Archive for the ‘Math Competitions’ Category.

Math Kangaroo’s Logic Puzzle

My AMSA students loved the following puzzle from the 2003 Math Kangaroo contest for grades 7-8:

The children A, B, C and D made the following assertions.

  • A: B, C and D are girls.
  • B: A, C and D are boys.
  • C: A and B are lying.
  • D: A, B and C are telling the truth.

How many of the children were telling the truth?
A) 0   B) 1   C) 2   D) 3   E) Impossible to determine

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Discussing a Problem from the Moscow Olympiad

I recently posted the following problem from the 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 acquaintances at this meeting.
  • Show that n can be greater than 4.

Here is the proof for the first bullet. Choose a person X. Take a pair of X‘s acquaintances. These two acquaintances have to share two acquaintances between themselves one of whom is X. In other words, we have described a function from all pairs of X‘s acquaintances to people who are not X. On the other hand, for every person who is not X, s/he and X share a pair of acquaintances. Hence, there is a bijection between people other than X and all pairs of X‘s acquaintances. If the number of X‘s acquaintances is a, and the total number of people is t, then we have shown that (a choose 2) = t−1. As this is true for any X, we see that everyone has the same number of acquaintances. Moreover, this situation can happen only if t−1 is a triangular number.

But wait. There is more work that needs to be done. The smallest triangular number is 1. That means that t might be 2. If there are two people at the meeting, then the condition holds: they have 0 common acquaintances. The next triangular number is 3. So we need to see what would happen if there are four people. In this case, if everyone knows each other, it works. This is why the second bullet asks us to find an example of the situation with more than four people, because four people is too easy.

Let’s look at larger triangular numbers. The situation described in the problem might also happen when there are:

  • 7 people total and everyone has 4 acquaintances,
  • 11 people total and everyone has 5 acquaintances,
  • 16 people total and everyone has 6 acquaintances,
  • and so on: a(a−1)/2 + 1 people total and everyone has a acquaintances.

The official Olympiad solution suggests the following example for 16 people total. Suppose we put 16 people in a square formation so that everyone knows people in the same row and column. I leave it to the reader to check that every two people share exactly two acquaintances.

Let me prove that there is no solution for a total of seven people. If there were a solution, then each person would have to know four people. My first claim is that the acquaintance graph can’t contain a four-clique. Suppose there is a four-clique. Then each person in the clique has to have another acquaintance outside of the clique to make it up to four. In addition, this extra acquaintance can’t be shared with anyone in the clique, because the clique contains all the acquaintances that they share. This means we need to have at least four more people.

Next, suppose two people a and b know each other and share an acquaintance c. Any two people in this group of three has to have another shared acquaintance, who is not shared with the third person. That is, there should be another person who is the acquaintance of a and b, a different person who is an acquaintance of a and c, and a third person who is acquainted with b and c. These three extra people are all the acquaintances of a, b, and c. Which means the last person who is not acquainted with a, b, or c, has less than for four acquaintances.

Let’s look at a more difficult problem that I offered at the same posting:

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 acquaintances at this meeting.
  • Is it possible that n can be greater than 2k?

As in the previous solution, we see that a, the number of acquaintances of a person and t, the total number of people, satisfy the following equation: (a choose k) = (t−1 choose k−1).

For example, if k = 3, the equation becomes (a choose 3) = (t−1 choose 2). This is a question of finding numbers that are both tetrahedral and triangular. They are known and their sequence, A027568, is finite: 0, 1, 10, 120, 1540, 7140. The corresponding number of acquaintances is 3, 5, 10, 22, 36 and the total number of people is 3, 6, 17, 57, 121. The first trivial example involves 3 people who do not know each other. The next example is also simple: it has 6 people and everyone knows everyone else.

What about non-trivial examples? If there are 17 people in the group, then each person has to know 10 people. Does the acquaintance graph exist so that every group of three people share 3 acquaintances?

We see that the problem consists of two different parts. First, we have to solve the equation that equates two binomial coefficients. And second, we need to build the acquaintance graph. Both questions are difficult. We see that for k = 2 we have an infinite number of solutions to the equation with binomial coefficients. For k = 3, that number is finite. What happens with other k? If there are 2k people and they all know each other, then this works. But are there other non-trivial solutions? I am grateful to Henry Cohn for directing me to the works of Singmaster who studied non-trivial repetitions of numbers in Pascal’s triangle. In particular, Singmaster showed that the equation (n+1 choose k+1) = (n choose k+2) has infinitely many solutions given by n = F2i+2F2i+3−1 and k = F2iF2i+2−1.

This sequence generates the following non-trivial examples (15 choose 5) = (14 choose 6), (104 choose 39) = (103 choose 40), and so on. That means it might be possible that there is a group of 16 people so that every 6 people share 6 acquaintances. In this situation every person must know everyone else except for one other person. That leads us to the structure of the acquaintance graph: it is a complement to the perfect matching graph. I leave it to my readers to check that the corresponding acquaintance graph doesn’t exist. Are there examples of two binomial coefficients that equal each other and that lead to an acquaintance graph that can be built?

