wolfram-mathematica

Inspired by Mike Bantegui's question on constructing a matrix defined as a recurrence relation, I wonder if there is any general guidance that could be given on setting up large block matrices in the least computation time. In my experience, constructing the blocks and then putting them together can be quite inefficient (thus my answer was actually slower than Mike's original code). Join and possibly ArrayFlatten are possibly less efficient than they could be. Obviously if the matrix is sparse, one can use SparseMatrix constructs, but there will be times when the block matrix you are

I've seen it's possible to make a connection between Mathematica and MySQL databases using Input Needs["DatabaseLink "] and conn = OpenSQLConnection[JDBC["MySQL(Connector/J)", "yourserver/yourdatabase"], "Username" -> "yourusername", "Password" -> "yourpassword"] (in case anyone wants to give it a try). Documentation of DatabaseLink here , by the way. Does anyone have experience using Mathematica in this way, probably to analyze data contained in the database? Are there obvious drawbacks (speed, memory needed, etc.)? I recently used databases to speed up a Manipulate[] block. Without the

I came across an old problem that you Mathematica/StackOverflow folks will probably like and that seems valuable to have on StackOverflow for posterity. Suppose you have a list of lists and you want to pick one element from each and put them in a new list so that the number of elements that are identical to their next neighbor is maximized. In other words, for the resulting list l, minimize Length@Split[l]. In yet other words, we want the list with the fewest interruptions of identical contiguous elements. For example: pick[{ {1,2,3}, {2,3}, {1}, {1,3,4}, {4,1} }] --> { 2, 2, 1, 1, 1 } (Or {3

I am thinking of process an image to generate in Mathematica given its powerful image processing capabilities. Could anyone give some idea as to how to do this? Thanks a lot. Brett Champion Here's one version, using a textures. It of course doesn't act as a real lens, just repeats portions of the image in an overlapping fashion. t = CurrentImage[]; (* square off the image to avoid distortion *) t = ImageCrop[t, {240,240}]; n = 20; Graphics[{Texture[t], Table[ Polygon[ Table[h*{Sqrt[3]/2, 0} + (g - h)*{Sqrt[3]/4, 3/4} + {Sin[t], Cos[t]}, {t, 0., 2*Pi - Pi/3, Pi/3} ], VertexTextureCoordinates ->

The St. Louis Federal Reserve Bank has a great set of data available on a variety of their web pages, such as: http://research.stlouisfed.org/fred2/series/OILPRICE/downloaddata?cid=32217 http://www.federalreserve.gov/releases/h10/summary/default.htm http://research.stlouisfed.org/fred2/series/DGS20 The data sets get updated, some as often as daily. I tend to have an interest in the daily data (see the above settings on the URLS) I'd like to import these kinds of price or rate data streams (accessible as CSV or Excel files at the above URLs) directly into Mathematica. I've looked at the

Google's Sketchup is a nice, simple 3D-object modeler. Moreover Google has an enormous warehouse of 3D objects so that you actually don't have to do much modeling yourself if you aren't particularly gifted in this area. Many of the 3D buildings in Google Earth are made with Sketchup. The capability to import Sketchup's SKP files in Mathematica would be very nice, but alas, it doesn't do that yet. The free version of Sketchup doesn't export to any other formats than the KMZ (Google Earth) and DAE (Collada) formats. Though MMA can read KMZ/KML files it doesn't read those containing 3D objects.

I'm building a large ParallelTable , and would like to maintain some sense of how the computation is going. For a non parallel table the following code does a great job: counter = 1; Timing[ Monitor[ Table[ counter++ , {n, 10^6}]; , ProgressIndicator[counter, {0, 10^6}] ] ] with the result {0.943512, Null} . For the parallel case, however, it's necessary to make the counter shared between the kernels: counter = 1; SetSharedVariable[counter]; Timing[ Monitor[ ParallelTable[ counter++ , {n, 10^4}]; , ProgressIndicator[counter, {0, 10^4}] ] ] with the result {6.33388, Null} . Since the value of