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The Universe of Discourse The Universe of Discourse (Mark Dominus Blog) tag:blog.plover.com,2005:/math Blosxom http://perl.plover.com/favicon.ico 2017-02-07T08:31:00Z tag:blog.plover.com,2017:/math/increasing-sequences How many 24 puzzles are there? 2017-02-07T08:31:00Z 2017-02-07T08:31:00Z Mark Dominus http://www.plover.com/ email@example.com
[ Note: The tables in this article are important, and look unusually crappy if you read this blog through an aggregator. The properly-formatted version on my blog may be easier to follow. ]

A few months ago I wrote about puzzles of the following type: take four digits, say 1, 2, 7, 7, and, using only +, -, ×, and ÷, combine them to make the number 24. Since then I have been accumulating more and more material about these puzzles, which will eventually appear here. But meantime here is a delightful tangent.

In the course of investigating this I wrote programs to enumerate the solutions of all possible puzzles, and these programs were always much faster than I expected at first. It appears as if there are 10,000 possible puzzles, from «0,0,0,0» through «9,9,9,9». But a moment's thought shows that there are considerably fewer, because, for example, the puzzles «7,2,7,1», «1,2,7,7», «7,7,2,1», and «2,7,7,1» are all the same puzzle. How many puzzles are there really?

A back-of-the-envelope estimate is that only about 1 in 24 puzzles is really distinct (because there are typically 24 ways to rearrange the elements of a puzzle) and so there ought to be around  puzzles. This is an undercount, because there are fewer duplicates of many puzzles; for example there are not 24 variations of «1,2,7,7», but only 12. The actual number of puzzles turns out to be 715, which I think is not an obvious thing to guess.

Let's write  for the set of sequences of length  containing up to  different symbols, with the duplicates removed: when two sequences are the same except for the order of their symbols, we will consider them the same sequence.

Or more concretely, we may imagine that the symbols are sorted into nondecreasing order, so that  is the set of nondecreasing sequences of length  of  different symbols.

Let's also write  for the number of elements of .

Then  is the set of puzzles where input is four digits. The claim that there are  such puzzles is just that . A tabulation of  reveals that it is closely related to binomial coefficients, and indeed that $$C(d,n)=\binom{n+d-1}{d-1}.\tag{$\heartsuit$}$$

so that the surprising  is actually . This is not hard to prove by induction, because  is easily shown to obey the same recurrence as : $$C(d,n) = C(d-1,n) + C(d,n-1).\tag{$\spadesuit$}$$

To see this, observe that an element of  either begins with a zero or with some other symbol. If it begins with a zero, there are  ways to choose the remaining  symbols in the sequence. But if it begins with one of the other  symbols it cannot contain any zeroes, and what we really have is a length- sequence of the symbols , of which there are .

 <colgroup><col style="background: lightblue"><col style="background: pink""></colgroup>
0 0 0 0 1 1 1
0 0 0 1 1 1 2
0 0 0 2 1 1 3
0 0 0 3 1 1 4
0 0 1 1 1 2 2
0 0 1 2 1 2 3
0 0 1 3 1 2 4
0 0 2 2 1 3 3
0 0 2 3 1 3 4
0 0 3 3 1 4 4
0 1 1 1 2 2 2
0 1 1 2 2 2 3
0 1 1 3 2 2 4
0 1 2 2 2 3 3
0 1 2 3 2 3 4
0 1 3 3 2 4 4
0 2 2 2 3 3 3
0 2 2 3 3 3 4
0 2 3 3 3 4 4
0 3 3 3 4 4 4

Now we can observe that  (they are both 35) so that . We might ask if there is a combinatorial proof of this fact, consisting of a natural bijection between  and . Using the relation  we have:

$$ \begin{eqnarray} C(4,4) & = & C(3, 4) + & C(4,3) \\ C(5,3) & = & & C(4,3) + C(5,2) \\ \end{eqnarray}$$

so part of the bijection, at least, is clear: There are  elements of  that begin with a zero, and also  elements of  that do not begin with a zero, so whatever the bijection is, it ought to match up these two subsets of size 20. This is perfectly straightforward; simply match up  (blue) with  (pink), as shown at right.

But finding the other half of the bijection, between  and , is not so straightforward. (Both have 15 elements, but we are looking for not just any bijection but for one that respects the structure of the elements.) We could apply the recurrence again, to obtain:

$$ \begin{eqnarray} C(3,4) & = \color{darkred}{C(2, 4)} + \color{darkblue}{C(3,3)} \\ C(5,2) & = \color{darkblue}{C(4,2)} + \color{darkred}{C(5,1)} \end{eqnarray}$$

and since $$ \begin{eqnarray} \color{darkred}{C(2, 4)} & = \color{darkred}{C(5,1)} \\ \color{darkblue}{C(3,3)} & = \color{darkblue}{C(4,2)} \end{eqnarray}$$

we might expect the bijection to continue in that way, mapping  and . Indeed there is such a bijection, and it is very nice.

To find the bijection we will take a detour through bitstrings. There is a natural bijection between  and the bit strings that contain  zeroes and  ones. Rather than explain it with pseudocode, I will give some examples, which I think will make the point clear. Consider the sequence . Suppose you are trying to communicate this sequence to a computer. It will ask you the following questions, and you should give the corresponding answers:

"Is the first symbol 0?" ("No")

"Is the first symbol 1?" ("Yes")

"Is the second symbol 1?" ("Yes")

"Is the third symbol 1?" ("No")

"Is the third symbol 2?" ("No")

"Is the third symbol 3?" ("Yes")

"Is the fourth symbol 3?" ("No")

"Is the fourth symbol 4?" ("Yes")

At each stage the computer asks about the identity of the next symbol. If the answer is "yes" the computer has learned another symbol and moves on to the next element of the sequence. If it is "no" the computer tries guessing a different symbol. The "yes" answers become ones and "no" answers become zeroes, so that the resulting bit string is 0 1 1 0 0 1 0 1.

It sometimes happens that the computer figures out all the elements of the sequence before using up its  questions; in this case we pad out the bit string with zeroes, or we can imagine that the computer asks some pointless questions to which the answer is "no". For example, suppose the sequence is :

"Is the first symbol 0?" ("Yes")

"Is the second symbol 0?" ("No")

"Is the second symbol 1?" ("Yes")

"Is the third symbol 1?" ("Yes")

"Is the fourth symbol 1?" ("Yes")

The bit string is 1 0 1 1 1 0 0 0, where the final three 0 bits are the padding.

We can reverse the process, simply taking over the role of the computer. To find the sequence that corresponds to the bit string 0 1 1 0 1 0 0 1, we ask the questions ourselves and use the bits as the answers:

"Is the first symbol 0?" ("No")

"Is the first symbol 1?" ("Yes")

"Is the second symbol 1?" ("Yes")

"Is the third symbol 1?" ("No")

"Is the third symbol 2?" ("Yes")

"Is the fourth symbol 2?" ("No")

"Is the fourth symbol 3?" ("No")

"Is the fourth symbol 4?" ("Yes")

We have recovered the sequence  from the bit string 0 1 1 0 1 0 0 1.

This correspondence establishes relation  in a different way from before: since there is a natural bijection between  and the bit strings with  zeroes and  ones, there are certainly  of them as  says because there are  bits and we may choose any  to be the zeroes.

We wanted to see why . The detour above shows that there is a simple bijection between

 and the bit strings with 4 zeroes and 3 ones

on one hand, and between

 and the bit strings with 3 zeroes and 4 ones

on the other hand. And of course the bijection between the two sets of bit strings is completely obvious: just exchange the zeroes and the ones.

