mathrecreation

liar-truther accusation graphs

2012-05-31T16:18:00.002-07:00

One of the puzzles concerning the island of liars and truthers (#3 in the first post about these problems) involved a bunch of islanders accusing each other of lying, leaving you to sort out who was telling the truth and who wasn't.

I decided to present the solution in a truth table (in this post), but it turns out that for this kind of puzzle the answer is presented better (and found more easily) if you use a graph. For example, the graph at the top of the page (K_24,12) could represent 24 truthers, each accusing a group of 12 liars of lying - or it might be 12 truthers and 24 liars - it's hard to tell :).

Let's say you have an "accusation" puzzle, where a bunch of islanders are directly accusing each other of lying. Let each islander be represented by a vertex, and let each accusation be represented by an edge.

Note that it amounts to the the same thing if A accuses B, B accuses A, or if they both accuse each other, so it doesn't matter who accuses who - our graph doesn't need to have directed edges. The important thing to note is that if there is an accusation between A and B, then one of them must be a liar and the other must be a truther.

We want to find out who among the islanders are liars are truthers, and maybe represent this by colouring the vertices for liars one colour, and the vertices for truthers another colour. If A and B are connected by an edge this means that one of them is accusing the other of being a liar, they can't both be liars or both be truthers - so the vertices cannot be the same colour. You may see where this is going: finding a solution to the puzzle is equivalent to finding a two-colouring of the graph.

What if in our puzzle we have some islanders "praising" other islanders - like if D says "A is telling the truth."? We might be tempted to add in a new kind of edge to represent this in our graph, but this isn't necessary. If D says that A is telling the truth, this means that both D and A must be the same - they must either both be liars or both be truthers. From the point of view of our graph, we can represent this by collapsing D and A and represent them both by the same vertex. Note that you can't have D and A accusing each other and praising each other at the same time - you get a liar paradox if you do.

Here is an example:

You meet a group of islanders and want to know whether they are liars or truthers. Alice says "Bob is a liar", Bob says "Carol is a liar" and Carol says "Bob is lying." At that moment, Dave walks up and says "Alice is telling the truth." Who are the liars and who are the truthers?

We can start by modeling the problem as a graph - with edges for "accusations" and to start with the "praise" as a dashed edge (1). Then we collapse the nodes that have a dashed edge between them (2) and finally find a 2-colouring of the graph (3).

This shows that either (a) Alice, Dave, and Carol are truthers and Bob is a liar or (b) Alice, Dave, and Carol are liars and Bob is a truther.

I think that seeing this as a coloring problem makes it more obvious what the solutions will generally look like. For example, if the puzzle hasn't "pinned" any of the islanders by saying explicitly that they are a liar or a truther, or hasn't fixed the number of liars or truthers (e.g. by saying "there are two liars" or something similar) any solution that you find will also give a "complementary" solution by reversing the colours - turning every liar into a truther and vice-versa. Also, in any puzzle where there is an accusation, the group of islanders cannot be all liars or all truthers.

In a problem like that doesn't have any islanders standing alone (i.e. not praising or accusing anyone and not being praised or accused by anyone else), if there is a solution, the graph will be bipartite. The picture at the top of this post is of a complete bipartite graph, which is what you get if all truthers are accusing all liars (or vice-versa). Here are some pictures of complete bipartite graphs, and here are some more.

A lot of liar-truther problems are not like these "accusation" scenarios. See here for more variations on the liar-truther theme.

return to the island of liars and truthers

2012-05-09T16:14:00.001-07:00

This post has the answers to the puzzles in the last post, so you might want to read that one first.

Jeff left a comment suggesting another question that I wish that I had worked into my little story of the islanders, so this late addition was put in as part 5 which you may have missed if you read the post early. Also there was an error in part 3 which meant that although you could solve the puzzle it didn't really work well, so this was fixed also. Sorry for any remaining errors.

Throughout we are assuming usual two valued logic and the law of the excluded middle, which you either believe or you don't (see here).

I seem to remember being stumped the first time that I came across my first liar and truther puzzle. If you are like me and didn't have the insight of how to solve these the first time, once you see how one is done you will still enjoy applying the same method to figuring out the others.

Before going further you may want to look at the problem in the original post.

1. Going to the Village

You can't just ask the islander what village is up ahead: she might be a liar. Instead you have to find a question whose response will (a) give the answer and (b) not depend on whether the islander is a liar or truther. One possilbe question to ask is "Is that your village?" If the answer is yes, you can rest assured that it is the village of the truthers.

