Saturday, September 15, 2018

Solving (some) Logic Puzzles with Sets

As you may have noticed, since around this time last year, I have been playing around with generating puzzles based on those found in some of Raymond Smullyan's books. This has included Knights and Knaves, Portia's Caskets, The Case Files of Inspector Craig, Tigers and Treasure, and The Isle of Dreams. Some of the differences between puzzles are superficial: A "Portia's Casket" puzzle can be recast as a "Knights and Knaves" puzzle, for example. Even though there is some common deep structure to these various puzzles, I've found that sometimes the puzzle types call out for different approaches when writing solvers or generators.

The latest puzzle type that I have been enjoying is based on some puzzles found in Smullyan's What is the Name of This Book?. The "Lion and the Unicorn" puzzles are built around characters from Lewis Carroll's Through the Looking-Glass, and What Alice Found Thereand for this logic puzzle variation, I found that using sets to model the puzzle (rather than, say, propositions, truth tables, or graphs) seemed to make the most sense.

The Lion and the Unicorn, posing
at the East Block 

As described in the chapter 47 Alice and the Forest of Forgetfulness,
When Alice entered the Forest of Forgetfulness, she did not forget everything, only certain things. She often forgot her name, and the one thing she was most likely to forget was the day of the week. Now, the Lion and the Unicorn were frequent visitors to the forest. These two are strange creatures. The Lion lies on Mondays, Tuesdays, and Wednesdays, and tells the truth on the other days of the week. The Unicorn, on the other hand, lies on Thursdays, Fridays, and Saturdays, but tells the truth on other days of the week.
One day, Alice met the Lion and the Unicorn resting under a tree. They made the following statements: 
Lion: Yesterday was one of my lying days. 
Unicorn: Yesterday was one of my lying days too.
Alice must know: What day is today? 
If you think you have a solution to this - test it out on the interactive version of the puzzle.

If we model this using sets, our universe of discourse for this problem is the days of the week.


We consider the set L of days for which the lion is lying, and the set U of days for which the unicorn is lying.

The days that the animals are telling the truth are listed in the complements of each set.


These two sets have an empty intersection - the two characters never lie at the same time. The intersection of their truth-telling days is non empty, however: both tell the truth on the same day once a week.
The set Days is a set with structure, the days are an ordered set - the lion and the unicorn can talk about 'yesterday' and 'tomorrow.' For any set of days we can ask for its 'tomorrows' - the set of next days, or its set of 'yesterdays', the set of preceding days. When the lion says "I told lies yesterday" this can be translated as "today is a tomorrow for one of my lying days." The set of days covered by Lion's statement would be:

But do any of the days covered by Lion's statement coincide with a day that he is telling the truth? To believe his statement about what day it is, it must describe a day that he is actually speaking truthfully. If Lion is telling the truth, it must be a day in the intersection of the days in Lion's statement and the set of Lion's truthful days.

But, Lion could be lying. If Lion is lying, then today is in the intersection of the days not in Lion's statement, and Lion's lying days.
Since we don't know whether the lion is telling the truth or lying, we have to consider both possibilities, so the set of days that it could be, based only on Lion's statement is:


Going through a similar process, we can get another set of days based on the Unicorn's statements.


Days that fall in both the set from the Lion and the set from the Unicorn are possible solutions for today's day - if the intersection is empty, then there is no solution, if there are several days in the intersection, then the puzzle is ambiguous, if there is a single day in the intersection, that is today:


The notation might make this way of thinking seem difficult - here is the process stated a bit more plainly (see that it lines up with the formula above...):

1. Consider the Lion. Which days does the Lion's statement refer to?
2. Of these days, which coincide with Lion's truthful days?
3. Which of the days are not covered by the Lion's statement? Do any of these coincide with Lion's lying days?
4. Combine these two lists of days from the Lion.
5. Follow steps 1 through 4 for the Unicorn to produce a list of possible days from the Unicorn.
6. If there is one day that that is in both the Lion's list and the Unicorn's list, that is the solution.

We can come up with variations on this puzzle by varying the statements made by the Lion and the Unicorn. Instead of saying "I told lies yesterday," we could have them say things like "I will tell truths tomorrow" or "today is a week day", or even "today is Wednesday." Some of these will generate good puzzles (one element in the final set), others may not.

