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  • PHP Math issue with negatives [closed]

    - by user1269625
    Possible Duplicate: PHP negatives keep adding I have this code here.... $remaining = 0; foreach($array as $value=>$row){ $remaining = $remaining + $row['remainingbalance']; } What its doing is that it is going through all the remaining balances in the array which are -51.75 and -17.85 with the code above I get -69.60 which is correct. But I am wondering how when its two negatives if they could subtract? Is that possible? I tried this $remaining = 0; foreach($clientArrayInvoice as $value=>$row){ $remaining = $remaining + abs($row['remainingbalance']); } but it gives me 69.60 without the negative. Anyone got any ideas? my goal is to take -51.75 and -17.85 and come up with -33.90 only when its a negative to do subtract. otherwise add

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  • Cannot install Fast debugger in Netbeans 6.8 for Ruby 1.9

    - by Bragaadeesh
    Hi, I am using Netbeans 6.8 version and Ruby 1.9.1 installed on Windows XP. I tried to install the fast debugger and I am getting the following error. Building native extensions. This could take a while... ERROR: Error installing ruby-debug-ide: ERROR: Failed to build gem native extension. D:/Ruby19/bin/ruby.exe mkrf_conf.rb Building native extensions. This could take a while... Gem files will remain installed in D:/Ruby19/lib/ruby/gems/1.9.1/gems/ruby-debug-ide-0.4.9 for inspection. Results logged to D:/Ruby19/lib/ruby/gems/1.9.1/gems/ruby-debug-ide-0.4.9/ext/gem_make.out Have someone else faced this problem before. Please help. Thanks.

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  • Fast permutation -> number -> permutation mapping algorithms

    - by ijw
    I have n elements. For the sake of an example, let's say, 7 elements, 1234567. I know there are 7! = 5040 permutations possible of these 7 elements. I want a fast algorithm comprising two functions: f(number) maps a number between 0 and 5039 to a unique permutation, and f'(permutation) maps the permutation back to the number that it was generated from. I don't care about the correspondence between number and permutation, providing each permutation has its own unique number. So, for instance, I might have functions where f(0) = '1234567' f'('1234567') = 0 The fastest algorithm that comes to mind is to enumerate all permutations and create a lookup table in both directions, so that, once the tables are created, f(0) would be O(1) and f('1234567') would be a lookup on a string. However, this is memory hungry, particularly when n becomes large. Can anyone propose another algorithm that would work quickly and without the memory disadvantage?

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  • Point inside Oriented Bounding Box?

