"Mathematician Charles L. Dodgson, better known as Lewis Carroll, created the game of doublets."
This page shows us how "Lewis Carroll Changes 'Four' to 'Five.'" Pappas explains to us the rules of the game:
"The object is to change the first word to the other word by forming other successive words of the same length. You may change only one letter at a time, trying to make the least number of transformations."
Here is how Carroll changes FOUR to FIVE:
Here are some others to try --
Cover EYE with LID
Prove PITY to be GOOD
Get WOOD from TREE
Change OAT to RYE
In trying to solve these doublets, it helps me to think about consonants and vowels. For example, let's look at the completed example of FOUR to FIVE. Fortunately, both the starting and target words start with the letter F, so we don't even need to change it. But the key steps to me are FOOT-FORT-FORE, since this is when we change the third letter from vowel to consonant and then the fourth letter from consonant to vowel.
In the example for you to solve, the first one is the easiest, while the third is the hardest. In general, the three-letter words are easier than the ones with four-letter words. That's the only hint you'll get.
Meanwhile, let's give the answers to yesterday's puzzle. As it turns out, the first two polyhex puzzles are possible to solve, while the third one is impossible to solve.
But how can I post the solutions to the two possible problems in ASCII? Well, let's go back to the website providing us with a list of the seven tetrahexes:
Let's label these tetrahexes from 1 to 7, where 1 represents the leftmost tetrahex and 7 represents the rightmost tetrahex. Then we can show where these seven tetrahexes belong:
3 3 3 3 5 4 4
7 6 5 5 4 2 4
7 6 6 5 2 1 1
7 7 6 2 2 1 1
7 4 4 6 6
7 4 6 6 1 1
7 5 4 0 2 1 1
7 5 5 2 2 2
5 3 3 3 3
The zero, of course, represents the hole given in the middle of the second puzzle.
Some people may prefer to give the tetrahexes more descriptive names -- for example, we could name them after the letter of the alphabet they most closely resemble. The third polyhex, for example, has the hexagons in a straight line, so we could call it the I tetrahex. The fourth tetrahex is shown as an upside-down U, so we can all it the U tetrahex. The fifth polyhex is the Y tetrahex -- look at it sideways to find out why.
For the other tetrahexes, we must think more abstractly to come up with the letters. We can call the sixth tetrahex the Z tetrahex -- but maybe you prefer N, or even S (if we reflect it). The seventh can be called the L tetrahex, while the second one is the T tetrahex. The first one is the hardest to name -- we could call it the X tetrahex, since a line joining the top and bottom hexes intersects another line joining the left and right hexes. But you might even prefer calling it O, since the four hexes surround a central point. For this post, though, let's stick to labeling them as 1 to 7.
As for today's doublets puzzle, I'll give the answers in my next post -- which will not be Monday. In the district whose calendar the blog is following, there is no school for students on Monday. It appears that this will be a PD day for teachers, but a day off for students and subs.
Notice that this off-day has nothing to do with Columbus Day. In previous years, I used the term "Columbus Day" to refer to certain days in October when schools were closed, but those were in fact different districts. And indeed, Columbus Day was observed on Monday, October 9th, but my current district will be closed a week later, on Monday, October 16th.
We also notice that the first day of the semester was of course Day 1, and I already gave last day of the semester as Day 83. And so Day 42 is exactly halfway between 1 and 83. This shows us a more likely reason as to why Monday is a no-school day -- it represents the start of the second quarter.
And so that means you get a full three-day weekend to solve the doublets puzzle. I won't give the answers until my next post on Tuesday. (Oh, and why did I post last Wednesday -- another day with no math class -- yet I won't this Monday? It's because Wednesday was Day 40, but Monday will remain an unnumbered day, with Day 43 on Tuesday.)
Oh, and before we leave polyhexes, notice that today is Friday the 13th. This marks the second such day this calendar year, with the first being back in January. At the risk of hurting sufferers of triskaidekaphobia, let me ask:
-- How many triskaidecominoes are there?
-- How many triskaidecahexes are there?
Well, according to the OEIS, there are 238,591 triskaidecominoes. Indeed, the number of n-ominos increases exponentially in n -- the number of n-ominoes for large n is about 4^n. Actually, the OEIS gives it as mu^n, with mu between 3.98 and 4.64. Notice that 4^13 is over 67 million -- which tells us that 13 isn't "large enough" for the limit mu^n to apply yet.
And the number of triskaidecahexes is greater -- it is 3,274,826. The number of n-hexes likely also grows exponentially in n, but the OEIS gives no clue regarding how fast it increases.
