Mathematics, applications of mathematics to life in general, and my life as a mathematician.
Archive for 2009
I am strongly opposed to questions of the type “which is the odd one out” during IQ tests. On the other hand, I do not mind them in different settings, especially when they are fun. Inspired by Martin Gardner, I spent a lot of time drawing this picture, and now I have to share it with the world. So, which is the odd one out?
I love the MIT Mystery Hunt. I like the adrenalin rush when solving problems under pressure. Plus, I like the togetherness of doing problems with other people. During the hunt I usually do not have time to look at all the puzzles: some of them are solved by my teammates while I’m sleeping and others are solved before I get to see them.
I’ve never tried to go back and check out the puzzles I missed nor the puzzles from the previous hunts, probably because without the goal of winning and without my team, I might find them boring. Often the solving process involves tedious Internet browsing to identify the images of different people or objects. I would only be motivated if the puzzle were related to something I am very interested in, such as Ballroom dancing. But I’m not thrilled at the thought of browsing through all the problems in order to find one that is relevant to the Tango.
In short, I need an index to the puzzles. For example, it would be nice to direct the lovers of square dancing to the Do Sa Do puzzle, or fans of Star Trek to the Alien Species puzzle. I hope that nobody blames me for hinting that those aliens are from Star Trek. I’m convinced that Trekkies who only want to solve Star Trek-related puzzles would immediately recognize them anyway. I do believe that I am not revealing too much by saying that the Facebook puzzle will appeal to the aficionados of the television show “24”.
It would be extremely useful to humanity to at least mark the MIT Mystery Hunt puzzles that are self-consistent, and do not require activities. For example, some of the puzzles involve interaction with headquarters, so you can’t solve them after the hunt. Some of the puzzles might expire, as for example the puzzle with pictures of different announcements in the infinite corridor.
Unfortunately, such an index doesn’t exist, and I do not have the time or expertise to create one myself. But I can fill this void at least partially by presenting a guide to math puzzles from the previous four hunts. I can’t promise that my guide is complete, as navigating the MIT Mystery Hunt website is very tiresome.
Before going into the math puzzles, I would like to list Sergei’s favorite type of puzzle: Duck Konundrums. The first Duck Konundrum puzzle appeared in 2000. It was created by Dan Katz, which is why his initials are in the title. One really needs to follow the instructions for this puzzle. This is very unusual as traditionally hunt puzzles do not have instructions at all. Do not be relieved: the instructions are really very complicated. The next Duck Konundrum puzzle appeared in 2002 and was considered to be even more amusing. People loved it, and this type of puzzle became a tradition in subsequent hunts. Here is my list of Duck Konundrums:
Many Mystery Hunt puzzles appeal to mathematicians. I have to warn you though. Puzzles often are divided into two stages. In the first stage, you need to solve a puzzle, like solving sudoku, a crossword or finding all the wedding dates of the people in the pictures. The second stage requires you to produce a word or a phrase that is the answer to the puzzle. The second stage might be as simple as listing the people in chronological order of their wedding dates and then taking the first letters of their last names. This second stage could also be quite difficult. Depending on your tastes one stage of the puzzle might be much more rewarding than the other. If you love solving sudokus, you might find it more fun to just stop with that solution, instead of continuing on to the second stage.
2006
2007
2008
2009
It would also be nice to have some ratings for puzzles. I am not sure how to persuade the webmasters of the MIT Mystery Hunt page to do the index and the rating. Feel free to send them an encouraging email.
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In the book “Mythematics: Solving the Twelve Labors of Hercules” Michael Huber adds details to Hercules’ labors so that in order that he can do each task, you need to help Hercules solve two or three math problems. For example, to defeat the Nemean Lion Hercules needs to solve the problem “Zeus Makes a Deal”, which is a Greek-myth version of the Monty Hall problem.
The problems in Mythematics are quite advanced. They range in topic from algebra, geometry and probability to differential equations and integral calculus. Plus, as a reward for helping Hercules, Huber gives you variations on Sudoku puzzles.
Solving some nice math problems might not be the only reason for people to buy this book. Here are some other reasons:
I like Huber’s approach. Future possibilities for more books are endless. Let’s write new math problems based on Harry Potter, Batman, the Bible or, maybe, The Joy of Sex.
In the book “Mythematics: Solving the Twelve Labors of Hercules
The problems in Mythematics are quite advanced. They range in topic from algebra, geometry and probability to differential equations and integral calculus. Plus, as a reward for helping Hercules, Huber gives you variations on Sudoku puzzles.