A Jupyter notebook that generates 132 puzzles like this can be found here, and you can the puzzles out over here.

Thursday, September 6, 2018

generating celtic knot patterns

This post describes an algorithm for generating celtic knot patterns - ornamental knots, links, and braids that are laid out in a grid, like the one below:


If you would rather skip reading about how these are generated and start playing around with creating patterns like the one above, please try out the editor and random knot-pattern generator that I've posted on my github pages.

I have tried out various strategies for generating these patterns (for example,  using tiles), but the method described here is closest to how I like to draw them by hand, as described in the book by Aidan Meehan, Celtic Design: Knotwork - The Secret Method of the Scribes. The variation offered here is intended to suggest how to write a program to generate these patterns based on a simplified version of the techniques in Meehan's book.

A knot pattern is made up of strands that represent string or chord, and the gaps between the woven strands. The technique described below actually involves drawing the gaps, with the strands emerging out of the negative space between the gaps. Essentially, a grid of dots are drawn, and lines are selectively drawn between adjacent dots - these become the gaps between the strands. Additional rules are applied to connect the dots to create a woven effect, and the dots are replaced with polygons to  create a more stylised effect.

1) define primary grid points
A knot pattern is laid out on a square coordinate system using a set of "primary" points that are set at one unit distances in the horizontal and vertical directions. We'll say that (0,0) is the top left corner of the grid, and the positive x direction is towards the right and positive y direction is down.  The dimensions of the primary grid must be odd (there must be a total odd number of dots in both the x and y directions). Because we are starting with (0,0) in the top left, the top right point (x, 0) must have x even (4 in the example below), and the bottom left point (0,y) must have y even (6 in the example below).

the primary grid

(Note: In Meehan's account, things are layered a little differently so what we are calling the primary grid is referred to as the tertiary grid.)

2) identify secondary grid points
Some of the points on the grid are special - these form a secondary grid. The special secondary grid points are those where both x and y values are even, or both are odd.

the secondary grid

In step 4 below, the secondary grid points where both x and y are even will be referred to as even nodes, and those that have both x and y odd will be referred to as odd nodes. The requirement to have the primary grid have odd dimensions (step 1) was needed to ensure that the corners of the pattern are all secondary points.

3. draw a quadrilateral around the nodes
Each node will become a gap in the node pattern - the basic shape of a gap is quadrilateral whose vertices lie 1/4 unit above, below, and to the right and left of each node.

the basic node polygon

With all of the polygons drawn for the nodes, we get a grid of 'diamonds' like this:

node polygons drawn for
secondary grid points


4. extend lines from node polygon vertices
To create a woven affect, we extend lines from the vertices of each node polygon


Doing this for all nodes creates an image like the one below.

lines extended from node
polygon vertices

If you exchange the rules for odd and even nodes, you end up with a correct "opposite" weave: strands that were going under instead go over, and vice-versa.

5. place barriers, drop lines
In the above image, the simple woven pattern seems to extend off the sides. To create an edge boundary for the pattern, and to create more interesting twists and turns, we follow some rules for drawing boundaries.

boundary rule 1: A boundary can connect any two non-diagonally adjacent nodes (secondary points), as long as rule 2 is not violated. The midpoint of a boundary segment will be a primary point.

boundary rule 2: A primary point cannot have more than one boundary going through it.

The example below shows boundaries drawn along the edge of the image, as well as some internal boundaries.
legal boundary examples, showing
primary and secondary points
(node polygons are hidden)

Now that we have introduced boundaries, we refine how lines are drawn coming out of the nodes (adjusting step 4):

node-line rule: Only draw a line from a node vertex if there is no boundary across from the vertex.

Applying the node-line rule, and drawing the polygons (and dropping the primary grid points) we get an image like the one below, where the weaving respects the boundaries - the strands (in white) that emerge seem to bounce off the edges and twist to avoid internal boundaries.

node polygons, boundaries, and lines 

6. refine node polygons
We can apply some styling rules to make the pattern look smoother - these changes to our original node polygon (step 3) will be based on whether or not there are boundaries next to the node.

node-style rule: Truncate (chop off) the vertex of a node polygon that is next to a boundary.