The table below shows the complete bijection between  and its descriptive bit strings (on the left in blue) and between  and its descriptive bit strings (on the right in pink) and that the two sets of bit strings are complementary. Furthermore the top portion of the table shows that the  subsets of the two families correspond, as they should—although the correct correspondence is the reverse of the one that was displayed earlier in the article, not the suggested  at all. Instead, in the correct table, the initial digit of the  entry says how many zeroes appear in the  entry, and vice versa; then the increment to the next digit says how many ones, and so forth.

 <colgroup> <col style="background: lightblue; border-right"> <col style="background: rgb(193,236,250);"> </colgroup> <colgroup> <col style="background: rgb(255,212,223);"> <col style="background: pink"> </colgroup>

(bits)(complement bits)
 0 0 0 0  1 1 1 1 0 0 0  0 0 0 0 1 1 1  4 4 4
 0 0 0 1  1 1 1 0 1 0 0  0 0 0 1 0 1 1  3 4 4
 0 0 0 2  1 1 1 0 0 1 0  0 0 0 1 1 0 1  3 3 4
 0 0 0 3  1 1 1 0 0 0 1  0 0 0 1 1 1 0  3 3 3
 0 0 1 1  1 1 0 1 1 0 0  0 0 1 0 0 1 1  2 4 4
 0 0 1 2  1 1 0 1 0 1 0  0 0 1 0 1 0 1  2 3 4
 0 0 1 3  1 1 0 1 0 0 1  0 0 1 0 1 1 0  2 3 3
 0 0 2 2  1 1 0 0 1 1 0  0 0 1 1 0 0 1  2 2 4
 0 0 2 3  1 1 0 0 1 0 1  0 0 1 1 0 1 0  2 2 3
 0 0 3 3  1 1 0 0 0 1 1  0 0 1 1 1 0 0  2 2 2
 0 1 1 1  1 0 1 1 1 0 0  0 1 0 0 0 1 1  1 4 4
 0 1 1 2  1 0 1 1 0 1 0  0 1 0 0 1 0 1  1 3 4
 0 1 1 3  1 0 1 1 0 0 1  0 1 0 0 1 1 0  1 3 3
 0 1 2 2  1 0 1 0 1 1 0  0 1 0 1 0 0 1  1 2 4
 0 1 2 3  1 0 1 0 1 0 1  0 1 0 1 0 1 0  1 2 3
 0 1 3 3  1 0 1 0 0 1 1  0 1 0 1 1 0 0  1 2 2
 0 2 2 2  1 0 0 1 1 1 0  0 1 1 0 0 0 1  1 1 4
 0 2 2 3  1 0 0 1 1 0 1  0 1 1 0 0 1 0  1 1 3
 0 2 3 3  1 0 0 1 0 1 1  0 1 1 0 1 0 0  1 1 2
 0 3 3 3  1 0 0 0 1 1 1  0 1 1 1 0 0 0  1 1 1
 1 1 1 1  0 1 1 1 1 0 0  1 0 0 0 0 1 1  0 4 4
 1 1 1 2  0 1 1 1 0 1 0  1 0 0 0 1 0 1  0 3 4
 1 1 1 3  0 1 1 1 0 0 1  1 0 0 0 1 1 0  0 3 3
 1 1 2 2  0 1 1 0 1 1 0  1 0 0 1 0 0 1  0 2 4
 1 1 2 3  0 1 1 0 1 0 1  1 0 0 1 0 1 0  0 2 3
 1 1 3 3  0 1 1 0 0 1 1  1 0 0 1 1 0 0  0 2 2
 1 2 2 2  0 1 0 1 1 1 0  1 0 1 0 0 0 1  0 1 4
 1 2 2 3  0 1 0 1 1 0 1  1 0 1 0 0 1 0  0 1 3
 1 2 3 3  0 1 0 1 0 1 1  1 0 1 0 1 0 0  0 1 2
 1 3 3 3  0 1 0 0 1 1 1  1 0 1 1 0 0 0  0 1 1
 2 2 2 2  0 0 1 1 1 1 0  1 1 0 0 0 0 1  0 0 4
 2 2 2 3  0 0 1 1 1 0 1  1 1 0 0 0 1 0  0 0 3
 2 2 3 3  0 0 1 1 0 1 1  1 1 0 0 1 0 0  0 0 2
 2 3 3 3  0 0 1 0 1 1 1  1 1 0 1 0 0 0  0 0 1
 3 3 3 3  0 0 0 1 1 1 1  1 1 1 0 0 0 0  0 0 0

Observe that since  we have in general that , which may be surprising. One might have guessed that since , the relation was  and that  would have the same structure as , but it isn't so. The two arguments exchange roles. Following the same path, we can identify many similar 'coincidences'. For example, there is a simple bijection between the original set of 715 puzzles, which was , and , the set of nondecreasing sequences of  of length .

[ Thanks to Bence Kodaj for a correction. ]
 tag:blog.plover.com,2016:/math/magic-square-puzzle Let's decipher a thousand-year-old magic square 2016-12-15T09:36:00Z 2016-12-15T09:36:00Z Mark Dominus http://www.plover.com/ firstname.lastname@example.com
The Parshvanatha temple in Madhya Pradesh, India was built around 1,050 years ago. Carved at its entrance is this magic square:

The digit signs have changed in the past thousand years, but it's a quick and fun puzzle to figure out what they mean using only the information that this is, in fact, a magic square.

A solution follows. No peeking until you've tried it yourself!

There are 9 one-digit entries
 and 7 two-digit entries
 so we can guess that the entries are the numbers 1 through 16, as is usual, and the magic sum is 34. The  appears in the same position in all the two-digit numbers, so it's the digit 1. The other digit of the numeral  is , and this must be zero. If it were otherwise, it would appear on its own, as does for example the  from  or the  from .

It is tempting to imagine that  is 4. But we can see it's not so. Adding up the rightmost column, we get

  +  +  +  =
  + 11 +  +  =
 (10 + ) + 11 +  +  = 34,

so that  must be an odd number. We know it isn't 1 (because  is 1), and it can't be 7 or 9 because  appears in the bottom row and there is no 17 or 19. So  must be 3 or 5.

Now if  were 3, then  would be 13, and the third column would be

  +  +  +  =
 1 +  + 10 + 13 = 34,

and then  would be 10, which is too big. So  must be 5, and this means that  is 4 and  is 8. ( appears only a as a single-digit numeral, which is consistent with it being 8.)

The top row has

  +  +  +  =
  +  + 1 + 14 =
  + (10 + ) + 1 + 14 = 34

so that  +  = 9.  only appears as a single digit and we already used 8 so  must be 7 or 9. But 9 is too big, so it must be 7, and then  is 2.

 is the only remaining unknown single-digit numeral, and we already know 7 and 8, so  is 9. The leftmost column tells us that  is 16, and the last two entries,  and  are easily discovered to be 13 and 3. The decoded square is:

712114
213811
163105
96154

I like that people look at the right-hand column and immediately see 18 + 11 + 4 + 8 but it's actually 14 + 11 + 5 + 4.

This is an extra-special magic square: not only do the ten rows, columns, and diagonals all add up to 34, so do all the four-cell subsquares, so do any four squares arranged symmetrically about the center, and so do all the broken diagonals that you get by wrapping around at the edges.

