Entries Tagged 'Math' ↓

Here’s a small puzzle/game that I made, using a simple geometric concept. You can try it here: disk intersection game. The game idea is loosely inspired by the game planarity, but it is also significantly different.

There are some disks, and you can move them around arbitrarily (drag with the mouse). The goal is to make all of them “good”. When a disk is “good”, its color turns to green. The other colors indicate different levels of “badness”, as shown in the legend to the right. With some experimentation you can probably discover what you need to do to achieve this, but here is an explanation of the idea:

There is a hidden “model”, created randomly, which you need to discover. The model determines how the disks need to be interconnected, meaning which pairs of disks have to intersect each other (overlap with each other). If you “realize” this model, by moving the disks in such a position as to intersect each other in the required way, you win the game. The colors of the disks indicate how far the current configuration is from the correct model. This is computed as follows: if a certain disk should intersect disk A and no other disks, but instead it intersects disk B and C, and it doesn’t intersect A, then its “badness” is 3, so it has the color corresponding to the number 3. In other words, “badness” shows how many errors there are due to a given disk. When the “badness” of all disks is 0, you have solved the puzzle.

If you find the color codes hard to recognize, you can click the “#” button, and the “badness” will appear as a number. You can also ask for help three times during the game. Clicking the help button will display the model that has to be realized (i.e. the connections between the disks, shown as lines). There are easy and hard games, in a hard game the model tends to be more complicated, therefore harder to discover.

Try the disk intersection game.

Remarks:

• I came up with the idea for the game, while working on a problem related to unit disk graphs. This is the first iteration of the game, so probably the gameplay and the design could be significantly improved. If you have any suggestions, please let me know in the comments.
• There is nothing special about disks here, maybe other intersection graphs would make more interesting (or more challenging) games.
• One reason I wrote the game, besides wanting to try out the idea, was to get familiar with HTML5 Canvas – in the future I’d like to make another, somewhat more complex game. As this was my first encounter with Canvas, and because I wrote the game in one sitting (~3 hours), the implementation is a bit rough around the edges. Please let me know if you find any bugs or if you have suggestions.

Some years ago I was reading about the game of Nim, then a bit more about the theory of impartial games, and as the rabbithole seemed to go deeper and deeper, I needed some pretext to continue. So I came up with the rules of a simple paper-pencil game that I could try to analyze with the theory that I was reading about. I never managed to fully analyze the game, but it seemed to be reasonably fun to play anyway. Later I found that some very similar games have already been described.

The rules are simple: two players take turns in marking squares on a grid (say, of size 8*8). You can only mark a square if all of its neighbors (including the diagonal neighbors) are empty. The first player unable to move loses.

More details of the game are here.

Recently, I had the pleasure of discovering a website that contains the descriptions of many paper and pencil games, most of them playable on the website against the computer. The site is maintained by David Johnson-Davies. In fact, I found out about the website from him, when he told me that he also included the game I described above. The game is filed under the name “Obstruction game” which very well describes what it is about. You can even play it against the computer. Here is the website with all the games and here is the link to the Obstruction game. There is even a description with some useful strategy-tips.

Happy playing!

Suppose there are n sports teams, some of them play each other, in each game one team wins and the other loses and in the end we want to order the teams such that if A beats B than A is before B in the order. If two teams A and B played each other multiple times but, say, A clearly dominated B, we consider as if they would have played once and A beat B. If they played each other several times and there is no clear winner, or if they played only one draw, we consider as if they would have played twice and once A won, once B.

We can represent the whole situation as a directed graph with n vertices {1, …, n} and an edge from i to j if i won against j. Clearly, there are no self-loops. If the graph has no cycles, we can efficiently compute a topological sorting of the vertices and we are done. What if the graph has cycles (two teams that played a draw, or a sequence of teams where A beat B, B beat C, …, X beat A), but we still want to output a reasonable ordering? What is a reasonable ordering?

