I have always liked puzzles. I really enjoy discovering a new puzzle. Sometimes when I have discovered a new puzzle, I enjoy it so much that I can’t seem to do enough of them. In such cases, I quickly run out of these puzzles to do and find myself looking for either a new puzzle or a way to generate them on my own.
Such was the case when I discovered the “Range Puzzles”. The rules are simple:
Every cell is marked either blue or gray
No two gray cells can be next to one another
The grid must be a connected (i.e. there is always a path from every cell to every other cell using horizontal and vertical connections.
Some cells have a number inside. This indicates the number of cells that can be viewed (horizontally and vertically in both directions) by this cell, including the cell itself.
In this post, I want to return to the idea of NP-Complete problems. There is a more technical, more formal definition that I can refer you to, but I like to refer to the images from Garey and Johnson’s “Computers and Intractability: A Guide to the Theory of NP Completeness”. The images helps to understand the difficulty of NP-Complete problems by presenting two images. The first image shows a single individual speaking to someone saying that he has been unable to solve the problem. The second image shows that same individual speaking to the same person behind the desk, but saying that not only was he unable to solve the problem but neither was a long line of people. The theory of NP Complete problems revolves around the concept that if an efficient algorithm exists for an NP-Complete problem, then an efficient algorithm exists for all problems in the class NP.
Today, I would like to present a puzzle I created to play with the Independent Set problem. This is the problem where we are given a graph G = (V, E) and are asked to find a maximum set of vertices S such that there is no edge in the graph G between any two vertices in S. The decision version (the problem asking whether there is an independent set of size k) of this problem is NP-Complete, so the known algorithms for problem either have a slow running time, or do not solve it exactly.
This problem is very related to another puzzle I posted last year called the clique problem. In fact, Karp originally proved that Clique was NP-Complete by showing that if Clique could be solved efficiently, then Independent Set could be solved efficiently. He did this by constructing a second graph [G bar], called the compliment of G (containing the same vertices in G, along with the edges that are not present in G. Any edge present in G will not be present in ). Then he showed that the nodes representing a maximum clique in G would represent a maximum independent set in . He had already shown that Independent Set was NP-Complete, which meant that both Independent Set and Clique were among the most difficult problems within the class known as NP.
The puzzle begins with an undirected graph and asks users to find a maximum independent set. Users should click on the numbers in the table below the graph indicating the nodes they wish to select in their independent set (purple indicates that the node is selected, gray indicates that it is not). Once a user have a potential solution, they can press the “Check” button to see if their solution is optimal. If a user is having trouble and simply wishes to see the maximum independent set, they can press the “Solve” button. And to generate a new problem, users can press the “New Problem” button.As a result of this relationship between the Clique problem and the Independent Set problem, the Bron-Kerbosch algorithm that was used to find maximum cliques previously can also be used here.
I still remember how I felt when I was first introduced to NP-Complete problems. Unlike the material I had learned up to that point, there seemed to be such mystery and intrigue and opportunity surrounding these problems. To use the example from Garey and Johnson’s book “Computers and Intractability: A Guide to the Theory of NP Completeness”, these were problems that not just one researcher found difficult, but that a number of researchers had been unable to find efficient algorithms to solve them. So what they did was show that the problems all had a special relationship with one another, and thus through this relationship if someone were to discover an algorithm to efficiently solve any one of these problems they would be able to efficiently solve all the problems in this class. This immediately got my mind working into a world where I, as a college student, would discover such an algorithm and be mentioned with the heavyweights of computer science like Lovelace, Babbage, Church, Turing, Cook, Karp and Dean.
Unfortunately I was a student so I did not have as much time to devote to this task as I would have liked. In my spare time though I would try to look at problems and see what kind of structure I found. One of my favorite problems was, The Clique Problem. This is a problem where we are given an undirected graph and seek to find a maximum subset of nodes in this graph that all have edges between them, i.e. a clique (Actually the NP-Complete version of this problem takes as input an undirected graph G and an integer k and asks if there is a clique in G of size k).
Although I now am more of the mindset that there do not exist efficient algorithms to solve NP-Complete problems, I thought it would be a nice project to see if I could re-create this feeling – both in myself and others. So I decided to write a program that generates a random undirected graph and asks users to try to find a maximum clique. To test users answers, I coded up an algorithm that works pretty well on smaller graphs, the Bron-Kerbosch Algorithm. This algorithm uses backtracking to find all maximal cliques, which then allows us to sort them by size and determine the largest.
Users should click on the numbers in the table below the canvas indicating the nodes they wish to select in their clique (purple indicates that the node is selected, gray indicates that it is not). Once they have a potential solution, they can press the “Check” button to see if their solution is optimal. If a user is having trouble and simply wishes to see the maximum clique, they can press the “Solve” button. And to generate a new problem, users can press the “New Problem” button.
