Archive for the ‘Puzzles’ Category.

Skyscrapers (Sum)

Skyscraper puzzles are one of my favorite puzzles types. Recently I discovered a new cute variation of this puzzle on the website the Art of Puzzles. But first let me remind you what the skyscraper rules are. There is an n by n square grid that needs to be filled as a Latin square: each number from 1 to n appears exactly once in each row and column. The numbers in the grid symbolize the heights of skyscrapers. The numbers outside the grid represent how many skyscrapers are seen in the corresponding columns/row from the direction of the number.

The new puzzle is called Skyscrapers (Sum), and the numbers outside the grid represent the sum of the heights of the skyscrapers you see from this direction. For example, if the row is 216354, then from the left you see 8(=2+6); and from the right you see 15(=4+5+6).

Skyscraper Sums

Here’s an easy Skyscrapers (Sum) puzzle I designed for practice.

The Art of Puzzles has four Skyscrapers (Sum) puzzles that are more difficult than the one above:

 

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ApSimon’s Mints Investigation

I recently wrote a post about the ApSimon’s Mints problem:

New coins are being minted at n independent mints. There is a suspicion that some mints might use a variant material for the coins. There can only be one variant material. Therefore, fake coins weigh the same no matter where they’ve been minted. The weight of genuine coins is known, but the weight of fake coins is not. There is a machine that can precisely weigh any number of coins, but the machine can only be used twice. You can request several coins from each mint and then perform the two weighings in order to deduce with certainty which mints produce fake coins and which mints produce real coins. What is the minimum total of coins you need to request from the mints?

The post was accompanied by my paper Attacking ApSimon’s Mints.

Unfortunately, both the post and the paper contain wrong information. They both state that the number of coins as a sequence of the number of mints is 1, 2, 4, 8, 15, 38, 74. This is wrong. I took this data from the sequence A007673 in the OEIS database. The sequence had incorrect data lying dormant for 20 years. I believe that the sequence was generated from the paper of R. K. Guy and R. J. Nowakowsky, ApSimon’s Mints Problem, published in Monthly in 1994. To the credit of Guy and Nowakowsky, they never claimed to find the best solution: they just found a solution, thus providing a bound for the sequence. Someone mistook their solution for the optimal one and generated the sequence in the database.

After my post, my readers got interested in the problem and soon discovered the mistake. First Konstantin Knop found a solution for 6 mints with 30 coins and for 7 mints with 72 coins. Konstantin is my long-time collaborator. I trust him so I was sure the sequence was flawed. Then someone located a reference to a paper in Chinese A New Algorithm for ApSimon’s Mints Problem. Although none of my readers could find the paper itself nor translate the abstract from Chinese. But judging from the title and the formulae it was clear that they found better bounds than the sequence in the database. My readers got excited and tried to fix the sequence. David Reynolds improved on Konstantin’s results with a solution for 6 mints with 29 coins and for 7 mints with 52 coins. David did even better on his next try with 28 and 51 coins correspondingly. He also found a solution with 90 coins for 8 mints. Moreover, his exhaustive search proved that these were the best solutions.

Now the sequence in the database is fixed. It starts 1, 2, 4, 8, 15, 28, 51, 90.

For future generations I would like to support each number of the sequence by an example. I use the set P(Q) to represent the sequence of how many coins are taken from each mint for the first(second) weighing. For one mint, only one coin and one weighing is needed. ApSimon himself calculated the first five values, so they were not in dispute.

  • a(2) = 2: P=(1,0) and Q=(0,1). Found by ApSimon.
  • a(3) = 4: P=(0,1,2) and Q=(1,1,0). Found by ApSimon.
  • a(4) = 8: P=(0,1,2,3) and Q=(1,0,2,2) or P=(0,1,1,4), Q=(2,0,1,1). Found by ApSimon.
  • a(5) = 15: P=(0,1,1,4,5) and Q=(2,1,2,5,0). Found by ApSimon.
  • a(6) = 28: P=(1,2,2,5,5,0) and Q=(0,1,2,1,8,10). Found by Robert Israel, Richard J. Mathar, and David Reynolds,
  • a(7) = 51. P=(2,3,7,2,8,12,0) and Q=(0,2,1,7,7,12,12). Found by David Applegate and David Reynolds.
  • a(8) = 90. P=(4,6,6,7,3,13,15,3) and Q=(4,0,1,6,12,12,1,27). Found by David Applegate and David Reynolds.