A truther answers "yes" if it is the truther village because they are telling the truth, a liar answers "yes" in the same situation because they are lying. A truther answers "no" if it is the liar village because that's the truth, and again the liar would answer "no" because they are lying. Nice, eh?

2. A Bunch of Islanders

This is probably the easiest of the questions, and there are a variety of ways to figure it out. I think the clearest way to see the solutions is to make a truth table, although presenting it this way is probably more work. Our table will have columns A, B, and C representing Alice, Bob, and Carol, but it will also have columns for the statements that Alice (A), Bob (B) and Carol (C) make about each other. Whether A, B, or C is T (truther) or F (liar) must be consistent with the statements made. For example, when Alice says "Bob is a liar" either A=T and B=F or A=F and B=T. Put another way:

Regardless of who the liars and truthers are, they have to be able to make the statements that they make. So, we know we've found a possible solution when all the columns in the truth table that represent these statements are true.

From the table, the only solutions are that either that Bob is a truther and Alice and Carol are liars, or that Bob is the liar and Alice and Carol are truthers.

3. Looking for the Ferry

Now, in my original post I messed up the wording of the puzzle a bit. I have changed it so you might want to check back. You can answer the puzzle in its original wording using the same method as part 5 (below). The change is that Xavier and Yvette are from different villages and don't want to talk about them, and this change in wording forces you to use a different approach.

In the spirit of the first puzzle, you need to implicate the islanders in the question, so that when the liar is lying then this is somehow conjuncted with their answer (negating their false answer). One way of doing this is to ask Xavier "Which way would Yvette tell me to take to get to the ferry?" Whatever Xavier answers, take the other path.

If X is a truther and Y is a liar, X will truthfully reply with what Y would tell you, which would be a lie. If X is a liar and Y is a truther, then X will lie and tell you the path that Y would not have chosen for you.

4. Leaving the Island

Isaac and Jane are giving you a two statement version of the liar paradox. If they are from the island, this can't be resolved (if I then not J, but then not I, etc.). So, these two must be from off the island, and when Isaac says "Jane is a liar" he doesn't mean "Jane is someone whose every statement is false" but rather means something along the lines "Jane sometimes lies" or maybe "don't trust Jane." In any case, probably best not to hang out with these two and instead spend more time with the islanders who are at least consistent.

5. Postscript: on the Ferry

So, how can you always get the right answer out of an islander? You can generalize the method used in question 1 about the village (unless, as in the reformulated question 3 the islander refuses to talk about their village).

Suppose you want to know if a statement A is true or false. You should ask the islander "would someone from your village say that A is true?" If the answer is yes, then you know, no matter whether the islander is a liar or a truther that A is true, but if they answer no, you know that A must be false. This relies on the double negation that will happen when a liar talks about their village.

So, you could ask the captain a question like "would someone from your village say that this is the ferry to the mainland?"

on the island of liars and truthers

2012-05-07T19:42:00.000-07:00

I've been looking at logic puzzles over the last couple of days - my favorites are the ones that involve the Island of Liars and Truthers. These logic puzzles are particularly appealing because they can be thought of as variations and elaborations of the famous liar paradox, known and loved by all.

I've come across these in various places - there are quite a few examples of these on the Internet, but if you just google "liars and truthers" you get lots of hits pertaining to conspiracies, so you have to go one further and google "liars and truthers logic puzzle" or "island of liars and truthers" to find them. I didn't find the ones below in my recent searches (although you might find them ... I didn't search too hard, after getting caught up reading about the melting point of steel, etc.) - they are adapted by memory from other sources that I can't recall - they are not original and variations on them are probably pretty common.

The Island of Liars and Truthers

Preamble

Imagine that you are visiting an island on which there are only two kinds of people (other than yourself): truthers, who always tell the truth, and liars, who always lie. There are two villages - one where all the truthers live, and another where all the liars live. Although they live in separate villages, liars and truthers frequently roam about the island together and generally get along just fine. Talking to islanders is a bit difficult because they all observe the peculiar custom of not answering more than one question in a conversation and generally don't elaborate on any statements they make. Another interesting feature of these islanders is that although outsiders can't distinguish between truthers and liars by how they look, liars and truthers can always tell each other apart.

1. Going to the Village

You are on the island and see a village on the road ahead of you, and you are not sure whether it is the truther village or the liar village. An islander, who may be a liar or a truther, is standing on the side of the road. What one question do you ask her to find out if the village is the truther village or the liar village?