So I hope users have fun with the clique problem puzzles, and who knows maybe someone will discover an algorithm that efficiently solves this problem and become world famous.
To help understand this problem, I want you to think about a common situation in many people’s lives. You have a road trip coming up today and you’ve overslept and are at risk of missing your flight. And to top matters off, you were planning to pack this morning but now do not have the time. You quickly get up and begin to get ready. You grab the first bag you see and quickly try to make decisions on which items to take. In your head you’re trying to perform calculations on things you’ll need for the trip versus things that you can purchase when you get there; things that you need to be able to have a good time versus things you can do without. And to top matters off, you don’t have time to look for your ideal luggage to pack these things. So you have the additional constraint that the items you pick must all fit into this first bag you found this morning.
The situation I described above is a common problem. Even if we ignore the part about the flight, and just concentrate on the problem of trying to put the most valuable set of items in our bag, where each item has its own value and its own size limitations, this is a problem that comes up quite often. The problem is known (in the math, computer science and operations research communities) as the knapsack problem. It is known to be difficult to solve (it is said to be NP-Hard and belongs to a set of problems that are thought to be the most difficult problems within its class). Because of this difficulty, this problem has been well studied.
What I provide in my script are two approaches to solving this problem. The first is a greedy approach, which selects a set of items by iteratively choosing the item with the highest remaining value to size ratio. This approach solves very fast, but can be shown to give sub-optimal solutions.
The second approach is a dynamic programming approach. This algorithm will solves the problem by ordering the items 0, 1, …, n and understanding that in order to have the optimal solution on the first i items, the optimal solution must have been first selected on the fist i-1 items. This algorithm will optimally solve the problem, but it requires the computation of many sub-problems which causes it to run slowly.
Update (4/2/2013): I enjoy this problem so much that I decided to implement two additional approaches to the problem: Linear Programming and Backtracking.
The Linear Programming approach to this problem comes from the understanding that the knapsack problem (as well as any other NP-Complete problem) can be formulated as an Integer Program (this is a mathematical formulation where we seek to maximize a linear objective function subject to a set of linear inequality constraints with the condition that the variables take on integer values). In the instance of the knapsack problem we would introduce a variable xi for each item i; the objective function would be to maximize the total value of items selected. This can be expressed as a linear objective function by taking the sum of the products of the values of each item vi and the variable xi; the only constraint would be the constraint saying that all items fit into the knapsack. This can be expressed as a linear inequality constraint by taking the sum of the products of the weights of each item wi and the variable xi. This sum can be at most the total size of the knapsack. In the integer programming formulation, we either select an item or we do not. This is represented in our formulation by allowing the variable xi = 1 if the item is selected, 0 otherwise.
The LP relaxation of an integer program can be found by dropping the requirements that the variables be integer and replacing them with linear equations. So in the case of the knapsack problem, instead of allowing the variables to only take on values of 0 and 1, we would allow the variables to take on any value in the range of 0 and 1, i.e 0 <= xi <= 1 for each item i. We can then solve this LP to optimality to get a feasible solution to the knapsack problem.
The second knapsack approach I implemented today is through backtracking. Similar to the Dynamic Programming approach to this problem, the backtracking approach will find an optimal solution to the problem, but these solutions generally take a long time to compute and are considered computationally inefficient. The algorithm I implemented here first orders the item by their index, then considers the following sub-problems for each item i "What is the best solution I can obtain with this initial solution?". To answer this question, the algorithm begins with an initial solution (initially, the empty set) and a set of unchecked items (initially, all items) and recursively calls itself on sub-problems with an additional item as a part of the initial solution and with this item removed from the unchecked items.
So check out my knapsack problem page. I think its a good way to be introduced to the problem itself, as well as some of the techniques that are used in the fields of mathematics, computer science, engineering and operations research.
Here in DC, we recently had an unexpected snow day. By the word unexpected, I don’t mean that the snow wasn’t forecast – it was definitely forecast. It just never came. However due to the forecast I decided to avoid traffic just in case the predictions were correct. So while staying at home, I began thinking about some things that I’ve been wanting to update on the site and one thing that came up was an update to my Sudoku program. Previously, it contained about 10000 sample puzzles of varying difficulty. However, I told myself that I would return to the idea of generating my own Sudoku puzzles. I decided to tackle that task last week.
The question was how would I do this. The Sudoku solver itself works through the dancing links algorithm which uses backtracking, so this was the approach that figured as most likely to get me a profitable result in generating new puzzles (I have also seen alternative approaches discussed where people start with an initial Sudoku and swap rows and columns to generate a new puzzle). The next question was how to actually implement this method.
Here is an overview of the algorithm. I went from cell to cell (left to right, and top to bottom starting in the top left corner) attempting to place a random value in that cell. If that value can be a part of a valid Sudoku (meaning that there exists a solution with the current cells filled in as is), then we continue and fill in the next cell. Otherwise, we will try to place a different value in the current cell. This process is continued until all cells are filled in.