Below is the same pattern above, but with the node polygons following the node-style rule. You can see the effects of the rule most clearly by looking at the nodes near the edge of the image, and particularly the corner nodes.

pattern using truncated
node polygons

It is possible to add further adjustments to how the nodes and lines are drawn to create smoother looking knot patterns. I have experimented a bit, but have not obtained great results. Here's an example of the same patter above that adjusts the node polygons and line thicknesses:

a slightly different style applied
to the knot pattern

I hope you enjoy playing around with this - either implementing the process described above yourself or playing around with my version: there is a simple editor available here, and one that allows for further adjustments of the size and dimensions here.


Wednesday, June 6, 2018

origami workshop again

A few years back, I posted some notes about an origami workshop that I had run with some middle school students. Last week I had the opportunity to run origami workshops with similar groups of students, using many of the same models I mentioned before (including the hopping frog).



One nice model that I used this time that is not mentioned in that other post is the multiform (aka windmill/pinwheel) -  a flexible hinged surface from which several simple models can be folded, including the windmill, the house, and the pajarita, and which can be extended to form the masu box and others.


Thursday, May 17, 2018

The Isle of Dreams

After a short break, we are returning to some logic puzzles inspired by those of Raymond Smullyan. Earlier we visited the island of knights and knaves, looked into Portia's caskets, explored the case files of Inspector Leslie Craig, and looked behind doors for tigers and treasure. In this post, we are visiting the Isle of Dreams. As Smullyan says in his book The Lady or The Tiger?:
I once dreamed that there was a certain island called the Isle of Dreams. The inhabitants of this island dream quite vividly; indeed, their thoughts while asleep are as vivid as while awake. Moreover, their dream life has the same continuity from night to night as their waking life has from day to day. As a result, some of the inhabitants sometimes have difficulty in knowing whether they are awake or asleep at a given time.  
Now, it so happens that each inhabitant is of one of two types: diurnal or nocturnal. A diurnal inhabitant is characterised by the fact that everything he believes while he is awake is true, and everything he believes while he is asleep is false. A nocturnal inhabitant is the opposite: everything a nocturnal person believes while asleep is true, and everything he believes while awake is false. 
On this island then, each islander has a type (diurnal or nocturnal), and a state (awake or asleep), and based on their type and state, you can assess the veracity of their thoughts (either true or false).

To play around with this, I decided to make some puzzles similar to ones found in The Lady or The Tiger?, but based on the thoughts of two islanders A and B. Each islander has two distinct thoughts: one about themselves (either about their state or their type), and one about the other (either their state or their type). Importantly, these thoughts occur to both A and B at exactly the same time. Here is an example:

  • Islander A has two distinct thoughts at the same moment: I am nocturnal. B is diurnal.
  • At the same moment, islander B has these distinct thoughts: I am awake. A is diurnal.
We want to know: what is the actual type and state of both A and B? Can we know everything about them, or is their something about them that we cannot tell? Or maybe these thoughts are impossible, and lead to contradictions.

To solve these kinds of puzzles, it helps to know the Four Laws of the Isle of Dreams:

  1. An islander while awake will believe they are diurnal.
  2. An islander while asleep will believe they are nocturnal.
  3. Diurnal inhabitants at all times believe they are awake.
  4. Nocturnal inhabitants at all times believe they are asleep.
Let's just establish the first law, and then you should try to convince yourself of the others. Consider the case of a diurnal awake islander: because they are diurnal and awake, they think true thoughts, so they will correctly think that they are diurnal. Second, consider the case of a nocturnal awake islander: because they are nocturnal and awake, they will think false thoughts, and will conclude that they are diurnal. So, no matter whether an islander is diurnal or nocturnal, when awake they will think they are diurnal (some correctly, some falsely). Using this along with rule 2, if an islander thinks they are diurnal, you should conclude that they are awake.

