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Binary Puzzles

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As you can probably tell, I’m a big fan of puzzles. On one hand you can say that a good puzzle is nothing but particular instance of a complex problem that we’re being asked to solve. What exactly makes a problem complex though?

To a large extent that depends on the person playing the puzzles. Different puzzles are based on different concepts and meant to highlight different concepts. Some puzzles really focus on dynamic programming like the Triangle Sum Puzzles or the Unidirectional TSP Puzzles.

Other puzzles are based on more complicated problems, in many cases instances of NP-complete problems. Unlike the puzzles mentioned above, there is generally no known optimal strategy for solving these puzzles quickly. Some basic examples of these are ones like Independent Set Puzzles, which just give a random (small) instance of the problem and ask users to solve it. Most approaches involve simply using logical deduction to reduce the number of possible choices until a “guess” must be made and then implementing some form of backtracking solution (which is not guessing since you can form a logical conclusion that if the guess you made were true, you reach either (a) a violation of the rules or (b) a completed puzzle).

One day a few months back i was driving home from work and traffic was so bad that i decided to stop at the store. While browsing the books, I noticed a puzzle collection. Among the puzzles I found in that book were the Range Puzzles I posted about earlier. However I also found binary puzzles.

Filled Binary puzzles are based on three simple rules
1. No the adjacent cells in any row or column can contain the same value (so no 000 or 111 in any row or column).
2. Every row must have the same number of zeros and ones.
3. Each row and column must be unique.

There is a paper from 2013 stating that Binary Puzzles are NP Complete. There is another paper that discusses strategies involved in Solving a Binary Puzzle. 

Once I finished the puzzles in that book the question quickly became (as it always does) where can I get more. I began writing a generator for these puzzles and finished it earlier this year. Now i want to share it with you. You can visit the examples section to play those games at Binary Puzzles.

Below I will go over a sample puzzle and how I go about solving it. First lets look at a 6 by 6 puzzle with some hints given:

 01   
0 10  
11 0  
 1  0 
   0  
1  1 0

We look at this table and can first look for locations where we have a “forced move”. An obvious choice for these moves wold be three adjacent cells in the same row or column where two have the same value. A second choice is that when we see that a row or column has the correct number of zeros or ones, the remaining cells in that row or column must have the opposite value.

So in the above puzzle, we can see that the value in cells (2, 2) and (2, 5) must also be a 0 because cells (2, 3) and (2, 4) are both 1. Now we see that column 2 has 5 of its 6 necessary values, and three 0’s. So the last value in this column (2, 6) must be a 1 in order for there to be an equal number of 0s and 1s.

For some easier puzzles these first two move types will get you far enough to completely fill in all the cells. For more advanced puzzles though, this may require a little more thorough analysis. 

As always, check it out and let me know what you think. 

Examples

Floyd-Warshall Shortest Paths

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The Floyd Warshall algorithm is an all pairs shortest paths algorithm. This can be contrasted with algorithms like Dijkstra’s which give the shortest paths from a single node to all other nodes in the graph.

Floyd Warshall’s algorithm works by considering first the edge set of the graph. This is the set of all paths of the graph through one edge. Node pairs that are connected to one another through an edge will have their shortest path set to the length of that edge, while all other node pairs will have their shortest path set to infinity. The program then runs through every triplet of nodes (i, j, k) and checks if the path from i to k and the path from k to j is shorter than the current path from i to j. If so, then the distance and the path is updated.

So lets consider an example on the graph in the image above. The edge set of this graph is E = {(0, 1), (0, 2), (0, 3), (1, 3), (3, 4)}. So our initial table is:

 01234
0inf(0, 1)(0, 2)(0, 3)inf
1(0, 1)infinf(1, 3)inf
2(0, 2)infinfinfinf
3(0, 3)(1, 3)infinf(3, 4)
4infinfinf(3, 4)inf

As we look to update the paths, we first look for routes that go through node 0:

Because node 0 connects to both node 1 and node 2, but node 1 does not connect to node 2, we have the following truth holding in the matrix above:
cost(0, 1) + cost(0, 2) < cost(1, 2), so we can update the shortest path from node 1 to node 2 to be (1, 0, 2).

Because node 0 connects to both node 2 and node 3, but node 2 does not connect to node 3, we have the following truth holding in the matrix above:
cost(0, 2) + cost(0, 3) < cost(2, 3), so we can update the shortest path from node 2 to node 3 to be (2, 0, 3).

Because node 3 connects to both node 0 and node 4, but node 0 does not connect to node 4, we have the following truth holding in the matrix above:
cost(0, 3) + cost(3, 4) < cost(0, 4), so we can update the shortest path from node 0 to node 4 to be (0, 3, 4).

Because node 3 connects to both node 1 and node 4, but node 1 does not connect to node 4, we have the following truth holding in the matrix above:
cost(1, 3) + cost(3, 4) < cost(1, 4), so we can update the shortest path from node 1 to node 4 to be (1, 3, 4).