[ Addendum: It has come to my attention that the digit symbols in the magic square are not too different from the current forms of the digit symbols in the Gujarati script. ]

[ Addendum 20161217: The temple is not very close to Gujarat or to the area in which Gujarati is common, so I guess that the digit symbols in Indian languages have evolved in the past thousand years, with the Gujarati versions remaining closest to the ancient forms, or else perhaps Gujarati was spoken more widely a thousand years ago. I would be interested to hear about this from someone who knows. ]

[ Addendum 20170130: Shreevatsa R. has contributed a detailed discussion of the history of the digit symbols. ]
 tag:blog.plover.com,2016:/math/even-odd Decomposing a function into its even and odd parts 2016-07-30T02:41:00Z 2016-07-30T02:41:00Z Mark Dominus http://www.plover.com/ email@example.com
As I have mentioned before, I am not a sudden-flash-of-insight person. Every once in a while it happens, but usually my thinking style is to minutely examine a large mass of examples and then gradually synthesize some conclusion about them. I am a penetrating but slow thinker. But there have been a few occasions in my life when the solution to a problem struck me suddenly out of the blue.

One such occasion was on the first day of my sophomore honors physics class in 1987. This was one of the best classes I took in my college career. It was given by Professor Stephen Nettel, and it was about resonance phenomena. I love when a course has a single overarching theme and proceeds to examine it in detail; that is all too rare. I deeply regret leaving my copy of the course notes in a restaurant in 1995.

The course was very difficult, But also very satisfying. It was also somewhat hair-raising, because of Professor Nettel's habit of saying, all through the second half "Don't worry if it doesn't seem to make any sense, it will all come together for you during the final exam." This was not reassuring. But he was right! It did all come together during the final exam.

The exam had two sets of problems. The problems on the left side of the exam paper concerned some mechanical system, I think a rod fixed at one end and free at the other, or something like that. This set of problems asked us to calculate the resonant frequency of the rod, its rate of damping at various driving frequencies, and related matters. The right-hand problems were about an electrical system involving a resistor, capacitor, and inductor. The questions were the same, and the answers were formally identical, differing only in the details: on the left, the answers involved length, mass and stiffness of the rod, and on the right, the resistance, capacitance, and inductance of the electrical components. It was a brilliant exam, and I have never learned so much about a subject during the final exam.

Anyway, I digress. After the first class, we were assigned homework. One of the problems was

Show that every function is the sum of an even function and an odd function.

(Maybe I should explain that an even function is one which is symmetric across the -axis; formally it is a function  for which  for every . For example, the function , shown below left. An odd function is one which is symmetric under a half-turn about the origin; formally it satisfies  for all . For example , shown below right.)

I found this claim very surprising, and we had no idea how to solve it. Well, not quite no idea: I knew that functions could be expanded in Fourier series, as the sum of a sine series and a cosine series, and the sine part was odd while the cosine part was even. But this seemed like a bigger hammer than was required, particularly since new sophomores were not expected to know about Fourier series.

I had the privilege to be in that class with Ron Buckmire, and I remember we stood outside the class building in the autumn sunshine and discussed the problem. I might have been thinking that perhaps there was some way to replace the negative part of  with a reflected copy of the positive part to make an even function, and maybe that  was always even, when I was hit from the blue with the solution:

$$ \begin{align} f_e(x) & = \frac{f(x) + f(-x)}2 \text{ is even},\\ f_o(x) & = \frac{f(x) - f(-x)}2 \text{ is odd, and}\\ f(x) &= f_e(x) + f_o(x) \end{align} $$

So that was that problem solved. I don't remember the other three problems in that day's homework, but I have remembered that one ever since.

But for some reason, it didn't occur to me until today to think about what those functions actually looked like. Of course, if  itself is even, then  and , and similarly if  is odd. But most functions are neither even nor odd.

For example, consider the function , which is neither even nor odd. Then we get

$$ \begin{align} f_e(x) & = \frac{2^x + 2^{-x}}2\\ f_o(x) & = \frac{2^x - 2^{-x}}2 \end{align} $$

The graph is below left. The solid red line is , and the blue and purple dotted lines are  and . The red line is the sum of the blue and purple lines. I thought this was very interesting-looking, but a little later I realized that I had already known what these graphs would look like, because  is just like , and for  the even and odd components are exactly the familiar  and  functions. (Below left, ; below right, .)

I wasn't expecting polynomials to be more interesting, but they were. (Polynomials whose terms are all odd powers of , such as , are always odd functions, and similarly polynomials whose terms are all even powers of  are even functions.) For example, consider , which is neither even nor odd. We don't even need the  and  formulas to separate this into even and odd parts: just expand  as  and separate it into odd and even powers,  and :

Or we could do  similarly, expanding it as  and separating this into  and :

I love looking at these and seeing how the even blue line and the odd purple line conspire together to make whatever red line I want.

I kept wanting to try familiar simple functions, like , but many of these are either even or odd, and so are uninteresting for this application. But you can make an even or an odd function into a neither-even-nor-odd function just by translating it horizontally, which you do by replacing  with . So the next function I tried was , which is the translation of . Here I got a surprise. I knew that  was undefined at , so I graphed it only for . But the even component is , which is undefined at both  and at . Similarly the odd component is undefined at two points. So the  formula does not work quite correctly, failing to produce the correct value at , even though  is defined there. In general, if  is undefined at some , then the decomposition into even and odd components fails at  as well. The limit $$\lim_{x\to -c} f(x) = \lim_{x\to -c} \left(f_o(x) + f_e(x)\right)$$ does hold, however. The graph below shows the decomposition of .

Vertical translations are uninteresting: they leave  unchanged and translate  by the same amount, as you can verify algebraically or just by thinking about it.

Following the same strategy I tried a cosine wave. The evenness of the cosine function is one of its principal properties, so I translated it and used . The graph below is actually for  to prevent the details from being too compressed:

This reminded me of the time I was fourteen and graphed  and was surprised to see that it was another perfect sinusoid. But I realized that there was a simple way to understand this. I already knew that . If you take  and multiply the whole thing by , you get $$\sqrt2\cos\left(x + \frac\pi4\right) = \sqrt2\sin x\cos\frac\pi4 + \sqrt2\cos x\sin\frac\pi4 = \sin x + \cos x$$ so that  is just a shifted, scaled cosine curve. The decomposition of  is even simpler because you can work forward instead of backward and find that , and the first term is odd while the second term is even, so that  decomposes as a sum of an even and an odd sinusoid as you see in the graph above.

Finally, I tried a Poisson distribution, which is highly asymmetric. The formula for the Poisson distribution is , for some constant . The  in the denominator is only defined for non-negative integer , but you can extend it to fractional and negative  in the usual way by using  instead, where  is the Gamma function. The  function is undefined at zero and negative integers, but fortunately what we need here is the reciprocal gamma function , which is perfectly well-behaved. The results are spectacular. The graph below has .

The part of this with  is the most interesting to me, because the Poisson distribution has a very distinctive shape, and once again I like seeing the blue and purple  functions working together to make it. I think it's just great how the red line goes gently to zero as  increases, even though the even and the odd components are going wild. ( increases rapidly with , so the reciprocal  function goes rapidly to zero. But the even and odd components also have a  part, and this is what dominates the blue and purple lines when .)

On the  side it has no meaning for me, and it's just wiggly lines. It hadn't occurred to me before that you could extend the Poisson distribution function to negative , and I still can't imagine what it could mean, but I suppose why not. Probably some statistician could explain to me what the Poisson distribution is about when .

You can also consider the function , which breaks down completely, because either  or  is undefined except when . So the claim that every function is the sum of an even and an odd function fails here too. Except perhaps not! You could probably consider the extension of the square root function to the complex plane, and take one of its branches, and I suppose it works out just fine. The geometric interpretation of evenness and oddness are very different, of course, and you can't really draw the graphs unless you have four-dimensional vision.

I have no particular point to make, except maybe that math is fun, even elementary math (or perhaps especially elementary math) and it's fun to see how it works out.

The beautiful graphs in this article were made with Desmos. I had dreaded having to illustrate my article with graphs from Gnuplot (ugh) or Wolfram|α (double ugh) and was thrilled to find such a handsome alternative.

