Caturday Post: Tamerlane the World-Conquering Sword and the Fifty Thousand Puffballs of Persia*

In the ongoing (we desperately hope not dropped) anime series Arslan Senki, the main tank character, when told by the strategist that with the addition of two more allies they will only have to defeat 50,000 Lusitanians apiece, considers this and says completely seriously that he thinks he can take 50,000. (And we think so too. He's invincible as only a legendary hero can be.)

Obviously Elisheba's little orange cat Tamerlane, as the World-Conquering Sword of hopelessly spoiled and pampered indoor cats, can defeat 50,000 of his puffballs. 

some of Tamerlane's puffballs

But how much room in our house would these 50,000 puffballs take up?  We did the math:

Assume that a typical puffball is 3.5 cm in diameter. We'll assume the puffballs are un-squishable, and, further, we'll assume they are cubical with a side of 3.5cm rather than spherical in order to do a quick and dirty space estimate rather than getting sucked into a minimal packing arrangement problem. With all our simplifying assumptions this becomes an exercise in dimensional analysis:

volume of 1 puffball * convert to meters for easier visualization of space required * number of puffballs \[\frac{(3.5cm)^3}{1 puffball}\cdot \left(\frac{1 m}{100 cm}\right)^3 \cdot \frac{50,000 puffballs}{1} \]
\[=\frac{3.5^3\cancel{cm^3}}{1 \cancel{puffball}}\cdot \frac{1 m^3}{100^3 \cancel{cm^3}} \cdot \frac{50,000 \cancel{puffballs}}{1} \approx 2.1 m^3 \]
Thus we need 2.1 cubic meters of space for the 50,000 puffballs, or a box \(2.1^{1/3} \approx 1.3\) meters on a side.

Considering that the puffballs actually squish considerably, we can probably get away with one extra-large moving box from Home Depot and drop Tamerlane in on top to wreak havoc.

"Cry havoc, and let slip the kittens of war!" -William Shakescat


*like the bowmen of Persia, but puffier and significantly less likely to fight back in any way.

Friday Fabulosity: Theoretical Muffin Topology

from the ever wonderful webcomic Questionable Content: http://questionablecontent.net/view.php?comic=3284

I would say that muffin topology is my new career goal, but actually I found algebraic topology straightforward but boring until we got to homology, at which point I became totally lost.  Fortunately that was at the very end of the introductory course, or I would not have passed that preliminary exam.  

Babylonian Algebra and an Uzbekistanian Tangent

Everyday Life in Babylonia and Assyria, by H.W.F. Saggs, is probably rather outdated at this point.  However, I found it an interesting and very readable overview.  And it has a math problem.  An actual Babylonian math problem, with solution, from a clay tablet.  The problem goes as follows:

I have added the surface of my two squares: 28;20.[The text uses sexagesimal numbers.  In our usual decimal system, \(28;20 = 20\cdot 1 + 28 \cdot 60 = 1700\).]
(The side of) one square is a quarter (the side of) the (other) square.
You put down 4 and 1.
You multiply 4 by 4: 16.
You multiply 1 by 1: 1.
You add 1 and 16: 17.
The reciprocal of 17 cannot be solved.
What must I put to 17 [i.e. What must I multiply 17 by] which will give me 28;20 [1700]? 1;40 [100].
This is the square of 10.
You raise 10 by 4 and 40 is (the side of) one square.
You raise 10 by I and 10 is (the side of) the second square.

A modern symbolic solution: Let x and y be the side lengths of the squares.  We know that

\[x^2 + y^2 = 1700 \] and (without loss of generality assume \(y < x\) ) \[ x = 4y \]

Substitute \(x = 4y\) into \(x^2 + y^2 = 1700\) and get
\begin{eqnarray*}
 16y^2 + y^2 & =& 1700 \\
17y^2 &=& 1700\\
y^2 &=& 100\\
y &=& \pm 10
\end{eqnarray*}

Since the problem is talking about physical squares, we discard the negative solution and get \(y = 10\), \(x=4y=40\).

You can find problems exactly like this in any present day basic algebra textbook, right down to the use of nice integers which tempt natural arithmeticians to not show their work and thus receive only partial credit on graded homework and tests. 

My number theory textbook says that the mathematician whose book Kitab al-jabr introduced Europe to algebra, Abu Ja'far Mohammed Ibn Musa Al-Khwarizmi, was from the town of Khiva (Kowarzizm) in modern day Uzbekistan.  I tried to do a short research report on Khiva my first year of college, and could (at the time) put my hands on almost zero substantive material.  The most anyone could seem to say about the town was that it (a) wasn't quite as notorious a slave-trading center as its neighbor town, Merv, and (b) has a lovely, though relatively recent, half-finished minaret:

image from Wikipedia

What Good Is Math?