Because node 3 connects to both node 2 and node 4, but node 2 does not connect to node 4, we have the following truth now holding:
cost(2, 3) + cost(3, 4) < cost(2, 4), so we can update the shortest path from node 2 to node 4 to be (2, 0, 3, 4).

The final table giving the list of shortest paths from every node to every other node is given below.

 01234
0inf(0, 1)(0, 2)(0, 3)(0, 3, 4)
1(0, 1)inf(1, 0, 2)(1, 3)(1, 3, 4)
2(0, 2)(1, 0, 2)inf(2, 0, 3)(2, 0, 3, 4)
3(0, 3)(1, 3)(2, 0, 3)inf(3, 4)
4(0, 3, 4)(1, 3, 4)(2, 0, 3, 4)(3, 4)inf

To see more examples and to help answer questions, check out the script in my examples section on the Floyd-Warshall algorithm

Examples

Longest Common Subsequence

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Suppose you and I each had an ordered list of items and we were interested in comparing how similar those lists are. One calculation we can perform on these two strings is the Longest Common Subsequence. A sequence X is an ordered list of elements <x1, …, xn>. A subsequence Z is another sequence where (1) Each element of Z is also an element of X and (2) The elements of Z occur in the same order (in Z) as they do in X.

Note that we do not say that the elements of Z need to be a continuous block of elements. If this were true we would be defining a substring. So as an example, suppose we have as an initial string,
X = C, O, M, P, U, T, E, R. 
Then the following are all subsequences: 
Z1 = C, M, U, T, R
Z2 = C, O, M, P
Z3 = U, T, E, R
Z4 = O, P, T, E

I will note that Z2 and Z3 are also substrings since they contain continuous sets of characters. 

The length of a substring is simply the number of characters it contains. So X has length 8, Z1 has length 5, Z2, Z3 and Z4 have length 4. 

Suppose now that we had a second string, Y = P, R, O, G, R, A, M, M, E, R and are interested in the longest common subsequence between the two. We can do that by observing that there is a bit of recursion going on with this question. What I mean by that is that asking the question of “What is the longest common subsequence between X and Y” is the same as asking “What is the longest common subsequence between X and Y once we have seen 8 characters of X and 10 characters of Y”

There are three possible ways to answer this question. 

If X<sub>8</sub> equals Y<sub>10</sub>, then we ask the same question about X<sub>7 and Y<sub>9</sub> and add 1 to the answer. 
If X<sub>8</sub> is not equal to Y<sub>10</sub>, then the answer to this will be the same as the maximum of the pair X<sub>7</sub>, Y<sub>10</sub> and the pair X<sub>8</sub>, Y<sub>9</sub>. 
If we reach a situation where we reach the beginning of either string, we are forced to answer 0 to that question. 

Then the function has the following look: 

LCS(Xi, Yj) =
0, if i is 0 or j is 0
1 + LCS(Xi-1, Yj-1) if Xi equals Yj
max(LCS(Xi-1, Yj), LCS(Xi, Yj-1))

Below is a table showing how we would solve the problem mentioned.

The strategy used to devise this concept is called dynamic programming. It is useful we can solve larger problems by solving overlapping subproblems, as was the case here. In this situation we generally can store the data in a table form and avoid re-solving subproblems for which many larger problems will be dependent.

You can see this algorithm in action at LEARNINGlover.com: Longest Common Subsequece. Let me know what you think.

Examples

Lets Learn About XOR Encryption

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One of the more common things about this generation is the constant desire to write up (type) their thoughts. So many of the conversations from my high school days were long lasting, but quickly forgotten. Today’s generation is much more likely to blog, tweet, write status updates or simply open up a notepad file and write up their thoughts after such a conversation.

When we feel that our thoughts are not ready for public eyes (maybe you want to run your idea by the Patent and Trademark Office before speaking about it) we may seek some form of security to ensure that they stay private. An old fashioned way of doing this was to write in a diary and enclosed it within a lock and key. The mathematical field of encryption also tries to grant privacy by encoding messages so that only people with the necessary information can read them.

The type of encryption I want to speak about today is called XOR encryption. It is based on the logical operation called “exclusive or” (hence the name XOR). The exclusive or operation is true between two logical statements if exactly one of the two statements is true, but not both statement. This can be represented with the following truth table

Input 1Input2XOR Result
TTF
TFT
FTT
FFF

XOR Encryption is particularly useful in this day and age because we every character we type is understood by the computer as a sequence of zeros and ones. The current standard encoding that is used is Unicode (also known as UTF-8). Under this encoding the letter ‘a’ is represented as the binary string ‘01100001’. Similarly every letter, number and special character can be represented as its own binary string. These binary strings are just an assignment of numbers to these characters so that we can to help represent them in the computer. The numbers can the be thought of in base 10, which is how we generally think about numbers, or in base 2 which is how computers generally work with numbers (or a number of other ways). The way we would use these binary strings in encoding is first by translating a text from human-readable text to machine readable text via its binary string. For example, the word “Invincible”, we would get the following binary strings:

LetterUnicode in base 10Unicode in base 2
I7301001001
n11001101110
v11801110110
i10501101001
n11001101110
c9901100011
i10501101001
b9801100010
l10801101100
e10101100101

To encrypt the message we need a key to encode the message and will simply perform an XOR operation on the key and every character in the string. Similarly, do decrypt the message we perform XOR operation on the key and every character in the encoded message. This means that the key (much like a normal key to a diary) must be kept private and only those whom the message is to be shared between have access to it.