2. A Bunch of Islanders

Leaving the village, you meet a group of three islanders and want to know whether they are liars or truthers. Alice says "Bob is a liar", Bob says "Carol is a liar" and Carol says "Bob is lying." After that, they don't say anything else. Suppose the group consists of one truther and two liars - who's the truther? Now suppose that the group consists of two truthers and one liar - who would the truthers be? Can this group be all truthers or all liars?

3. Looking for the Ferry

You've decided to leave the island and are trying to find the ferry that will take you back to the mainland. There is a fork in the road that splits off in two directions. Two islanders, Xavier and Yvette, are standing at the fork. Xavier and Yvette are from different villages; you don't know who is from the truther village and who is from the liar village, and Xavier and Yvette won't answer questions about their villages. What question do you ask one of them to find out how to get to the ferry?

4. Leaving the Island

At the ferry you meet Isaac and Jane. Isaac and Jane are either both from the island, or else have both just come off the ferry from the mainland. Isaac says "Jane is a liar" and Jane responds "Isaac is telling the truth." Are Isaac and Jane from the island?

5. Postscript: on the Ferry

It's a slow day, and you are the only passenger on the Ferry: it is just you and the captain. As it pulls out into the harbour you realize that you might have boarded the wrong ferry - is this really the boat that is going to the mainland? You can ask the captain, an islander himself, one question to find out.

some answers here

envelope doodle design family

2012-04-24T18:59:00.000-07:00

I'm still playing with some simple designs and also with the "envelope doodle" of a few posts back. I called it "envelope doodle" because, it's a doodle that I used to do (I think in middle school every time we were given graph paper) and it's made up of a bunch of lines that form the envelope of a curve.

Both the first (the one made of squares at the top of the post) and last (the one below on the right) design look like standard tiles that you might find on any floor. Some family resemblance might be clear, but it is a little surprising that both are stages in the same series.

some simple designs

2012-04-17T20:37:00.000-07:00

The images at the top and bottom of this post were made in Processing and were inspired by some of the exercises from the 1972 book Principles of Two-Dimensional Design by Wucius Wong.

Forty years ago this book introduced a language for thinking and talking about the then emerging world of modern design. Today it could also be a source of exercises in basic programming, particularly well-suited for Processing.

Wong's later book, Principles of Three-Dimensional Design includes really nice examples of geometric sculpture, many based on polyhedral forms.

phyllotaxis multiplication colouring pages

2012-04-05T19:19:00.000-07:00

I thought I would try to make some printable colouring pages of phyllotaxis spirals - thinking that they could be coloured-in using multiplication-table / skip-counting rules to make patterns like the ones shown in the previous post. I've put a two-page pdf here - page one has a spiral with the numbers filled in (as above), and page two has one without numbers (for unfettered colouring).

When colouring these in, you might just shade the multiples of five - and get the picture below.

As you colour, you'll see the spiral pattern formed by the sequence 5, 10, 15, ...

You also can't avoid noticing a radial pattern as the dots seem to line out on spokes pointing out from the center. Take a closer look at that, and you'll see that adjacent numbers on the radial arms always have a difference of 55. Neato! this was something I didn't see when I generated the same pattern via software - using a pencil slows things down a bit and helps you notice things.

I'm wondering what relationships might be found in the patterns that other multiplicative colouring rules produce.

* * *

Postscript: I shouldn't have been surprised! Any time you make a number spiral skip-counting by n with k spokes, the difference along the spokes is going to be nk - in this case, skip counting by 5 and generating 11 spokes gives a radial difference of 55.

modulo phyllo

2012-03-29T19:46:00.001-07:00

These pictures show some number patterns in phyllotaxis spirals. The one labeled modulo 17 above has points whose index is a multiple of 7 (these are equivalent to 0 modulo 7) shown in black and all other points shown in gray.

These are related to the pictures in the last post, as when you look at primes, what you see are all the numbers that are left over when you have removed all the integers greater than 1 that are not divisible by another integer - or all the x's not congruent to zero modulo some other y.

phyllotaxis & primes

2012-03-23T18:22:00.001-07:00

I was inspired to make some pictures of phyllotaxis with primes colored in by this post at Maxwell's Demon, which has not only nice pictures, but also explanations for why the primes appear to line up on certain curves. Like the spirals there, the ones in these pictures are not numbered in the "naturally" occurring way - in a phyllotaxis spiral that occurs in nature (as in a sunflower or pine-cone) the outer dots/seeds grow first and are pushed out by the younger seeds that emerge from the center.