The next step was to create a puzzle out of a filled in Sudoku. The tricky about this step is that if too many cells are removed then we wind up generating a puzzle that has multiple solutions. If too few cells are removed though, then the puzzle will be too easy to solve. Initially, I went repeatedly removed cells from the locations that were considered the most beneficial. This generally results in a puzzle with about 35-40 values remaining. To remove additional cells, I considered each of the remaining values and questioned whether hiding the cell would result in the puzzle having multiple solutions. If this was the case, then the cell value was not removed. Otherwise it was. As a result I now have a program that generates Sudoku puzzles that generally have around 25 hints.
Say we have a set of items. They could be practically anything, but for the sake of understanding lets think of something specific, like school supplies. In particular lets say that school is starting back and you have a checklist of things you need to buy:
a spiral tablet
a bookbag (or knapsack as I just learned people call them)
and some paper
Suppose also that as you’re shopping for these items you also see that stores sell the items as collections. In order to spend less money, you think it may be best to buy the items in collections instead of individually. Say these collections are:
a pencil, a notebook and a ruler
a pen, a calculator and a bookbag
a spiral tablet a lamp and some paper
The exact cover problem is a very important problem in computer science because many other problems can be described as problems of exact cover. Because the problem is in general pretty difficult to solve most strategies for solving these problems still generally take a long amount of time to solve. A common technique for solving these problems is called backtracking. This is basically a trial and error way of solving the problem. We try to construct a solution to the problem, and each time we realize our (partial) solution will not work, we realize the error, backup by eliminating part of the current solution and try to solve the problem by constructing another solution. This is basically how the backtracking procedure works. The main caveat to it is that we need to keep track of the partial solutions we have already tried so that we do not continuously retry the same solutions over and over again.
In particular Sudoku puzzles can be described as exact cover problems. Doing this involves translating the rules of Sudoku into exact cover statements. There are four basic rules of Sudoku that we need to take into consideration.
Each cell in the grid must receive a value
a value can only appear once in each row
a value can only appear once in each column
a value can only appear once in each pre-defined 3 by 3 subgrid.
In actuality these statements are joined together and what happens is that each position in the grid we (try to) fill in actually has effects in all four sections. We can set up the Suduko problem as an exact cover problem by noting these effects.
If a cell (i, j) in the grid receives the value v, then we also need to note that row i, column j, and the pre-defined subgrid have also received its proper value.
We can keep track of this by defining a table.
The first 81 (81 = 9*9) columns in the table will answer the question of does row i, column j have anything in that position. This will tell us whether the cell is empty or not, but will not tell us the value in the cell.
The next 81 columns in the table will answer the question of does row i (of the Sudoku grid) have the value v in it yet.
The next 81 columns in the table will answer the question of does column j (of the Sudoku grid) have the value v in it yet.
The final 81 columns in the table will answer the question of does the (3×3) subgrid containing (i, j) have the value v in it yet.
Notice that individually these columns do not give us much information about the grid, but because each value we place into the grid also has a precise location, a row, a column, a value and a subgrid we link together these columns of the new table and are able to understand more about the implications of each Sudoku operation. There are 9 possible rows, 9 possible columns and 9 possible values for each move, creating 729 possible moves. For each possible move, we create a row in the new table which links the columns of this new table together as follows:
row[i*9+j] = 1
(this says that cell (i, j) is nonempty)
row[81+i*9+v] = 1
(this says that cell row i has value v)
row[81*2+j*9+v] = 1
(this says that column j has value v)
row[81*3+(Math.floor(i/3)*3+Math.floor(j/3))*9+v] = 1
(this says that the 3 x 3 subgrid of (i, j) has value v)
Our goal is to choose a set of moves that fills in the grid. In the exact cover representation of the problem, this means we need to select a set of rows of the new table such that each column of the new table has exactly one 1 in our solution.
In my script I solve this problem using backtracking and the dancing links representation of this exact cover problem. The way dancing links works is that the exact cover version of this problem is translated from being a simple table to a graphical one where each cell becomes a node with pointers to the rows above and below it and the columns to the left and to the right of it. There is also a header row of column headers which contain information about that column, particularly the column number and the number of cells in that column that can cover it (basically the number of 1’s in the table form of the exact cover version of the problem). There is also a head node which points to the first column.
Once the graph version of the matrix is set up, the algorithm recursively tries to select a column and then a row in that column. Once these values have been chosen it tries to solve the new version of the problem with fewer cells remaining. If this is possible then it continues, if not then it goes back and chooses another row or another column and row.
I hope that was an understandable overview of the procedures I used to implement this algorithm. Otherwise I hope you enjoy the script and use it freely to help with your Sudoku puzzles.