this paper we discuss the connected components of underlying graphs of halving lines’ configurations. We show how to create a configuration whose underlying graph is the union of two given underlying graphs. We also prove that every connected component of the underlying graph is itself an underlying graph.
• Efficient Calculation of Determinants of Symbolic Matrices with Many Variables (with Ziv Scully). Efficient matrix determinant calculations have been studied since the 19th century. Computers expand the range of determinants that are practically calculable to include matrices with symbolic entries. However, the fastest determinant algorithms for numerical matrices are often not the fastest for symbolic matrices with many variables. We compare the performance of two algorithms, fraction-free Gaussian elimination and minor expansion, on symbolic matrices with many variables. We show that, under a simplified theoretical model, minor expansion is faster in most situations. We then propose optimizations for minor expansion and demonstrate their effectiveness with empirical data.

This is the promised solution to the puzzle Integers and Sequences that I posted earlier. The puzzle is attached below.

Today I do not want to discuss the underlying math; I just want to discuss the puzzle structure. I’ll assume that you solved all the individual clues and got the following lists of numbers.

• 12 42 18 40 30 24 20
• 2 1 132 42 429 14
• 7 9 1 8 5 3 10 4
• 92 117 70 145 35 1 22 12 5
• 137 1 37 13 107 1013 113
• 30 12 2 42 6
• 70 4030 836 7192

Since the title mentions sequences, it is a good idea to plug the numbers into the Online Encyclopedia of Integer Sequences. Here is what you will get:

• not clear
• Catalan numbers with 5 missing: 1, 1, 2, 5, 14, 42, 132, 429
• not clear
• Pentagonal numbers with 51 missing: 1, 5, 12, 22, 35, 51, 70, 92, 117, 145
• Primeval numbers with 2 missing: 1, 2, 13, 37, 107, 113, 137, 1013
• not clear
• Weird numbers with 5830 missing: 70, 836, 4030, 5830, 7192

Your first “aha moment” happens when you notice that the sequences are in alphabetical order and each has exactly one number missing. The alphabetical order is a good sign that you are on the right track; it can also narrow down the possible names of the sequences that you haven’t yet identified. Alphabetical order means that you have to figure out the correct order for producing the answer.

Did you notice that some groups above are as long as nine integers and some are as short as four? In puzzles, there is nothing random, so the lengths of the groups should mean something. Your second “aha moment” will come when you realize that, together with the missing number, the number of the integers in each group is the same as the number of letters in the name of the sequence. This means you can get a letter by indexing the index of the missing number into the name of the sequence.

So each group of numbers provides a letter. Now we need to identify the remaining sequences and figure out in which order the groups will produce the word that is the answer.

Let’s go back and try to identify the remaining sequences. We already know the number of letters in the name of each sequence, as well as the range within the alphabet. The third sequence might represent a challenge as its numbers are small and there might be many sequences that fit the pattern, but let’s try. The results are below with the capitalized letter being the one that is needed for the answer.

• abundAnt
• caTalan
• dEficient or iMperfect
• pentaGonal
• pRimeval
• proNic or proMic
• weiRd

What is going on? There are two sequences that fit the pattern of the third group and the sequence for the sixth group has many names, two of which fit the profile but produce different letters. Now we get to your third “aha moment”: you have already seen some of the sequence names before, because they are in the puzzle. This will allow you to disambiguate the names.

Now that we have all the letters, we need the order. Sequences are mentioned inside the puzzle. You were forced to notice that because you needed the names for disambiguation. Maybe there is something else there. On closer examination, all but one of the sequence names are mentioned. Moreover, with one exception the clues for one sequence mention exactly one other sequence. Once you connect the dots, you’ll have your last “aha moment:” the way the sequences are mentioned can provide the order. The first letter G will be from the pentagonal sequence, which was not mentioned. The clues for the pentagonal sequence mention the primeval sequence, which will give the second letter R, and so on.

The answer is GRANTER.

Many old-timers criticized the 2013 MIT Mystery Hunt. They are convinced that a puzzle shouldn’t have more than one “aha moment.” I like my “aha moments.”