Solving some nice math problems might not be the only reason for people to buy this book. Here are some other reasons:
I like Huber’s approach. Future possibilities for more books are endless. Let’s write new math problems based on Harry Potter, Batman, the Bible or, maybe, The Joy of Sex
Konstantin Knop sent me the following coins puzzle, which was created by Alexander Shapovalov and first appeared at the Regional round of the all-Russian math Olympiad in 2000.
Baron Münchhausen has 8 identical-looking coins weighing 1, 2, …, 8 grams. The Baron knows which coin is which and wants to demonstrate to his guests that he is right. He plans to conduct one weighing on a balance scale, so that the guests will be convinced about the weight of one of the coins. Can the Baron do this?
This being a sequence-lover blog, we want to create a sequence out of this puzzle. The sequence is the following: Let the Baron initially have n identical-looking coins that weigh exactly 1, 2, …, n grams. Then a(n) is the minimum number of weighings on a balance scale that the Baron needs in order to convince his guests about the weight of one of those coins.
The original puzzle can be restated as asking whether a(8) = 1. The sequence is defined starting from index 1 and the first several terms are easy to calculate: 0, 1, 1, 1, 2, 1, 1, 1. Can you continue this sequence?
Let’s look at where ones occur in this sequence:
Theorem. If the weight of a coin can be confirmed with one weighing, then one cup of that weighing must contain all the coins with weights from 1 to some i, and the other cup must contain all the coins with weights from some j to n. Furthermore, either the scale must balance, or the cup containing the 1-gram coin must be lighter.
Proof. What does it mean for the Baron to convince his guests about the weight of some coin with one weighing? From the perspective of the guests, a weighing is a number of coins in one cup, a number of coins in the other cup, and a number of coins not on the scale, together with the result the scale shows (one or the other cup heavier, or both the same weight). For the guests to be convinced of the weight of some particular coin, it must therefore be the case that all possible arrangements of coin weights consistent with that data agree on the weight of the coin in question. Our proof strategy, therefore, is to look for ways to alter a given arrangement of coin weights so as to change the weight given to the coin whose weight is being demonstrated, thus arriving at a contradiction.
First, obviously, the coin whose weight k the Baron is trying to confirm has to be alone in its group: either alone on some cup or the only coin not on the scale. After that we can divide the proof of the theorem into several cases.
Case 1. The coin k is on a cup and the scale is balanced. Then we want to show two things: k = n, and the coins on the other cup weigh 1, 2, …, i grams for some i. For the first part, observe that if k < n, then the coin with weight k+1 must not be on the scale (otherwise it would overbalance coin k). Therefore, we can substitute coin k+1 for coin k, and substitute a coin one gram heavier for the heaviest coin that was on the other cup, and produce thereby a different arrangement with the same observable characteristics but a different weight for the coin the Baron claims has weight k.
To prove the second part, suppose the contrary. Then it is possible to substitute a coin 1 gram lighter for one of the coins on the other cup. Now, if coin k-1 is not on the scale, we can also substitute k-1 for k, and again produce a different arrangement with the same observable characteristics but a different weight for the coin labeled k. On the other hand, if k-1 is on the scale but k-2 is not, then we can substitute k-2 for k-1 and then k-1 for k and the weighing is again unconvincing. Finally, if both k-1 and k-2 are on the scale, and yet they balance k, then k=3 and the theorem holds.
Consequently, k = n = 1 + 2 + … + i is a triangular number.
Case 2. The coin k is on the lighter cup of the scale. Then: first, k = 1, because otherwise we could swap k and the 1-gram coin, making the light cup lighter and the heavy cup heavier or unaffected; second, the 2-gram coin is on the heavy cup and is the only coin on the heavy cup, because otherwise we could swap k with the 2-gram coin and not change the weights by enough to affect the imbalance; and finally n = 2 because otherwise we could change the weighing 1 < 2 into 2 < 3.
Thus the theorem holds, and the only example of this case is k = 1, n = 2.
Case 3. The coin k is on the heavier cup of the scale. Then k = n and the lighter cup consists of some collection of the lightest available coins, by the same argument as Case 1 (but even easier, because there is no need to maintain the balance). Furthermore, k must weigh exactly 1 gram more than the lighter cup, because otherwise, k-1 is not on the lighter cup and can be substituted for k, making the weighing unconvincing.
