Archive for the ‘Weighings’ Category.

2015 Coin Problem Solution

A while ago I posted my second favorite problem from the 2015 All-Russian Math Olympiad:

Problem. A coin collector has 100 coins that look identical. He knows that 30 of the coins are genuine and 70 fake. He also knows that all the genuine coins weigh the same and all the fake coins have different weights, and every fake coin is heavier than a genuine coin. He doesn’t know the exact weights though. He has a balance scale without weights that he can use to compare the weights of two groups with the same number of coins. What is the smallest number of weighings the collector needs to guarantee finding at least one genuine coin?

Now it’s solution time. First we show that we can do this in 70 weighings. The strategy is to compare one coin against one coin. If the scale balances, we are lucky and can stop, because that means we have found two real coins. If the scale is unbalanced, the heavier coin is definitely fake and we can remove it from consideration. In the worst case, we will do 70 unbalanced weighings that allow us to remove all the fake coins, and we will find all the real coins.

The more difficult part is to show that 69 weighings do not guarantee finding the real coin. We do it by contradiction. Suppose the weights are such that the real coin weighs 1 gram and the i-th fake coin weighs 100i grams. That means whatever coins we put on the scale, the heaviest pan is the pan that has the fake coin with the largest index among the fake coins on the scale.

Suppose there is a strategy to find a real coin in 69 weighings. Given this strategy, we produce an example designed for this strategy, so that the weighings are consistent, but the collector cannot find a real coin.

For the first weighing we assign the heaviest weight, 10070 to one of the coins on the scale and claim that the pan with this coin is heavier. We continue recursively. If a weighing has the coins with assigned weights, we pick the heaviest coin on the pans and claim that the corresponding pan is heavier. If there are no coins with assigned weights on pans, we pick any coin on the pans, assigned the largest available weight to it and claim that the corresponding pan is heavier.

After 69 weighings, not more than 69 coins have assigned weights, while all the weighings are consistent. The rest of the coins can have any of the leftover weights. For example, any of the rest of the coins can weigh 100 grams. That means that there is no coin that is guaranteed to be real.

Share:Facebooktwittergoogle_plusredditpinterestlinkedinmail

A Random Scale Solution

I recently posted the following puzzle:

Puzzle. We have 32n identical-looking coins. One of the coins is fake and lighter than the other coins, which all are real. We also have three scales: two normal and one random. Find the fake coin in the smallest total number of weighings.

Here is my son Sergei’s solution. Divide the coins into nine groups of equal size and number the groups in ternary: 00, 01, 02, 10, 11, 12, 20, 21, and 22. On each scale we put three groups versus three groups. On the first scale we compare the three groups that start with 1 with the three groups that start with 2. For the second scale we do the same using the last digit instead of the first one, and for the third scale we use the sum of two digits modulo 3. Any pair of scales, if they are assumed to be normal, would point to one out of nine groups as the group containing the fake coin.

If all three pairs of scales agree on one group, then this is the group containing the fake coin. Thus in three weighings, we reduce the number of groups of coins by a factor of nine. If the pairs of scales do not agree, then the random scale produced a wrong weighing and thus can be found out. How do we do that? We have three out of nine groups of coins each of which might contain the fake coin. We compare two of the groups on all three scales. This way we know exactly which group contains the fake coin and, consequently, which scale generated a wrong weighing. If we know the random scale, we can speed up the rest of the process of finding the fake coin. Thus in the worst case we require 3n+3 weighings.

The big idea here is that as soon as the random scale shows a wrong weighing result it can be found out. So in the worst case, the random scale behaves as a normal scale and messes things up at the very end. Sergei’s solution can be improved to 3n+1 weighings. Can you do that?

The improved solution is written in a paper Взвешивания на «хитрых весах» (in Russian) by Konstantin Knop, that is published in Математика в школе 2009-2. The paper contains an even stronger solution that provides a better asymptotics.

Share:Facebooktwittergoogle_plusredditpinterestlinkedinmail

Alternators: People and Coins

If like me, you fancy Raymond Smullyan and his books, then you’ve heard about knights and knaves. Knights always tell the truth and knaves always lie. In addition to knights and knaves, there are normal people who sometimes tell the truth and sometimes lie. Here is a puzzle.