Meanwhile, let's return to Paul Hoffman's book. Today we'll read Chapter 2 -- and notice that in order to make it easier to remember, these chapters also follow the digit pattern, with Chapter 2 on the same day as Lesson 4-2. Chapter 3 will be on Tuesday, the same day as Lesson 4-3.
But I already wrote that Hoffman, like a few other math authors, numbers his chapters differently. We already read Chapter 0, and there is another chapter named for a constant -- Chapter e, where e of course refers to the constant 2.718281828....
Again, Chapter 3 will be posted along with Lesson 4-3 on Tuesday. Therefore, today we will read both Chapter 2 and Chapter e. I know -- I actually could post Chapter e over the three-day weekend, but I won't post for three days at all. Hey, at least you get a bonus chapter today! (And I can hear you saying, I'd rather get the answers to the doublets puzzle earlier than a bonus chapter today!)
Chapter 2 of Paul Hoffman's The Man Who Loved Only Numbers is called "Epszi's Enigma." As usual, we begin with a quote:
"Temetni tudunk." (How to bury people -- that is one thing we know.)
-- old Magyar saying
While Chapter 1 was a more mathematical chapter, this is more biographical and historical. Hoffman opens up with a story of the 17-year old Paul Erdos, who is invited on a "play date" (of sorts) to the home (actually, a shoe store) of 14-year old prodigy Andrew Vazsonyi. Of course, the only games Erdos wants to play are math games. He asks Vazsonyi to tell him a four-digit number, and so the younger teen replies with 2532. Erdos says, "The square of it is 6,411,024. Sorry. I am getting old and cannot tell you the cube." Again, cue the drens -- Erdos can square a four-digit number mentally, while we're lucky if our drens can square one-digit numbers mentally.
Hoffman now begins his biography of Erdos. Born on March 26th, 1913, Paul is the third child of two high school math teachers. Unfortunately, tragedy strikes the family on the day of Paul's birth -- his older sisters, three-year-old Klara and five-year-old Magda, both die of scarlet fever. It is because of this that the young Paul is homeschooled during his elementary years -- his parents fear that he may be exposed to germs, catch scarlet fever, and follow his sisters to an early grave.
At this point, the author switches to tell us the history of Paul's homeland, Hungary. The first settlers there were the Magyars, who arrived there in the ninth century. Throughout the centuries, the Magyar people are attacked by one tribe after another, beginning with Genghis Khan's Mongols in 1241. This explains the origin of the opening quote of the chapter.
Then exactly three centuries later, the Ottomans divided the land into three parts. One part went to the Hapsburgs, Transylvania to the Muslims (yes, the same Transylvania we speak of every year during this time, Halloween), and the twin cities of Buda and Pest for themselves. It wasn't until forty years before Paul's birth when Buda and Pest united to become Budapest. As Erdos often jokes, "The problem with Hungary is that in every war we choose the wrong side."
I also found it interesting that a holiday celebrated strongly in Hungary is All Soul's Day -- which is November 2nd on the calendar. Perhaps the only culture that observes the holiday more than Hungarians are Mexicans, including those of Mexican descent here in California. To them, All Soul's Day is known as the Day of the Dead.
I could continue with Hoffman's history, but I do wish to focus on the math. The one thing I do want to say is that I was always a strong math student who didn't care much for history. That's why it's fascinating to read about history in a book like Hoffman's -- history through the lens of math. Here in California, students learn about the history of modern Europe in sophomore World History. Most likely, there isn't much about Hungary per se in our textbooks. I suppose that much of what Hoffman writes may be part of the AP European History class, but I never took AP Euro.
Returning to Erdos himself, Hoffman tells us that his family was Jewish, particularly on his mother's side (which makes Erdos culturally Jewish himself). Hoffman writes that his mother was no longer a practicing Jew after his father asked her out on a date on Yom Kippur.
As Erdos grew up, he attends high school only sporadically. There are strong restrictions on Jewish university admissions, but fortunately Erdos is admitted to the University Pazmany Peter in Budapest at the age of 17, and four years later he completes his doctoral degree in math.
Hoffman writes that during this time, Erdos often meets with his fellow mathematicians at a park. He is afraid that they will be punished during these tough anti-Semitic years, and so this is when he develops many of those code words that appear throughout this book:
"on the long wavelength" = Communist
"on the short wavelength" = Fascist (the ruling class in Hungary at the time)
"epsilons" = children
"poison" = alcohol
"bosses" = women
That last word explains why Erdos never marries -- he never becomes a "slave" to a "boss." (And yes, when growing up some assume that Erdos is gay.) Naturally, Erdos disapproves strongly of fascism, and so anyone he doesn't like he calls "fascist" -- including God, the Supreme Fascist (SF).