Consequently, k = n = (1 + 2 + … + i) + 1 is one more than a triangluar number.
Case 4. The coin k is not on a cup and the scale is not balanced. Then, since k must be off the scale by itself, all the other coins must be on one cup or the other. Furthermore, all coins heavier than k must be on the heavier cup, because otherwise we could make the lighter cup even lighter by substituting k for one of those coins. Likewise, all coins lighter than k must be on the lighter cup, because otherwise we could make the heavier cup even heavier by substituting k for one of those coins. So the theorem holds; and furthermore, the cups must again differ in weight by exactly 1 gram, because otherwise we could swap k with either k-1 or k+1 without changing the weights enough to affect the result on the scale.
Consequently, the weight of the lighter cup is k(k-1)/2, the weight of the heavier cup is 1 + k(k-1)/2. Thus the total weight of all the coins is n(n+1)/2 = k2+1. In other words, case 4 is possible iff n is the index of a triangular number that is one greater than a square.
Case 5. The coin k is not on a cup and the scale is balanced. This case is hairier than all the others combined, so we will take it slowly (noting first that all the coins besides k must be on some cup).
Lemma 1. The two coins k-1 and k-2 must be on the same cup, if they exist (that is, if k > 2). Likewise k-2 and k-4; k+1 and k+2; and k+2 and k+4.
Proof. Suppose they’re not. Then we can rotate k, k-1, and k-2, that is, put k on the cup with k-1, put k-1 on the cup with k-2, and take k-2 off the scale. This makes both cups heavier by one gram, producing a weighing with the same outward characteristics as the one we started with, but a different coin off the scale. The same argument applies to the other three pairs of coins we are interested in, mutatis mutandis.
Lemma 2. The four coins k-1, k-2, k-3 and k-4 must be on the same cup if they exist (that is, if k ≥ 5).
Proof. By Lemma 1, the three coins k-1, k-2, and k-4 must be on the same cup. Suppose coin k-3 is on the other cup. Then we can swap k-1 with k-3 and k with k-4. Each cup becomes heavier by 2 grams without changing the number of coins present, the balance is maintained, and the Baron’s guests are not convinced.
Lemma 3. If coin k-4 exists, that is if k ≥ 5, all coins lighter than k must be on the same cup.
Proof. By Lemma 2, the four coins k-1, k-2, k-3 and k-4 must be on the same cup. Suppose some lighter coin is on the other cup. Call the heaviest such coin c. Then, by choice of c, the coin with weight c+1 is on the same cup as the cluster k-1, …, k-4, and is distinct from coin k-2 (because c is on a different cup from k-3). We can therefore swap c with c+1 and swap k with k-2. This increases the weight on both cups by 1 gram without changing how many coins are on each, but moves k onto the scale. The Baron’s guests are again unconvinced.
Lemma 4. The theorem is true for k ≥ 5.
Proof. By Lemma 3, all coins lighter than k must be on the same cup. Further, if a coin with weight k+4 exists, then by the symmetric version of Lemma 3, all coins heavier than k must also be on the same cup. They must be on the other cup from the coins lighter than k because otherwise the scale wouldn’t balance, and the theorem is true.
If no coin with weight k+4 exists, that is, if n ≤ k+3, how can the theorem be false? All the coins lighter than k must be on one cup, and their total weight is k(k-1)/2. Further, in order to falsify the theorem, at least one of the coins heavier than k must also be on that same cup. So the minimum weight of that cup is now k(k-1)/2 + k+1. But we only have at most two coins for the other cup, whose total weight is at most k+2 + k+3 = 2k + 5. For the scale to even have a chance of balancing, we must have
k(k-1)/2 + k+1 ≤ 2k + 5 ⇔ k(k-1)/2 ≤ k + 4 ⇔ k(k-1) ≤ 2k + 8 ⇔ k2 – 3k – 8 ≤ 0.
Finding the largest root of that quadratic we see that k < 5.
So for k ≥ 5, the collection of all coins lighter than k is heavy enough that either one needs all the coins heavier than k to balance them, or there are enough coins heavier than k that the theorem is true by symmetric application of Lemma 3.
Completion of Case 5. It remains to check the case for k < 5. If n > k+3, then coin k+4 exists. If so, all the coins heavier than k must be on the same cup. Furthermore, since k is so small, they will together weigh more than half the available weight, so the scale will be unbalanceable. So k < 5 and n ≤ k+3 ≤ 7.
