Saturday, January 13, 2018

bipartite art

A bipartite graph consists of two sets of nodes, N and M, where every node in N is connected to every node in M by an edge.

If you set up the nodes N and M on a pair of circles centred around the same point, you can get quite a variety of nice looking diagrams. As part of playing around with generating svg images using d3js, I set up a page that allows you to create some 'bipartite art.' Some of the diagrams turned out nicely, particularly when you hide the nodes completely:

Try it out at:

Tuesday, December 19, 2017

Hello Phyllotaxis

Phyllotaxis spirals are a favorite of recreational math - often explored in connection to other perennial topics such as the golden mean and Fibonacci numbers.

When I am trying out a new data visualization platform or programming language, I like to try to draw phyllotaxis spirals as a sort of "Hello World." My latest "Hello Phyllotaxis" came about during the early stages of learning how to use D3js. You can try it out here.

Sketched with D3js (see here)

Here is a "Hello Phyllotaxis" for Desmos (blog post here). Over the years I have also pointed to phyllotaxis-like sketches using Fathom here, Processing here, and R here.

In the image above, the dark circles are chosen by skip-counting out from the center - in this case, skip counting by 11 (see #6 in this post).

Below is a brief visual roundup of how this little "program" displays on the various platforms I have tried it on.

Sketched with Fathom (see here)

Sketched with Processing (see here)

Sketched with R (see here)

Sketched with Desmos (see here and here)

Thursday, December 7, 2017

Constructing Portia's Caskets

In Shakespeare's The Merchant of Venice, Portia tested her suitors by asking them to discover which of three caskets concealed her portrait. Inscriptions on the caskets presented riddles that would challenge the virtue of her potential mates. In his classic What is the Name of this Book? logician Raymond Smullyan, imagined several generations of clever Portias, who presented potential suitors with caskets inscribed with logic puzzles that provided the key to finding her hidden portrait.

On this page I have set up interactive puzzle generators that correspond (roughly) to Smullyan’s first three generations of Portias. In this post I thought I would describe what I find interesting about the way the puzzles are solved and  how they are generated. The code that generates and presents these puzzles is found here.

Here's the first Portia's Caskets puzzle in "What is the Name of this Book?":

One way to solve problems like this is to work backwards. For each casket, imagine that the portrait is concealed within that casket, then count how many statements would be true. When you find the portrait placement that matches the requirement that "at most one is true" you have your solution.

Building our first 348 Portia Puzzles

Working backwards also gives us a way to generate puzzles like this one. To generate these puzzles, consider all possible statements on the caskets. Each set of statements gives you three possible puzzles: a puzzle where the portrait is in the gold casket, a puzzle where the portrait is in the silver casket, and a puzzle where it is in the lead casket. To find out if any (or all) of these are valid solvable puzzles, check to see if the portrait placement gives you a 'truth count' that unique within that group of potential puzzles.

But what about generating 'all possible statements'? What are the statements that can be included? The statements on the caskets can be thought of as pointers to caskets, pointers that are either positive ("the portrait is in the silver casket") or negative ("the portrait is not in the gold casket"), and that may be directed at the current casket itself ("the portrait is in this casket").

For three caskets, this can be represented as an array of three integers. For example the statements for the puzzle above would be represented as [1, -2, -1], the 1 in the first position is saying that the first casket is pointing positively to itself ("the portrait is in this casket"). The -2 in the second position is saying  that the second casket is pointing negatively to itself ("the portrait is not in this casket"), and the -1 in the third position is saying that the third casket is pointing negatively at the first casket ("the portrait is not in the gold casket"). We can analyse this puzzle using a chart like the one below:

For the puzzle that says "there is at most one true statement," this means that the portrait is in casket 2 (silver), the only placement which gives at most one true statement. The chart also tells us that we cannot use this statement arrangement to make up other puzzles like "there are no true statements" or "there are exactly two true statements," as the other casket positions (p=1, p=3) both make two statements true.

So, generating this sort of Portia Casket problem is easy: first generate all lists of length three (for the three caskets) made up of the values -1, -2, -3, 1, 2, 3 (for positive and negative statements about the three caskets). Each of these lists gives a family of three possible puzzles - one for the portrait being in each casket. Check each possible solution to see how many statements become true with that solution. Any solution that gives a truth count that is unique in that family corresponds to a valid puzzle.

For three caskets, following this method of generating puzzles, we get a total of 348 puzzles. Some of these are trivially easy, as in the case where the clue is "all statements are true" and one of the caskets says "the portrait is in here."

Smullyan's first Portia puzzle, puzzle 67a, corresponds to the puzzle with identifier portia1-44 in our Portia I generator. Smullyan presents another puzzle by the first generation of Portias, puzzle 67b. This puzzle has statements [-2, -2, 3], and corresponds to portia1-271 in the Portia I generator.

All the Portia I puzzles are listed in this json file. Here is an example of one from the page:

Hopefully, you can see that we would represent the statements as [-1, -2, 2] and that the solution to this puzzle would have to be casket 1.