Now back to the puzzle:

  • Islander A has two distinct thoughts at the same moment: I am nocturnal. B is diurnal.
  • At the same moment, islander B has these distinct thoughts: I am awake. A is diurnal.
Applying the four laws of the Isle of Dreams to the first thoughts of the islanders in the puzzle above, we know that A must be asleep (law 2) , and that B must be diurnal (law 3). Now turning to A's second thought: because they are asleep and thinking something that is true (B is diurnal) A must be nocturnal. B's second thought is not true, so since they are diurnal they must be asleep. So, A is nocturnal and asleep, while B is diurnal and asleep. Sweet dreams, A and B.

How many puzzles can we make like this, where we have two islanders, each thinking something about their state or type and something about the state or type of the other? Well, there are 4 possible thoughts an islander could have about themselves (I am awake/asleep/nocturnal/diurnal) and 4 possible thoughts about the other (They are awake/asleep/nocturnal/diurnal), giving us 16 pairs of thoughts. Since there are two islanders involved, this gives 256 puzzles (really only 128 truly different puzzles, as A and B are interchangeable).

You can try out all 256 of them, or as many as you like, here. They look something like this:



This collection of puzzles has some interesting features. There are 192 that are completely solvable: you can find the type and state of both A and B from the thoughts that they think (like the example above). There are 32 partially solvable puzzles, where the first thoughts of A and B (their thoughts about themselves) tell us something about their state and type, but their second thoughts (about the other islander) are inconclusive. Finally, there are 32 puzzles included in the set where the thoughts of A and B are contradictory, and therefore impossible. We can include these contradictory items in the set, as the question page gives you the chance to identify these nasty puzzles.



It turns out that the distribution of the partially solvable and impossible puzzles display an interesting pattern. Let's consider all 16 pairs of thoughts, and make a graph showing which combinations are (a) completely solvable, (b) partially solvable, or (c) impossible.

Here are the 16 pairs of thoughts an islander might have:

1: I am awake. The other is awake.
2: I am awake. The other is asleep.
3: I am awake. The other is diurnal.
4: I am awake. The other is nocturnal.
5: I am asleep. The other is awake.
6: I am asleep. The other is asleep.
7: I am asleep. The other is diurnal.
8: I am asleep. The other is nocturnal.
9: I am diurnal. The other is awake.
10: I am diurnal. The other is asleep.
11: I am diurnal. The other is diurnal.
12: I am diurnal. The other is nocturnal.
13: I am nocturnal. The other is awake.
14: I am nocturnal. The other is asleep.
15: I am nocturnal. The other is diurnal.
16: I am nocturnal. The other is nocturnal.

Let's create a graph where the horizontal axis represents A's thoughts and the vertical axis represents B's thoughts. A white square on the graph represents a completely solvable puzzle for that x/y combination of thoughts, a grey square on the graph represents a partially solvable puzzle, and a black square represents an unsolvable puzzle.

solvable, partially solvable, and unsolvable
Isle of Dreams puzzles

This is really neat: the partially solvable and unsolvable combinations form an interesting pattern dispersed through the whitespace of the completely solvable puzzles. There are 16 "problem squares" of 4 puzzles that have a distinct symmetric pattern, and these 16 problem squares are arranged in 4 sets of 4 puzzles that also have an interesting symmetry.

We'll zoom in on one of the "problem squares" to give a better picture of what the graph is displaying:


Let's look at one of the contradictory puzzles - the one in the bottom left of this "problem square."

A is thinking #1: I am awake. The other is awake.
B is thinking #5: I am asleep. The other is awake.

From their first thoughts, we know that A is diurnal and B is nocturnal. If A is awake, they they will think true thoughts and consequently B is awake. If B is awake, they must be thinking false thoughts, requiring A to be asleep - a contradiction. On the other hand, if A is asleep, they will be thinking false thoughts, so B will be asleep. B will then be thinking true thoughts, requiring A to be awake, again a contradiction.

But why to these partial/contradictory puzzles form the patterns that they do? Maybe we will return again to the Isle of Dreams someday to find an answer.

Try the puzzles out here: https://dmackinnon1.github.io/inspectorCraig/dreamers.html

Sunday, May 6, 2018

more bipartite art


Playing around with some of the images created by connecting two sets of dots. In this case, every dot from the second set is connected to every dot in the first set, and the two sets are arranged in concentric circles. In the picture above, the first set of dots has 12 equally spaced dots in a circle, and the second set has 48, but the second set is arranged on a circle whose radius is much, much larger than the first, so the lines from the second set to the first come in from a great distance.

