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  • How do you calculate the reflex angle given to vectors in 3D space?

    - by Reimund
    I want to calculate the angle between two vectors a and b. Lets assume these are at the origin. This can be done with theta = arccos(a . b / |a| * |b|) However arccos gives you the angle in [0, pi], i.e. it will never give you an angle greater than 180 degrees, which is what I want. So how do you find out when the vectors have gone past the 180 degree mark? In 2D I would simply let the sign of the y-component on one of the vectors determine what quadrant the vector is in. But what is the easiest way to do it in 3D?

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  • NSPredicate as a constraint solver?

    - by Felixyz
    I'm working on a project which includes some slightly more complex dynamic layout of interface elements than what I'm used to. I always feel stupid writing complex code that checks if so-and-so is close to such-and-such and in that case move it x% in some direction, etc. That's just not how programming should be done. Programming should be as declarative as possible! Precisely because what I'm going to do is fairly simple, I thought it would be a good opportunity to try something new, and I thought of using NSPredicate as a simple constraints solver. I've only used NSPredicate for very simple tasks so far, but I know that it capable of much more. Are there any ideas, experiences, examples, warnings, insights that could be useful here? I'll give a very simple example so there will be something concrete to answer. How could I use NSPredicate to solve the following constraints: viewB.xmid = (viewB.leftEdge + viewB.width) / 2 viewB.xmid = max(300, viewA.rightEdge + 20 + viewB.width/2) ("viewB should be horizontally centered on coordinate 300, unless its left edge gets within 20 pixels of viewB's right edge, in which case viewA's left edge should stay fixed at 20 pixels to the right of viewB's right edge and viewA's horizontal center get pushed to the right.") viewA.rightEdge and viewB.width can vary, and those are the 'input variables'. EDIT: Any solution would probably have to use the NSExpression method -(id)expressionValueWithObject:(id)object context:(NSMutableDictionary *)context. This answer is relevant.

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  • SQL Server view: how to add missing rows using interpolation

    - by Christopher Klein
    Running into a problem. I have a table defined to hold the values of the daily treasury yield curve. It's a pretty simple table used for historical lookup of values. There are notibly some gaps in the table on year 4, 6, 8, 9, 11-19 and 21-29. The formula is pretty simple in that to calculate year 4 it's 0.5*Year3Value + 0.5*Year5Value. The problem is how can I write a VIEW that can return the missing years? I could probably do it in a stored procedure but the end result needs to be a view.

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  • Finding coordinates of a point between two points?

    - by Nicros
    Doing some 3D stuff in wpf- want to use a simpler test to see if everything is working (before moving to curves). The basic question is given two points x1,y1,z1 and x2,y2,z2 I have calculated the distance between the points. But how to find the coordinates of another point (x3,y3,z3) that lies on that line at some distance? I.e. if my line is 100 long between -50,0,0 and 50,0,0 what are the coordinates of the point at 100 * 0.1 along the line? I think this is a simple formula but I haven't found it yet....

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  • Alternative to as3isolib?

    - by tedw4rd
    Hi everyone, I've been working on a Flash game that involves an isometric space. I've been using as3isolib for a while now, and I'm less than impressed with how easy it is to use. Whether I'm approaching it the wrong way or it's just not that great to use is a question for another post. Anyways, I've been thinking of a different way to approach the problem of isometric positions, and I think I've got an idea that might work. Essentially, each object that is to be rendered to the iso-space maintains a 3-coordinate position. Those items are then registered with a camera that projects that 3-coordinate position to a 2-coordinate point on the screen according to the math on this Wikipedia article. Then, the MovieClip is added to the stage (or to the camera's MovieClip, perhaps) at that point, and at a child index of the point's y-value. That way, I figure objects that are closer to the camera will be "above" the objects further away, and will get rendered over them. So my question, then, is two-fold: Do you think this idea will work the way I think it will? Are there any existing 3D matrix/vector packages that I should look at? I know there's a Matrix3 class in Flex 3, but we're not using Flex for this game. Thanks!

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  • rotate a plane around a diagonal

    - by compie
    I would like to rotate a plane, not around a single (X or Y) axis, but around the diagonal (45 degrees between X and Y). How do I calculate the Rx and Ry given the Rdiagonal? (Rdiagonal is the amount of rotation I would like to achieve around the diagonal axis).