*****

• (the largest integer n such that there exists a Platonic solid with n vertices, a Platonic solid with n edges, and a Platonic solid with n faces)
• (the largest two-digit tetrahedral number)/(the smallest value the second smallest angle of a convex hexagon with all integer degrees can have)
• (the number of positive integers less than 2013 that are divisible by 100, but not divisible by 70)
• (the number of two-digit numbers that produce a square when summed up with their reverse) ⋅ (the smallest number of weighings on a balance scale that guarantees to find the only fake coin out of 100 identical coins, where the fake coin is lighter than other coins)
• (the only two digit number n such that 2ⁿ ends with n) − (the second smallest, and conjectured to be the largest, triangular number such that its square is also triangular)
• (the smallest non-trivial compositorial number that is also a factorial)
• (the sum of the smallest three positive pronic numbers)

*****

• (the digit you get when you sum up the digits of 2013²⁰¹³ repeatedly until you get a single digit) − (the greatest common factor of the indices of the Fibonacci numbers divisible by 13)
• (the largest common divisor of numbers of the form p² − 1 for primes p greater than three) − (the largest sum of digits that can appear on a 12-hour digital clock starting from 1:00 up to 12:59)
• (the largest Fibonacci number, such that it and all positive Fibonacci numbers less than it are deficient) + (the difference between the sum of all even numbers up to 100 and the sum of all odd numbers up to 100) − (the first digit of a four-digit square that has the first two digits the same and the last two digits the same)
• (the smallest composite Jacobsthal number) ⋅ (the only digit needed to express the number of diagonals of a convex hendecagon)/(the smallest prime divisor of 13²⁰¹³ + 1)
• (the smallest integer the fate of whose aliquot sequence is unknown) + (the largest amount of money in cents you can have in American coins without having change for 2 dollars) − (the repeated number in the aliquot cycle of 95) ⋅ (the second-smallest integer n such that the Russian word for n has n letters)
• (the smallest positive even integer that’s not a totient)

*****

• (the number of letters in the last name of a famous Russian writer whose year of birth many Russians use to help them memorize the digits of e)
• (the number of pluses you need to insert in a row of 20 fives so that the sum is 1000)
• (the number of positive integers less than 2013 such that not all their digits are distinct) − (the number of four-digit numbers with only odd digits) − (the largest Fibonacci square)
• (the number of positive integers n for which the sum of the n smallest positive integers evenly divides 18n)
• (the number of trailing zeroes of 2013!) − (the number of sets in the game of Set such that every feature is different on all three cards) − (an average speed in miles per hour of a person who drives somewhere with a speed of 420 miles per hour, then drives back using the same route with a speed of 210 miles per hour)
• (the smallest fortunate triangular number)
• (the smallest weird number)/(the only prime one less than a cube)
• (the third most probable product of the numbers showing when two standard six-sided dice are rolled)

*****

• (the largest integer number of dollars you can’t pay if you have an unlimited supply of 9-dollar bills and 13-dollar bills) − (the positive difference between the two prime numbers that do not share a unit digit with any other prime number)
• (the largest three-digit primeval number) − (the largest number of distinct SET cards without a set)
• (the number conjectured to be the second-largest number such that two to its power has no zeroes) − (the largest number whose cube has at most two distinct digits and no zeroes)
• (the number of 5-digit palindromic integers in base 5) + (the only positive integer that is five times the sum of its digits)
• (the only Fibonacci number that is a double of a prime) + (the only prime p such that p! has p digits) − (the only fixed point of look-and-say operation)
• (the only number whose concatenation with itself is prime)
• (the only positive integer that that differs by 1 from a square and a nonsquare cube) − (the largest number such that its divisors are each 1 less than a prime)
• (the smallest admirable number)
• (the smallest evil untouchable number)

*****

• (the alphanumeric value of MANIC SAGES) + (the sum of all three-digit numbers you can get by permuting digits 1, 2, and 3) + (the number of two-digit integers divisible by 9) − (the number of rectangles whose sides are composed of edges of squares of a chess board)
• (the integer whose standard Roman numeral representation is alphabetically later than all others) − (the number you get if you divide a three digit number with identical digits by the sum of the digits)
• (the largest even integer that is not a sum of two abundant numbers) − (the digit in the first position where e and π have the same digit)
• (the number formed by the last two digits of the sum: 1! + 2! + 3! + 4! + . . . + 2013!)
• (the only positive integer such that if you sum the digits and the squares of the digits, you get the original number back) + (the largest prime factor of the smallest Carmichael number)
• (the smallest multi-digit hyperperfect number such that more than half of its digits are the same) − (the sum of digits that cannot be the last digits of squares) ⋅ (the largest base n in which 8n is not written like 80) ⋅ (the smallest positive integer that leaves a remainder of 2 when divided by 3, 4, and 5)
• (the smallest three-digit brilliant number) − (the first decimal digit of the number that in hexadecimal gives the house number of Sherlock Holmes)