[ Addendum: I've just discovered that in Desmos you can include a parameter in the functions that it graphs, and attach the parameter to a slider. So for example you can arrange to have it display  or , with the value of  controlled by the slider, and have the graph move left and right on the plane as you adjust the slider, with its even and odd parts changing in real time to match. ]

[ For example, check out travelling Gaussians or varying sinusoid. ]
 tag:blog.plover.com,2016:/math/17-puzzle A simple but difficult arithmetic puzzle 2016-07-12T19:13:00Z 2016-07-12T19:13:00Z Mark Dominus http://www.plover.com/ email@example.com
Lately my kids have been interested in puzzles of this type: You are given a sequence of four digits, say 1,2,3,4, and your job is to combine them with ordinary arithmetic operations (+, -, ×, and ÷) in any order to make a target number, typically 24. For example, with 1,2,3,4, you can go with $$((1+2)+3)×4 = 24$$ or with $$4×((2×3)×1) = 24.$$

We were stumped trying to make 6,6,5,2 total 24, so I hacked up a solver; then we felt a little foolish when we saw the solutions, because it is not that hard. But in the course of testing the solver, I found the most challenging puzzle of this type that I've ever seen. It is:

Given 6,6,5,2, make 17.

There are no underhanded tricks. For example, you may not concatenate 2 and 5 to make 25; you may not say  and  and concatenate 1 and 7 to make ; you may not interpret the 17 as a base 12 numeral, etc.

I hope to write a longer article about solvers in the next week or so.
 tag:blog.plover.com,2016:/math/horse-puzzle The sage and the seven horses 2016-04-20T02:11:00Z 2016-04-20T02:11:00Z Mark Dominus http://www.plover.com/ firstname.lastname@example.com
A classic puzzle of mathematics goes like this:

A father dies and his will states that his elder daughter should receive half his horses, the son should receive one-quarter of the horses, and the younger daughter should receive one-eighth of the horses. Unfortunately, there are seven horses. The siblings are arguing about how to divide the seven horses when a passing sage hears them. The siblings beg the sage for help. The sage donates his own horse to the estate, which now has eight. It is now easy to portion out the half, quarter, and eighth shares, and having done so, the sage's horse is unaccounted for. The three heirs return the surplus horse to the sage, who rides off, leaving the matter settled fairly.

(The puzzle is, what just happened?)

It's not hard to come up with variations on this. For example, picking three fractions at random, suppose the will says that the eldest child receives half the horses, the middle child receives one-fifth, and the youngest receives one-seventh. But the estate has only 59 horses and an argument ensues. All that is required for the sage to solve the problem is to lend the estate eleven horses. There are now 70, and after taking out the three bequests,  horses remain and the estate settles its debt to the sage.

But here's a variation I've never seen before. This time there are 13 horses and the will says that the three children should receive shares of  and . respectively. Now the problem seems impossible, because . But the sage is equal to the challenge! She leaps into the saddle of one of the horses and rides out of sight before the astonished heirs can react. After a day of searching the heirs write off the lost horse and proceed with executing the will. There are now only 12 horses, and the eldest takes half, or six, while the middle sibling takes one-third, or 4. The youngest heir should get three, but only two remain. She has just opened her mouth to complain at her unfair treatment when the sage rides up from nowhere and hands her the reins to her last horse.
 tag:blog.plover.com,2015:/math/se/2015-08 Math.SE report 2015-08 2015-12-18T21:50:00Z 2015-12-18T21:50:00Z Mark Dominus http://www.plover.com/ email@example.com
I only posted three answers in August, but two of them were interesting.

In why this  keeps apearing in my group theory book? (cycle decomposition) the querent asked about the "conjugation" operation that keeps cropping up in group theory. Why is it important? I sympathize with this; it wasn't adequately explained when I took group theory, and I had to figure it out a long time later. Unfortunately I don't think I picked the right example to explain it, so I am going to try again now.

Consider the eight symmetries of the square. They are of five types:

Rotation clockwise or counterclockwise by 90°.

Rotation by 180°.

Horizontal or vertical reflection

Diagonal reflection

The trivial (identity) symmetry

What is meant when I say that a horizontal and a vertical reflection are of the same 'type'? Informally, it is that the horizontal reflection looks just like the vertical reflection, if you turn your head ninety degrees. We can formalize this by observing that if we rotate the square 90°, then give it a horizontal flip, then rotate it back, the effect is exactly to give it a vertical flip. In notation, we might represent the horizontal flip by , the vertical flip by , the clockwise rotation by , and the counterclockwise rotation by ; then we have

$$ \rho H \rho^{-1} = V$$

and similarly

$$ \rho V \rho^{-1} = H.$$

Vertical flips do not look like diagonal flips—the diagonal flip leaves two of the corners in the same place, and the vertical flip does not—and indeed there is no analogous formula with  replaced with one of the diagonal flips. However, if  and  are the two diagonal flips, then we do have

$$ \rho D_1 \rho^{-1} = D_2.$$

In general, When  and  are two symmetries, and there is some symmetry  for which

$$xax^{-1} = b$$

we say that  is conjugate to . One can show that conjugacy is an equivalence relation, which means that the symmetries of any object can be divided into separate "conjugacy classes" such that two symmetries are conjugate if and only if they are in the same class. For the square, the conjugacy classes are the five I listed earlier.

This conjugacy thing is important for telling when two symmetries are group-theoretically "the same", and have the same group-theoretic properties. For example, the fact that the horizontal and vertical flips move all four vertices, while the diagonal flips do not. Another example is that a horizontal flip is self-inverse (if you do it again, it cancels itself out), but a 90° rotation is not (you have to do it four times before it cancels out.) But the horizontal flip shares all its properties with the vertical flip, because it is the same if you just turn your head.

Identifying this sameness makes certain kinds of arguments much simpler. For example, in counting squares, I wanted to count the number of ways of coloring the faces of a cube, and instead of dealing with the 24 symmetries of the cube, I only needed to deal with their 5 conjugacy classes.

The example I gave in my math.se answer was maybe less perspicuous. I considered the symmetries of a sphere, and talked about how two rotations of the sphere by 17° are conjugate, regardless of what axis one rotates around. I thought of the square at the end, and threw it in, but I wish I had started with it.

How to convert a decimal to a fraction easily? was the month's big winner. OP wanted to know how to take a decimal like  and discover that it can be written as . The right answer to this is of course to use continued fraction theory, but I did not want to write a long treatise on continued fractions, so I stripped down the theory to obtain an algorithm that is slower, but much easier to understand.

The algorithm is just binary search, but with a twist. If you are looking for a fraction for , and you know , then you construct the mediant  and compare it with . This gives you a smaller interval in which to search for , and the reason you use the mediant instead of using  as usual is that if you use the mediant you are guaranteed to exactly nail all the best rational approximations of . This is the algorithm I described a few years ago in your age as a fraction, again; there the binary search proceeds down the branches of the Stern-Brocot tree to find a fraction close to .

I did ask a question this month: I was looking for a simpler version of the dogbone space construction. The dogbone space is a very peculiar counterexample of general topology, originally constructed by R.H. Bing. I mentioned it here in 2007, and said, at the time:

[The paper] is on my desk, but I have not read this yet, and I may never.

I did try to read it, but I did not try very hard, and I did not understand it. So my question this month was if there was a simpler example of the same type. I did not receive an answer, just a followup comment that no, there is no such example.
 tag:blog.plover.com,2015:/math/se/2015-07 Math.SE report 2015-07 2015-08-16T16:38:00Z 2015-08-16T16:38:00Z Mark Dominus http://www.plover.com/ firstname.lastname@example.com
My overall SE posting volume was down this month, and not only did I post relatively few interesting items, I've already written a whole article about the most interesting one. So this will be a short report.