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  • implementation of interp1 function of MATLAB in J2ME

    - by Jeeka
    Hi, i am looking to implement the interp1, 1-D data interpolation (table lookup), function available in MATLAB in J2ME or JAVA. here is the link http://www.mathworks.com/access/helpdesk/help/techdoc/ref/interp1.html Is there any library available in J2ME or JAVA which has already implemented the same function ? If not can anybody help me in implementing interp1 function in J2ME or JAVA ?

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  • Solve equation from string to result in C

    - by Alexandre Cassagne
    Hi, I would like to know if anyone has info or experience on how to do something which sounds simple but doesn't look like it when trying to program it. The idea is : give a string containing an equation, such as : "2*x = 10" for example (this is simple, but it could get very complex, such as sqrt(54)*35=x^2; and so on....) and the program would return x = 5 and possibly give a log of how he got there. Is this doable ? If so, does anyone have a lead ? For info there is this site (http://www.numberempire.com/equationsolver.php) which does the same thing in PHP, but isn't open source. Thanks for any help !

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  • How to Create a Duotone Interpolation Color in GDI+ (.NET)?

    - by Otaku
    I have two colors that I need to linearly interpolate to create a duotone color based of another color's luminance value. Dim clr1 = Color.Black Dim clr2 = Color.FromArgb(224, 255, 165) Dim lum = Color.FromArgb(226, 189, 136).GetBrightness 'lum = 0.71 I know the result of this should be R=171, G=192, B=125, but I can't figure out how to do this. I've searched through SO quite a bit and all over Google, but have not discovered a way to produce this result. Can anyone help?

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  • Simple encryption - Sum of Hashes in C

    - by Dogbert
    I am attempting to demonstrate a simple proof of concept with respect to a vulnerability in a piece of code in a game written in C. Let's say that we want to validate a character login. The login is handled by the user choosing n items, (let's just assume n=5 for now) from a graphical menu. The items are all medieval themed: eg: _______________________________ | | | | | Bow | Sword | Staff | |-----------|-----------|-------| | Shield | Potion | Gold | |___________|___________|_______| The user must click on each item, then choose a number for each item. The validation algorithm then does the following: Determines which items were selected Drops each string to lowercase (ie: Bow becomes bow, etc) Calculates a simple string hash for each string (ie: `bow = b=2, o=15, w=23, sum = (2+15+23=40) Multiplies the hash by the value the user selected for the corresponding item; This new value is called the key Sums together the keys for each of the selected items; this is the final validation hash IMPORTANT: The validator will accept this hash, along with non-zero multiples of it (ie: if the final hash equals 1111, then 2222, 3333, 8888, etc are also valid). So, for example, let's say I select: Bow (1) Sword (2) Staff (10) Shield (1) Potion (6) The algorithm drops each of these strings to lowercase, calculates their string hashes, multiplies that hash by the number selected for each string, then sums these keys together. eg: Final_Validation_Hash = 1*HASH(Bow) + 2*HASH(Sword) + 10*HASH(Staff) + 1*HASH(Shield) + 6*HASH(Potion) By application of Euler's Method, I plan to demonstrate that these hashes are not unique, and want to devise a simple application to prove it. in my case, for 5 items, I would essentially be trying to calculate: (B)(y) = (A_1)(x_1) + (A_2)(x_2) + (A_3)(x_3) + (A_4)(x_4) + (A_5)(x_5) Where: B is arbitrary A_j are the selected coefficients/values for each string/category x_j are the hash values for each string/category y is the final validation hash (eg: 1111 above) B,y,A_j,x_j are all discrete-valued, positive, and non-zero (ie: natural numbers) Can someone either assist me in solving this problem or point me to a similar example (ie: code, worked out equations, etc)? I just need to solve the final step (ie: (B)(Y) = ...). Thank you all in advance.

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  • mystified by qr.Q(): what is an orthonormal matrix in "compact" form?