It is a lot easier to simulate a phyllotaxis spiral as a function of the radius (described here) rather than modeling the pushing-out from the center that happens in plants (a good description of this is found in The Book of Numbers by Conway and Guy).

Drawing number spirals with primes on them is some sort of recreational mathematics meme - the ulam spiral is the most famous, they are also drawn on quadratic spirals and others.

another way to draw simple labyrinths

2012-03-21T19:07:00.001-07:00

A while back there was a post that mentioned one way to draw labyrinths (via the method pictured above).

Here's another way. For this second method, you can use some square graph paper, and start with drawing an axis that has two units in each direction (you might call these 4 radial spokes).

From the end of each axis you draw a path as pictured below, extending the end of each axis in an alternating fashion.

If you want, you can finish off the labyrinth by extending the path you drew in step 4 down to the bottom of the grid (step 5).

Once you see the alternating pattern for extending the end of each axis, you can try without the grid, and might find that the free-hand version looks nicer.

If you wanted to extend this process to make larger labyrinths, you would need to start with more radial spokes (this example used four radial spokes - can you do it with three, or with five?).

If you'd like to have a labyrinth that brings you back to the beginning, instead of leaving you stranded in the middle, you can draw a curve through center of the labyrinth's path - you go in on one side of this new curve, and out on the other.

origami envelope redux

2012-03-17T11:22:00.000-07:00

A while ago I posted about how the traditional origami envelope could make a good object for simple math investigations. You can find instructions for the envelope on Origami USA's website. Like I mentioned, there are a lot of nice aspects to the envelope, such as the shapes in its crease pattern and its rotational symmetry. One thing to look at is the relationship between the size of the envelope and the paper you start with. It turns out that there is a nice relationship between the interior rectangle in the crease pattern and original paper (the area of the front/back of the envelope is just this rectangle with two of its corners trimmed off).

If you assume that the sides of the original paper are of length a and b, with a greater than b, you can begin to work your way towards finding out the dimensions of the resulting rectangle by following along with the folds, and using the Pythagorean theorem (actually, you just need a very special case of Pythagoras - for triangles with 45 degree angles)

Following the folds, subtracting and adding from some lengths what you deduce from other folds leads to some slightly intimidating looking expressions.

Thankfully, these simplify down to what I found to be a surprising result. The interior rectangle has an area equal to one quarter the area of the original rectangle. More surprising to me was that each side of the smaller rectangle depended only on one of the sides of the original - the longer side of the interior rectangle is 1/sqrt(2) of the shorter original side, and the shorter side of the interior rectangle is 1/(2sqrt(2)) of the longer original.

I'm sure there are probably easier ways to see this relationship. If you construct it in GSP or other dynamic geometry package you can experiment easily with increasing the side lengths.

envelope doodle

2012-03-09T18:25:00.001-08:00

prisoner's dilemma

2012-03-07T14:41:00.000-08:00

"The essence of mathematics resides in its freedom."
- Georg Cantor

In a recent article in Prospect magazine, David McConnell writes of his experience teaching basic mathematics to prison inmates. He explains, "I once decided to teach maths to prisoners. Surprisingly, many of them embraced prison discipline to study for the General Educational Development (GED) test...." Like his students, McConnell's own learning path wasn't a direct one:

I’d re-learned maths myself as a kind of tourist or traveller. Instead of the disconnected and seemingly arbitrary techniques I’d studied in the odd moments between recess and leaf collecting and dinosaur books, I found maths a unified country, sober and dazzling, when I returned to it as an adult.

Both in his return to math as an adult, and in his reflections on math's appeal for those that have lost their freedom, McConnell's piece reminded me of an article about Wole Soyinka written by Anushree Majumdar in the Indian Express a couple of years ago:

As a school student, Nigerian author Wole Soyinka loathed mathematics... Little did the Nobel Laureate know at the time, that mathematics would save him from losing his mind, during his imprisonment in 1997. “When I was imprisoned, I was thrown into solitary confinement. I had been placed under trial but it was a barren existence. I invented games in my head. I began doing mathematics again. I’d scratch on the floor of the cell with a stone, working out permutations and combinations, using different formulae. Hours would pass but it nearly drove me crazy too ... I had to create an interior life to survive”

A literary example of using mathematics as a means of hanging on to sanity (unsuccessfully) under imprisonment (and torture) is found in George Orwell's 1984. In 1984, Winston Smith clings to the one truth that he feels cannot be unmade, that 2 + 2 = 4.