At this time, during the 1930's, Erdos meets a talented student named Esther Klein. He calls her "epsilon," or "Epszi" for short -- so she's the Epszi mentioned in the chapter title. And her "enigma" is a geometry problem that she gives the other mathematicians:
-- Given any five points on a plane, with no three of them collinear, prove that four of these points will always form a convex quadrilateral. (Recall that "convex polygon" is defined in Lesson 2-7 of the U of Chicago text."
To prove this, she divides this into three cases. In Case I, the five points are the vertices of a convex pentagon, then any four points are the vertices of a convex quadrilateral. In Case II, one point is interior to the other four points, which are the vertices of a convex quadrilateral. In Case III, two points are interior to the triangle formed by the other three points. Then we draw a line through those interior points. Two of the triangle's vertices must lie on one side of this line (with the last point on the other side), and so these two plus the two on the line are the vertices of a convex quadrilateral.
After learning about this, Erdos and the others want to generalize this. They find that nine points are necessary to guarantee a convex pentagon. In this case, eight isn't enough -- a counterexample is the vertices of two rectangles, with one completely interior to the other. They conjecture that to guarantee a convex n-gon requires 2^(n - 2) + 1 points, but they can't even prove this for the hexagon. Like many problems in Ramsey theory, they are able to find an upper bound of 71 points, but that's a long way from the conjectured 17 points. Hoffman writes that since then, the upper bound has been reduced to 37 points.
Hoffman spends the rest of Chapter 2 writing about Erdos and his later acquaintances, including the English mathematician G.H. Hardy. I mentioned Hardy in my May 1st post in conjunction with the movie The Man Who Knew Infinity -- namely Ramanujan. In fact, Erdos and Hardy often spoke about Ramanjuan and his discoveries.
Of course, they begin with the taxicab number -- the number on Ramanujan's taxicab, 1729:
1729 = 12^3 + 1^3 = 10^3 + 9^3
Both Erdos and Ramanujan enjoyed looking for asymptotic formulas, which estimate how many natural number below a certain number n have a certain property. Hoffman briefly mentions the Prime Number Theorem, which we've seen in Ogilvy's book as well.
Ramanujan -- and later on Erdos -- finds an asymptotic formula for the number of highly composite numbers -- numbers with more divisors than any number below it. They also notice that the prime factors of these record-breakers follow patterns:
6 = 2^1 * 3^1
12 = 2^2 * 3^1
24 = 2^3 * 3^1
332,640 = 2^5 * 3^3 * 5^1 * 7^1 * 11^1
43,243,200 = 2^6 * 3^3 * 5^2 * 7^1 * 11^1 * 13^1
2,248,776,129,600 = 2^6 * 3^3 * 5^2 * 7^2 * 11^1 * 13^1 * 17^1 * 19^1 * 23^1
In every case, the largest exponent is on the prime 2, and then they decrease from there.
All three mathematicians, Hardy, Ramanujan, and Erdos, also write about "round numbers," which have more divisors than just the numbers around them (rather than all the naturals below them starting with 1). Hoffman explains that 1 million is "round," since it has 12 prime factors (six 2's and six 5's) while all the other numbers from 999,991 to 1,000,010 have at most eight (1,000,008), and on average about four.
The final example in Chapter 2 is partitions. For example, the number 5 has seven partitions:
5 = 5
5 = 4 + 1
5 = 3 + 2
5 = 3 + 1 + 1
5 = 2 + 2 + 1
5 = 2 + 1 + 1 + 1
5 = 1 + 1 + 1 + 1 + 1
The partition numbers appear in the OEIS:
Hardy and Ramanujan find an asymptotic formula for them, while Erdos is able to give an "elementary proof" of this formula -- one which uses only real numbers, not complex numbers. (We saw an example of a proof in Ogilvy where complex numbers appeared out of nowhere.)
Oh, and technically, the formula 4^n for the number of n-ominoes is an asymptotic formula as well.
Chapter e of Paul Hoffman's The Man Who Loved Only Numbers is "Problems with Sam and Joe." I begin with the opening quote:
"There is a much quoted story about David Hilbert, who one day noticed that a certain student had stopped attending class. When told the student had decided to drop mathematics to become a poet, Hilbert replied, 'Good -- he did not have enough imagination to become a mathematician.'"
