Tag Archives: JavaScript

Blog, Examples

Prim’s Algorithm

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I have just written a script that executes Prim’s Algorithm that finds the minimum spanning tree on a randomly generated graph.

Given a weighted graph, many times we are interested in finding a minimum spanning tree (MST) for that graph. This has many applications including the very important network simplex method. Prim’s algorithm is a greedy method which does finds this MST. A spanning tree is a subset of the edges of a graph that connects every vertex, but contains no cycles. This spanning tree is called a minimum spanning tree if in addition the sum of the weights of the edges included in this tree is less than or equal to the sum of the weights of the edges of any other spanning tree for this graph.

Prim’s algorithm works by the following procedure.
1. Let Treev be the set of vertices included in the tree, and TreeE be the set of edges included in the tree. Initially Treev and TreeE are empty.
2. Add an arbitrary vertex to Treev (TreeE is still empty).
3. Find the edge e of minimum weight such that one vertex is in Treev and vertex is not in Treev. Add the associated vertex to Treev, and add e to TreeE.
4. If edge was found in step 3, goto 5, else go to 6.
5. If the number of vertices in Treev is less than the number of vertices in the original graph, then the graph is not connected and thus does not contain a minimum spanning tree. Goto 8.
6 If the number of vertices in Treev is less than the number of vertices in the original graph, go to 2, else go to 7.
7. Output “The Minimum Spanning Tree is “, TreeE.
8. Output “This graph does not have a minimum spanning tree because it is not connected. ”

For example, consider the graph represented by the following adjacency matrix:

01234
0–––1312
1––––16
2––––24
313–––13
412162413–

Graph with 5 nodes

Initially our tree (Tv is empty). The first step says to choose a random vertex and add it to the tree, so lets choose vertex 2.

Iteration 1: Now our tree contains the vertex 2 (i.e. Tv = {2}) and likewise TE contains the edges coming from Tv. Thus TE = {(2, 4)}.
We want to choose the cheapest edge that has one endpoint in Tv and one endpoint not in Tv. These edges are represented by TE. Notice that TE only contains one edge, so we select this
edge, which has a cost of 24.

Iteration 2: Our tree thus contains the vertices 2 and 4 (i.e Tv = {2, 4}) and likewise TE contains the edges coming from Tv. Thus TE = {(0, 4), (1, 4), (3, 4)}.
Again, we want to choose the cheapest edge that has one endpoint in Tv and one endpoint not in Tv. This will be the edge (0, 4) which has a cost of 12.

Iteration 3: Now Tv = {0, 2, 4} and TE = {(1, 4), (3, 4), (0, 3)}. The cheapest of these three edges is the edge (0, 3) with a cost of 13, which means we will add it to our tree.

Iteration 4: Now Tv = {0, 2, 3, 4} and TE = {(1, 4)}. Since (1, 4) is the only edge connected to our tree we add it and it has a cost of 16.

Iteration 5: Now Tv = {0, 1, 2, 3, 4} and TE = {}. Because our tree contains all the vertices of the graph it is now spanning tree. The cost of this spanning tree is 24 + 12 + 13 + 16 = 65.

To learn more and see more examples, view Prim’s Algorithm at LEARNINGlover.com

Blog, Examples

Gaussian Elimination

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I have just written a script which executes the Gaussian Elimination Algorithm.

When we have a collection of lines we wish to know if they all intersect at some point. Many times we are interested in determining what that point is. In order to calculate this information, we first need an understanding of the lines themselves. The way the Gaussian Elimination Algorithm works is that the collection of lines are input using a notation of Ax = b, where the matrix A is called the coefficient matrix, as the nth row of it corresponds to the coefficients for the nth line being considered. The vector b represents the right hand side vector (in two dimensions, we would call these constants the y-intercepts of the lines. In higher dimensions they hold a similar property). The vector x represents the point where the lines intersect. It is this quantity which Gaussian Elimination seeks to determine.

The basic procedure of Gaussian Elimination is to use \”elementary row operations\” on the matrix (A|b), which is called the augmented matrix, to transform A into upper triangular form. Once this is done, a procedure called back-substitution can find the solution (x) to this problem.

The elementary row operations that we are allowed to perform are:

  • Interchange two rows.
  • Multiply a row by a nonzero number.
  • Add a row to another one multiplied by a number.

    • For the last property listed above, we will determine this number by dividing the coefficient of the term we which to eliminate by the negative of the coefficient of the element on the main diagonal of the same column of the matrix. This will have the property of cancelling out, or producing a desired zero in the resulting row.

    If this algorithm produces an upper triangular matrix from which we can solve for x using back-substitution. This procedure of back-substitution is simply solving for the vector x from the bottom of the matrix to the top. If the algorithm does not produce an upper triangular matrix (because somewhere along the line, we are unable to obtain a ratio because we have zero’s on the diagonal and all zeros below the diagonal), then we say the matrix is singular. This means that there is no unique point where the lines all intersect.

    To learn more and see more examples, check out My Script on Gaussian Elimination.