    - by Milo
    I have an OBB2D class based on SAT. This is my point in OBB method: public boolean pointInside(float x, float y) { float newy = (float) (Math.sin(angle) * (y - center.y) + Math.cos(angle) * (x - center.x)); float newx = (float) (Math.cos(angle) * (x - center.x) - Math.sin(angle) * (y - center.y)); return (newy > center.y - (getHeight() / 2)) && (newy < center.y + (getHeight() / 2)) && (newx > center.x - (getWidth() / 2)) && (newx < center.x + (getWidth() / 2)); } public boolean pointInside(Vector2D v) { return pointInside(v.x,v.y); } Here is the rest of the class; the parts that pertain: public class OBB2D { private Vector2D projVec = new Vector2D(); private static Vector2D projAVec = new Vector2D(); private static Vector2D projBVec = new Vector2D(); private static Vector2D tempNormal = new Vector2D(); private Vector2D deltaVec = new Vector2D(); private ArrayList<Vector2D> collisionPoints = new ArrayList<Vector2D>(); // Corners of the box, where 0 is the lower left. private Vector2D corner[] = new Vector2D[4]; private Vector2D center = new Vector2D(); private Vector2D extents = new Vector2D(); private RectF boundingRect = new RectF(); private float angle; //Two edges of the box extended away from corner[0]. private Vector2D axis[] = new Vector2D[2]; private double origin[] = new double[2]; public OBB2D(float centerx, float centery, float w, float h, float angle) { for(int i = 0; i < corner.length; ++i) { corner[i] = new Vector2D(); } for(int i = 0; i < axis.length; ++i) { axis[i] = new Vector2D(); } set(centerx,centery,w,h,angle); } public OBB2D(float left, float top, float width, float height) { for(int i = 0; i < corner.length; ++i) { corner[i] = new Vector2D(); } for(int i = 0; i < axis.length; ++i) { axis[i] = new Vector2D(); } set(left + (width / 2), top + (height / 2),width,height,0.0f); } public void set(float centerx,float centery,float w, float h,float angle) { float vxx = (float)Math.cos(angle); float vxy = (float)Math.sin(angle); float vyx = (float)-Math.sin(angle); float vyy = (float)Math.cos(angle); vxx *= w / 2; vxy *= (w / 2); vyx *= (h / 2); vyy *= (h / 2); corner[0].x = centerx - vxx - vyx; corner[0].y = centery - vxy - vyy; corner[1].x = centerx + vxx - vyx; corner[1].y = centery + vxy - vyy; corner[2].x = centerx + vxx + vyx; corner[2].y = centery + vxy + vyy; corner[3].x = centerx - vxx + vyx; corner[3].y = centery - vxy + vyy; this.center.x = centerx; this.center.y = centery; this.angle = angle; computeAxes(); extents.x = w / 2; extents.y = h / 2; computeBoundingRect(); } //Updates the axes after the corners move. Assumes the //corners actually form a rectangle. private void computeAxes() { axis[0].x = corner[1].x - corner[0].x; axis[0].y = corner[1].y - corner[0].y; axis[1].x = corner[3].x - corner[0].x; axis[1].y = corner[3].y - corner[0].y; // Make the length of each axis 1/edge length so we know any // dot product must be less than 1 to fall within the edge. for (int a = 0; a < axis.length; ++a) { float l = axis[a].length(); float ll = l * l; axis[a].x = axis[a].x / ll; axis[a].y = axis[a].y / ll; origin[a] = corner[0].dot(axis[a]); } } public void computeBoundingRect() { boundingRect.left = JMath.min(JMath.min(corner[0].x, corner[3].x), JMath.min(corner[1].x, corner[2].x)); boundingRect.top = JMath.min(JMath.min(corner[0].y, corner[1].y),JMath.min(corner[2].y, corner[3].y)); boundingRect.right = JMath.max(JMath.max(corner[1].x, corner[2].x), JMath.max(corner[0].x, corner[3].x)); boundingRect.bottom = JMath.max(JMath.max(corner[2].y, corner[3].y),JMath.max(corner[0].y, corner[1].y)); } public void set(RectF rect) { set(rect.centerX(),rect.centerY(),rect.width(),rect.height(),0.0f); } // Returns true if other overlaps one dimension of this. private boolean overlaps1Way(OBB2D