Against this narrative of mathematics as liberation or resistance stands another that sees the rigor and certainty of math as an oppressive force. As Dostoyevsky's underground man says: " twice two makes five is sometimes a very charming thing too." (See the Wikipedia article on 2+2=5 for more - and also this.)

three views of K(6,12)

2012-03-03T19:39:00.001-08:00

some complete bipartite graphs

2012-03-02T18:37:00.001-08:00

a fractal family

2012-02-15T12:50:00.000-08:00

These fractals were generated in a way similar to the Mandelbrot set. For the Mandelbrot set, you use the recursive formula z_{n+1} = z_n^2 +c, where z_0 is 0 and c is an element of C. As you input c values, and perform the recursion, if the magnitudes of the results get big they are not in the set, if they stay small, they are.

For these, the same recursive formula z_{n+1} = z_n^2 + c is used, except z_0 is your input value and c is a constant complex number. Different values for c yield different fractals. As with the Mandelbrot set, initial values (z_0 in this case) are in the set if the recursion stays bounded.

My favorite so far is this last one - when I look at it, there seems to be a slight optical illusion in play that makes the dark centers seem to grow slightly as you look at the picture. In all these images, points that stayed small for more iterations are darker.

fractal island

2012-02-10T16:25:00.000-08:00

This strange blob that looks a bit like an island was generated by the same process as the Mandelbrot set described here, except instead of using z_0 = 0 in the z_n = (z_{n-1})^2 +c recursive function, it uses z_0 = (-1,1/2).

Also, following Alexandre Muñiz's suggestion in his nice post Children of Julia Sets, I used 255 iterations, and assigned the grayscale colour to the points based on which iteration they failed to converge. This gives the wispy archipelagos in the south of the island.

better late than never: Mandelbrot Set

2012-02-03T10:37:00.000-08:00

Although I have a math and computer science background, this (above) is my first attempt to draw the Mandelbrot set. It seems overdue since plotting the set is such a well established computer-math project that it's almost cliche (there is so much online about this, but this Math Munch article has some nice pointers). But if I am just finally getting around to this, then it's not too late for you too.

So, if you haven't written a little program to draw the set yet, I strongly recommend it. There are lots of nice things to think about as you explore this, and very little of it has much to do with fractals - although having that strange inkblot-like image appear at the end is the carrot, or cauliflower, that will motivate you along.

I'll share my little program and explanations below. You can find or create much more slick and efficient code if you try. For example, you don't really need to know the magnitude of the complex points (the square of the magnitude is enough), and I'm sure my point generation isn't the best.

what are you plotting?
I would guess that most students on the pre-calc train tend to think of plotting points almost exclusively in terms of single variable functions of real numbers. This is not one of those kinds of plots. Instead of (x,y) indicating y = f (x), we are plotting points belonging to C, and making them one colour if they belong to a particular set M, and a different colour if they don't.

To figure out whether a point is in M, for every point that you consider, you need to construct a particular sequence - if that sequence stays bounded, then it is in M, but if the sequence is unbounded it isn't in M. The further you go in calculating your sequence, the more sure you are that your points are actually in the set.

some complex number stuff
You don't need to write much code to implement the complex number operations that you need for this, but I'm partial to encapsulating this sort of thing in a general purpose class like this Processing example:

class CPoint {
float xvalue;
float yvalue;
float pointsize = 0.2;
CPoint(float x, float y){
xvalue = x;
yvalue = y;
}
CPoint sum(CPoint p){
return new CPoint(xvalue +p.xvalue, yvalue+ p.yvalue);
}
CPoint prod(CPoint p){
float newx = xvalue*(p.xvalue) - yvalue*(p.yvalue);
float newy = xvalue*(p.yvalue) + yvalue*(p.xvalue);
return new CPoint(newx, newy);
}
CPoint squared(){
return this.prod(this);
}
float magnit(){
float part = pow(xvalue,2) + pow(yvalue,2);
return sqrt(part);
}
void display(float xshift, float yshift, float zoom) {
fill(255,255);
ellipse(xvalue*zoom + xshift, yvalue*zoom + yshift, pointsize, pointsize);
}
}

The key item is the complex multiplication - this is what distinguishes a point in C from one in R^2.