    - by gappy
    R has a qr() function, which performs QR decomposition using either LINPACK or LAPACK (in my experience, the latter is 5% faster). The main object returned is a matrix "qr" that contains in the upper triangular matrix R (i.e. R=qr[upper.tri(qr)]). So far so good. The lower triangular part of qr contains Q "in compact form". One can extract Q from the qr decomposition by using qr.Q(). I would like to find the inverse of qr.Q(). In other word, I do have Q and R, and would like to put them in a "qr" object. R is trivial but Q is not. The goal is to apply to it qr.solve(), which is much faster than solve() on large systems.

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  • Body CSS Gradient that Stops and Continues as Solid Color

    - by Alfo
    Something similar to this question has been asked here - HTML/CSS Gradient that stops at a perticular height and continues further with a solid color, but as far as I can see this doesn't work when using it on the body for a background color - which is what I want to achieve. Specifically, I would like it to be light blue at the top of the page, gradient into dark blue 200px further down, and then continue in dark blue for ever. Thanks to anybody who can help, Alfo.

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  • Rejigging a floating point equation ...

    - by Jamie
    I'd like to know if there is a way to improve the accuracy of calculating a slope. (This came up a few months back here). It seems by changing: float get_slope(float dXa, float dXb, float dYa, float dYb) { return (dXa - dXb)/(dYa - dYb); } to float get_slope(float dXa, float dXb, float dYa, float dYb) { return dXa/(dYa - dYb) - dXb/(dYa - dYb); } might be an improvement. Suggestions? Edit: It's precision I'm after, not efficiency.

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  • Where can I find BLAS example code (in Fortran)?

    - by Feynman
    I have been searching for decent documentation on blas, and I have found some 315 pages of dense material that ctrl-f does not work on. It provides all the information regarding what input arguments the routines take, but there are a LOT of input arguments and I could really use some example code. I am unable to locate any. I know there has to be some or no one would be able to use these libraries! Specifically, I use ATLAS installed via macports on a mac osx 10.5.8 and I use gfortran from gcc 4.4 (also installed via macports). I am coding in Fortran 90. I am still quite new to Fortran, but I have a fair amount of experience with mathematica, matlab, perl, and shell scripting. I would like to be able to initialize and multiply a dense complex vector by a dense symmetric (but not hermitian) complex matrix. The elements of the matrix are defined through a mathematical function of the indices--call it f(i,j). Could anyone provide some code or a link to some code?

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  • How to solve generic algebra using solver/library programmatically? Matlab, Mathematica, Wolfram etc?

    - by DevDevDev
    I'm trying to build an algebra trainer for students. I want to construct a representative problem, define constraints and relationships on the parameters, and then generate a bunch of Latex formatted problems from the representation. As an example: A specific question might be: If y < 0 and (x+3)(y-5) = 0, what is x? Answer (x = -3) I would like to encode this as a Latex formatted problem like. If $y<0$ and $(x+constant_1)(y+constant_2)=0$ what is the value of x? Answer = -constant_1 And plug into my problem solver constant_1 > 0, constant_1 < 60, constant_1 = INTEGER constant_2 < 0, constant_2 > -60, constant_2 = INTEGER Then it will randomly construct me pairs of (constant_1, constant_2) that I can feed into my Latex generator. Obviously this is an extremely simple example with no real "solving" but hopefully it gets the point across. Things I'm looking for ideally in priority order * Solve algebra problems * Definition of relationships relatively straight forward * Rich support for latex formatting (not just writing encoded strings) Thanks!

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  • Why is glpk-java.jar not the same across all platforms?

    - by pruefsumme
    I'm using glpk-java in one of my projects. It provides a JNI to GLPK, a free LP/MIP solver. To use it under different platforms, you need different shared libraries (libglpk.so under Linux, libglpk.dylib under Mac OS X, glpk.dll under Windows) which is fully understandable. In addition, a java library is needed: glpk-java.jar. It's created as part of the build process. I was wondering why this JAR file (which only contains .class files) is different under different platforms? I.e., I cannot use the same glpk-java.jar under Linux and Mac OS X which means that I cannot put this file under revision control.

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  • Outer product using CBLAS

    - by The Dude
    I am having trouble utilizing CBLAS to perform an Outer Product. My code is as follows: //===SET UP===// double x1[] = {1,2,3,4}; double x2[] = {1,2,3}; int dx1 = 4; int dx2 = 3; double X[dx1 * dx2]; for (int i = 0; i < (dx1*dx2); i++) {X[i] = 0.0;} //===DO THE OUTER PRODUCT===// cblas_dgemm(CblasRowMajor, CblasNoTrans, CblasTrans, dx1, dx2, 1, 1.0, x1, dx1, x2, 1, 0.0, X, dx1); //===PRINT THE RESULTS===// printf("\nMatrix X (%d x %d) = x1 (*) x2 is:\n", dx1, dx2); for (i=0; i<4; i++) { for (j=0; j<3; j++) { printf ("%lf ", X[j+i*3]); } printf ("\n"); } I get: Matrix X (4 x 3) = x1 (*) x2 is: 1.000000 2.000000 3.000000 0.000000 -1.000000 -2.000000 -3.000000 0.000000 7.000000 14.000000 21.000000 0.000000 But the correct answer is found here: https://www.sharcnet.ca/help/index.php/BLAS_and_CBLAS_Usage_and_Examples I have seen: Efficient computation of kronecker products in C But, it doesn't help me because they don't actually say how to utilize dgemm to actually do this... Any help? What am I doing wrong here?

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  • Is it possible to convert striped logical volume to linear logical volume?

    - by JooMing
    I've a logical volume that is striped across three physical volumes. I had to move this logical volume to another physical volume. This worked nicely with pvmove command. However, I discovered later that the logical volume is still striped and now all three stripes are on the same physical volume. Is there any way to convert striped logical volumes to linear logical volumes? I'm using LVM2 on linux. I figured that the obvious possibility is to rename the striped logical volume, create a new linear logical volume, and then copy data over, but that requires taking the filesystem system offline for some time. Unfortunately, I can't do that before the next week. Is there any better alternative?

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  • Is Linear Tape File System (LTFS) Best For Transportable Storage?

    - by rickramsey
    Those of us in tape storage engineering take a lot of pride in what we do, but understand that tape is the right answer to a storage problem only some of the time. And, unfortunately for a storage medium with such a long history, it has built up a few preconceived notions that are no longer valid. When I hear customers debate whether to implement tape vs. disk, one of the common strikes against tape is its perceived lack of usability. If you could go back a few generations of corporate acquisitions, you would discover that StorageTek engineers recognized this problem and started developing a solution where a tape drive could look just like a memory stick to a user. The goal was to not have to care about where files were on the cartridge, but to simply see the list of files that were on the tape, and click on them to open them up. Eventually, our friends in tape over at IBM built upon our work at StorageTek and Sun Microsystems and released the Linear Tape File System (LTFS) feature for the current LTO5 generation of tape drives as an open specification. LTFS is really a wonderful feature and we’re proud to have taken part in its beginnings and, as you’ll soon read, its future. Today we offer LTFS-Open Edition, which is free for you to use in your in Oracle Enterprise Linux 5.5 environment - not only on your LTO5 drives, but also on your Oracle StorageTek T10000C drives. You can download it free from Oracle and try it out. LTFS does exactly what its forefathers imagined. Now you can see immediately which files are on a cartridge. LTFS does this by splitting a cartridge into two partitions. The first holds all of the necessary metadata to create a directory structure for you to easily view the contents of the cartridge. The second partition holds all of the files themselves. When tape media is loaded onto a drive, a complete file system image is presented to the user. Adding files to a cartridge can be as simple as a drag-and-drop just as you do today on your laptop when transferring files from your hard drive to a thumb drive or with standard POSIX file operations. You may be thinking all of this sounds nice, but asking, “when will I actually use it?” As I mentioned at the beginning, tape is not the right solution all of the time. However, if you ever need to physically move data between locations, tape storage with LTFS should be your most cost-effective and reliable answer. I will give you a few use cases examples of when LTFS can be utilized. Media and Entertainment (M&E), Oil and Gas (O&G), and other industries have a strong need for their storage to be transportable. For example, an O&G company hunting for new oil deposits in remote locations takes very large underground seismic images which need to be shipped back to a central data center. M&E operations conduct similar activities when shooting video for productions. M&E companies also often transfers files to third-parties for editing and other activities. These companies have three highly flawed options for transporting data: electronic transfer, disk storage transport, or tape storage transport. The first option, electronic transfer, is impractical because of the expense of the bandwidth required to transfer multi-terabyte files reliably and efficiently. If there’s one place that has bandwidth, it’s your local post office so many companies revert to physically shipping storage media. Typically, M&E companies rely on transporting disk storage between sites even though it, too, is expensive. Tape storage should be the preferred format because as IDC points out, “Tape is more suitable for physical transportation of large amounts of data as it is less vulnerable to mechanical damage during transportation compared with disk" (See note 1, below). However, tape storage has not been used in the past because of the restrictions created by proprietary formats. A tape may only be readable if both the sender and receiver have the same proprietary application used to write the file. In addition, the workflows may be slowed by the need to read the entire tape cartridge during recall. LTFS solves both of these problems, clearing the way for tape to become the standard platform for transferring large files. LTFS is open and, as long as you’ve downloaded the free reader from our website or that of anyone in the LTO consortium, you can read the data. So if a movie studio ships a scene to a third-party partner to add, for example, sounds effects or a music score, it doesn’t have to care what technology the third-party has. If it’s written back to an LTFS-formatted tape cartridge, it can be read. Some tape vendors like to claim LTFS is a “standard,” but beauty is in the eye of the beholder. It’s a specification at this point, not a standard. That said, we’re already seeing application vendors create functionality to write in an LTFS format based on the specification. And it’s my belief that both customers and the tape storage industry will see the most benefit if we all follow the same path. As such, we have volunteered to lead the way in making LTFS a standard first with the Storage Network Industry Association (SNIA), and eventually through to standard bodies such as American National Standards Institute (ANSI). Expect to hear good news soon about our efforts. So, if storage transportability is one of your requirements, I recommend giving LTFS a look. It makes tape much more user-friendly and it’s free, which allows tape to maintain all of its cost advantages over disk! Note 1 - IDC Report. April, 2011. “IDC’s Archival Storage Solutions Taxonomy, 2011” - Brian Zents Website Newsletter Facebook Twitter