other) { for (int a = 0; a < axis.length; ++a) { double t = other.corner[0].dot(axis[a]); // Find the extent of box 2 on axis a double tMin = t; double tMax = t; for (int c = 1; c < corner.length; ++c) { t = other.corner[c].dot(axis[a]); if (t < tMin) { tMin = t; } else if (t > tMax) { tMax = t; } } // We have to subtract off the origin // See if [tMin, tMax] intersects [0, 1] if ((tMin > 1 + origin[a]) || (tMax < origin[a])) { // There was no intersection along this dimension; // the boxes cannot possibly overlap. return false; } } // There was no dimension along which there is no intersection. // Therefore the boxes overlap. return true; } public void moveTo(float centerx, float centery) { float cx,cy; cx = center.x; cy = center.y; deltaVec.x = centerx - cx; deltaVec.y = centery - cy; for (int c = 0; c < 4; ++c) { corner[c].x += deltaVec.x; corner[c].y += deltaVec.y; } boundingRect.left += deltaVec.x; boundingRect.top += deltaVec.y; boundingRect.right += deltaVec.x; boundingRect.bottom += deltaVec.y; this.center.x = centerx; this.center.y = centery; computeAxes(); } // Returns true if the intersection of the boxes is non-empty. public boolean overlaps(OBB2D other) { if(right() < other.left()) { return false; } if(bottom() < other.top()) { return false; } if(left() > other.right()) { return false; } if(top() > other.bottom()) { return false; } if(other.getAngle() == 0.0f && getAngle() == 0.0f) { return true; } return overlaps1Way(other) && other.overlaps1Way(this); } public Vector2D getCenter() { return center; } public float getWidth() { return extents.x * 2; } public float getHeight() { return extents.y * 2; } public void setAngle(float angle) { set(center.x,center.y,getWidth(),getHeight(),angle); } public float getAngle() { return angle; } public void setSize(float w,float h) { set(center.x,center.y,w,h,angle); } public float left() { return boundingRect.left; } public float right() { return boundingRect.right; } public float bottom() { return boundingRect.bottom; } public float top() { return boundingRect.top; } public RectF getBoundingRect() { return boundingRect; } public boolean overlaps(float left, float top, float right, float bottom) { if(right() < left) { return false; } if(bottom() < top) { return false; } if(left() > right) { return false; } if(top() > bottom) { return false; } return true; } public static float distance(float ax, float ay,float bx, float by) { if (ax < bx) return bx - ay; else return ax - by; } public Vector2D project(float ax, float ay) { projVec.x = Float.MAX_VALUE; projVec.y = Float.MIN_VALUE; for (int i = 0; i < corner.length; ++i) { float dot = Vector2D.dot(corner[i].x,corner[i].y,ax,ay); projVec.x = JMath.min(dot, projVec.x); projVec.y = JMath.max(dot, projVec.y); } return projVec; } public Vector2D getCorner(int c) { return corner[c]; } public int getNumCorners() { return corner.length; } public boolean pointInside(float x, float y) { float newy = (float) (Math.sin(angle) * (y - center.y) + Math.cos(angle) * (x - center.x)); float newx = (float) (Math.cos(angle) * (x - center.x) - Math.sin(angle) * (y - center.y)); return (newy > center.y - (getHeight() / 2)) && (newy < center.y + (getHeight() / 2)) && (newx > center.x - (getWidth() / 2)) && (newx < center.x + (getWidth() / 2)); } public boolean pointInside(Vector2D v) { return pointInside(v.x,v.y); } public ArrayList<Vector2D> getCollsionPoints(OBB2D b) { collisionPoints.clear(); for(int i = 0; i < corner.length; ++i) { if(b.pointInside(corner[i])) { collisionPoints.add(corner[i]); } } for(int i = 0; i < b.corner.length; ++i) { if(pointInside(b.corner[i])) { collisionPoints.add(b.corner[i]); } } return collisionPoints; } }; What could be wrong? When I getCollisionPoints for 2 OBBs I know are penetrating, it returns no points. Thanks

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  • More Fun With Math

    - by PointsToShare
    More Fun with Math   The runaway student – three different ways of solving one problem Here is a problem I read in a Russian site: A student is running away. He is moving at 1 mph. Pursuing him are a lion, a tiger and his math teacher. The lion is 40 miles behind and moving at 6 mph. The tiger is 28 miles behind and moving at 4 mph. His math teacher is 30 miles behind and moving at 5 mph. Who will catch him first? Analysis Obviously we have a set of three problems. They are all basically the same, but the details are different. The problems are of the same class. Here is a little excursion into computer science. One of the things we strive to do is to create solutions for classes of problems rather than individual problems. In your daily routine, you call it re-usability. Not all classes of problems have such solutions. If a class has a general (re-usable) solution, it is called computable. Otherwise it is unsolvable. Within unsolvable classes, we may still solve individual (some but not all) problems, albeit with different approaches to each. Luckily the vast majority of our daily problems are computable, and the 3 problems of our runaway student belong to a computable class. So, let’s solve for the catch-up time by the math teacher, after all she is the most frightening. She might even make the poor runaway solve this very problem – perish the thought! Method 1 – numerical analysis. At 30 miles and 5 mph, it’ll take her 6 hours to come to where the student was to begin with. But by then the student has advanced by 6 miles. 6 miles require 6/5 hours, but by then the student advanced by another 6/5 of a mile as well. And so on and so forth. So what are we to do? One way is to write code and iterate it until we have solved it. But this is an infinite process so we’ll end up with an infinite loop. So what to do? We’ll use the principles of numerical analysis. Any calculator – your computer included – has a limited number of digits. A double floating point number is good for about 14 digits. Nothing can be computed at a greater accuracy than that. This means that we will not iterate ad infinidum, but rather to the point where 2 consecutive iterations yield the same result. When we do financial computations, we don’t even have to go that far. We stop at the 10th of a penny.  It behooves us here to stop at a 10th of a second (100 milliseconds) and this will how we will avoid an infinite loop. Interestingly this alludes to the Zeno paradoxes of motion – in particular “Achilles and the Tortoise”. Zeno says exactly the same. To catch the tortoise, Achilles must always first come to where the tortoise was, but the tortoise keeps moving – hence Achilles will never catch the tortoise and our math teacher (or lion, or tiger) will never catch the student, or the policeman the thief. Here is my resolution to the paradox. The distance and time in each step are smaller and smaller, so the student will be caught. The only thing that is infinite is the iterative solution. The race is a convergent geometric process so the steps are diminishing, but each step in the solution takes the same amount of effort and time so with an infinite number of steps, we’ll spend an eternity solving it.  This BTW is an original thought that I have never seen before. But I digress. Let’s simply write the code to solve the problem. To make sure that it runs everywhere, I’ll do it in JavaScript. function LongCatchUpTime(D, PV, FV) // D is Distance; PV is Pursuers Velocity; FV is Fugitive’ Velocity {     var t = 0;     var T = 0;     var d = parseFloat(D);     var pv = parseFloat (PV);     var fv = parseFloat (FV);     t = d / pv;     while (t > 0.000001) //a 10th of a second is 1/36,000 of an hour, I used 1/100,000     {         T = T + t;         d = t * fv;         t = d / pv;     }     return T;     } By and large, the higher the Pursuer’s velocity relative to the fugitive, the faster the calculation. Solving this with the 10th of a second limit yields: 7.499999232000001 Method 2 – Geometric Series. Each step in the iteration above is smaller than the next. As you saw, we stopped iterating when the last step was small enough, small enough not to really matter.  When we have a sequence of numbers in which the ratio of each number to its predecessor is fixed we call the sequence geometric. When we are looking at the sum of sequence, we call the sequence of sums series.  Now let’s look at our student and teacher. The teacher runs 5 times faster than the student, so with each iteration the distance between them shrinks to a fifth of what it was before. This is a fixed ratio so we deal with a geometric series.  We normally designate this ratio as q and when q is less than 1 (0 < q < 1) the sum of  + … +  is  – 1) / (q – 1). When q is less than 1, it is easier to use ) / (1 - q). Now, the steps are 6 hours then 6/5 hours then 6/5*5 and so on, so q = 1/5. And the whole series is multiplied by 6. Also because q is less than 1 , 1/  diminishes to 0. So the sum is just  / (1 - q). or 1/ (1 – 1/5) = 1 / (4/5) = 5/4. This times 6 yields 7.5 hours. We can now continue with some algebra and take it back to a simpler formula. This is arduous and I am not going to do it here. Instead let’s do some simpler algebra. Method 3 – Simple Algebra. If the time to capture the fugitive is T and the fugitive travels at 1 mph, then by the time the pursuer catches him he travelled additional T miles. Time is distance divided by speed, so…. (D + T)/V = T  thus D + T = VT  and D = VT – T = (V – 1)T  and T = D/(V – 1) This “strangely” coincides with the solution we just got from the geometric sequence. This is simpler ad faster. Here is the corresponding code. function ShortCatchUpTime(D, PV, FV) {     var d = parseFloat(D);     var pv = parseFloat (PV);     var fv = parseFloat (FV);     return d / (pv - fv); } The code above, for both the iterative solution and the algebraic solution are actually for a larger class of problems.  In our original problem the student’s velocity (speed) is 1 mph. In the code it may be anything as long as it is less than the pursuer’s velocity. As long as PV > FV, the pursuer will catch up. Here is the really general formula: T = D / (PV – FV) Finally, let’s run the program for each of the pursuers.  It could not be worse. I know he’d rather be eaten alive than suffering through yet another math lesson. See the code run? Select  “Catch Up Time” in www.mgsltns.com/games.htm The host is running on Unix, so the link is case sensitive. That’s All Folks

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  • Kalculate = math + fun

    - by Devin A. Rychetnik
    Kalculate is a you vs. the Internet style game for math lovers. The rules are simple: answer as many math problems as you can in 90 seconds. At the end of each round, Kalculate will tally up all the scores and show you where you ranked relative to others currently playing.Tip: answering 3 questions in 10 seconds earns you a score multiplier      If you prefer to just practice and stay out of the competition, there's an offline mode that allows you to play solo.Kalculate is free (ad-supported) and can be downloaded here.