I already wrote up Building a box from smaller boxes on the blog here. But maybe I have a couple of extra remarks. First, the other guy's proposed solution is awful. It's long and complicated, which is forgivable if it had answered the question, but it doesn't. And the key point is "blah blah blah therefore code a solver which visits all configurations of the search space". Well heck, if this post had just been one sentence that ended with "code a solver which visits all configurations of the search space" I would not have any complaints about that.

As an undergraduate I once gave a talk on this topic. One of my examples was the problem of packing 31 dominoes into a chessboard from which two squares have been deleted. There is a simple combinatorial argument why this is impossible if the two deleted squares are the same color, say if they are opposite corners: each domino must cover one square of each color. But if you don't take time to think about the combinatorial argument you could waste a lot of time on computer search learning that there is no solution in that case, and completely miss the deeper understanding that it brings you. So this has been on my mind for a long time.

I wrote a few posts this month where I thought I gave good hints. In How to scale an unit vector  in such way that  where  is a scalar I think I did a good job identifying the original author's confusion; he was conflating his original unit vector  and the scaled, leading him to write . This is sure to lead to confusion. So I led him to the point of writing  and let him take it from there. The other proposed solution is much more rote and mechanical. ("Divide this by that…")

In Find numbers  so that  the OP got stuck partway through and I specifically addressed the stuckness; other people solved the problem from the beginning. I think that's the way to go, if the original proposal was never going to work, especially if you stop and say why it was never going to work, but this time OP's original suggestion was perfectly good and she just didn't know how to get to the next step. By the way, the notation  here means the number .

In Help finding the limit of this series  it would have been really easy to say "use the formula" or to analyze the series de novo, but I think I almost hit the nail on the head here: it's just like , which I bet OP already knows, except a little different. But I pointed out the wrong difference: I observed that the first sequence is one-fourth the second one (which it is) but it would have been simpler to observe that it's just the second one without the . I had to review it just now to give the simpler explanation, but I sure wish I'd thought of it at the time. Nobody else pointed it out either. Best of all, would have been to mention both methods. If you can notice both of them you can solve the problem without the advance knowledge of the value of , because you have  and then solve for .

In Visualization of Rhombus made of Radii and Chords it seemed that OP just needed to see a diagram ("I really really don't see how two circles can form a rhombus?"), so I drew one.

 tag:blog.plover.com,2015:/math/ounce-of-theory-2 Another ounce of theory 2015-07-28T14:48:00Z 2015-07-28T14:48:00Z Mark Dominus http://www.plover.com/ email@example.com
A few months ago I wrote an article here called an ounce of theory is worth a pound of search and I have a nice followup.

When I went looking for that article I couldn't find it, because I thought it was about how an ounce of search is worth a pound of theory, and that I was writing a counterexample. I am quite surprised to discover that that I have several times discussed how a little theory can replace a lot of searching, and not vice versa, but perhaps that is because the search is my default.

Anyway, the question came up on math StackExchange today:

John has 77 boxes each having dimensions 3×3×1. Is it possible for John to build one big box with dimensions 7×9×11?

OP opined no, but had no argument. The first answer that appeared was somewhat elaborate and outlined a computer search strategy which claimed to reduce the search space to only 14,553 items. (I think the analysis is wrong, but I agree that the search space is not too large.)

I almost wrote the search program. I have a program around that is something like what would be needed, although it is optimized to deal with a few oddly-shaped tiles instead of many similar tiles, and would need some work. Fortunately, I paused to think a little before diving in to the programming.

 Order
 How to Solve It

 with kickback
 no kickback

For there is an easy answer. Suppose John solved the problem. Look at just one of the 7×11 faces of the big box. It is a 7×11 rectangle that is completely filled by 1×3 and 3×3 rectangles. But 7×11 is not a multiple of 3. So there can be no solution.

Now how did I think of this? It was a very geometric line of reasoning. I imagined a 7×11×9 carton and imagined putting the small boxes into the carton. There can be no leftover space; every one of the 693 cells must be filled. So in particular, we must fill up the bottom 7×11 layer. I started considering how to pack the bottommost 7×11×1 slice with just the bottom parts of the small boxes and quickly realized it couldn't be done; there is always an empty cell left over somewhere, usually in the corner. The argument about considering just one face of the large box came later; I decided it was clearer than what I actually came up with.

I think this is a nice example of the Pólya strategy "solve a simpler problem" from How to Solve It, but I was not thinking of that specifically when I came up with the solution.

For a more interesting problem of the same sort, suppose you have six 2×2x1 slabs and three extra 1×1×1 cubes. Can you pack the nine pieces into a 3×3x3 box?
 tag:blog.plover.com,2015:/math/se/2015-04 Math.SE report 2015-04 2015-07-19T00:28:00Z 2015-07-19T00:28:00Z Mark Dominus http://www.plover.com/ firstname.lastname@example.com
[ Notice: I originally published this report at the wrong URL. I moved it so that I could publish the June 2015 report at that URL instead. If you're seeing this for the second time, you might want to read the June article instead. ]

A lot of the stuff I've written in the past couple of years has been on Mathematics StackExchange. Some of it is pretty mundane, but some is interesting. I thought I might have a little meta-discussion in the blog and see how that goes. These are the noteworthy posts I made in April 2015.

Languages and their relation : help is pretty mundane, but interesting for one reason: OP was confused about a statement in a textbook, and provided a reference, which OPs don't always do. The text used the symbol . OP had interpreted it as meaning , but I think what was meant was .

I dug up a copy of the text and groveled over it looking for the explanation of , which is not standard. There was none that I could find. The book even had a section with a glossary of notation, which didn't mention . Math professors can be assholes sometimes.

Is there an operation that takes  and , and returns  is more interesting. First off, why is this even a reasonable question? Why should there be such an operation? But note that there is an operation that takes  and  and returns , namely, multiplication, so it's plausible that the operation that OP wants might also exist.

But it's easy to see that there is no operation that takes  and  and returns : just observe that although , the putative operation (call it ) should take  and yield , but it should also take  and yield . So the operation is not well-defined. And you can take this even further:  can be written as , so  should also take  and yield .

They key point is that the representation of a number, or even an integer, in the form  is not unique. (Jargon: "exponentiation is not injective".) You can raise , but having done so you cannot look at the result and know what  and  were, which is what  needs to do.

But if  can't do it, how can multiplication do it when it multiplies  and  and gets ? Does it somehow know what  is? No, it turns out that it doesn't need  in this case. There is something magical going on there, ultimately related to the fact that if some quantity is increasing by a factor of  every  units of time, then there is some  for which it is exactly doubling every  units of time. Because of this there is a marvelous group homomophism $$\log : \langle \Bbb R^+, \times\rangle \to \langle \Bbb R ,+\rangle$$ which can change multiplication into addition without knowing what the base numbers are.

In that thread I had a brief argument with someone who thinks that operators apply to expressions rather than to numbers. Well, you can say this, but it makes the question trivial: you can certainly have an "operator" that takes expressions  and  and yields the expression . You just can't expect to apply it to numbers, such as  and , because those numbers are not expressions in the form . I remembered the argument going on longer than it did; I originally ended this paragraph with a lament that I wasted more than two comments on this guy, but looking at the record, it seems that I didn't. Good work, Mr. Dominus.

how 1/0.5 is equal to 2? wants a simple explanation. Very likely OP is a primary school student. The question reminds me of a similar question, asking why the long division algorithm is the way it is. Each of these is a failure of education to explain what division is actually doing. The long division answer is that long division is an optimization for repeated subtraction; to divide  you want to know how many shares of three cookies each you can get from  cookies. Long division is simply a notation for keeping track of removing  shares, leaving  cookies, then  further shares, leaving none.