If both sets have 3 dots, both sets are on concentric circles, and one of the sets is on a much larger circle, you might get something like this:



The second set is so far out, that it looks like the lines from a point the second set are parallel. If the first set has 3 points and the second far-out set has 6 points, you might get something like this:



Increasing size of the far-out set to 18 points:



Can you figure out the number of points in each set that would generate an image like this? You can test out your guesses here.



Friday, April 27, 2018

some Chessboard Puzzle solutions

In the previous post I mentioned some mathematical chessboard puzzle puzzles, created as part of working through the book Across the Board, by John J. Watkins. This post provides some possible solutions to the puzzles on that puzzle page.

Queens on a 5 by 5 board

The puzzle "Place 3 queens on a 5x5 chessboard. The board must be dominated," is asking you to find the minimal dominating set for queens on a 5x5 board (3 is the queen's domination number for 5x5 boards). Here are two solutions:

The large dots show where the queens are placed, and a small dot appears on every square that is reachable by a queen. In the solution on the left, all three queens can be attacked, but in the solution on the right, the queen in the corner is uncovered.

The puzzle "Place 5 queens on a 5x5 chessboard. The board must be dominated. The pieces must be independent," is asking you to find the maximal independent set for queens on a 5x5 board (5 is the queens independence number for 5x5 boards). Here's a solution:


There is also a puzzle that asks you to find an arrangement between the domination and independence numbers, "Place 4 queens on a 5x5 chessboard. The board must be dominated. The pieces must be independent." Here is one solution for that:



Queens on other boards

On a 6x6 board, our queen puzzles will be bounded by the domination number of 3 and the independence number of 6. Here are solutions for those:


In between these, we have "Place 5 queens on a 6x6 chessboard. The board must be dominated. The pieces must be independent;" here's a solution for that one:


Just to get a sense of what solutions to these might look like in general, let's jump up to 8x8. In this case, the domination number for queens is 5, so the puzzle in our set with the fewest queens on 8x8 is "Place 5 queens on a 8x8 chessboard. The board must be dominated. The pieces must be independent." Here is one solution:


This particular solution is of interest because the pieces are in a pattern known as the Spencer-Cockayne construction, which can be used to find coverings of square boards of side length 9, 10, 11, and 12 as well. More interesting details can be found in Across the Board.

Knights on a 5x5 board

There are plenty of "independence and domination" problems for the knight on a 5x5 board, because the gap between the domination number (5) and the independence number (13) is so large (compared to queens on the 5x5, at least). Finding solutions for some of the intermediate numbers is a bit tricky, you may find. 


For example, here is a solution to "Place 9 knights on a 5x5 chessboard. The board must be dominated. The pieces must be independent":


Knights on other boards

All puzzles based on the maximum number of independent knights on a board have the same solution: put a knight on every square of the colour that has the most squares (on odd boards, one colour has more squares than the other). 


Here is an example of a puzzle based on a "sub-optimal" dominating set that is also independent: "Place 11 knights on a 6x6 chessboard. The board must be dominated. The pieces must be independent." And a solution:


Bishops on 5x5

Of the remaining pieces that we have puzzles for, bishops, kings, and rooks, the bishop is the most interesting, and the 5x5 board gives a good idea of how to construct the puzzle solutions.

Consider these two puzzles:

"Place 5 bishops on a 5x5 chessboard. The board must be dominated. The pieces must be independent."

"Place 8 bishops on a 5x5 chessboard. The board must be dominated. The pieces must be independent."


The minimum dominating set for bishops on a 5x5 board has 5 pieces, and the maximum independent set has 8. In between these, we can also form puzzles based on non optimal dominating sets (that are also independent), such as:

"Place 6 bishops on a 5x5 chessboard. The board must be dominated. The pieces must be independent."

"Place 7 bishops on a 5x5 chessboard. The board must be dominated. The pieces must be independent."


Solutions for finding similar solutions for bishops on boards of other sizes follow the same patterns as those on the 5x5 board.

Hopefully, these examples will help you out if you get stuck on any of the puzzles. As mentioned earlier, if you are interested in learning more about the mathematics behind these puzzles, check out Across the Board.