Here is a link to the script where you can check out XOR Encrpytio. Try it out and let me know what you think.

Examples

Topological Sort

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One of the things I generally say about myself is that I love learning. I can spend hours upon hours reading papers and algorithms to better understand a topic. Some of these topics are stand alone segments that I can understand in one sitting. Sometimes, however, there is a need to read up on some preliminary work in order to fully understand a concept.

Lets say that I was interested in organizing this information into a new course. The order I present these topics is very important. Knowing which topics depend on one another allows me to use the topological sorting algorithm to determine an ordering for the topics that respects the preliminary work.

The input for the topoligical sorting algorithm is a Directed Acyclic Graph (DAG). This is a set of relationships between pairs of topics, where if topic 1 must be understood before topic 2, we would add the relationship (topic 1, topic 2) to the graph. DAGs can be visualized by a set of nodes (points) representing the topics. Relationships like the one above (topic 1, topic 2) can then represented by a directed arc originating at topic 1 and flowing in the direction of topic 2. We say that the graph is “Acyclic” because there cannot be a cycle in the topic preliminaries. This amounts to us saying that a topic cannot be a prerequisite for itself. An example of a DAG is shown in the image above.

With the topics represented as a DAG, the topologial ordering algorithm works by searching the set of nodes for the one with no arcs coming into it. This node (or these nodes is multiple are present) represents the topic that can be covered next without losing understanding of the material. Such a node is guaranteed to exist by the acyclic property of the DAG. Once the node is selected, we can remove this node as well as all arcs that originate at this node from the DAG. The algorithm then repeats the procedure of searching for a nod with no arcs coming into it. This process repeats until there are no remaining nodes from which to choose.

Now lets see how the topological sort algorithm works on the graph above. We will first need to count the in-degree (the number of arcs coming into) each node.

Node | Indegree
—————-
0 | 2
1 | 2
2 | 0
3 | 2
4 | 2
5 | 2
6 | 0
7 | 2
8 | 3

Node to be removed (i.e. node with the minimum indegree): Node 2.
Arcs connected to node 2: (2, 5), (2, 3)
Resulting Indegree Count:
Node | Indegree
—————-
0 | 2
1 | 2
3 | 1
4 | 2
5 | 1
6 | 0
7 | 2
8 | 3

Node to be removed: Node 6:
Arcs connected to node 6: (6, 1), (6, 3), (6, 4), (6, 5), (6, 7), (6, 8)
Resulting Indegree Count:
Node | Indegree
—————-
0 | 2
1 | 1
3 | 0
4 | 1
5 | 0
7 | 1
8 | 2

Node to be removed: Node 3
Arcs connected to node 3: (3, 0), (3, 8)
Resulting Indegree Count:
Node | Indegree
—————-
0 | 1
1 | 1
4 | 1
5 | 0
7 | 1
8 | 1

Node to be removed: Node 5
Arcs connected to node 5: (5, 0), (5, 8)
Resulting Indegree Count:
Node | Indegree
—————-
0 | 0
1 | 1
4 | 1
7 | 1
8 | 0

Node to be removed: Node 0:
Arcs connected to node 0: (0, 1), (0, 4)
Resulting Indegree Count:
Node | Indegree
—————-
1 | 0
4 | 0
7 | 1
8 | 0

Node to be removed: Node 1
Arcs connected to node 1: none
Resulting Indegree Count:
Node | Indegree
—————-
4 | 0
7 | 1
8 | 0

Node to be removed: Node 4
Arcs connected to node 4: none
Resulting Indegree Count:
Node | Indegree
—————-
7 | 1
8 | 0

Node to be removed: Node 8
Arcs connected to node 8: (8, 7)
Resulting Indegree Count:
Node | Indegree
—————-
7 | 0

Node to be removed: Node 7
Arcs connected to node 7: none
Resulting Indegree Count:
Node | Indegree
—————-

Since there are no nodes remaining, we have arrived at a topological ordering. Going through this iteration, we can see that we arrived at the ordering (2, 6, 3, 5, 0, 1, 4, 8, 7). There were several occasions where there were multiple nodes with indegree of 0 and we could have selected an alternative node. This would have given us a different topological ordering of the nodes, but it would still be valid.

There are more learning opportunities and an interactive demonstration of the algorithm at Topological Sort Examples at LEARNINGlover.