In this question there was a similar answer.  is  because if you have one cookie, and want to give each kid a share of  cookies, you can get out two shares. Simple enough.

I like division examples that involve giving cookies to kids, because cookies are easy to focus on, and because the motivation for equal shares is intuitively understood by everyone who has kids, or who has been one.

There is a general pedagogical principle that an ounce of examples are worth a pound of theory. My answer here is a good example of that. When you explain the theory, you're telling the student how to understand it. When you give an example, though, if it's the right example, the student can't help but understand it, and when they do they'll understand it in their own way, which is better than if you told them how.

How to read a cycle graph? is interesting because hapless OP is asking for an explanation of a particularly strange diagram from Wikipedia. I'm familiar with the eccentric Wikipedian who drew this, and I was glad that I was around to say "The other stuff in this diagram is nonstandard stuff that the somewhat eccentric author made up. Don't worry if it's not clear; this author is notorious for that."

In Expected number of die tosses to get something less than 5, OP calculated as follows: The first die roll is a winner  of the time. The second roll is the first winner  of the time. The third roll is the first winner  of the time. Summing the series  we eventually obtain the answer, . The accepted answer does it this way also.

But there's a much easier way to solve this problem. What we really want to know is: how many rolls before we expect to have seen one good one? And the answer is: the expected number of winners per die roll is , expectations are additive, so the expected number of winners per  die rolls is , and so we need  rolls to expect one winner. Problem solved!

I first discovered this when I was around fifteen, and wrote about it here a few years ago.

As I've mentioned before, this is one of the best things about mathematics: not that it works, but that you can do it by whatever method that occurs to you and you get the same answer. This is where mathematics pedagogy goes wrong most often: it proscribes that you must get the answer by method X, rather than that you must get the answer by hook or by crook. If the student uses method Y, and it works (and if it is correct) that should be worth full credit.

Bad instructors always say "Well, we need to test to see if the student knows method X." No, we should be testing to see if the student can solve problem P. If we are testing for method X, that is a failure of the test or of the curriculum. Because if method X is useful, it is useful because for some problems, it is the only method that works. It is the instructor's job to find one of these problems and put it on the test. If there is no such problem, then X is useless and it is the instructor's job to omit it from the curriculum. If Y always works, but X is faster, it is the instructor's job to explain this, and then to assign a problem for the test where Y would take more time than is available.

I see now I wrote the same thing in 2006. It bears repeating. I also said it again a couple of years ago on math.se itself in reply to a similar comment by Brian Scott:

If the goal is to teach students how to write proofs by induction, the instructor should damned well come up with problems for which induction is the best approach. And if even then a student comes up with a different approach, the instructor should be pleased. ... The directions should not begin [with "prove by induction"]. I consider it a failure on the part of the instructor if he or she has to specify a technique in order to give students practice in applying it.

 tag:blog.plover.com,2015:/math/logic/annoying-boxes-solution The annoying boxes puzzle: solution 2015-07-03T12:15:00Z 2015-07-03T12:15:00Z Mark Dominus http://www.plover.com/ email@example.com I presented this logic puzzle on Wednesday:

 There are two boxes on a table, one red and one green. One contains a treasure. The red box is labelled "exactly one of the labels is true". The green box is labelled "the treasure is in this box."
 Can you figure out which box contains the treasure?

 It's not too late to try to solve this before reading on. If you want, you can submit your answer here:
 <form method=GET action="http://perl.plover.com/annoying-boxes.cgi"> <input type=radio name="r" value="red">The treasure is in the red box
 <input type=radio name="r" value="green">The treasure is in the green box
 <input type=radio name="r" value="wut">There is not enough information to determine the answer
 <input type=radio name="r" value="other">Something else: <input type=text name="o" size=50>
 <input type=submit value="Submit Solution"> </form>
Results

 There were 506 total responses up to Fri Jul 3 11:09:52 2015 UTC; I kept only the first response from each IP address, leaving 451. I read all the "something else" submissions and where it seemed clear I recoded them as votes for "red", for "not enough information", or as spam. (Several people had the right answer but submitted "other" so they could explain themselves.) There was also one post attempted to attack my (nonexistent) SQL database. Sorry, Charlie; I'm not as stupid as I look.

 66.52% 300 red 25.72 116 not-enough-info 3.55 16 green 2.00 9 other 1.55 7 spam 0.44 2 red-with-qualification 0.22 1 attack 100.00 451 TOTAL

One-quarter of respondents got the right answer, that there is not enough information given to solve the problem, Two-thirds of respondents said the treasure was in the red box. This is wrong. The treasure is in the green box.

What?
 Let me show you. I stated:

There are two boxes on a table, one red and one green. One contains a treasure. The red box is labelled "exactly one of the labels is true". The green box is labelled "the treasure is in this box."

 The labels are as I said. Everything I told you was literally true.
 The treasure is definitely not in the red box.

 No, it is actually in the green box.

 (It's hard to see, but one of the items in the green box is the gold and diamond ring made in Vienna by my great-grandfather, which is unquestionably a real treasure.)
 So if you said the treasure must be in the red box, you were simply mistaken. If you had a logical argument why the treasure had to be in the red box, your argument was fallacious, and you should pause and try to figure out what was wrong with it.
 I will discuss it in detail below.

Solution
 The treasure is undeniably in the green box. However, correct answer to the puzzle is "no, you cannot figure out which box contains the treasure". There is not enough information given. (Notice that the question was not "Where is the treasure?" but "Can you figure out…?")

(Fallacious) Argument A
 Many people erroneously conclude that the treasure is in the red box, using reasoning something like the following:

Suppose the red label is true. Then exactly one label is true, and since the red label is true, the green label is false. Since it says that the treasure is in the green box, the treasure must really be in the red box.
Now suppose that the red label is false. Then the green label must also be false. So again, the treasure is in the red box.
Since both cases lead to the conclusion that the treasure is in the red box, that must be where it is.

What's wrong with argument A?
 Here are some responses people commonly have when I tell them that argument A is fallacious:

"If the treasure is in the green box, the red label is lying."
 Not quite, but argument A explicitly considers the possibility that the red label was false, so what's the problem?

"If the treasure is in the green box, the red label is inconsistent."
 It could be. Nothing in the puzzle statement ruled this out. But actually it's not inconsistent, it's just irrelevant.

"If the treasure is in the green box, the red label is meaningless."
 Nonsense. The meaning is plain: it says "exactly one of these labels is true", and the meaning is that exactly one of the labels is true. Anyone presenting argument A must have understood the label to mean that, and it is incoherent to understand it that way and then to turn around and say that it is meaningless! (I discussed this point in more detail in 2007.)

"But the treasure could have been in the red box."
 True! But it is not, as you can see in the pictures. The puzzle does not give enough information to solve the problem. If you said that there was not enough information, then congratulations, you have the right answer. The answer produced by argument A is incontestably wrong, since it asserts that the treasure is in the red box, when it is not.

"The conditions supplied by the puzzle statement are inconsistent."
 They certainly are not. Inconsistent systems do not have models, and in particular cannot exist in the real world. The photographs above demonstrate a real-world model that satisfies every condition posed by the puzzle, and so proves that it is consistent.