    Blog, Examples

    The Euclidean Algorithm

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    I remember being in school and learning about fractions. In particular, I remember the problems we had when trying to add and subtract fractions. This problem also presented itself when we tried to multiply fractions, although we still received partial credit if we couldn’t reduce fractions to their lowest terms.

    What I’m referring to is the problem of finding the Greatest Common Divisor (GCD) of two numbers. The GCD of two numbers is the largest number that divides into both numbers with a remainder of zero each time. This problem has many applications, but for most of us we can relate to it because of our frustrations with fractions.

    The algorithm that I’m writing about and have written a script for is called the Euclidean Algorithm, which solves precisely this problem. To be more precise, the Euclidean Algorithm finds the greatest common divisor between two integer numbers.

    There are different versions of the algorithm, but the one I have implemented finds the GCD by subtracting the smaller number from the larger number, and if the result is greater than 0, the procedure repeats itself with the two lower numbers. Otherwise, the result (the GCD) is the final number that was greater than 0.

    Lets see an example:
    Consider the number 9 and 21.
    21 – 9 = 12. 12 is greater than 0, so we repeat the procedure with 9 and 12.
    12 – 9 = 3. 3 is greater than 0, so we repeat the procedure with 3 and 9.
    9 – 3 = 6. 6 is greater than 0, so we repeat the procedure with 3 and 6.
    6 – 3 = 3. 3 is greater than 0, so we repeat the procedure with 3 and 3.
    3 – 3 = 0. 0 is not greater than 0, so can exit the loop portion of the algorithm.
    Since 3 was the last positive number that we arrived at in this procedure, we see that the GCD of 9 and 12 is 3.

    To learn more and see more examples, check out My Script on The Euclidean Algorithm.

    Examples

    Sieve of Eratosthenes

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    Prime numbers are an important concept in Number Theory and Cryptography which often uses the difficulty of finding prime numbers as a basis for building encryption systems that are difficult to break without going through all (or a very large number of) possible choices.

    Remember that a prime number is a number greater than 1 whose only divisors are 1 and that number itself. One of the most famous algorithms for searching for prime numbers is the Sieve of Eratosthenes. I added a script which implements the Sieve of Eratosthenes to my Examples page.

    This algorithm prints out all prime numbers less than a given number by first canceling out all multiples of 2 (the smallest prime), then all multiples of 3 (the second smallest prime), then all multiples of 5 (the third smallest prime – multiples of 4 do not need to be considered because they are also multiples of 2), etc until we have reached a number which cannot be a divisor of this maximum number.

    So if we are given a number, n, the first step of the algorithm is write out a table that lists all the numbers that are less than n. For example lets run this Sieve on 50. So all numbers less than 50 are

    1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50.

    So since 1 is not a prime number (by the definition of prime numbers), we cancel that number out.

    1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50.

    Next, we look at the list and the first number that is not crossed out is a prime. That number is 2. We will put a mark by 2 and cancel out all of 2’s multiples.

    1, 2*, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50.

    Again, we look at the list and the first number that is not marked or crossed out is 3, so that number is prime. We will put a mark by 3 and cancel out all of 3’s multiples.

    1, 2*, 3*, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50.

    Once again, we look at the list and the first number that is not marked or crossed out is 5, so that number is prime. We will put a mark by 5 and cancel out all of 5’s multiples.

    1, 2*, 3*, 4, 5*, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50.

    We look at the list and the first number that is not marked or crossed out is 7, so that number is prime. We will put a mark by 7 and cancel out all of 7’s multiples.

    1, 2*, 3*, 4, 5*, 6, 7*, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50.

    Now, we look and the first number that is not crossed out is 11. However, since 11 is greater than sqrt(50) we know that each of 11’s multiples that are less than 50 will have been cancelled out by a previous prime number. So we have finished the algorithm.

    Check out my script which implements the Sieve of Eratosthenes for more examples.

    Blog

    Hello, World!

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    The ideas for this site have been bouncing around inside my head and on my computer for years, and this site is an acknowledgement that it was time to finally act on these ideas.

    The site will feature a collection of scripts I have written to help illustrate different concepts. A large part of this will be a flash cards section which will provide an avenue to study or to refresh one’s memory on various subjects. I recognize, though, that all subjects are not easily understood through flash cards and so I also have an examples section where various algorithms are implemented on example problems (problem sets) to provide users with a more hands on experience.

    The site will be updated regularly, generally with either new subject areas added to the flash cards database, new scripts added to the examples. I leave open the possibility, though, of entirely new sections being introduced as ideas continue to develop and the site continues to grow.

    EDIT:
    There are several areas that I would like to take the site, but with each area I have the problems of (a) showing the concept, (b) visualizing the concept, c) showing the intuition behind the concept. Sometimes, I will think of (what I call) more clever ways to teach or learn an idea I’ve already discussed. This may lead to more than one page on the same concept. I encourage you to try all such pages to see if you like any of them.

    I hope you enjoy.