This is the vertex ordering problem and we can formulate many different criteria. I’ll go through some variants and mention for each of them, whether they are NP-hard or polynomially solvable. It turns out that almost all of them are NP-hard, and those which are not are quite trivial (plus the topological sorting mentioned before, which is also polynomially – in fact, linearly – computable). I’ll try to refer to sources where I found some, where I didn’t find any, the result is simple enough to be considered folklore. Not all the criteria that I list are reasonable – a reasonable criterion should actually lead to the topological sorting when there are no cycles. Did I forget some sensible criteria? Let me know what you think.

The general set-up is the following: we search for a permutation f:[n] -> [n], such as to minimize or maximize some quantity as follows:

1. The first criterion is that we want to minimize/maximize some reasonable function of the number of wins and the number of losses of a team. This includes the methods used in most sport tournaments where wins and loses are worth some number of points and in the end the teams are ranked by their accumulated number of points. This criterion is easy, because we can just compute the indegree/outdegree of each vertex, compute the necessary function values and sort vertices according to it.

Now we move to some more interesting criteria.

• We either minimize (MIN) or maximize (MAX) some quantity.
• The quantity is either the SUM, the MAXimum, the MINimum or the COUNT of something.
• That something is either the length of EDGEs, the length of BACK-EDGEs, or SIGNED length of EDGEs.

This gives 2*4*3=24 cases.

For example, if i->j means that there is an edge from i to j:

MIN COUNT EDGE refers to    

MAX SUM EDGE refers to    

MIN SUM BACK-EDGE refers to    

MAX MIN SIGNED-EDGE refers to     .

These are all for directed graphs, as mentioned earlier. Before starting to enumerate all these cases, let’s define the same problem for undirected graphs. For undirected graphs there are less cases, as we don’t have to consider separately any edge or back-edge or signed edge. Thus there are just 2*4=8 possibilities here.

Now let’s see what we’ve got, numbering continuously all the cases.

Undirected graphs

2. MIN SUM: This is called Minimum Linear Arrangement, a known NP-hard problem.

3. MIN MAX: This is called Minimum Bandwidth, a known NP-hard problem.

4. MIN MIN: makes no sense, it can always be 1.

5. MIN COUNT: makes no sense, it is always the number of edges.

6. MAX SUM: this is equivalent to 2. on the complement graph, therefore NP-hard.

7. MAX MAX: makes no sense, it can always be n-1.

8. MAX MIN: This is the anti-bandwidth problem, studied in literature, see here, or here (leads to paywall). It is NP-hard. This can be seen via the following reduction: “Does G have a Hamiltonian path?” is the same as “can the vertices of G be arranged such that there are edges between neighbors?” is the same as “can the vertices of G be arranged such that the complement of G has anti-bandwith at least 2?”. If we could answer the last question efficiently, we could answer the first.

9. MAX COUNT: makes no sense, it is always the number of edges.

Now we move on to the original question:

Directed graphs

10. MIN SUM EDGE: NP-hard. From 2. if we replace every undirected edge with a directed edge.

11. MIN SUM BACK-EDGE: NP-hard. From 2. if we replace every undirected edge with two directed edges.

12. MIN SUM SIGNED-EDGE: this is quite interesting. Notice that we can replace a path a->b->c with a single edge a->c without changing the sum. If we do this sufficiently many times we will be left with vertices with only incoming or only outgoing edges. To minimize the sum we clearly have to put vertices with outgoing edges first, then vertices with incoming edges. Otherwise we could make a simple replacement that would decrease the sum. Furthermore, vertices with outgoing edges have to be ordered in decreasing order of degree, vertices with incoming edges ordered in increasing order of degree. Again, otherwise, a simple swap would decrease the sum. That’s all.. Observe that what we’ve got is the same as ordering the original vertices in decreasing order of [outderee-indegree]. In this sense, this is a special case of 1.