Finding 16152 more!

In "What is the Name of this Book?" a second generation Portia makes more difficult puzzles by putting two statements on each casket. Our Portia II generator is based on Smullyan's puzzle 68b.

It seems clear that these puzzles are very similar to the Portia I puzzles except: (1) there are two lists of statements, and (2) instead of giving a total count of the true statements, we describe the distribution ("on one lid both statements are true, on another both statements were false, and on a third one statement was true and one false").

If we model the statements as two lists of integers, such as [-1, -1, -3], [2 ,3, 1] for the above, and then for each pair of lists include puzzles where a portrait position generates a unique distribution, we can create puzzles like this one. It makes sense to exclude statement lists where the second set of statements is the same as the first, and to consider puzzles the same if we just exchange the first list of statements with the second list.

Following this approach we get a large number (16152) of puzzles (all of them are here). These are generally quite a bit trickier to solve than Portia I puzzles, although the method for solving them is essentially the same. Puzzle 68b above corresponds to portia2-6854, although the Portia II algorithm we used swaps the first statement list for the second (it generates as  [2 ,3, 1],  [-1, -1, -3]).

Here's an example of a generated puzzle:

For this particular puzzle we don't have to work backwards, we can take a direct approach aided by the contradictory statements on casket 2: We know this must be the one casket with one true statement. Since the statements on casket 1 cannot both be true, we know that this must be the casket with no true statements. This implies that casket 3 is the casket with two true statements. One of the statements on casket 3 says the portrait is in casket 2... so that is where it must be.

Bellini and Cellini give us 3600 more puzzles

Instead of telling us how many of the caskets have true statements, or how true statements are distributed across the caskets, what if solving another riddle was the key to figuring out which statements are true and which are false?

When attempting to generate "Bellini and Cellini" puzzles, I decided on an approach that does not quite match up with any of the puzzles in "What is the Name of this Book," but is a combination of Portia I, Portia II, and the "Knights and Knaves" puzzles that Smullyan describes in an earlier chapter (see this post).

For Portia III, we imagine that the caskets have inscriptions similar to Portia I, but also have an extra inscription that says something about the provenance of the caskets. For example, casket 1 might have an inscription "casket 2 was made by Bellini." You might think "great, that means that whatever is written on casket 2 must be true." Well, that is correct only if casket 1 was also made by Bellini, if it was made by Cellini, then whatever is written on casket 2 is actually false.

So, our Portia III algorithm is actually a set of Portia I statements with a "three-islander Knights and Knaves puzzle" layered on top of it. You can learn more about these here. For simplicity, a specific kind of "three islander" problem was chosen: the three caskets (islanders) point at each other using 2 "accusation/affirmations" and one "similarity/difference" statement. These are always (pretty easily) solvable. Maybe a future Portia puzzle generator will involve different varieties of Bellini & Cellini statements.

All the Portia III puzzles that the current algorithm generates are listed in this json file. Let's try one:

Casket 2 includes the inscription "Fashioned by a different maker than 1." If casket 2 is made by Bellini, this means that casket 1 is made by Cellini, since we can trust casket  2 in this case. If on the other hand, casket 2 is made by Cellini, then we should disbelieve the statement, and conclude that casket 1 is made by the same maker (Cellini). Hence, regardless of who made casket 2, we know that casket 1 is made by Cellini, and its inscriptions will be false. Now casket 1 makes the (false) assertion that casket 3 is made by Cellini - so we know that actually casket 3 is made by Bellini and must contain only true inscriptions. Casket 3 says that the portrait is in casket 2, so we believe it:

Working backwards is not the approach to take with Portia III: these puzzles require you to first solve the Bellini/Cellini riddle and figure out which caskets are stating the truth, and which are lying. Next you can use this information to follow the true (or false) statements to the portrait.

When generating these puzzles, the method is to first generate all possible Portia I statements. For a given set of statements, each possible solution (casket 1, casket 2, or casket 3) makes each statement either true or false. To verify if the puzzle is solvable, we need to check whether the portrait can be discovered once the truth or falsehood of each statement is known. There is a simple rule for this: if p is the position of the portrait, then either we need a statement with value +/-p or we need a statement for each of the remaining caskets. An example of a puzzle that does not work would be [1, 1, 1] where p = 2. The three caskets would all be lying (they would all be saying the portrait was in casket 1), but there would not be enough information to tell us if the portrait was in casket 2 or 3. The same statement list, [1,1,1] is a valid puzzle if p = 1. Once we have a solvable puzzle, we can overlay a few different Bellini & Cellini riddles that would say which caskets are lying and which are stating truths.

So, all told, these three Portias have given us 20100 puzzles, enough to keep us busy for a while. I had a lot of fun digging into them, and I hope you have fun with them as well - please try them out: 

Thus hath the candle singed the moth.
O, these deliberate fools! when they do choose,
They have the wisdom by their wit to lose. 
- Act II, Scene VII, The Merchant of Venice 

Saturday, November 18, 2017

the island of knights and knaves

There is classic type of logic problem where we are asked to imagine an island consisting of two types of people: those that always tell the truth (knights), and those that always tell lies (knaves). In puzzles based on this trope, the islanders make statements, and we have to figure out which islanders are knights and which islanders are knaves.