I have a like/dislike relationship with Processing - I like how quickly things can be created and that it makes nice pictures without much effort, but I dislike how I can't seem to help breaking fundamental programming rules when using it (I end up using global variables, and always break model-view separation) - likely a personal problem, rather than a problem with Processing itself.

determining if a point is in the set M
You are going to be finding a sequence of points based on a special rule - if applying the special rule repeatedly causes the points to get too big, then they are out of the set. In addition to knowing the special rule, you also need to know "how big is too big" and "how many times will I apply the rule."

The calculation is encapsulated in this other Processing class Map. Can you figure out what the rule is, and how you specify the "how big" and "how many" parts of the calculation?

class Map {
CPoint c;
CPoint first;
CPoint current;

Map(CPoint initial, CPoint cValue){
first = initial;
current = initial;
c = cValue;
}

void iterate(){
current = current.squared().sum(c);
}

boolean iterate(int iterations, int bound){
for(int i=0; i< iterations; i++){
this.iterate();
if(current.magnit()>bound) break;
}
return (current.magnit() < bound);
}

void display(float xshift, float yshift, float zoom){
c.display(xshift, yshift, zoom);
}
}

If you only calculate for a few iterations, some points that should not be in the set get included - after 5 iterations, you get something that looks like a skate.

Having a somewhat less precise plot gives you funkier looking pictures than the crisper image that you get with more iterations. For this reason, people often include points not quite in the set in their pictures of the Mandelbrot set, and often use colour to show at which iteration a given point failed the membership test. These fuzzy pictures are really good examples of fuzzy sets - where instead of set membership being true or false, it is a range of values indicating by what level the point failed to be in the actual set. The picture below was generated using 20 iterations.

plotting the set
The last thing you need to do is to set up your window, and generate your points. Here's the main Processing file that I used for this (no zoom or panning in this one - exercise left to the reader):

//various magic numbers
int windowX = 600;
int windowY = 500;
int iterations = 200;
int zoom = 250;
int disk = 2;
int numberPoints = 100000;
int randomRange = 100;
float centerX = -0.5;
float centerY = 0;

//init
void setup() {
size(windowX, windowY);
noStroke();
smooth();
background(0);
}

void draw()
{
loop();
CPoint zero = new CPoint(0,0);
float x;
float y;
int signx = -1;
int signy = -1;
Map newMap;
CPoint cPoint;
for( int i=0; i < numberPoints; i++){
cPoint = randomCPoint(1.5, randomRange);
newMap = new Map(zero,cPoint);
if(newMap.iterate(iterations, disk)){
newMap.display(windowX/2 - centerX*zoom, windowY/2 - centerY*zoom, zoom );
}
}
}

CPoint randomCPoint(float bound, float depth){
float x;
float y;
int signx = -1;
int signy = -1;
x = bound*random(depth)/depth;
y = bound*random(depth)/depth;
if(random(100)<50) signx *= (-1);
if(random(100)<50) signy *= (-1);
return new CPoint(signx*x,signy*y);
}

If you implement it like this, the set will slowly emerge as more and more points are tested. This crisper version below used 200 iterations (maybe a bit excessive).

primes on a log spiral

2012-01-11T11:29:00.000-08:00

Since looking again at Theodore Andrea Cook's The Curves of Life a few posts back I've been planning on playing with logarithmic spirals, which are identified in that book as the type of spiral that you often encounter in nature and in architecture. I was inspired to finally spend some time with them after reading a recent post on Math Hombre.

For fun I treated the curve like a number line and plotted prime numbers on it using Processing. It seems to me that a nice thing about curling up the number line is that it allows you to take in more of the line at a glance. You can notice both the (seemingly) increasing gaps that occur between primes, as well as the (apparent) persistent occurrence of twin primes.

the best of 2011

2012-01-10T14:45:00.000-08:00

Once in a while I get sent books to review and recommend - this is very nice, but unfortunately I haven't had the chance to post many reviews. It is not only in the book review department that I'm failing - I seem to be having a more general problem finding time to do any recreational mathematics (and then to post about it here).

So it was a treat to receive a copy of The Best Writing on Mathematics, 2011 (Mircea Pitici, ed.), a book that solves both problems: it is a book that I really have to recommend, and it is also certain to inspire me in more mathematical recreations.

The anthology gets off to a good start: In the forward, eminent physicist Freeman Dyson proclaims that "Recreational mathematics is a splendid hobby which young and old can equally enjoy... To enjoy recreational mathematics you do not need to be an expert." A great statement that I should probably take as the motto for this blog.