13. MAX SUM EDGE: NP-hard. From 6. if we replace every edge with a directed edge.

14. MAX SUM BACK-EDGE: NP-hard. From 6. if we replace every edge with two directed edges.

15. MAX SUM SIGNED-EDGE: solution is the reverse order of 12.

16. MIN MAX EDGE: NP-hard. From 3. if we replace every undirected edge with a directed edge.

17. MIN MAX BACK-EDGE: NP-hard. From 3. if we replace every undirected edge with two directed edges.

18. MIN MAX SIGNED-EDGE: NP-hard. From 3. if we replace every undirected edge with two directed edges. Known as Minimum Directed Bandwidth.

19. MAX MAX EDGE: makes no sense, it can always be n-1.

20. MAX MAX BACK-EDGE: makes no sense, it can always be n-1.

21. MAX MAX SIGNED-EDGE: makes no sense, it can always be n-1.

22. MIN MIN EDGE: makes no sense, it can always be 1.

23. MIN MIN BACK-EDGE: makes no sense, it can always be 1 and has to be at least 1 if there are cycles.

24. MIN MIN SIGNED-EDGE: makes no sense, it can always be -(n-1).

25. MAX MIN EDGE: NP-hard. From 8. if we replace every undirected edge with a directed edge.

26. MAX MIN BACK-EDGE: NP-hard. From 8. if we replace every undirected edge with two directed edges.

27. MAX MIN SIGNED-EDGE: NP-hard. From 8. if we replace every undirected edge with two directed edges.

28. MIN COUNT EDGE: makes no sense, it is always the number of edges.

29. MIN COUNT BACK-EDGE: NP-hard, known as Minimum Feedback Arc Set.

30. MIN COUNT SIGNED-EDGE: makes no sense, it is always the number of edges.

31. MAX COUNT EDGE: makes no sense, it is always the number of edges.

32. MAX COUNT BACK-EDGE: NP-hard, it is the reverse order of 29.

33. MAX COUNT SIGNED-EDGE: makes no sense, it is always the number of edges.

Phew :) Did I forget any interesting criteria?

From the point of view of the original motivation (a ranking of teams) the criteria that seem to some extent reasonable are: 11, 12, 17, 18, 29.

Let’s revisit the topic of the previous post on self-counting sentences. We looked at sentences like this:

In this sentence the number of occurrences of 1 is __, of 2 is __, ..., of n is __.

Let’s go up one more step on the ladder of abstraction, to arrive at sentences of the type:

In this sentence, there are 3 numbers appearing 1 time, 1 number appearing 2 times, 1 number appearing 3 times, 0 numbers appearing 4 times.

In this sentence, there are 2 numbers appearing 1 time, 3 numbers appearing 2 times, 0 numbers appearing 3 times, 0 numbers appearing 4 time.

Observe, that both say the truth about themselves. Below are three puzzles. In all cases the task is to fill in the gaps such as to make the sentence say the truth and the questions are: for what values of n does a solution exist and what are the solutions (if they exist).

In this sentence there are __ numbers occurring 1 time, __ numbers occurring 2 times, ..., __ numbers occurring n times.

In this sentence, from the numbers 0,1, ..., n-1, there are __ numbers occurring underlined 0 time, __ numbers occurring underlined 1 time, ..., __ numbers occurring underlined n-1 times.

In this sentence there are __ numbers occurring underlined 1 time, __ numbers occurring underlined 2 times, ..., __ numbers occurring underlined n times.

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The second problem is equivalent to the first problem. There exists a solution to problem 1, if and only if the same sequence of numbers is a solution to problem 2 (note that in problem 2 the counting starts from 0). Therefore we only look at the first and the third problems.

Solution of problem 1:

For small values of n, we can find the solutions by trial and error or with a small program. We denote the numbers of the solution as f(1), f(2), …, f(n).

For n=1, there is a single solution: f(1)=2.

For n=2, there is a single solution: f(1)=4, f(2)=0.

For n=3, there is a single solution: f(1)=4, f(2)=1, f(3)=0.

For n=4, there are two solutions (see the beginning of the post).

For n=5, n=6, there are no solutions.

For n=7, there are two solutions: f(1)=4, f(2)=3, f(3)=0, f(4)=1, f(5)=0, f(6)=0, f(7)=0, and f(1)=5, f(2)=1, f(3)=1, f(4)=1, f(5)=0, f(6)=0, f(7)=0.

For n=8, there are two solutions: f(1)=5, f(2)=2, f(3)=1, f(4)=1, f(5)=0, f(6)=0, f(7)=0, f(8)=0, and f(1)=5, f(2)=3, f(3)=0, f(4)=0, f(5)=1, f(6)=0, f(7)=0, f(8)=0.

For every n>=9 there are exactly three solutions:
a) f(1)=n-3, f(2)=3, f(n-3)=1, and the remaining values 0,
b) f(1)=n-2, f(2)=1, f(4)=1, f(n-4)=1, and the remaining values 0,
c) f(1)=n-3, f(2)=2, f(3)=1, f(n-4)=1, and the remaining values 0.

These solutions clearly work. Let’s prove that no other solutions exist.

We can observe two properties of the sequence f(i):

Property 1: ∑ i * f(i) = 2n. (where i goes from 1 to n)
Proof:
There are 2n numbers in the sentence and each i*f(i) term counts how many numbers are there with multiplicity i. To have a multiplicity greater than n would mean that all gaps are filled with the same value, which cannot yield a true sentence for any value. Thus the possible multiplicities are between 1 and n, therefore ∑ i * f(i) gives the total number of numbers in the sentence, which is 2n.

Property 2: ∑ f(i) = n+1. (where i goes from 1 to n)
Proof:
Clearly, for all i, f(i)>=0 and f(i)<=n.
If for all i, f(i)>0 would hold, ∑ i * f(i) would be larger than 2n, violating Property 1. Therefore, some of the f(i) values have to be equal to 0, which means that all numbers 0,1,…,n appear in the sentence. Furthermore, all numbers that appear in the sentence have multiplicity between 1 and n. Thus, ∑ f(i) gives the total number of distinct values, which is n+1.

Now we prove that a) b) c) are the only solutions for n>=9.

If f(1)>n-2,
it must be that f(1)=n-1, but now n-1 occurs twice, so to maintain that n-1 values occur once, we need to write n-1 in every other position as well. This is not a correct solution.

If f(1)=n-2,
we must have exactly one other nonzero value (besides n-2) written in the gaps. We cannot write n-2 anywhere else, as that would violate Property 2. To get a sum of n+1, the possibilities are writing one time 3, or three times 1. Writing 3 anywhere doesn’t make the sentence correct, so it remains to write three 1′s. But then the number of 0′s will be n-4, the number of 1′s will be 4 and the number of (n-2)’s will be 2. This gives us solution b).

If f(1)=n-3,
we have two other nonzero values (besides n-3) written in the gaps. They also have to add up to 4 (Property 2). That can be as 3+1 or as 2+1+1. The first case means that 1,3 and n-3 have multiplicities 2 and 0 has multiplicity n-3. The second case means that 2 and n-3 have multiplicity 2, 0 has multiplicity n-4 and 1 has multiplicity 3. These cases uniquely lead to solutions a) and c).

If f(1)<n-3,
suppose f(1)=n-k (where k>3). The value 0 has to appear more than one time (otherwise the sum of Property 1 would be too large). Therefore, among the values 1, …, n, there are k distinct values which are written somewhere in the gaps (to make their multiplicities larger than one in the whole sentence). We know that n-k appears in the first gap, so we have k-1 other values in the gaps starting from the second. Let’s denote these values by a₁ < a₂ < . . .  < a_k − 1. Since they are written in the gaps after the first one, it follows that there are at least a₁ + a₂ + . . .  + a_k − 1 distinct numbers that appear more than once in the sentence. Even if one of them is n-k, and one of them is 0, that still leaves S = a₁ + a₂ + . . .  + a_k − 1 − 2 distinct numbers appearing more than once (therefore, also in the gaps). Since a₁ ≥ 1, we have S >= k(k-1)/2 – 2. Even if the smallest of these numbers is 1, we get that ∑f(i) >> n+1, for k>3. This violates Property 2, therefore no solution of this kind is possible.

Solution of problem 3:

For n=1, we have f(1)=1, unique solution.
For n=2, we have f(1)=2, f(2)=0, unique solution.
For n=3, we have f(1)=1, f(2)=1, f(3)=0, unique solution.
For n=4, we have f(1)=2, f(2)=1, f(3)=0, f(4)=0, unique solution.

For n>=5, we have exactly two solutions:
a) f(2)=1, f(n-2)=1, the remaining values 0
b) f(1)=2, f(n-2)=1, the remaining values 0.

Proof:

Similarly to the first problem, now we have the property:
∑ i * f(i) = n.

Clearly f(n)=0, otherwise all values would have to be equal, which is not a correct solution.
Also, f(n-1)=0. The only alternative would be f(n-1)=1, and all other values equal to zero, but this is not a correct solution either.

If f(n-2)=1 we need that ∑ i * f(i) over all other values is 2. This leads to solutions a) and b) only.
f(n-2)>=2 is not possible, as it would violate the property for n>=5.

Now suppose f(n-k) is the last nonzero value, for k>=3. This means that there are k numbers that are not equal to one of the most frequent values. If 0 is one of the most frequent values, then there are k nonzero values. All of them cannot be equal to 1, as that would not be a correct solution, so ∑ i * f(i) >= k(k-1)/2 + n-k + 1, which is larger than n for k>=3, violating the required property. On the other hand, if less than half of the elements are 0, the property is clearly violated, which shows that there are no other solutions.

In this sentence the number of occurrences of 1 is 2, of 2 is 3, of 3 is 2, of 4 is 1.

This self-referential, true sentence is a variant of a puzzle attributed to the logician Raphael Robinson, popularized in Douglas Hofstadter’s book Metamagical Themas. In the book the puzzle appears in the following form:

Fill in the gaps such as to make the sentence true:

In this sentence the number of occurrences of the digit 0 is __, of digit 1 is __, of digit 2 is __, ..., of digit 9 is __.

With some trial and error you can find two different solutions, and it has been shown that no other solutions are possible. The problem has also been studied using some theory from the field of dynamical systems. Here are two interesting links, both containing the solution of the puzzle.

In this post, however, I would like to look at a slightly different puzzle. Instead of counting digits, I am counting the numbers as indivisible units. If, for instance, “15″ appears in the sentence, I count it as “15″, not as one “1″ and one “5″. In this way the problem is independent of the base in which the numbers are represented.

In both of the following two puzzles the task is to fill in the gaps such as to make the sentences true. In the first sentence the values can be between 1 and n (including 1 and n), in the second sentence the values can be between 0 and n-1 (both included). For what values of n are the problems solvable? How many solutions are there?

In this sentence the number of occurrences of 1 is __, of 2 is __, ..., of n is __.

In this sentence the number of underlined occurrences of 0 is __, of 1 is __, of 2 is __, ..., of n-1 is __.

Think about the puzzles for some time, or scroll down for the solution. Let me know if you find a simpler solution. I also wrote a follow-up post, with a different variant of the problem.

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It can be shown that the two problems are equivalent: let’s denote the numbers written in the gaps of the first sentence by f(1), …, f(n). It can be seen easily that f(1), …, f(n) form a solution for the first problem if and only if f(1)-1, …, f(n)-1 form a solution for the second problem. So it’s enough to look at the first puzzle only.