Puzzle. How, in one sentence, can a normal person prove that they are normal?

We can draw a parallel between people and coins. We can say that knights correspond to real coins, and knaves to fake coins that are lighter than real ones. Inspired by normal people, my coauthor Konstantin Knop invented chameleon coins. Chameleon coins can change their weight and behave like real or fake coins. I just wrote a post about chameleon coins.

Normal people are too unpredictable: they can consistently pretend to be knights or knaves. So logicians invented a simpler type of person, one who switches from telling the truth in one sentence to a lie in the next and then back to the truth. Such people are called alternators. Here is another puzzle:

Puzzle. You meet a person who is one of the three types: a knight, a knave, or an alternator. In two questions, find out which type they are.

Continuing a parallel between people and coins we can define alternator coins: the coins that switch their weight each time they are on the scale from weighing as much as real ones to weighing as much as fake ones. For the purposes of this essay, we assume that the fake coins are lighter than real ones. Unlike the chameleon coin, which might never reveal itself by always pretending to be real, the alternators can always be found. How do you find a single alternator among many real coins? There is a simple strategy: repeat every weighing twice. This strategy allows us to find an alternator among 9 coins in four weighings. Can we do better?

I used the alternator coins as a research project for my PRIMES STEP program where we do math research with students in seventh and eighth grade. The students started the alternator project and immediately discovered the strategy above. The next step is to describe a better strategy. For example, what is the maximum number of coins containing one alternator such that the alternator can always be found in four weighings?

But first we count possible outcomes. Suppose there is a strategy that finds an alternator. In this strategy we can’t have two unbalanced weighings in a row. To prove that, let us suppose there was an unbalanced weighing. Then the alternator switches its weight to a real coin and whether or not the alternator is on the scale, the next weighing must balance. The beauty of it is that given a strategy each outcome has to point to a particular coin as an alternator. That means the number of outcomes bounds the total number of coins that can be processed.

Counting the number of possible outcomes that do not have two unbalances in row is a matter of solving a recurrence, which I leave to the readers to find. The result is Jacobshtal numbers: the most beautiful sequence you might never have heard of. For example, the total number of possible outcomes of four weighings is 11. Since each outcome of a strategy needs to point to a coin, the total number of coins that can be processed in four weighings is not more than 11. But 11 is better than 9 in our previous strategy. Can we process 11 coins in four weighings? Yes, we can. I will describe the first part of the strategy.

So we have 11 coins, one of which is an alternator. In the first weighing we compare 5 coins against 5 coins. If the weighing unbalances, the alternator is on a lighter pan. Our problem is reduced to finding the alternator among five coins when we know that it is in the real state. If the weighing balances, then we know that if the alternator is among the coins on the scale it must now be in the light state. For the second weighting, we pick two sets of three coins out of this ten coins and compare them against each other. Notice that 3 is a Jacobsthal number, and 5, the number of coins outside the scale, is also a Jacobsthal number. If the second weighing balances, the alternator must be among 5 coins outside the scale. All but one of these coins are in the light state, and I leave it to the readers to finish the strategy. If the weighing unbalances, we need to find the alternator among 3 coins that are in the real state now. This can be done in two weighings, and again the readers are to the rescue.

It appears that Jacobsthal numbers provide the exact lower bound of the number of coins that can be processed. This is what my middle-schoolers discovered and proved. We wrote a paper on the subject. The strategy in the paper is adaptive. That means it changes depending on the results of the previous weighings. Can we find an oblivious strategy? I will tell you in later posts.

Share:Facebooktwittergoogle_plusredditpinterestlinkedinmail

A Random Scale

Suppose we have 3n identical-looking coins. One of the coins is fake and lighter than the other coins which all are real. We also have a random scale. That is a scale that at each weighing behaves randomly. Find the fake coin in the smallest number of weighings. Oops! That won’t work! It is impossible to find the fake coin. The scale can consistently misbehave in such a way as to blame a specific real coin for being fake.

Let’s try something else. Suppose we have two scales: one normal and one random. Find the fake coin.

What am I thinking? The normal scale can point to one coin and the random scale can point to another coin and we are in a “she said, he said” situation which we can’t resolve.