From Kiery King, who was deprived of a basic education, the reason why they need math:

http://www.kieryking.com/2016/01/on-math-programming/

and how being denied an education based on sex is still very much alive and well in the United States of America:

http://www.kieryking.com/2014/02/in-which-my-genitals-mean-i-dont-learn-math-or-science/

If you know someone who wants to do (math, science, sewing, whatever) don't tell them they can't because they are (female, black, male, whatever).  Don't stand there and let that person's parents tell them they don't need to/are not allowed to learn.  Deliberately sabotaging someone else's choices because of arbitrary and nonsensical gender restrictions is a vile thing to do.

Do I Really Have To Find The Conveniently Lying Around Puzzle Hint?: Basic Combinatorics and Computer Games

Screen text: I don't know the right combination, and trying to find it will take too long.

Situation 1: combination lock.  3 buttons (levers? thingamajigs?) , each button has 5 sides.  Without any further information to rule out possible combinations, I have 5 choices for the first button, 5 choices for the second button, and 5 choices for the third button, for 5*5*5 = 125 total possibilities to try for a brute force test-everything solution.  Assume I need 2 seconds to set each button to a possibility.  Then the maximum time I need to test a combination is 6 seconds (less for combinations where I need to change only one or two slots).  6 seconds * 125 tries = 750 seconds or 12.5 minutes.  Assume I stop for a break and to make notes to keep track of my attempts.  I can still reasonably force the lock in under 30 minutes.  Even in a real-time game, unless you are in a down-to-the-minute crunch, this is doable.  Tedious (very tedious!), but straightforward.  I don't truly need someone's conveniently left out password here. 

How about this one?

Screen text: I can't figure this out without a plan.

Situation 2: A clearly magical harp sitting in a swamp (if it weren't magical, we would have serious water damage to this poor instrument).  We do have a limited number of pitches: 17, one for each colored segment.  (My mouse pointer is covering up one such segment). If we assume the magical melody is

• exactly 17 notes long, no repeats, we have 17 choices for the first note, then 16 choices for the second note, 15 choices for the third note, etc., for 17*16*15*14*...*3*2*1 = 17! = 355687428096000 combinations.  Way too many to test without a computer!  (Read the "!" as "factorial" in this context.)
• exactly 17 notes long, repeats allowed, we have  17*17*17*...*17 = 17^17 = 827240261886336764177 possible choices.  Way too many to test!  (With a computer?  I'm not sure.  Let me do some more research.)
• Less than or more than 17 notes? Less than, and you could probably still force it with a computer eventually.  Small enough and you could do it by hand.  More than 17?  Well, how much more?  At some point you can always make the problem too big.  Even the integers are infinite.  (And the real numbers are infinitely more infinite than the integers.  In a mathematically well-defined and amazingly mind-blowing way.)
• If timing and tempo matter?  Forget it.  

I do need a conveniently lying around hint for this one, even with an untimed game.

Edit: turned out situation 2 was just about tuning the strings, rather than a combination.  Since tuning is an aesthetic choice (individual or cultural), yes, I did need a guide. 

Friday Fabulousity: The Endless Entertainment of Electromagnets

For all your walking on the ceiling needs, see Hackaday: http://hackaday.com/2014/05/23/electromagnetic-boots-for-all-your-upside-down-needs/

The craftsmanship seems to be at a high level, although I am not trained in the use of power tools.  But all the pieces fit neatly together and the all the connections are tidy, just like a good piece of clothing.  I like watching craftspeople at work.   I might need some ceiling walking in my training exercises.

Here's the main making-of video from the Hackaday site:



As Oersted, Ampere, and Maxwell said  
\[ \nabla \times \mathbf{B} = \frac{4\pi}{c}\mathbf{J} +\frac{1}{c} \frac{\partial \mathbf{E}}{\partial t} \] That is, the magnetic field B generated around a wire (the curl \(\nabla \times \mathbf{B} \)) is proportional to the sum of the current density J and to the change in time of the electric field E!  This equation encapsulates a whole boatload of exciting math, as well as being almost unbelievably beautiful and tidy!   We have no a priori reason to believe that the universe is (sometimes) so tidy, but sometimes it is.   Oh, and if you have enough J you can walk on the ceiling!   (The \(\partial \mathbf{E}/\partial t \) effect is usually pretty small compared to gravity on Earth.)

Mathematical Monday: In Puerto Rico. With Pigeons.

There were many pigeons in San Juan.  Tourists paid to feed them.  This I do not understand since pigeons are ugly disease-ridden rats with wings.  I wouldn't even feed them to my cats, since I would fear a flea infestation.

However!  I have a however!  There was a wall filled with pigeonholes!  There were more pigeons than pigeonholes!  Several pigeonholes had more than one pigeon!

In case it's not obvious why I'm excited, my absolute favorite mathematical principle is that of the pigeonhole, which simply states that if there are n pigeonholes and >n pigeons, at least 1 pigeonhole will contain more than 1 pigeon.  And it's true!  I don't think that one could have a cat with cardboard box principle the same way, since in the case of my cats, if I have more than one cat, they will fight over one box, regardless of how many boxes there are.