This page will generate knight and knave puzzles of varying difficulty. Here’s an example:

An "easy" puzzle from

The grandfather of all these puzzles is an actual islander who referenced an actual island  - around 600 BCE, the Cretan Epimenides is credited with the statement “All Cretans are liars.” The fun has not stopped since.

The authoritative source for island puzzles is “What is the Name of this Book?” by Raymond M. Smullyan, which builds these seemingly simple puzzles from humble beginnings (knights and knaves alone on islands) to much more complicated ones that include normals (they sometimes lie, and sometimes tell the truth), the insane (they think lies are truths, and vice versa), monkeys, clubs of knights and knaves, vampires, werewolves and other characters. Using this motley crew, the book slyly leads us to a puzzle-based statement of Godel’s incompleteness theorem.

We will stick to the first type of puzzles: knights and knaves alone on an island. How do we solve the puzzle above? Here's one way to reason it out:

A sample solution for a puzzle generated

The puzzle generation page mentioned above limits itself to knights and knaves making three kinds of statements:

Accusations and Affirmations
In an accusation, an islander A says something like "B is a knave" or an equivalent statement like "B always lies." In an affirmation, islander A says something like "B is a knight" or "B always tells the truth."

Unfortunately, we don't know if A is telling the truth or not, so how can we know if what they are saying about B is true or not? Even without knowing the type of A, we can learn something very helpful about A and B from these statements. If A and B are linked by an accusation, they must be of different types: either A is a knave and B is a knight, or vice versa. If A and B are linked by an affirmation, they must be of the same type: either both are knights, or both are knaves. See if you can reason out why this must be so.

One approach to use when solving puzzles that feature several accusations and affirmations is to draw a diagram (as described in an old post).

Knave Conjunctions
An example of a knave conjunction, is when A says "B is a knight, or I am a knave," or  "C is a knave and I am a knave."

These are very helpful statements, as they always tell us the type of both the speaker and the spoken-of. Any islander who says "or I am a knave" will be making a statement that must be, overall, truthful and is therefore a knight, while any islander who says "and I am a knave" will by lying, and must be knave. Try and convince yourself of that.

An interesting thing about "knave conjunctions" is that their usefulness comes from how close they are to the liar paradox. The liar paradox is a more self-directed refinement of that original statement of Epimenides, where we imagine that someone says: "I am lying" and wonder if they are telling the truth or not.

An islander cannot make the statement "I am lying," or any equivalent statement such as "I am a knave." If such a statement is true, then the speaker is lying, making the statement also false; and if it is false, the speaker is lying, making the statement also true. So such a statement is contradictory - neither false or true, and our islanders are restricted to statements that are either true or false.

Islanders, however, can get close to generating the liar paradox by including its statement as a clause along with another statement. To see how they can do this, you have to be clear on how statements that include "and" and "or" work. If someone on the island says "X or Y" then only X or Y has to be true in order for the whole statement to be true. So a knight can say "X or Y" when only one of the statements is true. However, if a knave says "X or Y" then both X and Y must be false (if one was true, the whole statement would be true, and a knave cannot make a true statement). If an islander says "X and Y" then both X and Y have to be true in order for the statement to be true, and only one needs to be false for the statement to be false. So a knave can say "X and Y" when only one of the statements is false.

Similarity and Difference Statements
Sometimes an islander A might say "B is my type" or maybe "C is not my type." We might not know if A is a knight or knave, but we can infer the type of B and C right away. No matter whether A is lying or telling the truth, if A claims "B is my type," then B must be a knight. On the other hand, if A claims that "C is not my type" we know that C is a knave. Do you agree?

It is interesting to compare these statements with our first type of statements, the "accusations/affirmations," these two types of statements are reciprocal in a way. When an islander directly says what type another islander is (an accusation or affirmation), all we learn is that the source and target of the statement are similar or different, without learning the actual type. However, when an islander makes a statement about similarity or difference, we learn exactly what type the target is, without learning if this is similar or different than the source.

There are many more interesting things that islanders could say, but once you have reasoned out how these three kinds of statements work, you can beat any puzzle on the page.

Friday, November 10, 2017

trihexagonal & rhombille tilings

The image above is the superposition of two tessellations. The dark bold lines show a tiling of the plane that is made up of regular hexagons and triangles (the  trihexagonal tiling).

The light lines show portions of the reciprocal (or dual) of the dark-lined tessellation.

To create a reciprocal tessellation, for every two adjacent tiles in the original tessellation, join the centers of the two tiles by a line segment perpendicular to their shared side. This line segment becomes the edge of one of the tiles in the reciprocal tessellation.

The reciprocal of trihexagonal tiling is made up entirely of rhombs, the rhombille tiling.