Now that I’ve tackled the solution to this Olympiad problem, I see that I generated more questions than I answered.

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Next Tanya Khovanova

Many years ago at Gelfand’s seminar in Moscow, USSR, someone pointed out a young girl and told me: “This is Natalia Grinberg. In her year in the math Olympiads, she was the best in the country. She is the next you.”

We were never introduced to each other and our paths never crossed until very recently.

Several years ago I became interested in the fate of the girls of the IMO (International Math Olympiad). So, I remembered Natalia and started looking for her. If she was the best in the USSR in her year, she would have been a gold medalist at the IMO. But I couldn’t find her in the records! The only Grinberg I found was Darij Grinberg from Germany who went to the IMO three times (2004, 2005, and 2006) and won two silver medals and one gold.

That was clearly not Natalia. I started doubting my memory and forgot about the whole story. Later I met Darij at MIT and someone told me that he was Natalia’s son.

I was really excited when I received an email from Natalia commenting on one of my blog posts. We immediately connected, and I asked her about past events.

Natalia participated in the All-Soviet Math Olympiads three times. In 1979 as an 8th grader she won a silver medal, and in 1980 and 1981 she won gold. That indeed was by far the best result in her year. So she was invited to join the IMO team.

That year the IMO was being held in the USA, which made Soviet authorities very nervous. At the very last moment four members of the team did not get permission to travel abroad. Natalia was one of them. The picture below, which Natalia sent to me, was taken during the Soviet training camp before the Olympiad. These four students were not allowed to travel to the IMO: Natalia Grinberg, Taras Malanyuk, Misha Epiktetov, and Lenya Lapshin.

1981 IMO training camp

Because of the authorities’ paranoia, the Soviet team wasn’t full-sized. The team originally contained eight people, but as they rejected four, only six traveled to the USA, including two alternates.

I have written before how at that time the only way for a Jewish student to get to study mathematics at Moscow State University was to get to the IMO. I wrote a story about my friend Sasha Reznikov who trained himself to get to the IMO, but because of some official machinations, still was not accepted at MSU.

Natalia’s story surprised me in another way. She didn’t get to the IMO, but she was accepted at MSU. It appears that she was accepted at MSU as a member of the IMO team, because that decision was made before her travel documents were rejected.

Natalia became a rare exception to the rule that the only way for a Jewish person to attend MSU was to participate in the IMO. It was a crack in the system. They had to block visas at the last moment, so that people wouldn’t have time to make a fuss and do something about it. Natalia slipped through the crack and got to study at the best university in the Soviet Union.

Unfortunately, the world lost another gold IMO girl. Three Soviet team members won gold medals that year. Natalia, being better then all of them, would have also won the gold medal.

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My New Favorite Hat Puzzle

My new favorite hat puzzle was invented by Konstantin Knop and Alexander Shapovalov. It appeared (in a different wording) in March 2013 at the Tournament of the Towns:

A sultan decides to give 100 of his sages a test. The sages will stand in line, one behind the other, so that the last person in the line sees everyone else. The sultan has 101 hats, each of a different color, and the sages know all the colors. The sultan puts all but one of the hats on the sages. The sages can only see the colors of the hats on people in front of them. Then, in any order they want, each sage guesses the color of the hat on his own head. Each hears all previously made guesses, but other than that, the sages cannot speak. They are not allowed to repeat a color that was already announced. Each person who guesses his color wrong will get his head chopped off. The ones who guess correctly go free. The rules of the test are given to them one day before the test, at which point they have a chance to agree on a strategy that will minimize the number of people who die during this test. What should that strategy be?

I loved it so much that I wrote a paper about it. You can find the solution there.

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A Problem from the Moscow Olympiad

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?
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Four Papers in Three Weeks

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.
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Integers and Sequences Solution

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)
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Three out of Three

Davidson Institute for Talent Development announced their 2012 Winners. Out of 22 students, three were recognized for their math research. All three of them are ours: that is, they participated in our PRIMES and RSI programs:

  • David Ding’s project, “Infinitesimal Cherednik Algebras of gln,” came out of his participation in the PRIMES program.
  • Sitan Chen’s project, “On the Rank Number of Grid Graphs,” came out of his participation in the RSI program.
  • Xiaoyu He’ project, “On the Classification of Universal Rotor-Routers,” came out of his participation in the PRIMES program.

I already wrote about Xiaoyu’s project. Today I want to write about Sitan’s project and what I do as the math coordinator for RSI.

RSI students meet with their mentors every day and I meet with students once a week. On the surface I just listen as they describe their projects. In reality, I do many different things. I cheer the students up when they are overwhelmed by the difficulty of their projects. I help them decide whether they need to switch projects. I correct their mistakes. Most projects involve computer help, so I teach them Mathematica. I teach them the intricacies of Latex and Beamer. I explain general mathematical ideas and how their projects are connected to other fields of mathematics. I never do their calculations for them, but sometimes I suggest general ideas. In short, I do whatever needs to be done to help them.