"But that's not fair! You could have made up any random garbage at all, and then told me afterwards that you had been lying."
 Had I done that, it would have been an unfair puzzle. For example, suppose I opened the boxes at the end to reveal that there was no treasure at all. That would have directly contradicted my assertion that "One [box] contains a treasure". That would have been cheating, and I would deserve a kick in the ass.
 But I did not do that. As the photograph shows, the boxes, their colors, their labels, and the disposition of the treasure are all exactly as I said. I did not make up a lie to trick you; I described a real situation, and asked whether people they could diagnose the location of the treasure.
 (Two respondents accused me of making up lies. One said:
There is no treasure. Both labels are lying. Look at those boxes. Do you really think someone put a treasure in one of them just for this logic puzzle?
 What can I say? I did put a treasure in a box just for this logic puzzle. Some of us just have higher standards.)

"But what about the labels?"
 Indeed! What about the labels?

The labels are worthless
 The labels are red herrings; the provide no information. Consider the following version of the puzzle:

 There are two boxes on a table, one red and one green. One contains a treasure.
 Which box contains the treasure?

 Obviously, the problem cannot be solved from the information given.
 Now consider this version:

 There are two boxes on a table, one red and one green. One contains a treasure. The red box is labelled "gcoadd atniy fnck z fbi c rirpx hrfyrom". The green box is labelled "ofurb rz bzbsgtuuocxl ckddwdfiwzjwe ydtd."</s>
 Which box contains the treasure?

 One is similarly at a loss here.
 (By the way, people who said one label was meaningless: this is what a meaningless label looks like.)

 There are two boxes on a table, one red and one green. One contains a treasure. The red box is labelled "exactly one of the labels is true". The green box is labelled "the treasure is in this box."
 But then the janitor happens by. "Don't be confused by those labels," he says. "They were stuck on there by the previous owner of the boxes, who was an illiterate shoemaker who only spoke Serbian. I think he cut them out of a magazine because he liked the frilly borders."
 Which box contains the treasure?

 The point being that in the absence of additional information, there is no reason to believe that the labels give any information about the contents of the boxes, or about labels, or about anything at all. This should not come as a surprise to anyone. It is true not just in annoying puzzles, but in the world in general. A box labeled "fresh figs" might contain fresh figs, or spoiled figs, or angry hornets, or nothing at all.

 Order
 What is the Name of this Book?

 with kickback
 no kickback

Why doesn't every logic puzzle fall afoul of this problem?
 I said as part of the puzzle conditions that there was a treasure in one box. For a fair puzzle, I am required to tell the truth about the puzzle conditions. Otherwise I'm just being a jerk.
 Typically the truth or falsity of the labels is part of the puzzle conditions. Here's a typical example, which I took from Raymond Smullyan's What is the name of this book? (problem 67a):

 … She had the following inscriptions put on the caskets:

GoldSilverLead
 THE PORTRAIT IS IN THIS CASKET THE PORTRAIT IS NOT IN THIS CASKET THE PORTRAIT IS NOT IN THE GOLD CASKET
 Portia explained to the suitor that of the three statements, at most one was true.
 Which casket should the suitor choose [to find the portrait]?

 Notice that the problem condition gives the suitor a certification about the truth of the labels, on which he may rely. In the quotation above, the certification is in boldface.
 A well-constructed puzzle will always contain such a certification, something like "one label is true and one is false" or "on this island, each person always lies, or always tells the truth". I went to What is the Name of this Book? to get the example above, and found more than I had bargained for: problem 70 is exactly the annoying boxes problem! Smullyan says:

 Good heavens, I can take any number of caskets that I please and put an object in one of them and then write any inscriptions at all on the lids; these sentences won't convey any information whatsoever.
 (Page 65)
 Had I known ahead of time that Smullyan had treated the exact same topic with the exact same example, I doubt I would have written this post at all.

But why is this so surprising?
 I don't know.

Final notes
 16 people correctly said that the treasure was in the green box. This has to be counted as a lucky guess, unacceptable as a solution to a logic puzzle.
 One respondent referred me to a similar post on lesswrong.
 I did warn you all that the puzzle was annoying.
 I started writing this post in October 2007, and then it sat on the shelf until I got around to finding and photographing the boxes. A triumph of procrastination!
 [ Addendum 20150911: Steven Mazie has written a blog article about this topic, A Logic Puzzle That Teaches a Life Lesson. ] tag:blog.plover.com,2015:/math/logic/annoying-boxes The annoying boxes puzzle 2015-07-01T15:13:00Z 2015-07-01T15:13:00Z Mark Dominus http://www.plover.com/ firstname.lastname@example.com Here is a logic puzzle. I will present the solution on Friday.

 There are two boxes on a table, one red and one green. One contains a treasure. The red box is labelled "exactly one of the labels is true". The green box is labelled "the treasure is in this box."
 Can you figure out which box contains the treasure?

 <form method=GET action="http://perl.plover.com/annoying-boxes.cgi"> <input type=radio name="r" value="red">The treasure is in the red box
 <input type=radio name="r" value="green">The treasure is in the green box
 <input type=radio name="r" value="wut">There is not enough information to determine the answer
 <input type=radio name="r" value="other">Something else: <input type=text name="o" size=50>
 <input type=submit value="Submit Solution"> </form> Starting on 2015-07-03, the solution will be here.<p tag:blog.plover.com,2015:/math/se/2015-05 Math.SE report 2015-05 2015-06-19T03:01:00Z 2015-06-19T03:01:00Z Mark Dominus http://www.plover.com/ email@example.com
A lot of the stuff I've written in the past couple of years has been on math.StackExchange. Some of it is pretty mundane, but some is interesting. My summary of April's interesting posts was well-received, so here are the noteworthy posts I made in May 2015.

What matrix transforms  into  and tranforms  into ? was a little funny because the answer is $$\begin{pmatrix}2 & 4 \\ 6 & 8 \end{pmatrix}$$ and yeah, it works exactly like it appears to, there's no trick. But if I just told the guy that, he might feel unnecessarily foolish. I gave him a method for solving the problem and figured that when he saw what answer he came up with, he might learn the thing that the exercise was designed to teach him.

Is a "network topology'" a topological space? is interesting because several people showed up right away to say no, it is an abuse of terminology, and that network topology really has nothing to do with mathematical topology. Most of those comments have since been deleted. My answer was essentially: it is topological, because just as in mathematical topology you care about which computers are connected to which, and not about where any of the computers actually are.

Nobody constructing a token ring network thinks that it has to be a geometrically circular ring. No, it only has to be a topologically circular ring. A square is fine; so is a triangle; topologically they are equivalent, both in networking and in mathematics. The wires can cross, as long as they don't connect at the crossings. But if you use something that isn't topologically a ring, like say a line or a star or a tree, the network doesn't work.

The term "topological" is a little funny. "Topos" means "place" (like in "topography" or "toponym") but in topology you don't care about places.

Is there a standard term for this generalization of the Euler totient function? was asked by me. I don't include all my answers in these posts, but I think maybe I should have a policy of including all my questions. This one concerned a simple concept from number theory which I was surprised had no name: I wanted  to be the number of integers  that are no larger than  for which . For  this is the famous Euler totient function, written .

But then I realized that the reason it has no name is that it's simply  so there's no need for a name or a special notation.

As often happens, I found the answer myself shortly after I asked the question. I wonder if the reason for this is that my time to come up with the answer is Poisson-distributed. Then if I set a time threshold for how long I'll work on the problem before asking about it, I am likely to find the answer to almost any question that exceeds the threshold shortly after I exceed the threshold. But if I set the threshold higher, this would still be true, so there is no way to win this particular game. Good feature of this theory: I am off the hook for asking questions I could have answered myself. Bad feature: no real empirical support.

how many ways can you divide 24 people into groups of two? displays a few oddities, and I think I didn't understand what was going on at that time. OP has calculated the first few special cases:

1:1 2:1 3:3 4:3 5:12 6:15

which I think means that there is one way to divide 2 people into groups of 2, 3 ways to divide 4 people, and 15 ways to divide 6 people. This is all correct! But what could the 1:1, 3:3, 5:12 terms mean? You simply can't divide 5 people into groups of 2. Well, maybe OP was counting the extra odd person left over as a sort of group on their own? Then odd values would be correct; I didn't appreciate this at the time.