This anthology offers a lot for recreational mathematicians, mathematics educators, professional math practitioners, and hopefully others as well. A couple of the articles in the collection were "old favorites" that inspired posts on this blog when they appeared in their original contexts. Doris Schattschneider's article on Escher and Coxeter prompted this post, and Dana Mackenzie's article on Apollonian gaskets motivated this one and another. These articles remain among my favorites in the collection, but there are many others that make interesting reading, including others like these that focus on aesthetic aspects of mathematics (in ribbed sculptures, in bronze and stone, and in strange-attractors).

Some of the articles are against the grain of our prevailing zeitgeist - Melvyn B. Nathanson strikes a somewhat contrarian tone against the promises of polymath, and Martin Campbell-Kelly wistfully recalls the now obsolete numerical table. I particularly liked how Underwood Dudley asks "What is Mathematics For" and takes aim at an assumption that is now almost sacrosanct: that we teach mathematics because it is useful.

Dudley's thesis, that mathematics (particularly school algebra) may not be used very often but helps us learn to think and reason, although not currently popular, is actually one of the oldest arguments in favor of learning algebra. The very first English-language algebra textbook (published by Robert Recorde in 1557) was titled "The Whetstone of Whitte" precisely because algebra was considered by its author to be like a knife-sharpener for the brain. Of algebra, it said:

Its use is great, and more than one.

Here if you lift your wits to wet,

Much sharpness thereby shall you get.

Dull wits hereby do greatly mend,

Sharp wits are fined to their full end.

I think that many who appreciate the appeal to the aesthetics and cultural significance of mathematics in Lockhart's Lament will agree with Dudley's call for a more subtle (and accurate) understanding of what mathematics education gives us beyond the merely utilitarian.

Although there is a broad appeal to these articles, I'm guessing that the audience that will most appreciate this collection are those involved in mathematics education. Of particular interest to teachers are two career retrospectives by eminent math-education- theorists Alan Schoenfeld and John Mason, the previously mentioned paper by Underwood Dudley, two other papers specifically about mathematics education, as well as a paper on the cognitive aspects of perceiving numbers.

Thinking about these kinds of articles, I was reminded that when Martin Gardner died in 2010, many wrote about how his columns inspired them to take up mathematics as a hobby and as a profession. With Gardner as an example, it is clear that the authors of these and other popular mathematics articles are doing something worthwhile.

more window patterns in gsp

2011-11-20T18:40:00.001-08:00

You'd be right in saying 'hey, these are just a bunch of overlapping squares.' Yes. The only redeeming thing that I can point to is that they are made by following a rule, and the rule is one that is easy to reproduce without using any external measuring device (like a ruler or protractor), only the squares themselves. Think origami: you find midpoints by folding, etc. In this case, GSP is used, but only simple constructions like mid-point finding and segment creating.

The trick is to find a rule that allows you to start with a square and then construct two points that you can base another square on, and then repeat.

These were made from the same window-pattern instructions mentioned here.

A4 window patterns and special triangles

2011-11-11T20:30:00.001-08:00

A short while ago I mentioned that A4 paper has nice proportions - it's a silver rectangle, which means that the ratio of its long side to its short side is sqrt(2). Because of their nice proportions, silver rectangles can be used to construct special triangles that we know and love from trigonometry.

One nice way to note the angles in these triangles is to form window patterns based on them - these are shapes made from overlapping pieces of paper that have been rotated according to a rule. The term window pattern comes from William Gibbs - so named because if you put them up in a window, the light shining through the different layers of paper reveals additional patterns and shapes.

Here's one example of the special-triangle-window-pattern process. Start with an A4 or similarly proportioned rectangle, and find the midpoint of one of the shorter sides (by folding the paper, for example).

Now take a second rectangle the same size, and place it so that one vertex lines up with the midpoint drawn, and the other vertex along the same short side of the second rectangle touches the long side of the first. It's easier to see this in a picture:

By doing this, you've constructed the tricky length of sqrt(3)/2 and built the 30-60-90 (pi/6, pi/3, pi/2) triangle. You can confirm that the angle that you've formed a 60 degree triangle by repeating the process and finding that you come "full circle" after 6 pieces of paper (360/6 = 60).

If you change the first placement a bit so that the second rectangle lies mostly across the interior of the first, you get the pattern at the top of the post.