I had a lot of fun working with Sitan. His project was about the rank number of grid graphs. A vertex k-ranking is a labeling of the vertices of a graph with integers from 1 to k so that any path connecting two vertices with the same label passes through a vertex with a greater label. The rank number of a graph is the minimum possible k for which a k-ranking exists for that graph. When Sitan got the project, the ranking numbers were known for grid graphs of sizes 1 by n, 2 by n, and 3 by n. So Sitan started working on the ranking number of the 4 by n graph.

His project was moving unusually fast and my job was to push him to see the big picture. I taught him that the next step, once he finishes 4 by n graphs is not to do 5 by n graphs, as one might think. After the first step, the second step should be bigger. He should use his insight and understanding of 4 by n graphs to try to see what he can do for any grid graphs.

This is exactly what he did. After he finished the calculation of the rank number of the 4 by n graphs, he found a way to improve the known bounds for the ranking number of any grid graph. His paper is available at the arXiv.

I just looked at my notes for my work with Sitan. The last sentence: “Publishable results, a potential winner.”

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A Median Coin

Baron Münchhausen is famous for his tall tales. My co-author Konstantin Knop wants to rehabilitate him and so invents problems where the Baron is proven to be truthful from the start. We already wrote a paper about one such problem. Here is a new problem by Konstantin:

Kostya has a black box, such that if you put in exactly 3 coins of distinct weights, the box will expose the coin of median weight. The Baron gave Kostya 5 coins of distinct weights and told him which coin has the median weight. Can Kostya check that the Baron is right, using the box not more than 3 times?

Actually, Konstantin designed a more complicated problem that was given at the Euler Olympiad, 2012 in Russia.

Let n be a fixed integer. Kostya has a black box, such that if you put in exactly 2n+1 coins of distinct weights, the box will expose the coin of median weight. The Baron gave Kostya 4n+1 coins of distinct weights and told him which coin has the median weight. Can Kostya check that the Baron is right, using the box not more than n+2 times?

Note that Kostya can’t just put 4n+1 coins in the box. The box accepts exactly 2n+1 coins. The problem that I started with is for n = 1. Even such a simple variation was a lot of fun for me to solve. So, have fun.

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Guessing the Suit

I recently published my new favorite math problem:

A deck of 36 playing cards (four suits of nine cards each) lies in front of a psychic with their faces down. The psychic names the suit of the upper card; after that the card is turned over and shown to him. Then the psychic names the suit of the next card, and so on. The psychic’s goal is to guess the suit correctly as many times as possible.
The backs of the cards are asymmetric, so each card can be placed in the deck in two ways, and the psychic can see which way the top card is oriented. The psychic’s assistant knows the order of the cards in the deck; he is not allowed to change the order, but he may orient any card in either of the two ways.
Is it possible for the psychic to make arrangements with his assistant in advance, before the latter learns the order of the cards, so as to ensure that the suits of at least (a) 19 cards, (b) 23 cards will be guessed correctly?
If you devise a guessing strategy for another number of cards greater than 19, explain that too.

If the psychic is only allowed to look at the backs of the cards, then the amount of transmitted information is 236, which is the same amount of information as suits for 18 cards. This number of guesses is achievable: the backs of every two cards can clue in the suit of the second card in the pair. This way the psychic can guess the suits of all even-numbered cards in the deck. So the problem is to improve on that. Using the info from the cards that the psychic is permitted to turn over can help too.

The problem is from the book Moscow Mathematical Olympiads, 2000-2005. The book and Russian blog discussions provide many different ideas on how to guess more than half of the deck.

Here is the list of ideas.

Idea 1. Counting cards. If you count cards you will know the suits of the last cards.

Idea 2. Trading. As we discussed before, the psychic can correctly guess the suits of even-numbered cards. By randomly guessing the odd-numbered cards she can correctly guess on average the suits of 4.5 additional cards. Unfortunately, this is not guaranteed. But wait. What if we trade the knowledge of the second card’s suit for the majority suit among odd-numbered cards?

Idea 3. Three cards. Suppose we have three cards. Three bits can provide the following knowledge: the majority color, plus the suit of the first and of the second cards in the majority color. Thus, three bits of information will allow the psychic to guess the suits of two cards out of three.

Idea 4. Which card. Suppose the assistant signals the suits of even-numbered cards. With no loss, the psychic can guess the even-numbered card and repeat the same suit for the next card. If this is the plan, the assistant can choose which of the two cards to describe. Which card of the two matches the psychic’s guess provides an additional bit of information.

Idea 5. Surprise. Suppose we have a strategy to inform the psychic about some cards. Suppose the assistant deliberately fails on one of the cards. Then the index of this card provides info to the psychic.

I leave it to my readers to use these ideas to find the solution for 19, 23, 24 and maybe even for 26 cards.

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