*****

• (the number of evil minutes in an hour)
• (the number of fingers on ten hands) − (the smallest number such that its square has a digit repeated three times)
• (the number of ways you can rearrange letters of MANIC)/(the number of ways you can rearrange letters of SAGES)
• (the only multi-digit Catalan number with digits in strictly decreasing order)
• (the smallest perfect number)

*****

• (the largest product of positive integers that sum up to 10) + (the smallest perimeter of a rectangle with integral sides of area 120) − (the day of the month of the second Thursday in a January that has exactly 4 Mondays and 4 Fridays)
• (the second-largest number with all distinct digits, such that all the words in its American English representation start with the same letter) + (the largest square-free composite number that contains each of the digits 1, 2, 3, 4 exactly once in its prime factorization) + (the number of ways you can flip a coin 10 times so that the number of heads is the same as the number of tails) + (the smallest positive integer such that 2 to its power contains 2013 as a substring) + (the sum of five prime numbers formed from the digits 2, 3, 5, 7, 8, 9 where each digit is used exactly once) + (the number of days in a year where the day of the month is odious) + (the sum of the digits each of which spelled out has an alphanumeric value equal to the meaning of life, the universe, and everything) ⋅ (the sum of all prime numbers p such that p + 20 and p + 40 are also prime) + (the first digit of the total number of legal moves of the Black king in chess)
• (the second-largest three-letter palindrome in Roman numerals)/((the smallest composite number not divisible by any of its digits)/(the last digit of 2013²⁰¹³) − (the digit in position 2013 of the string formed by concatenation of all integers into one stream: 123456789101112…)) − (the number of days in a year such that the month and the day of the month are simultaneously composite)
• (the second-smallest cube with only prime digits) ⋅ (the smallest perimeter of a Pythagorean triangle)/(the last digit to appear in the units place of a Fibonacci number) + (the greatest common divisor of the sums in degrees of the interior angles of convex polygons with an even number of sides) + (the number of subsets that you can form from the set {1,2,3,4,5,6,7,8,9} that do not contain two consecutive numbers) − (the only common digit of 2013 base 8 and base 9)

Should I eat this piece of cake or not? I will certainly enjoy it very much. What harm will it do? Will this piece increase my weight? Maybe not. The next piece might, but this particular one looks harmless. Even if my weight increases by half a pound, it could be muscle weight. Yes, it probably would be due to muscle weight: I just went out of my house to throw away my garbage and this has to count as exercise.

Do you see the problem? Eating the cake provides an immediate reward, but the punishment is vague and in the far distant future. That is why I got excited when my son Alexey sent me the link to Beeminder, a company that creates an artificial non-vague and not far-in-the-future punishment for eating that piece of cake.

Here is how it works. You give them your target number — in my case my desired weight, but it could be any measurable goal — and the date by which you want to hit it. They draw a yellow path on a weight chart. You must weigh yourself every day. Whenever your weight is above your path, you have to pay real money to the company. Five dollars!

This is a great idea. Suddenly that piece of cake looks threatening. The only problem with using their system is that I have no clue how to lose weight. The company doesn’t provide tools to lose weight: it just provides a commitment device. So it is difficult to stick with the weight-loss commitment without having a proven weight-loss plan.

The truth is that my son sent me the link, I laughed, and forgot about it. Besides, if I ever want to pay money for failing in my commitments, I would rather choose the beneficiary myself. Then I realized that I can use the yellow-road idea to try to lose weight while figuring out what works for me. I call my new plan the Adaptive Diet.

Starting from my actual weight on Day One, I drew a line that represents my target weight, assuming a daily decrease of 0.1 pounds. A deviation of one pound from my target weight on my daily weigh-in is what I call my Yellow Zone. When I am in the Yellow, I continue doing what I was doing before: trying to build new, healthier habits.

If I am more than one pound below my target weight, then I have entered what I call the Green Zone. When I am in the Green, I can allow myself to indulge my cravings. However, when I am one pound above my target weight, I call that the dreaded Red Zone. This Zone has different shades of red. If I am between 1 and 2 pounds above my target weight, I have to eat only apples after 8:00pm. If I am 2 to 3 pounds above my target weight, only-apples time starts at 6:00pm. And so on. Every extra pound above my target weight moves the cut-off time by two hours. That means that if I am 7 pounds above my target weight, I would have to eat apples all day long.

The system has to work: I do not like apples.