Now, in my final try, I’ll make it right. We actually have three scales, one of which is random. So here we go, with thanks to my son Sergei for giving me this puzzle:

Puzzle. We have 3n identical-looking coins. One of the coins is fake and lighter than the other coins, which all are real. We also have three scales: two normal and one random. Find the fake coin in the smallest total number of weighings.

Let’s start with this strategy: repeat every weighing on all three scales and have a majority vote. At least two of the scales will agree, thus pointing to the true result. This way we can use a divide-into-three-equal-groups strategy for one scale to find the fake coin. It will require 3n weighings.

Can we do better? Of course, we can. We can repeat every weighing on two scales. If they agree we do not need the third scale. If they do not agree, one of the scales is random and lying, and we can repeat the weighing on the third scale to “out” the random scale. After we identify one normal scale, the process goes faster. In the worst case we will need 2n + 1 weighings.

Can we do even better? Yes, we can. I will leave it to the readers to find a beautiful solution that is asymptotically better than the previous one.

Update on Dec 24, 2016. The total number of coins should be 32n, not 3n. We are looking at the worst case scenario, when the random scale is adversarial.

Share:Facebooktwittergoogle_plusredditpinterestlinkedinmail

Chameleon Coins

We all have played with problems in which we had real coins and fake (counterfeit) coins. For this post I assume that the fake coins are always lighter than the real coins. My coauthor Konstantin Knop invented a new type of a coin: a chameleon coin. This coin can mimic a fake or a real coin. It can also choose independently which coin to mimic for each weighing on a balance scale.

You cannot find the chameleon coin in a mix with real coins if it does not want to be found, because it can consistently behave as a real coin. Let’s add classic fake coins to the mix, the ones that are lighter. Still the task of identifying the chameleons using a balance scale cannot be achieved: the chameleons can pretend to be fake coins. We can’t identify the fake coins either, as the chameleons can mess things us up by consistently pretending to be fake.

What we can do is to find a small number of coins some of which are guaranteed to be fake. Consider the simplest setup, when we have one fake coin and one chameleon in our mix of N coins. That is we have N − 2 real coins. Our task now is to find TWO coins, one of which has to be fake. As usual we want to do it in the smallest number of weighings that guarantees that we’ll find the two coins. Let me give you a fun problem to solve:

Puzzle. The total number of coins is four. And as above we have one chameleon and one classic fake. In two weighings find two coins so that one of them is guaranteed to be fake.

If you want to learn more, we just wrote a paper titled Chameleon Coins.

Share:Facebooktwittergoogle_plusredditpinterestlinkedinmail

A Faulty Scale

Today I have two new coin puzzles that were inspired by my son, Alexey Radul:

You have N > 2 identical-looking coins. All but one of them are real and weigh the same. One coin is fake and is lighter than the real ones. You also have a balance scale which might or might not be faulty. A faulty scale differs from a normal one in reversing the sense of unbalanced weighings—it shows a lighter pan as heavier and vice versa (but still shows equal-weight pans as weighing the same). What is the smallest number of weighings you need in order to figure out whether the scale is faulty?

If you think about it, this problem is isomorphic to a known problem I wrote about before:

You have N ≥ 2 identical-looking coins. All but one of them are real and weigh the same. One coin is fake and is either lighter than the real ones or heavier than the real ones. You also have a normal balance scale. What is the smallest number of weighings you need in order to figure out whether the fake coin is lighter or heavier?

To make things more interesting let’s mix the problems up.

You have N > 2 identical-looking coins. All but one of them are real and weigh the same. One coin is fake and is lighter than the real ones. You also have M > 1 identical-looking balance scales. All but one of them are normal and one is faulty. The faulty scale differs from a normal one in reversing the sense of unbalanced weighings—it shows a lighter pan as heavier and vice versa (but still shows equal-weight pans as weighing the same). What is the smallest number of total weighings needed to figure out which scale is faulty?

Share:Facebooktwittergoogle_plusredditpinterestlinkedinmail

Two Fake Coins

You are given N coins such that two of them are fake and the other coins are genuine. Real coins weigh the same. Fake coins weigh the same and are lighter than real ones. You need to find both fake coins using a balance scale in the smallest number of weighings.

It is easy to estimate the number of weighings from below using information theory. Given N coins you will have N choose 2 different answers. The number of possible answers has to be not more than 3w, where w is the number of weighings. This generates a trivial information-theoretic bound (ITB) on the number of coins that can be processed in a given number of weighings.