For lack of any better creativity, we will tackle the remaining portion of the problem by complete enumeration of the possible cases, except for the one observation that, to balance the scale with just the coin k off it, the total weight of the remaining coins, that is, n(n+1)/2 – k must be even. This observation cuts our remaining work in half. Now to it.
Case 5. Seven Coins. n = 7. Then 5 > k ≥ n – 3 = 4, so k = 4. Then the weight on each cup must be 12. One of the cups must contain the 7 coin, and no cup can contain the 4 coin, so the only two weighings the Baron could try are 7 + 5 = 1 + 2 + 3 + 6, and 7 + 3 + 2 = 1 + 5 + 6. But the first of those is unconvincing because k+1 = 5 is not on the same cup as k+2 = 6, and the second because it has the same shape as 7 + 3 + 1 = 2 + 4 + 5 (leaving out the 6-gram coin instead of the asserted 4-gram coin).
Case 5. Six Coins. n = 6. Then 5 > k ≥ n – 3 = 3, and n(n+1)/2 = 21 is odd, so k must also be odd. Therefore k=3, and the weight on each cup must be 9. The 6-gram coin has to be on a cup and the 3-gram coin is by presumption out, so the Baron’s only chance is the weighing 6 + 2 + 1 = 4 + 5, but that doesn’t convince his skeptical guests because it looks too much like the weighing 1 + 3 + 4 = 6 + 2.
Case 5. Five Coins. n = 5. Then 5 > k ≥ n – 3 = 2, and n(n+1)/2 = 15 is odd, so k must also be odd. Therefore k=3, and the weight on each cup must be 6. The only way to do that is the weighing 5 + 1 = 2 + 4, which does not convince the Baron’s guests because it looks too much like 1 + 4 = 2 + 3.
Case 5. Four Coins. n = 4. Then the only way to balance a scale using all but one coin is to put two coins on one cup and one on the other. The only two such weighings that balance are 1 + 2 = 3 and 1 + 3 = 4, but they leave different coins off the scale.
The remaining cases, n < 4, are even easier. That concludes the proof of Case 5.
Consequently, by the argument similar to the one in case 4 we can show that the number of coins in case 5 must be the index of a square triangular number.
This concludes the proof of the theorem.
Now we can describe all possible numbers of coins that allow the Baron to confirm a coin in one weighing, or, in other words, the indices of ones in the sequence a(n). The following list corresponds to the five cases above:
If we have four coins, then the same weighing 1+2 < 4 identifies two coins: the coin that weighs three grams and is not in a cup and the coin weighing four grams that is in a cup. The other case like this is for two coins. Comparing them to each other we can identify each of them. It is clear that there are no other cases like this. Indeed, for the same weighing to identify two different coins, it must be the n-gram on the cup, and the n-1 coin off the scale. From here we can see that n can’t be big.
As usual we want to give something to think about to our readers. We have given you the list of sequences describing all the numbers for which the Baron can prove the weight of one coin in one weighing. Does there exist a number greater than four that belongs to two of these sequences? In other words, does there exist a total number of coins such that the Baron can have two different one-weighing proofs for two different coins?
To conclude this essay we would like to note that the puzzle we are discussing is related to the puzzle in one of Tanya’s previous posts:
You have 6 coins weighing 1, 2, 3, 4, 5 and 6 grams that look the same, except for their labels. The number (1, 2, 3, 4, 5, 6) on the top of each coin should correspond to its weight. How can you determine whether all the numbers are correct, using the balance scale only twice?
The latter puzzle appeared at the last round of Moscow math Olympiad in 1991. The author of this problem was Sergey Tokarev.
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I would like to tell you a story from my childhood and how I started on my math path.
When I was in elementary and middle school, I was very good with mathematics. Actually, I was by far the best math student in my class and my math teacher didn’t know what to do with me. Our algebra book had 2,000 problems and was intended to cover three years of study. But I worked out those problems, one after another, whenever I had a free minute in my math class. As I result, I got way ahead of everyone else.
One day a new boy named Lenya Kostyukov joined our class. He had extraordinarily long eye-lashes that covered his eyes, and all the girls envied him. He was a nice smart kid, but other than his lashes, I didn’t notice him very much. After a year or two, he announced that he was leaving our school, because he had been accepted to a math school for gifted children.