Related posts and pages
domination and independence puzzles 
post introducing chessboard puzzles
chess tour puzzles
post on chess tour puzzles


Tuesday, April 24, 2018

Mathematical Chessboard Puzzles

Chess problems are compositions where a set of pieces are arranged as if in a game and a specific goal is set - the problem is to determine how to get from the arrangement to the end goal. An interesting variation on the traditional chess problem are the retrograde analysis chess problems of Raymond Smullyan, where instead of a goal being set, a question is asked about the conditions that may have lead to the arrangement (a backwards looking problem, rather than the traditional forwards looking type). Mathematical chessboard problems are completely different than these traditional chess problems, and bear little connection to the actual game of chess - they are more concerned with the structure of how particular pieces can move on the board, and ask questions about how a single piece can move about the board, or about what positions are reachable by collections of the same type of piece. These problems are questions in graph theory in (thin) disguise, and have attracted the attention of both professional and recreational mathematicians.

A useful and very readable guide to mathematical chessboard problems is Across the Board, by John J. Watkins. I’ve been playing around with knight tours for a few years, and since picking up this book a while back, I have been returning to it again and again to learn new and interesting things about them. Although I had heard about other mathematical chessboard problems, like the eight queens problem, Across the Board introduced me to the general category of chess independence and domination problems and encouraged me to learn more about them.

A group of chess pieces of the same type is said to dominate a board if every square is either occupied or a neighbour (reachable in one move) of an occupied square. A group of chess pieces of the same type is said to be independent if no piece is a neighbour of any other piece. Domination (sometimes called covering) problems are, generally, to find a minimal dominating set, for example, the smallest number of queens required to dominate a board. Independence (or un-guarding) problems generally require you to find a maximal set of pieces that can be placed and remain independent; the greatest number of knights, for example, that can be placed so that no knight attacks another.

Uninteresting examples of dominating queens and independent knights.
A minimal dominating set and a maximal independent set would be more interesting

As part of working through Across the Board and understanding chess domination and independence, I tried to create an interactive ‘mathematical chessboard puzzle’ set (try it out here). Here is a screenshot example:

An example puzzle from the online set.
 The solver is not off to a good start.

What is the difference between a mathematical chessboard problem and a mathematical chessboard puzzle? When considering the problem of queens independence, we would expect a serious treatment: a solution which finds the maximum number of independent queens for boards up to a certain size, an algorithm or method for generating maximal independent arrangements, and for some cases that remain unsolved by methods provided, some way of placing bounds on the independence number. A puzzle based on the idea of queens independence is a much simpler thing: merely an instruction like “Find a way to place 8 queens on an 8x8 board so that the board is dominated and the queens are independent.” Across the Board provides a great survey of results that mathematicians have used in tackling the problems of finding dominating sets, independent sets, and tours. The rest of this post is about puzzles (like the example above) that are generated from those results.

In puzzles inspired by the problems of domination and independence, we want to ask the solver to come up with arrangements of pieces of a single type, constrained so that the pieces either dominate the board, are independent, or both. Recall that the domination problem is looking for a minimum number of pieces required to cover the board (either by placement or by attack), while the independence problem is looking for a maximum number of pieces that can be placed independently. For example, for queens on a 5x5 board, the domination number is 3, but the independence number is 5. So for queens on a 5x5 board, our puzzles will require placements of sets ranging from 3 to 5 queens.

There is an asymmetry between domination and independence that we have to keep in mind: A solution to the domination problem might not be independent, but the maximal independent set will always be dominating. The example of 3 queens on the 5x5 board shows that you cannot always make your dominating set independent. On the other hand, a maximal independent set will always dominate: if the set does not dominate the board, that means there is a square that cannot be attacked by any of the current pieces - you can therefore add one more piece to the board at that spot, contradicting the fact that you already had a maximal independent set.

For our puzzles, we’ll just consider boards from 4x4 to 8x8 (so that they fit reasonably on the screen). In the table below, the lowest number in each cell represents the domination number for that piece on the given board size, and the largest represents the independence number. The letters next to each number indicate whether the set of that size should be said to be independent (i) and/ or dominant (d) - some of this information is redundant, but all indicators are included for completeness. The numbers between the least and greatest represents other possible arrangements. For example, for queens on a 4x4 board, the domination number is 2 (dominant set is not independent in this case), and the independence number is 4, but it is possible to find a dominating independent set of size 3, giving us the entries 2d, 3di, and 4di.


The independence and domination numbers in the table above are from the results described in Across the Board; the values between were found looking at the solutions for either domination or independence and perturbing them slightly. For example, to fill in the values for queens on an  8x8 board, start with one of the solutions to the queens domination problem for 8x8, which consists of an arrangement of 5 pieces, and move one of the pieces to a reachable square with fewer neighbours, and fill in the gaps with additional pieces. Proceeding by trial and error, this leads to dominating independent sets of 6 and 7 pieces. Finding additional dominating and independent sets for knights is a little more challenging than others - there are some gaps in the table (maybe you can fill them). Most of these possible puzzles were written out in a format for displaying online, which you can view here.

If you interested in exploring the mathematical chessboard problems through the playful medium of chessboard puzzles, please give these a try; if you are interested in learning more about the mathematics behind these puzzles, check out Across the Board.

domination and independence puzzles: https://dmackinnon1.github.io/chessdom/puzzles.html
chess tour puzzles: https://dmackinnon1.github.io/kixote/

some puzzle solutions

Wednesday, March 7, 2018

Symmetry and Asymmetry in Tigers and Treasure

Tiger and treasure logic puzzles, like ones you can try out here,  offer you a choice between two doors that might lead to treasure, or a tiger. Statements on "door 1" are true only if they lead to treasure, and statements on "door 2" are true only if they lead to a tiger.

The previous post gave an overview of the different "tiger and treasure" logic puzzles that could be formed from a starting list of 14 statements:
  1. this room has treasure
  2. the other room has treasure
  3. at least one room has treasure
  4. both rooms have treasure
  5. this room has a tiger
  6. the other room has a tiger
  7. at least one room has a tiger
  8. both rooms have a tiger
  9. this room has treasure or the other room has a tiger
  10. the other room has treasure or this room has a tiger
  11. this room has treasure and the other room has a tiger
  12. the other room has treasure and this room has a tiger
  13. both rooms have treasure or both rooms have a tiger
  14. one room has treasure and the other has a tiger
All possible puzzles are listed here (statement numbers in the file start at 0, rather than 1).

When exploring all the different possible puzzles we can make from these, we won't include puzzles that lead to a contradiction, or puzzles where the clues don't allow you to identify either door. This leaves 96 good puzzles out of the 196 combinations of statements, shown in black below:


Symmetry among Puzzles

For each statement in the list of fourteen, you can find the negation of that statement in the list - for example, statement 1 "this room has treasure" has its negation in statement 5 "this room has a tiger." If we plot the list of statements on door 1 vs. the the list of statements on door 2, but re-arrange the statements on the door 2 axis so they are the negations of our original list (the negated list would be statements 5, 6, 8, 7, 1 ,2, 4, 3, 12, 11, 10, 9, 14, 13), we can see the symmetry in these puzzles more clearly:


This graph is  telling us that if we have a puzzle that works, then if we swap the signs on the doors and negate them, we will also get a working puzzle. If you explore this a bit further, you'll see that the symmetry goes deeper. Let's get a  little mathy with this.

If a and b are statements in our list, a puzzle P can be described by the ordered pair (a,b). Every puzzle P also has a solution, (s, t) where s and t are either "tiger", "treasure", or "unknown."

For any statement in the list x, we can write the negation of x as -x. If we form a new puzzle by putting the negation of a on door 2 and the negation of b on door 1, we get a new puzzle -P = (-b,-a).

The symmetry of our tiger treasure puzzles can be expressed as this little theorem:
If P = (a,b) is a tiger treasure puzzle with solution (s,t), then its negation, -P = (-b, -a), will have the solution (t, s).
Here is an example. Consider the example puzzle from the previous post.

Puzzle P
Door 1 says: Both rooms have a tiger. (statement 8)
Door 2 says: The other room has treasure and this room has a tiger. (statement 12)

In the previous post, we worked out that door 1 has a tiger and door 2 has treasure.

Statement 8's negation is statement 3, and statement 12's negation is statement 9. So the puzzle -P looks like this:

Puzzle -P
Door 1 says: This room has treasure or the other room has a tiger (statement 9) 
Door 2 says: At least one room has treasure (statement 3)

If you have tried out a few of these, you may be able to find the solution to -P, which is that door 1 has a tiger and door 2 has treasure, which is the reverse of Puzzle P, where door 1 had treasure and door 2 had the tiger - the contents behind the doors have switched.

The symmetric graph above helped point us towards a nice symmetry that holds true for all of the tiger and treasure puzzles, but our original non-symmetric graph can point to another interesting things too.


Using asymmetry to make puzzles more interesting

In Raymond Smullyan's book "The Lady or the Tiger?" he presented a nice twist on the usual presentation of this kind of puzzle:
"There are no signs above the doors!" exclaimed the prisoner. "Quite true," said the king. "The signs were just made, and I haven't had time to put them up yet." "Then how do you expect me to choose?" demanded the prisoner. "Well, here are the signs," replied the king. That's all well and good," said the prisoner anxiously, "but which sign goes on which door?" The king thought awhile. "I needn't tell you," he said. "You can solve this problem without that information."
Let's look around for puzzles like this. One question to ask: Which of our problems would allow us to interchange the signs on the doors without affecting the solution to the problem? We'd like to know when does P = (a,b) give the same solution as Q = (b,a)? To explore which puzzles might work in this way, we can start looking at our first graph above, but limit our attention to those puzzles who's reflection in the line door1= door2 is also a puzzle. These are shown in black below (grey shows other puzzles whose reflection is not also a puzzle).


But we don't just want puzzles whose reflection gives us a puzzle, but ones whose reflections have the same solution as the original. It turns out that this gives an uninspiring set of six puzzles:


Only the trivial cases work: puzzles where the statements on the doors are exactly the same end up giving us puzzles that have exactly the same solution when the statements are interchanged. So what about the problem from "The Lady or the Tiger?", it uses these statements:
  • this room contains a tiger  (statement 5)
  • both rooms contain tigers (statement 8)
Why does the solver not need to know which door each goes on? Well, if statement 5 goes on door 1 we get a contradiction, so it must go on door 2. This is why the solver does not need to be told how to label the doors: there is only one possible way to do so without getting a contradiction. So, a way to find more problems like this is to look for puzzles P = (a,b) where P is a legitimate puzzle, but Q = (b, a) is not a puzzle, as that labelling of the doors leads to a contradiction.



There are 32 puzzles in this category (shown in black above): puzzles when you exchange the statements on the doors, you obtain a contradiction. Adding to this the 6 trivial cases where both doors have the same statement, we have 38 puzzles where we simply present the statements without explaining which statement is on each door.

For example, the puzzle (13, 10) falls into this set.

Let's try putting 13 on door 1 and 10 on door 2:

door 1: both rooms have treasure or both rooms have a tiger (13)

door 2: the other room has treasure or this room has a tiger (10)

Because inscriptions on door 1 are only true of door 1 leads to treasure, statement 13 implies that door 2 must lead to treasure. Door 2 leading to treasure makes its statement (10) false, requiring door 1 to lead to a tiger.

However, if we switch the statements, we run into trouble:

door 1: the other room has treasure or this room has a tiger (10)

door 2: both rooms have treasure or both rooms have a tiger (13)

If statement 10 on door 1 is false, then door 1 would lead to a tiger. However, the statement says that it leads to a tiger, so this cannot be. Door 1 must lead to treasure, making statement 10 true, requiring door 2 to also lead to a treasure. If door 2 leads to treasure, then its statement (13) must be false. However, statement 13 is true (both rooms have treasure), a contradiction.

So, presented with statements 10 and 13, there is only one way to arrange them on the doors: put statement 13 on door 1 and statement 10 on door 2, leading to a tiger behind door 1 and treasure behind door 2.

Please try out the tiger-treasure puzzles here and ask yourself: "What would the negation of this puzzle (-P) be?" and "Is this one of the 38 puzzles that can be answered without being told which sign is on which door?"