There is a solution for nine mints using 193 coins that is not confirmed to be optimal. It was found by David Reynolds: P=(1,2,4,12,5,4,20,39,43) and Q=(0,1,3,3,25,33,34,18,27). In addition, David Reynolds provided a construction that reduces the upper bound for n mints to (3(n+1)−2n−3)/4. The following set of coins work: P=(1,3,7,15,…,2n−1) and Q=(1,4,13,40,…,(3n−1)/2).

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No Averages Solution

I recently posted the following old Olympiad problem:

Prove that you can choose 2k numbers from the set {1, 2, 3, …, 3k−1} in such a way that the chosen set contains no averages of any two of its elements.

Let me show how to find 2k − 1 such numbers. We can pick all the numbers that have 0 or 1 in their ternary representation. Let me prove that this set doesn’t contain averages. Summing two such numbers doesn’t involve carry, and the sum will contain a 1 in each place where the digits differ. On the other hand, a double of any number in this set doesn’t contain ones.

This solution is pretty, but it is not good enough: we need one more number. We can add the number 3k−1. I will leave it to the reader to prove that the largest number in the group whose binary representation consists only of twos can be added without any harm.

There are other ways to solve this problem. It is useful to notice that multiplying a no-average set by a constant or adding a constant to it, doesn’t change the no-average property.

If we were allowed to use 0, then the problem would have been solved. As zero doesn’t belong to the initial range, we can add zero and shift everything by 1. The resulting sequence is sequence A3278. This sequence is the lexicographically first non-averaging sequence.

Another solution was suggested by devjoe in my livejournal mirror blog site. If we multiply our non-averaging set (the one that doesn’t have twos in their ternary representation) by 2, we get a set of all numbers that do not contain ones in their ternary representation. By linearity, such a set doesn’t contain averages either. We can add 1 to this set.

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Crypto Word Search II

I recently published a crypto word search puzzle: a word search puzzle where all the letters are encrypted by a substitution cypher. The answer to such a puzzle is a word or a phrase formed by those decrypted letters that are not in the hidden words.

Crypto Word Search 2

The puzzle I posted was easy. It can be solved by analyzing the repeated letters in the hidden words. The new puzzle is more difficult. No hidden word has repeated letters.

The hidden words are: FUN, HUNT, SOLVE, STARK, TABLE, THINK.

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Crypto Word Search

The puzzle below can be defined as a Crypto Word Search. Guess what needs to be done in this puzzle. The answer is a word.

Crypto Word Search

The hidden words are: DUKE, EYES, RUSE, WORD, WUSS.

The idea of a crypto word search came to me from a beautiful, but devilishly difficult, puzzle In the Details, designed by Derek Kisman. In the Details can be described as a fractal word search; it contains a crypto word search as one of the simpler steps.

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No Averages

Here is an old Olympiad problem:

Prove that you can choose 2k numbers from the set {1, 2, 3, …, 3k−1} in such a way that the chosen set contains no averages of any two of its elements.

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A Logic Quiz

This is a variation on an old quiz. Can you answer the last question?

—An airplane carries 500 bricks. One of the bricks falls out. How many bricks are left in the airplane?
—This is easy: 499!
—Correct. Next question. How do you put a giraffe into a refrigerator?
—Open the refrigerator, put in the giraffe, and close the refrigerator door.
—Good, next. How do you put an elephant into a refrigerator?
—Open the refrigerator, take out the giraffe, put in the elephant and close the door.
—Correct. The Lion King is hosting his birthday party. All the animals come to congratulate him—except one. Why?
—The elephant couldn’t come because it is in the refrigerator.
—Fantastic, next. A man needs to cross a river inhabited by crocodiles and he doesn’t have a boat. What should he do?
—He can just swim: all the crocodiles are attending Lion King’s birthday party.
—Amazing! The last question: The man swims across the river, and dies. What happened?