This chapter opens at the onset of World War II. By now Erdos has left his homeland and moves to England and ultimately the United States. In the chapter title, the problems referred to occur after the war, when Erdos tries to return to Hungary. Sam and Joe are two more Erdos nicknames -- "Sam" refers to the United States (as in "Uncle Sam") and Joe is the USSR ("Joseph Stalin").
This post is already long enough as I try to write about one chapter, so squeezing in a second chapter would make this post unbearably long if I write about all of the adventures of Erdos. (Again, if you really want these details, just read the book!)
Stanislaw Ulam is a set theorist whom Erdos meets in England in 1940. He is most famous for the Ulam spiral:
Hoffman shows us an Ulam spiral with 17 at the center and stopping at 272. Then there is an entire diagonal of primes, from 227 in one corner to 257 in the opposite corner. But it can be proved that the pattern doesn't go forever. All the primes in this diagonal are values of the polynomial n^2 + n + 17, and as we learned from Ogilvy, no polynomial can produce only primes.
Meanwhile, Erdos meets many other mathematicians during this period. Hoffman writes about the logician Kurt Godel. Actually, the author does a great job explaining Godel's Incompleteness Theorem and why it's a huge blow for mathematicians. David Kung also speaks about Kurt Godel in his lectures on paradoxes -- see my posts of January 2016 for more information.
Godel's Theorem tells us that no set of axioms can prove all of mathematics. Axioms are postulates -- and so Hoffman begins with the geometrical postulates of Euclid, and the subsequent discovery that it's impossible to prove Euclid's fifth postulate from the first four. We then move on to Russell and Whitehead, who tried to prove all of number theory from a few axioms. Again, we can go back to my January 2016 posts for the whole story here, including Russell's Paradox.
Ultimately, Hoffman ends the chapter by writing about another contemporary -- Einstein. Indeed, Einstein and Erdos speak to each other about the nature of time as the fourth dimension.
OK, let's finally get to the U of Chicago text!
Lesson 4-2 of the U of Chicago text is called "Reflecting Figures." This is what I wrote two years ago about today's lesson:
Let me state the Reflection Postulate, because it is so important, directly from the U of Chicago text:
Under a reflection:
a. There is a 1-1 correspondence between points and their images.
This means that each preimage has exactly one image, and each image comes from exactly one preimage.
b. If three points are collinear, then their images are collinear.
Reflections preserve collinearity. The image of a line is a line.
c. If B is between A and C, then the image of B is between the images of A and C.
Reflections preserve betweenness. The image of a line segment is a line segment.
d. The distance between two preimages equals the distance between their images.
Reflections preserve distance.
e. The image of an angle is an angle of the same measure.
Reflections preserve angle measure.
This postulate corresponds to Dr. Franklin Mason's "Rigid Motion Postulate," in the old version of his Lesson 3.1 last year. Since then, Dr. M has completely changed his Chapter 3 -- this is almost certainly because isometries ("rigid motions") aren't emphasized on the Common Core texts nearly as much as either of us thought they would when we first read the standards. Nowadays, Dr. M uses the classical definition of congruent polygons (i.e., equality of corresponding measures). He assumes SAS as a postulate (just as the mathematician Hilbert did a century ago), and uses Euclid's ancient proof to derive ASA. But for SSS, Dr. M still uses rigid motions to move one of the triangles into place (similar to the start of the U of Chicago proof) before using SAS and Isosceles Triangle Theorem to prove SSS.
Part a is a very important part of the Reflection Postulate. Without it, a point A could have two reflection images -- there could be two distinct points B and C such that the reflecting line m is the perpendicular bisector of both
According to the text, reflections preserve:
a nice little mnemonic for the students.
The first theorem of this chapter is the Figure Reflection Theorem:
If a figure is determined by certain points, then its reflection image is the corresponding figure determined by the reflection images of those points.
This theorem is used to conclude, for example, that if A' is the image of A, and B' is the image of B, then
But that is the sort of proof that I don't want to confuse students with. It's best just to use the informal proof given in the text and save formal proofs for later.
The text states that when a figure intersects the reflecting line, the image must intersect the reflecting line in the same point or points. This follows immediately from the fact that the image of a point on the reflecting line is the point itself.