These are nice patterns, but they don't actually use the special properties of A4 (you could do a similar thing with square or letter paper). A little more complicated placing of one rectangle over the other can allow you to create a right triangle with one leg equal to 1 and the other equal to sqrt(2)-1. This is not one of your "standard" special triangles, but it is special in that it allows you to calculate exact values of certain angles (which angles, we'll find out when we complete our pattern).

Here's what the placement looked like that constructed this triangle. I'm afraid that text instructions for the placement would be just too much for this post - maybe you can figure out how it is done from the diagram :).

If you continue placing the rectangles, you will find that it takes 16 of them to come back to the start, which tells us that our triangle contains an angle of pi/8 or 22.5 degrees - the others are pi/2 (90) and 3pi/8 (67.5).

So.. our new special triangle tells us, for example, that tan(pi/8) is equal to sqrt(2)-1 (what other exact values do we get?).

animated iterations - too much fun

2011-11-11T10:38:00.001-08:00

Not much to this post - just playing with the GSP sketch that I pointed to earlier. These are just iterations, plus animation, plus tracing, with what I think are some nice results.

The 'colored Pythagoras tree' fractal below is a classic that I learned in a GSP workshop years ago, and it's based on one of the projects in the free booklet 101 Project Ideas for GSP. I'm sure there are some instructions for the whole thing floating around somewhere. [Update: See the Nov 15th blog post at sine of the times for some instructions on the basic tree.]

The image below is a later stage of the one at the top of the post - an iteration made up of pentagons and curves - the bottom image shows what the first generation of this iteration looks like.

some gsp fractal sketches

2011-11-05T19:32:00.000-07:00

I found an old GSP file with a bunch geometric fractals in them - I thought that some of them looked nice, so I've posted them here. If you'd like to try them out, you can get the GSP file here - for the most part, they involve pretty standard use of the "iterate" feature.

Animating them in random ways creates some strange looking forms - the same sketch that produces the pentagon fractal above also gives the one below.

The same sketch that gives the snowflake-like pattern at the top of the post gives this odd looking sponge:

butterflies, bus transfers, cotangents

2011-11-02T19:16:00.000-07:00

Because it is my habit to do paper-folding while using public
transportation, people sometimes turn their heads and cast pitying
eyes upon me; but because of my strong concentration
at these times, their looks do not bother me.

- Kazuo Haga, Origamics

The origami model that I fold most frequently is Nick Robinson's A4 butterfly. You can find this model in Nick's book The Origami Bible (unfortunately I don't think that the instructions are posted on his website). Being in North America, A4 paper is not so easily obtained, but luckily I get handed a little piece of almost-the-same-ratio-as-A4-paper every workday morning in the form of a bus transfer.

Rectangles that have the same proportions as A4 paper have nice geometric properties - they are silver rectangles (named in contrast to golden rectangles), and the niceness of these silver rectangles is due the fact that the ratio of the long side to the short side is sqrt(2). If you don't have an appropriately proportioned bus transfer, or you want to make your own A4-style silver rectangle, Nick Robson provides some helpful instructions here.

Really, you don't need a perfect silver rectangle for the butterfly model - it is pretty forgiving, and tends to work well for bus transfers, ticket stubs, and magazine-subscription inserts (golden rectangles, for example, work too). However, if you look at the simplified crease pattern you can see that the model completely breaks down if the ratio of long to short side is too small or too large. To make things precise, things don't work at all if the ratio of long-side to short-side, r, approaches r = cotan(pi/4) = 1 on the low end, or r = cotan(pi/8) ~= 2.414 on the high end.

The reason that these ratios are as they are is that the fold that creates the outer edge of the wing has an angle of pi/4 with the midline of the paper, while the fold that creates the inner edge of the wing has an angle of pi/8. If either of these lines hit the corner of the rectangle, the model no longer works. That is why the ratio of long-side to short-side (or in trig-ratio speak, adjacent to opposite) is bounded by the cotan of these angles.

Butterflies attempted with almost-square paper have large bodies and almost no wings, while the long paper produces butterflies that have too-long wings and undersized bodies. Although it seems that the model enforces sharply defined boundaries on the range of paper can be used, finding the size of paper that produces the optimal butterfly is another problem. Are silver-rectangle butterflies the best, golden ones, or maybe ones with r = cotan(3pi/16)? This might be a question of personal origami-aesthetics rather than mathematics.

tintin - the calculus affair

2011-10-20T14:29:00.000-07:00

There is a larger version for printing here - I suggest that you just print the first 4 pages.