With some case-based analysis we can show that there is no solution for the cases n=1, n=2, n=3 and n=6.
For n=4, there are two solutions, the one given in the beginning of the post and the following: f(1)=3, f(2)=1, f(3)=3, f(4)=1.

For n=5, the solution is given by the sequence: f(1)=3, f(2)=2, f(3)=3, f(4)=1, f(5)=1 and it is unique. These can be verified manually.

For every n>=7, there is a solution, given by f(1)=n-3, f(2)=3, f(3)=2, f(n-3)=2, and f(k)=1, for every k other than 1,2,3, and n-3. Let’s prove that this is the only solution:

Suppose there is some other solution.
If f(1)=n-3, then f(n-3)>=2.
—> If f(n-3)=2, then we need f(2)>=3.
——> If f(2)=3, then we get the same solution.
——> If f(2)>3, then there exist distinct x and y (also different from n-3, 1 and 2), for which f(x)=f(y)=2, so we have a value different from 1 at places 1,2,x,y,n-3, which means that we don’t have enough places to put 1′s, to satisfy f(1)=n-3.
—> If f(n-3)>2, then we have some x (different from 1 and n-3), for which f(x)=n-3. But then there are n-4 places where we need to put x, again not leaving enough places to satisfy f(1)=n-3.

If f(1)>n-3, then f(f(1))>=2.
—> If f(f(1))=2, then we need f(2)>=3 and f(f(2))>=2. We ran out of places for 1′s, since 1,2,f(2) and f(1) are distinct and they map to values different from 1.
—> If f(f(1))>2, then there is some x (other than 1) with f(x)=f(1)>=n-2, which means that we need to fill at least n-3 places with the value x, not leaving enough place for the 1′s.

The interesting case remains when f(1)<n-3. This condition means that in at least 5 places we need to write a number different from 1.
Suppose there are exactly k places, corresponding to the indices x1, …, xk, where we have an entry different from 1. This means that f(x1), …, f(xk) are different from 1 and k>=5. Let x1=1, since we know that in place 1 we have an entry different from 1.

Observe that among f(x1), …, f(xk) there have to be exactly k-1 distinct elements. This is because we need a value different from 1 in all places 1, f(x1), …, f(xk), and unless there are exactly k such places, we reached a contradiction. Since we have k-1 distinct elements all different from 1, the maximum among them is at least k and the second largest element is at least k-1. This means that there is a number t different from 1, such that f(t)>=k-1, therefore, t appears as a count in at least k-2 places. If out of k counts we have k-2 equal, we cannot have k-1 distinct values, so we reached a blatant contradiction. Therefore this case is impossible and the solution given above is unique.

Bongard problems are a type of puzzle invented by Russian computer scientist M. M. Bongard, intended as challenges for pattern recognition algorithms. Regardless of the possible relevance to computer science, they are much fun simply as puzzles created by people for other people to solve. They were popularized in the Gödel, Escher, Bach book and Hofstadter himself designed many Bongard problems. Harry Foundalis has several pages dedicated to these problems, with an extensive index of example Bongard problems.

The idea of a Bongard problem is the following: given a set A of six figures (examples) and another set B of six figures (counter-examples), discover what is the rule that the figures in A obey and figures in B violate.

Here are some very nice examples of varying difficulty and a much more detailed description:
http://www.foundalis.com/res/diss_research.html

Inspired by these examples I tried my hand at making some Bongard puzzles myself. The modest result is below. If you can solve them, leave the answer in the comments.

Puzzle 1.

Puzzle 2.

Puzzle 3.

Puzzle 4.

solutions

Inequalities are the bread and butter of mathematical proofs, especially those with an “approximate” flavor. I found many useful sources that describe the most important inequalities, there exist even some books dedicated exclusively to inequalities. However, I couldn’t find a concise “cheat-sheet” style summary of the most important ones, so I wrote one myself, collecting information from the sources I found. Check out my useful inequalities cheat sheet.