“Why is he going to a math school? I am the math star here. Why aren’t I going to a math school?” I knew about math schools, and I knew that I was good at math; I just never made the connection. I never felt that I was supposed to apply. Despite enjoying my reputation, I just passively went with the flow. Lenya figuratively kicked me in the butt. If he can, why can’t I?
So I applied to the same school on the last permissible day and was accepted. It turned out that I accidentally went to the room where they were giving the test for a grade higher than mine. I passed it with flying colors. My parents, though, were scared of a long commute and didn’t really want me to go so far away. They found a different math school closer to home, and used my extraordinary results to convince that school to accept me, even though their application date had passed.
For many years I continued to be a very passive person. Applying to a math school was the single big step I took for myself, but it was a defining step. I am grateful to Lenya for that. Or more likely to his parents, who were actively looking around trying to find the best place for their gifted son, and as a byproduct found a place for me. Once I was on the path of mathematics, I had the guidance of teachers and supervisors, for better or for worse, which allowed me to continue to be passive.
I have described my defining moment to you, but I don’t want to leave you in the dark about Lenya’s fate. Here’s what happened to him.
As I mentioned, Leonid (Lenya’s formal name) and I ended up in different math schools, so I lost track of him. Four years later I went to study math at Moscow State University and stumbled upon a guy with very long eye-lashes. We recognized each other immediately and eventually became friends.
He was doing logic and was very good at it. He was recommended for graduate school. But by that time our MSU administration noticed his lashes too. The lashes were obviously very suspicious; they hinted at the existence of non-Russian blood in his veins. As it was the period of brutal anti-semitism at MSU, they didn’t allow him to go to graduate school.
Leonid Kostyukov dropped mathematics and became a famous writer.
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Here is a strange puzzle that was inspired by the palindrome problem. Suppose you have a sequence of words in some alphabet with the initial term a and all the other terms b: a, b, b, b, b, etc. Suppose this sequence generates palindromes every time you concatenate the first several terms, not counting the first term itself. So, ab, abb, abbb, and so on — are all palindromes. We call words b “papaya” words, when a exists, such that a and b generate the sequence with this palindrome property. Can you describe papayas?
Theorem. The word b is a papaya word if and only if b is a substring of Reverse(b)Reverse(b).
Proof. After we have added b so many times that the initial part a is much smaller than half of the concatenated string, the middle part of the concatenated string would consist of several words copies of the word b. The middle of the reverse string consists of several concatenations of Reverse(b). So the word b must be a substring of Reverse(b)Reverse(b). On the other hand, suppose b is a substring of Reverse(b)Reverse(b). Then Reverse(b)Reverse(b) is of the form xby, and we can choose a = y.
Theorem. Papaya words are either palindromes or concatenations of two palindromes.
Proof. Suppose our word consists of two palindromes cd. Then the reverse of it is dc and its double is dcdc. The word cd is a substring of dcdc, thus according to the first theorem, cd is a papaya word. Let’s do the other direction. Suppose the word b is a substring of Reverse(b)Reverse(b). Then Reverse(b)Reverse(b) is of the form xby. Then b = yx, and Reverse(b) = xy, which equals Reverse(x)Reverse(y). From here, Reverse(x) = x and Reverse(y) = y. If x or y has zero length, then our word is a palindrome. QED.
Hey, do you already know why we call these words papayas?
Just for fun we would like to study the structure of papaya words. Any one-character or two-character word is a papaya word, so the patterns are: a, aa, ab. For three-character words there are four patterns: aaa, aab, aba, abb. For four-character words there are 10 patterns: aaaa, aaab, aaba, abaa, aabb, abab, abba, abbb, abac, abcb. In this manner we get the sequence of the number of n-character papaya patterns: 1, 2, 4, 10, 21, 50 etc, which is sequence A165137 in the OEIS. This sequence depends on the number of letters in your alphabet. But the first n terms of these sequences are the same for all alphabets that have at least n letters.
Let us assume that we are working with an infinite alphabet. The complementary sequence would be the number of patterns for non-papaya words. The total number of patterns is described by sequence A000110 — Bell numbers: the number of word structures of length n using an infinite alphabet. So the beginning of this complementary sequence A165610 is: 0, 0, 1, 5, 31, 153, etc. The list of corresponding patterns is abc, aabc, abbc, abcc, abca, abcd, etc.
Historically, we first invented the corresponding sequence for numbers, not for words. We call a number a papaya number if it is a palindrome or a concatenation of two palindromes. If we use numbers instead of words in the problem, we need to carefully look at what happens if we encounter initial zeroes. Let’s take the papaya number 2200100, and see if we can find a number a, such that adding 22010 repeatedly to this sequence starting with a will always generate a palindrome. The number a must be 00100. But this is not a number. We have two choices: to say that we are working with strings of digits, or to allow several numbers to start the sequence before we add b repetitively and before getting to palindromes. In the latter case our sequence can start 0, 0, 100, 22010, 22010, and so on.
As we mentioned before, the number of patterns of papaya numbers will start the same as the number of patterns of papaya words. Later the sequence of patterns of numbers A165136 will be smaller than the corresponding sequence for words. As the sequence of Bell numbers is much more famous than the sequence A164864 of patterns of numbers, we expect the papaya patterns sequence corresponding to the infinite alphabet to be more interesting and important than the sequence of papaya patterns for numbers.
Though papaya numbers might be less important than papaya words with an infinite alphabet, they have an advantage in that we can generate more sequences with them. For example we can calculate the number of positive papaya number with n digits, as in the sequence A165135: 9, 90, 252, 1872, 4464, 29250, etc. And we can also calculate the sequence A165611 of n-digit non-papaya numbers: 0, 0, 648, 7128, 85536 etc.
Share: This geometric problem was given to me by Arkady Berenstein:
There are n points on the plane, but not all of them are on one line. Prove that a line exists that passes through exactly two points of this set.
Arkady gave me a beautiful solution to this problem. First, draw a line through each pair of points. Suppose you calculate all the distances from each point to all the lines that the point doesn’t belong to. Pick the smallest distance. The corresponding line will be the one with two points. To finish the solution you need to fill in the details. That process is usually left to the reader.
I suspect that there might also be a solution using linear algebra. Can you find one?
I would like to reformulate this problem without using geometry. Suppose there is a set of n elements. Let’s call a family of subsets line-like if any two distinct subsets of this family can have as an intersection not more than one element. Then the geometry problem above has a set-theoretical analogue:
You have a set of n elements and a line-like family of subsets of this set such that any two elements of the set belong to a subset from this family, and that the family doesn’t contain the whole set. Is it true that there always exists a subset in this family consisting of two elements?
Usually I give such problems as homework for the reader, but this time I decided to change my habit, so I’m including the picture which contains the solution of this problem by my son Alexey Radul.
Conclusion: geometry is important.
I remember this question from my childhood:
Why is the South Pole colder than the North Pole?
Indeed, the average winter temperature at the North Pole of -34°C is the same as the temperature at the South Pole at the beginning and end of its summer. The South Pole is only warmer than the North Pole 40 days per year. So the South Pole is a much, much colder place. According to Wikipedia there are three major reasons for this:
I remember when I was a child my father gave me a completely different explanation.
The Earth’s orbit is not a circle, but rather an ellipse. According to Kepler’s second law: “A line joining a planet and the sun sweeps out equal areas during equal intervals of time.” This means that the earth has a slower angle motion around the aphelion — in its furthest point — than around the perihelion — in its closest point to the Sun. Consequently, the summer is longer than the winter for the North Pole, whereas the opposite is true for the South Pole.
Something in my father’s explanation bothered me. Now I understand what: though the summer is longer at the North Pole, it should get less sunshine as the North Pole is further away from the Sun than the South Pole during its summer. So the effects might cancel each other out. In any case, as the earth’s orbit is almost circular, the contribution of the shape of the orbit should be minor, compared to the effects of elevation, the water underneath and continentality.
On the other hand, it is possible that my father wasn’t talking about the poles, but rather about the difference in hemispheres. I wonder if someone can calculate if there is a difference in the amount of sunlight the poles get due to the fact that the Earth’s orbit is not circular. Is the temperature different for the places that are equidistant from the equator, and have similar elevation and continentality, but which are located in different hemispheres?
I remember a funny article explaining why the northern hemisphere has more land. They said that continents drifted into the northern hemisphere because they wanted a nicer climate.
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I bought the book “Heard on The Street: Quantitative Questions from Wall Street Job Interviews” by Timothy Falcon Crack several years ago when I was looking for a job and felt that working in finance was a possibility. Despite having bought it simply to prepare for employment interviews, I actually enjoyed the math problems in the book.
The book has problems in logic, probability, statistics and finance, as well as a very useful chapter of general interview questions. If you’re interested in buying this book, I should mention that some questions require calculus and knowledge of financial terms.