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ApSimon’s Mints

Hugh ApSimon described the following coin puzzle in his book Mathematical Byways in Ayling, Beeling and Ceiling.

New coins are being minted at n independent mints. There is a suspicion that some mints might use a variant material for the coins. There can only be one variant material: fake coins weigh the same independently of the mint. The weight of genuine coins is known, but the weight of fake coins is not. There is a machine that can precisely weigh any number of coins, but the machine can only be used twice. You can request several coins from each mint and then perform the two weighings so that you can deduce with certainty which mints produce fake coins and which mints produce real coins. What is the minimum total of coins you need to request from the mints?

I will follow ApSimon’s notation. Suppose Pr and Qr is the number of coins from the mint r used in the first and the second weighing correspondingly. That is, we are minimizing Σmax(Pr,Qr). (All my summations are over all the mints. I skip the summation limits because it is difficult to write math in html.) Let us denote by W the weight of the genuine coin and by W(1 + ε) the weight of the fake coin. We do not know ε, except that it is not zero.

Let dr be either 0 or 1, depending on what material the r-th mint uses. Thus, the coin from the r-th mint weighs W(1 + drε). We know the results of these two weighings and the weight of the genuine coin. Therefore, we can calculate the following two values: a = ΣPrdrε and b = ΣQrdrε.

It is clear that we need to request at least one coin from each mint and use it in at least one weighing: Pr + Qr > 0. If both sums a and b are zero, then all the mints are producing genuine coins. Neither of the two values gives us much information as we do not know ε. We can get rid of ε by dividing a by b.

There are 2n − 1 combinations of possible answers: these are subsets of the set of mints producing fake coins given that there is at least one. Thus we need to select numbers Pr and Qr, so that a/b produces 2n − 1 possible answers for different sets of values of dr.

Let us consider cases in which the total number of mints is small. If there is one mint we can take one coin and we won’t even need a second weighing. For two mints we need one coin from each mint for a total of 2. For three mints, one coin from each mint is not enough. I leave this statement as an exercise. It is possible to test three mints with four coins: one each from the first and second mints and two from the third mint. The coins from each mint for the first and second weighings are (0,1,2) and (1,1,0) respectively.

To prove that this works we need to calculate (d2 + 2d3)/(d1 + d2) for seven different combinations of dr. I leave this as an exercise.

This puzzle seems to be very difficult. We only know the answer if the number of mints is not more than seven. The corresponding sequence A007673 in the OEIS is: 1, 2, 4, 8, 15, 38, 74. It is possible to give bounds for this sequence, but they are so far apart. The lower bound is n. And the ApSimon’s book offers a construction for two weighings were Pr = r! and Qr = 1.

You can try to find a better construction, or you can try calculating more terms of the sequence. You can also read more about this problem in my short paper Attacking ApSimon’s Mints.

I do not want to leave the readers with the puzzle that might end up being intractable. So I suggest the following easy puzzle. Solve the ApSimon’s Mints problem assuming that the weight of the fake coin is known.

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Hat Puzzle: Create a Distribution

Here is a setup that works for the several puzzles that follow it:

The sultan decides to test his hundred wizards. Tomorrow at noon he will randomly put a red or a blue hat—from his inexhaustible supply—on every wizard’s head. Each wizard will be able to see every hat but his own. The wizards will not be allowed to exchange any kind of information whatsoever. At the sultan’s signal, each wizard needs to write down the color of his own hat. Every wizard who guesses wrong will be executed. The wizards have one day to decide together on a strategy.