But having calculated 6 special cases correctly, why can't OP calculate the seventh? Perhaps they were using brute force: the next value is 48, hard to brute-force correctly if you don't have a enough experience with combinatorics.

I tried to suggest a general strategy: look at special cases, and not by brute force, but try to analyze them so that you can come up with a method for solving them. The method is unnecessary for the small cases, where brute force enumeration suffices, but you can use the brute force enumeration to check that the method is working. And then for the larger cases, where brute force is impractical, you use your method.

It seems that OP couldn't understand my method, and when they tried to apply it, got wrong answers. Oh well, you can lead a horse to water, etc.

The other pathology here is:

I think I did what you said and I got 1.585times 10 to the 21

for the  case. The correct answer is $$23\cdot21\cdot19\cdot17\cdot15\cdot13\cdot11\cdot9\cdot7\cdot5\cdot3\cdot1 = 316234143225 \approx 3.16\cdot 10^{11}.$$ OP didn't explain how they got  so there's not much hope of correcting their weird error.

This is someone who probably could have been helped in person, but on the Internet it's hopeless. Their problems are Internet communication problems.

Lambda calculus typing isn't especially noteworthy, but I wrote a fairly detailed explanation of the algorithm that Haskell or SML uses to find the type of an expression, and that might be interesting to someone.

I think Special representation of a number is the standout post of the month. OP speculates that, among numbers of the form  (where  are prime), the choice of  is unique. That is, the mapping  is reversible.

I was able to guess that this was not the case within a couple of minutes, replied pretty much immediately:

I would bet money against this representation being unique.

I was sure that a simple computer search would find counterexamples. In fact, the smallest is  which is small enough that you could find it without the computer if you are patient.

The obvious lesson to learn from this is that many elementary conjectures of this type can be easily disproved by a trivial computer search, and I frequently wonder why more amateur mathematicians don't learn enough computer programming to investigate this sort of thing. (I wrote recently on the topic of An ounce of theory is worth a pound of search , and this is an interesting counterpoint to that.)

But the most interesting thing here is how I was able to instantly guess the answer. I explained in some detail in the post. But the basic line of reasoning goes like this.

Additive properties of the primes are always distributed more or less at random unless there is some obvious reason why they can't be. For example, let  be prime and consider . This must have exactly one of the three forms  or  for some integer . It obviously has the form  almost never (the only exception is ). But of the other two forms there is no obvious reason to prefer one over the other, and indeed of the primes up to 10,000, 611 are of the type  and and 616 are of the type .

So we should expect the value  to be distributed more or less randomly over the set of outputs, because there's no obvious reason why it couldn't be, except for simple stuff, like that it's obviously almost always even.

So we are throwing a bunch of balls at random into bins, and the claim is that no bin should contain more than one ball. For that to happen, there must be vastly more bins than balls. But the bins are numbers, and primes are not at all uncommon among numbers, so the number of bins isn't vastly larger, and there ought to be at least some collisions.

In fact, a more careful analysis, which I wrote up on the site, shows that the number of balls is vastly larger—to have them be roughly the same, you would need primes to be roughly as common as perfect squares, but they are far more abundant than that—so as you take larger and larger primes, the number of collisions increases enormously and it's easy to find twenty or more quadruples of primes that all map to the same result. But I was able to predict this after a couple of minutes of thought, from completely elementary considerations, so I think it's a good example of Lower Mathematics at work.

This is an example of a fairly common pathology of math.se questions: OP makes a conjecture that  never occurs or that there are no examples with property , when actually  almost always occurs or every example has property .

I don't know what causes this. Rik Signes speculates that it's just wishful thinking: OP is doing some project where it would be useful to have  be unique, so posts in hope that someone will tell them that it is. But there was nothing more to it than baseless hope. Rik might be right.

[ Addendum 20150619: A previous version of this article included the delightful typo "mathemativicians". ]


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Re: Good Hostel in Queenstown
I would recommend you have a look on Rankers.co.nz - people can review and rank different operators based on their experience with them. There is a section for accommodation which, you should have a look at.
Be the first to rate this answer!Was this helpful? Quote & Answer
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== Search results for kinoma download ==
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I understand everything that's being said. Dana is correct on the difficulty photographing ABs, not just the variation in color, but the form generally has a progression over a 3 day life span. Yes, colors vary according to growing conditions and light. Certainly a photo captures a single moment. Nonetheless I think some effort should be made for accuracy as they grow in my environment. Photos are for my use in the off season and a blue lavender bloom that's only represented as true blue doesn't cut it. I want an 8x10 glossy. Hail damage and wind distortion (my worst enemy taking photos) are not part of the genetic makeup, so I want the bloom at its best and the most accurate for I see. Odd thing about wind distortion. Often impossible to avoid taking a photo, the actual impression is usually not that of the windblown version, but closer to the ideal of no wind. Some colors tend to be harder. I'm usually happy with yellows, pinks and oranges and mostly of the browns. Many reds aren't satisfactory but with a lot of patience and a lot of photos you can make them work. Blue lavenders and blue purples are worse. A few cultivars work, but most are really hard to impossible. Worst of all are blooms with near white standards in any color. They tend to photograph white even though the falls look correct. Very aggravating because it really gives an incorrect impression.
Lowell, if you saved one of those layers of the cut-out bloom, I'd like to see it posted with different colors for the background. Sort of a test of my theory that you can trick the eye into seeing a different color in the bloom by use of background color.
Donald Eaves
Texas Zone 7b, USA


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- 08-31-2014, 05:33 AM Thread Author #1
Unable to download the attachement file
I have the samsung galaxy s4 mini.Currently I M unable to open the attachement form the mail received.Kindly help in resolving the problems.While clicking on the attachment i m receiving a message that " you must first download the message to download the attachment".
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Re: Unable to download the attachement file
Welcome to Android Central! That system message makes it sound like you're dealing the the texting app rather than email. Which one is it?
If you're on your Messaging app, make sure you have mobile data (3G/4G) on, since I'm not sure attachments can be downloaded over voice signal or wi-fi. This also applies to sending photos via Messaging, as well as sending texts to groups.
If you're on email, which email app are you using? The stock Email app, or Gmail, or some other? Make sure you're connected to the web, either via wi-fi or mobile data. Do you still get that error?
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ABC and Finke Desert Race to the Rescue. Finke 2007.
Brad Loadsman was sure living up to his name.
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Travelling from Shepparton in Victoria along the Finke track, Loadsman found he's packed a little too much weight on the back of his bike.
Hitting the heavy sands halfway between Finke and Alice Springs, his 200kg bike toppled over, along with 70kg of luggage he was carting around.
"I met up with these blokes from Oodnadatta last night in Finke, and we were riding the track together. I guess the didn't realise I'd fallen".
While Loadsman realised he was carrying a pretty heavy load for a motorbike - there's not much that can be sacrificed, as he is carrying everything he needs to travel Australia for the next couple of years.
Carrying the large load can be a problem - especially when there's nobody there to help pick the bike up.
"I'd been waiting here half an hour until you guys showed up. There's plenty of bikes doing their training run here, but not much else".
The "you guys" Loadsman was referring to were Finke Desert Race President Antony Yoffa, along with ABC reporters Charlie King and Kate Humphris.
An unexpected diversion in our drive down to Finke, we were happy to help Loadsman lift his bike out of the sand and push him off - hopefully on a safe journey towards Alice Springs.
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Blogs // 3 Articles
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PRINCE William is keeping one secret from his bride — their honeymoon
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