I love the author’s taste in problems, and here are some sample questions from the book.
Question 2.7: How many degrees (if any) are there in the angle between the hour and minute hands of a clock when the time is a quarter past three?
Question 5.1.14: Welcome to your interview. Sit in this chair. Excuse me while I tie your arms and legs to the chair. Thank you. Now we are going to play “Russian roulette.” I have a revolver with six empty chambers. Watch me as I load the weapon with two contiguous rounds (i.e., two bullets side-by-side in the cylindrical barrel). Watch me as I spin the barrel. I am putting the gun against your head. Close your eyes while I pull the trigger. This is your lucky day: you are still alive! Our game differs from regular Russian roulette because I am not going to add any bullets to the barrel before we continue, and I am not going to give you the gun.
My question for you: I am going to shoot at you once more before we talk about your résumé. Do you want me to spin the barrel once more, or should I just shoot?
Question 6.1.16: Tell me something you tried but ended up quitting on.
I can tell you what I would have answered to the last question: I tried smoking, but ended up quitting.
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Mathematically we can describe a marriage by a graph. People are vertices and two spouses are connected by an edge.
Mathematical models tend to oversimplify life, so let us assume that a person can only be one of two genders. Therefore, the vertices of a graph are colored in two colors: pink and blue. In this article I explore the graph theory of different types of marriages.
A monogamous couple is represented as a complete K2 graph: two vertices connected by an edge. The graph is bipartite, no matter how you color it. But actually our vertices are already colored from the start. If we are considering traditional marriage, one vertex is pink and the other is blue.
Historically, the second most common type of marriage is polygyny, in which one man has several wives. Less common in history, but a mathematically equivalent type, is polyandry, in which one woman has several husbands. Both these types of marriages emphasize inequality, as husbands and wives have completely different sets of rights.
From a mathematical point of view, polygyny and polyandry are described by star graphs. Star graphs are bipartite graphs and the natural coloring is the one that proves bipartiteness.
The final type of marriage is polygynandry, which refers to a group marriage, where more than one man and more than one woman create a family. Everyone can have sex with everyone else of the other gender. Mathematically this type of marriage corresponds to a complete bipartite graph Kn,m. Actually, in this case I can imagine that a particular pair of people of different genders wouldn’t like each other and might not consummate their marriage. So this graph is not necessarily complete.
How can same-sex marriages change the graph theory of marriages? As a graph, a monogamous same-sex marriage is the same bipartite K2 graph as a heterosexual marriage. It will just be less colorful, as both vertices will be of the same color.
But what happens if we add the same-sex idea to polygamous marriages?
Suppose a homosexual man wants to live with several spouses at the same time. What name can we give to a family unit of more than two homosexual men? Homopolygamy? Their marriage graph will be a star graph in which all the vertices are of the same color.
If a man can have several spouses, what about his spouses? Can they form multiple marriages too? If only one person is allowed to engage in several marriages, then we will see inequality within the same gender. If any spouse is allowed to form other marriages, then we will have a situation in which several men are all spouses to each other. So mathematically we will see complete graphs with more than two vertices to represent a marriage. If two people in a group do not like each other and do not want to be married, then the corresponding graph doesn’t need to be complete.
By symmetry we can describe a marriage of several women, and mathematically it will be similar to a marriage of several men.
Another interesting aspect is the idea of mixed types of marriages involved in polygamy. Suppose a husband has several wives. Some of them might get bored waiting for his attention, and start spending so much time with each other that they end up developing feelings for each other. Suppose two wives of the same man decide to marry each other. What name would we give to this type of marriage? I am afraid that we do not have enough words to cover all the potential situations.
Suppose we have a heterosexual married couple and the man decides to bring another woman into their house. Thus the transition from a traditional marriage to a polygyny is created. If they got along so well that the first wife decides to marry the second wife, this would require a transition to a new type of marriage. Oh, I see that my essay just went in another direction — how different types of marriages might evolve into each other. For now, I’ll leave this for future research.
Talking about different directions. I recently wrote a piece about condoms. Now I have a new generalization for the classic condom puzzle. Suppose we have a mixed-type marriage defined by a graph. Suppose tonight every couple of people corresponding to the edge of this graph wants to have sex with each other. What is the smallest number of condoms they can use? In my condom essay, I didn’t define the condom usage for the sex of two women. I will leave it to your imagination and definition.
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