Previous authors used computers to completely analyze small numbers of weighings: from 2 to 5.

My colleagues from Russia, Konstantin Knop and Oleg Polubasov, performed some fantastic programming accompanied with some subtle heuristic and found optimal solutions for up to 10 weighings. For 11 and 12 weighings, the program is not efficient enough to find the optimal solutions: it found some solutions. Surprisingly, for up to 10 weighings, the optimal solutions match the information-theoretic bound (ITB). The results are in the table below. The first row represents the number of weighings w. The second row is the largest number of coins N for which a solution is found. The last row is the information-theoretic bound (ITB) we explained above.

w 5 6 7 8 9 10 11 12
N 22 38 66 115 198 344 585 1026
ITB 22 38 66 115 198 344 595 1031

Their paper, Two counterfeit coins revisited, is available in Russian. Lucky for us, 70 of 73 pages are pseudo-code of solutions for which you do not need any Russian. You just need to understand the pseudo-code. Here is the explanation.

Each line begins with its number. After it they have the weighing in the format 1 10 v 4 5 meaning coins 1 and 10 are weighed versus coins 4 and 5. Each weighing is followed by a colon, after which they describe in order actions for three different outcomes of the weighing: equality, the first pan is lighter, and the second pan is lighter. Each action is one of the following:

  • => L means go to line L.
  • (a, b) means the fake coins are a and b.
  • – means this branch is impossible and there is no output.
  • => 23. sym indicates the symmetry of the weighing and its result, therefore the resulting go-to line, 23, is omitted as being equivalent to another line.
  • – . sym means the output is symmetric to another one.

The line numbers after => in line L are always 3L+1, 3L+2 and 3L+3. The sym symbol implies that line 3L+3 is omitted as a symmetric version of line 3L+2.

For example, here is their solution for 2 weighings and 4 coins in their pseudo-code. An empty line separates different weighings:

0. 1 v 2 : => 1, => 2, => 3. sym

1. 1 v 3 : – , (1, 2), (3, 4).
2. 3 v 4 : – , (1, 3), – . sym

Another solution for 3 weighings and 7 coins:

0. 1 2 v 3 4 : => 1, => 2, => 3. sym

1. 1 v 2 : => 4, => 5, => 6. sym
2. 1 v 2 : (1, 2), => 8, => 9. sym

4. 5 v 6 : (5, 6), (5, 7), – . sym
5. 3 v 4 : – , (1, 3), – . sym
8. 5 v 6 : (1, 7), (1, 5), – . sym

If you want to see an optimal solution for 10 weighings, look at the paper: the algorithm takes 36 pages.

Share:Facebooktwittergoogle_plusredditpinterestlinkedinmail

A New Question about Old Coins

I want to come back to a middle-school Olympiad problem I posted a while ago.

Streamline School Olympiad 2000 (8th grade). You have six bags of coins that look the same. Each bag has an infinite number of coins and all the coins in the same bag weigh the same amount. Each different bag contains coins of a different weight, ranging from 1 to 6 grams exactly. There is a label (1, 2, 3, 4, 5, 6) attached to each bag that is supposed to correspond to the weight of the coins in that bag. You have only a balance scale. What is the least number of times you need to weigh the coins in order to confirm that the labels are correct?

The answer is unpretentious: one weighing is enough. We can take one 5-gram coin, two 4-gram coins, three 3-gram coins, four 2-gram coins and five 1-gram coins for the total of 35 grams. This number is not divisible by 6, so we can add one more 1-gram coin and weigh all of them against six 6-gram coins. I leave it to the reader to show that this solution works and to extrapolate the solution for any number of bags.

My new challenge is to find a weighing for the above problem using the smallest number of coins. What is the number of coins in such a weighing for a given number of bags?

I manually calculated this number for a small number of bags, but I would like to get a confirmation from my readers. Starting from 6 bags, I don’t know the answer. Can you help me?

Share:Facebooktwittergoogle_plusredditpinterestlinkedinmail

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).

Share:Facebooktwittergoogle_plusredditpinterestlinkedinmail

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.

Share:Facebooktwittergoogle_plusredditpinterestlinkedinmail