But what I find interesting is the related statement -- if a figure intersects its reflection image, then it must intersect the reflecting line in the same point or points. This statement is false in general, but it's true if the figure to be reflected is itself a line. This fact helps us greatly -- for example, consider Question 21 from the text:
The reflection image of Triangle ABC is Triangle XYZ. Now Lines AC and XZ intersect at a point -- which we now know must lie on the reflecting line. And Lines BC and YZ intersect at a point -- which we now know must lie on the reflecting line as well. And those two points determine exactly one line -- the reflecting line! So all the student has to do is draw the line through the two points of intersection.
And as a corollary, it follows that if a line is parallel to the reflecting line, it must be parallel to its reflection image, Last year, I called this the "Line Parallel to Mirror Theorem" (where "mirror" refers to the reflecting line) But I will wait a few days before introducing that theorem to students. (Notice that the Common Core Standards state that a line must be parallel to its dilation image, so why not give the conditions when a line is parallel to its reflection, rotation, or translation images?)
Today is an activity day. Two years ago, I posted several activities after Lesson 4-4, and so I might as well reblog those activities today, in addition to today's Reflection Postulate worksheet. This is what I wrote two years ago about the activities:
Meanwhile, I created worksheets for the proof of reflection over the axes. This is all in addition to the original worksheet for today, the Centauri challenge.
The following activity is another one from Michael Serra's text -- it appears in his Chapter 15, since this one often goes with two-column proofs (and proofs appear in his book late). In this activity, strings consisting of the letters P, Q, R, and S are converted into others using four rules (that end up being our postulates):
Rule 1. Any two adjacent letters in a string can change places with each other. (PQ=>QP)
Rule 2. If a string ends in the same two letters, then you may substitute a Q for those two letters. (RSS=>RQ)
Rule 3. If a string begins in the same two letters, then you may add an S in front of those two letters. (PPR=>SPPR)
Rule 4. If a string of letters starts and finishes with the same letter, then you may substitute an R for all the letters between the first and last letters. (PQRSP=>PRP).
Then the text gives the following theorem, PQQRSS=>QRQ. (Notice that I chose to write "=>" where the text writes ">>" since, after all, we've already used the former to denote the hypothesis and conclusion of a conditional.)
1. PQQRSS 1. Given
2. PQQRQ 2. By Rule 2
3. QPQRQ 3. By Rule 1
4. QRQ 4. By Rule 4
So for the students, this is a puzzle which gets them thinking about the logical structure of proofs without having to think about geometry.
The text calls this the "Centauri challenge," which I assume refers to Alpha Centauri, the closest star system to the sun. Notice that many of the Cooperative Problem Solving challenges in Serra are said to take place in a futuristic lunar colony. For this one, the inhabitants of this colony are trying to communicate with aliens from (Alpha) Centauri, but apparently, the Centaurian alphabet consists of only four letters.
My worksheet contains all of Challenge 1, then adds Challenge 2 as a Bonus. In case you're curious, here are my answers to Challenge 2.
1. Can you produce a string of five or more letters that cannot be reduced to RQ?
My answer is that I can't -- but now I must prove it. Here is my proof, in paragraph form:
Our string has five letters, but there are only four letters available. So one of those letters must appear at least twice! (This is called the Pigeonhole Principle.) Let's call the letter that appears twice X. (I know, it's actually P, Q, R, or S, not X, but here I'm using X as a variable to stand for one of the letters P, Q, R, or S, since I want this proof to be as general as possible.)
[2017 update -- Erdos used the Pigeonhole Principle in many of his proofs in graph or Ramsey theory as well. Indeed, Case III of the convex quadrilateral proof uses it -- there are three points and two sides of the line, so two points must lie on the same side of the line.]
Using Rule 1, we take the first appearance of X and change it with the letter on its left. Now take that X and change it with the new letter on its left. Keep doing this until the X is the first letter. Now take the last appearance of X and change it with the letter on its left. Keep doing this until this other X is the last letter.
Now our string begins with X and ends with X. So by Rule 4, it becomes XRX. Using Rule 1, we can change the first X and the R to obtain RXX. Finally, this string ends in the same two letters (whatever letter the X stands for), so by Rule 2, it becomes RQ. QED
After doing this, the second question becomes obvious.
2. One of the rules of Centauri can be removed without losing any of the first five theorems proved in Challenge 1. Which of the rules can be removed?
Look at which rules have been used in the post so far, and notice which one's missing. That's the rule that can be removed. I like the similarity between determining which theorems are provable with or without a certain rule, and finding out which theorems in geometry require a Parallel Postulate. [2017 update -- this is what Russell, Whitehead, Godel, Erdos etc. did in general -- try to figure out what proofs require which axioms, or postulates, or rules.]
Here is the activity: