# A puzzle puzzle

Jigsaw puzzles that say they have 1,000 pieces have approximately 1,000 pieces, but probably not exactly 1,000. Jigsaw puzzle pieces are typically arranged in a grid, so the number of pieces along a side has to be a divisor of the total number of pieces. This means there aren’t very many ways to make a puzzle with exactly 1,000 pieces, and most have awkward aspect ratios. Since jigsaw pieces are irregularly shaped, it may be surprising to learn that the pieces are actually arranged in a regular grid. At least they usually are. There are exceptions such as circular puzzles or puzzles that throw in a couple small pieces that throw off the grid regularity.

How many aspect ratios can you have with a rectangular grid of 1,000 points? Which ratio comes closest to the golden ratio? More generally, answer the same questions with 10^n points for positive integer n.

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# Sum-free subset challenge

A set of integers is called sum-free if no element of the set is the sum of any other pair of elements in the set. For example, {1, 10, 100} is sum-free.

Let’s look at pulling out a sum-free subset of a larger set. For example, if we start with {1, 2, 3, …, 10}, then {1, 5, 10} as a sum-free subset. So is {1, 2, 4, 7}. Notice in this case 1 + 2 + 4 = 7, but that’s OK because we’re only concerned with whether an element is the sum of two other elements.

[Update: Thanks to Sjoerd Visscher for pointing out that the definition of sum-free does not require that the elements of a sum be distinct. So when I said that the set {1, 2, 4, 7} is sum-free, this was wrong because 2 + 2 = 4. The set A is sum-free if the intersection of A+A with A is empty.]

Now let A be a set of integers with n elements. How large of a sum-free subset does A contain? It could be as large as n if the set A were sum-free to begin with, so that’s an upper bound. But what is a lower bound on the size of the largest sum-free subset?

There is a theorem that gives a number k such that every set of n non-zero integers contains a sum-free subset of size at least kn. You could let k be zero, but that’s no fun. Can you find a larger value of k? I’ll tell you later what value of k the theorem has. Until then, maybe you could try to find your own value.

Suppose you want to write a program to explore this empirically. For a given set, how would you find a maximal sum-free subset? Brute force examination of all subsets would take 2n steps, so hopefully you could do better than that.

What are some sets that have relatively small maximal sum-free subsets?

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# Probability question

Suppose you have a large number of buckets and an equal number of balls. You randomly pick a bucket to put each ball in one at a time. When you’re done, about what proportion of buckets will be empty?

One line of reasoning says that since you have as many balls as buckets, each bucket gets one ball on average, so nearly all the buckets get a ball.

Another line of reasoning says that randomness is clumpier than you think. Some buckets will have several balls. Maybe most of the balls will end up buckets with more than one ball, and so nearly all the buckets will be empty.

Is either extreme correct or is the answer in the middle? Does the answer depend on the number n of buckets and balls? (If you look at the cases n = 1 and 2, obviously the answer depends on n. But how much does it depend on n if n is large?) Hint: There is a fairly simple solution.

What applications can you imagine for the result?

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# A knight’s random walk

Here’s a puzzle I ran across today:

Start a knight at a corner square of an otherwise-empty chessboard. Move the knight at random by choosing uniformly from the legal knight-moves at each step. What is the mean number of moves until the knight returns to the starting square?

There’s a slick mathematical solution that I’ll give later.

You could also find the answer via simulation: write a program to carry out a knight random walk and count how many steps it takes. Repeat this many times and average your counts. Related post: A knight’s tour magic square

# An algebra problem from 1798

The Lady’s Diary was a popular magazine published in England from 1704 to 1841. It contained mathematical puzzles such as the following, published in 1798.

What two numbers are those whose product, difference of their squares, and the ratio or quotient of their cubes, are all equal to each other?