I wrote about puzzles with this setup before in my essay The Wizards’ Hats. My first request had been to maximize the number of wizards who are guaranteed to survive. It is easy to show that you cannot guarantee more than 50 survivors. Indeed, each wizard will be right with probability 0.5. That means whatever the strategy, the expected number of wizards guessing correctly is 50. My second request had been to maximize the probability that all of them will survive. Again, the counting argument shows that this probability can’t be more than 0.5.

Now here are some additional puzzles, including the first two mentioned above, based on the same setup. Suggest a strategy—or prove that it doesn’t exist—in which:

  1. 50 wizards will be guaranteed to survive.
  2. 100 wizards will survive with probability 0.5.
  3. 100 wizards will survive with probability 0.25 and 50 wizards will survive with probability 0.5.
  4. 75 wizards will survive with probability 1/2, and 25 wizards survive with probability 1/2.
  5. 75 wizards will survive with probability 2/3.
  6. The wizards will survive according to a given distribution. For which distributions is it possible?

As I mentioned, I already wrote about the first two questions. Below are the solutions to those questions. If you haven’t seen my post and want to think about it, now is a good time to stop reading.

To guarantee the survival of 50 wizards, designate 50 wizards who will assume that the total number of red hats is odd, and the rest of the wizards will assume that the total number of red hats is even. The total number of red hats is either even or odd, so one of the groups is guaranteed to survive.

To make sure that all of them survive together with probability 0.5, they all need to assume that the total number of red hats is even.

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Kolmogorov Student Olympiad in Probability

There are too many Olympiads. Now there is even a special undergraduate Olympiad in probability, called Kolmogorov Student Olympiad in Probability. It is run by the Department of Probability Theory of Moscow State University. I just discovered this tiny Olympiad, though it has been around for 13 years.

A small portion of the problems are accessible for high school students. These are the problems that I liked. I edited them slightly for clarity.

Second Olympiad. Eight boys and seven girls went to movies and sat in the same row of 15 seats. Assuming that all the 15! permutations of their seating arrangements are equally probable, compute the expected number of pairs of neighbors of different genders. (For example, the seating BBBBBBBGBGGGGGG has three pairs.)

Third Olympiad. One hundred passengers bought assigned tickets for a 100-passenger railroad car. The first 99 passengers to enter the car get seated randomly so that all the 100! possible permutations of their seating arrangements are equally probable. However, the last passenger decides to take his reserved seat. So he arrives at his seat and if it is taken he asks the passenger in his seat to move elsewhere. That passenger does the same thing: she arrives at her own seat and if it is taken, she asks the person to move, and so on. Find the expected number of moved passengers.

Third Olympiad. There are two 6-sided dice with numbers 1 through 6 on their faces. Is it possible to “load” the dice so that when the two dice are thrown the sum of the numbers on the dice are distributed uniformly on the set {2,…,12}? By loading the dice we mean assigning probabilities to each side of the dice. You do not have to “load” both dice the same way.

Sixth Olympiad. There are M green and N red apples in a basket. We take apples out randomly one by one until all the apples left in the basket are red. What is the probability that at the moment we stop the basket is empty?

Seventh Olympiad. Prove that there exists a square matrix A of order 11 such that all its elements are equal to 1 or −1, and det A > 4000.

Twelfth Olympiad. In a segment [0,1] n points are chosen randomly. For every point one of the two directions (left or right) is chosen randomly and independently. At the same moment in time all n points start moving in the chosen direction with speed 1. The collisions of all points are elastic. That means, after two points bump into each other, they start moving in the opposite directions with the same speed of 1. When a point reaches an end of the segment it sticks to it and stops moving. Find the expected time when the last point sticks to the end of the segment.

Thirteenth Olympiad. Students who are trying to solve a problem are seated on one side of an infinite table. The probability that a student can solve the problem independently is 1/2. In addition, each student will be able to peek into the work of his or her right and left neighbor with a probability of 1/4 for each. All these events are independent. Assume that if student X gets a solution by solving or copying, then the students who had been able to peek into the work of student X will also get the solution. Find the probability that student Vasya gets the solution.

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