Tanya Khovanova and Joel Brewster Lewis

In skyscraper puzzles you have to put an integer from 1 to n in each cell of a square grid. Integers represent heights of buildings. Every row and column needs to be filled with buildings of different heights and the numbers outside the grid indicate how many buildings you can see from this direction. For example, in the sequence 213645 you can see 3 buildings from the left (2,3,6) and 2 from the right (5,6).

In mathematical terminology we are asked to build a Latin square such that each row is a permutation of length n with a given number of left-to-right and right-to-left-maxima. The following 7 by 7 puzzle is from the Eighth World Puzzle Championship.

Latin squares are notoriously complicated and difficult to understand, so instead of asking about the entire puzzle we discuss the mathematics of a single row. What can you say about a row if you ignore all other info? First of all, let us tell you that the numbers outside have to be between 1 and n. The sum of the left and the right numbers needs to be between 3 and n+1. We leave the proof as an exercise.

Let’s continue with the simplest case. Suppose the two numbers are n and 1. In this case, the row is completely defined. There is only one possibility: the buildings should be arranged in the increasing order from the side where we see all of them.

Now we come to the question we are interested in. Given the two outside numbers, how many permutations of the buildings are possible? Suppose the grid size is n and the outside numbers are a and b. Let’s denote the total number of permutations by f_n(a, b). We will assume that a is on the left and b is on the right.

In a previous example, we showed that f_n(n, 1) = 1. And of course we have f_n(a, b) = f_n(b, a).

Let’s discuss a couple of other examples.

First we want to discuss the case when the sum of the border numbers is the smallest — 3. In this case, f_n(1, 2) is (n−2)!. Indeed, we need to put the tallest building on the left and the second tallest on the right. After that we can permute the leftover buildings anyway we want.

Secondly we want to discuss the case when the sum of the border numbers is the largest — n+1. In this case f_n(a,n+1-a) is (n-1) choose (a-1). Indeed, the position of the tallest building is uniquely defined — it has to take the a-th spot from the left. After that we can pick a set of a-1 buildings that go to the left from the tallest building and the position is uniquely defined by this set.

Before going further let us see what happens if only one of the maxima is given. Let us denote by g_n(a) the number of permutations of n buildings so that you can see a buildings from the left. If we put the shortest building on the left then the leftover buildings need to be arrange so that you can see a-1 of them. If the shortest building is not on the left, then it can be in any of the n-1 places and we still need to rearrange the leftover buildings so that we can see a of them. We just proved that the function g_n(a) satisfies the recurrence:

Actually g_n(a) is a well-known function. The numbers g_n(a) are called unsigned Stirling numbers of the first kind (see https://oeis.org/A132393); not only do they count permutations with a given number of left-to-right (or right-to-left) maxima, but they also count permutations with a given number of cycles, and they appear as the coefficients in the product (x + 1)(x + 2)(x + 3)…(x + n), among other places. (Another pair of exercises.)

We are now equipped to calculate f_n(1, b). The tallest building must be on the left, and the rest could be arranged so that, in addition to the tallest building, b-1 more buildings are seen from the right. That is f_n(1, b) = g_n-1(b-1).

Here is the table of non-zero values of f_n(1, b).

Now we have everything we need to consider the general case. In any permutation of length n, the left-to-right maxima consist of n and all left-to-right maxima that lie to its left; similarly, the right-to-left maxima consist of n and all the right-to-left maxima to its right. We can take any permutation counted by f_n(a, b) and split it into two parts: if the value n is in position k + 1 for some 0 ≤ k ≤ n-1, the first k values form a permutation with a – 1 left-to-right maxima and the last n – k – 1 values form a permutation with b – 1 right-to-left maxima, and there are no other restrictions. Thus:

Let’s have a table for f₇(a,b), of which we already calculated the first row:

We see that the first two rows of the puzzle above correspond to the worst case. If we ignore all other constrains there are 675 ways to fill in each of the first two rows. By the way, the sequence of the number of ways to fill in the most difficult row for n from 1 to 10 is: 1, 1, 2, 6, 22, 105, 675, 4872, 40614, 403704. The maximizing pairs (a,b) are (1, 1), (1, 2), (2, 2), (2, 2), (2, 2), (2, 3), (2, 3), (2, 3), (3, 3).

The actual skyscraper puzzles are designed so that they have a unique solution. It is the interplay between rows and columns that allows to reduce the number of overall solutions to one.

I just compared two searches on Google Trends:

• How to become a vampire is in blue.
• How to become a mathematician is in red.