From Benjamin Wardhaugh’s new book A Wealth of Numbers: An Anthology of 500 Years of Popular Mathematics Writing.  # Monday morning math puzzle

This morning Futility Closet mentioned that How common is this? In other words, what is the probability that a random set of digits ai will satisfy the following? How does the probability depend on n?

# Roman numeral puzzle

I noticed an ad for Super Bowl XLVI on a pizza box this morning. The Roman numeral XLVI does not repeat any character. This brought up a couple questions.

• How many Roman numerals are possible if you’re not allowed to repeat a character?
• Could you write a (reasonably short) regular expression to find all such numbers?

There has never been universal agreement on the rules for constructing Roman numerals, so your solution would depend on your choice of rules. For our purposes here, assume the valid characters are I, V, X, L, C, D, and M. Also, assume any character can be subtracted from a larger character. For example, you can assume IL is a valid representation of 49.

For a more challenging problem, you can use the more restrictive subtraction rules.

1. I can be subtracted from V and X only.
2. X can be subtracted from L and C only.
3. C can be subtracted from D and M only.
4. V, L, and D can never be subtracted.

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# A Renaissance math puzzle

In 1471, Johannes Müller asked where you should stand so that a vertical bar appears longest.

To be more precise, suppose a vertical bar is hanging so that the top of the bar is a distance a above your eye level and the bottom is a distance b above your eye level. Let x be the horizontal distance to the bar. For what value of x does the bar appear longest?

Note that the apparent length of the bar is determined by the size of the angle between your lines of sight to the top and bottom of the bar.

Please don’t give solutions in the comments. I’ll post my solution tomorrow, and you can give your solutions in the comments to that post if you’d like.

Source

Update: See this post for more historical background.

# A little math puzzle

Futility Closet posted the other day that log 237.5812087593 = 2.375812087593.

Make a formal statement of the problem that 2.375812087593 solves and show that there’s exactly one other solution.

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# Counterfeit coins and rare diseases

Here’s a puzzle I saw a long time ago that came to mind recently.

You have a bag of 27 coins. One of these coins is counterfeit and the rest are genuine. The genuine coins all weigh exactly the same, but the counterfeit coin is lighter. You have a simple balance. How can you find the counterfeit coin while using the scale the least number of times?

The surprising answer is that the false coin can be found while only using the scales only three times. Here’s how. Put nine coins on each side of the balance. If one side is lighter, the counterfeit is on that side; otherwise, it is one of the nine not on the scales. Now that you’ve narrowed it down to nine coins, apply the same idea recursively by putting three of the suspect coins on each side of the balance. The false coin is now either on the lighter side if the scales do not balance or one of the three remaining coins if the scales do balance. Now apply the same idea one last time to find which of the remaining three coins is the counterfeit. In general, you can find one counterfeit in 3n coins by using the scales n times.

The counterfeit coin problem came to mind when I was picking out homework problems for a probability class and ran into the following (problem 4.56 here):

A large number, N = mk, of people are subject to a blood test. This can be administered in two ways.

1. Each person can be tested separately. In this case N tests are required.
2. The blood samples of k people can be pooled and analyzed together. If the test is negative, this one test suffices for k people. If the test is positive, each of the k people must be tested separately, and, in all, k+1 test are required for the k people.

Suppose each person being tested has the disease with probability p. If the disease is rare, i.e. p is sufficiently small, the second approach will be more efficient. Consider the extremes. If p = 0, the first approach takes mk tests and the second approach takes only m tests. At the other extreme, if p = 1, the first approach still takes mk tests but the second approach now takes m(k+1) tests.

The homework problem asks for the expected number of tests used with each approach as a function of p for fixed k. Alternatively, you could assume that you always use the second method but need to find the optimal value of k. (This includes the possibility that k=1, which is equivalent to using the first method.)

I’d be curious to know whether these algorithms have names. I suspect computer scientists have given the coin testing algorithm a name. I also suspect the idea of pooling blood samples has several names, possibly one name when it is used in literally testing blood samples and other names when the same idea is applied to analogous testing problems.

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