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  • Are there IDEs in which the code-completion widget is not a linear list?

    - by Uri
    Most mainstream IDEs use code-completion in the form of a linear list of suggestions (typically methods). Are there any IDEs (mainstream or not) or IDE plugins that use a non-linear widget, such as a tree? (e.g., pick category first, then the actual recommendation)? I'm working on an IDE feature and want to make sure I'm not reinventing the wheel or infringing some patent.

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  • Linear Interpolation. How to implement this algorithm in C ? (Python version is given)

    - by psihodelia
    There exists one very good linear interpolation method. It performs linear interpolation requiring at most one multiply per output sample. I found its description in a third edition of Understanding DSP by Lyons. This method involves a special hold buffer. Given a number of samples to be inserted between any two input samples, it produces output points using linear interpolation. Here, I have rewritten this algorithm using Python: temp1, temp2 = 0, 0 iL = 1.0 / L for i in x: hold = [i-temp1] * L temp1 = i for j in hold: temp2 += j y.append(temp2 *iL) where x contains input samples, L is a number of points to be inserted, y will contain output samples. My question is how to implement such algorithm in ANSI C in a most effective way, e.g. is it possible to avoid the second loop? NOTE: presented Python code is just to understand how this algorithm works. UPDATE: here is an example how it works in Python: x=[] y=[] hold=[] num_points=20 points_inbetween = 2 temp1,temp2=0,0 for i in range(num_points): x.append( sin(i*2.0*pi * 0.1) ) L = points_inbetween iL = 1.0/L for i in x: hold = [i-temp1] * L temp1 = i for j in hold: temp2 += j y.append(temp2 * iL) Let's say x=[.... 10, 20, 30 ....]. Then, if L=1, it will produce [... 10, 15, 20, 25, 30 ...]

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  • Can a Highcharts range selector use non-date linear ranges?

    - by Simon
    I am using the HighStock JS lib to produce a chart that uses a linear series (not a time-series) for the xAxis. I'd still like to use the range-selector in order to zoom to pre-determined ranges within my linear series. Is this possible? For example; say my xAxis has a series: [[121,616],[122,600],[123,605],[124,585.5],[125,575.5],[126,580.5],[127,582],[128,582],[129,584],[130,583]] I'd like to use the range selector to zoom to the last n in the series.

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