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  • SQLAuthority News – Fast Track Data Warehouse 3.0 Reference Guide

    - by pinaldave
    http://msdn.microsoft.com/en-us/library/gg605238.aspx I am very excited that Fast Track Data Warehouse 3.0 reference guide has been announced. As a consultant I have always enjoyed working with Fast Track Data Warehouse project as it truly expresses the potential of the SQL Server Engine. Here is few details of the enhancement of the Fast Track Data Warehouse 3.0 reference architecture. The SQL Server Fast Track Data Warehouse initiative provides a basic methodology and concrete examples for the deployment of balanced hardware and database configuration for a data warehousing workload. Balance is measured across the key components of a SQL Server installation; storage, server, application settings, and configuration settings for each component are evaluated. Description Note FTDW 3.0 Architecture Basic component architecture for FT 3.0 based systems. New Memory Guidelines Minimum and maximum tested memory configurations by server socket count. Additional Startup Options Notes for T-834 and setting for Lock Pages in Memory. Storage Configuration RAID1+0 now standard (RAID1 was used in FT 2.0). Evaluating Fragmentation Query provided for evaluating logical fragmentation. Loading Data Additional options for CI table loads. MCR Additional detail and explanation of FTDW MCR Rating. Read white paper on fast track data warehousing. Reference: Pinal Dave (http://blog.SQLAuthority.com)   Filed under: Business Intelligence, Data Warehousing, PostADay, SQL, SQL Authority, SQL Documentation, SQL Download, SQL Query, SQL Server, SQL Tips and Tricks, SQL White Papers, SQLAuthority News, T SQL, Technology

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  • Make the Web Fast: Google Web Fonts - making pretty, fast!

    Make the Web Fast: Google Web Fonts - making pretty, fast! Join us for a technical deep-dive on Web Fonts: how they work, the data formats, performance optimizations, and tips and tricks for making your site both fast and pretty at the same time - turns out, these two goals are not mutually exclusive! From: GoogleDevelopers Views: 468 69 ratings Time: 01:11:43 More in Science & Technology

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  • Advice on learning programming languages and math.

    - by Joris Ooms
    I feel like I'm getting stuck lately when it comes to learning about programming-related things; I thought I'd ask a question here and write it all down in the hope to get some pointers/advice from people. Perhaps writing it down helps me put things in perspective for myself aswell. I study Interactive Multimedia Design. This course is based on two things: graphic design on one hand, and web development on the other hand. I have quite a decent knowledge of web-related languages (the usual HTML/JS/PHP) and I'll be getting a course on ASP.NET next year. In my free time, I have learnt how to work with CodeIgniter, aswell as some diving into Ruby (and Rails) and basic iOS programming. In my first year of college I also did a class on Java (19/20 on the end result). This grade doesn't really mean anything though; I have the basics of OOP down but Java-wise, we learnt next to nothing. Considering the time I have been programming in, for example, PHP.. I can't say I'm bad at it. I'm definitely not good or great at it, but I'm decent. My teachers tell me I have the programming thing down. They just tell me I should keep on learning. So that's what I do, and I try to take in as much as possible; however, sometimes I'm unsure where to start and I have this tendency to always doubt myself. Now, for the 'question'. I want to get into iOS programming. I know iOS programming boils down to programming in Cocoa Touch and Objective-C. I also know Obj-C is a superset of C. I have done a class on C a couple of years ago, but I failed miserably. I got stuck at pointers and never really understood them.. Until like a month ago. I suddenly 'got' it. I have been working through a book on Objective-C for a week or so now, and I understand the basics (I'm at like.. chapter 6 or so). However, I keep running into similar problems as the ones I had when I did the C class: I suck at math. No, really. I come from a Latin-Modern Languages background in high school and I had nearly no math classes back then. I wanted to study Computer Science, but I failed there because of the miserable state of my mathematics knowledge. I can't explain why I'm suddenly talking about math here though, because it isn't directly related to programming.. yet it is. For example, the examples in the book I'm reading now are about programming a fraction-calculator. All good, I can do the programming when I get the formulas down.. but it takes me a full day or more to actually get to that point. I also find it hard to come up with ideas for myself. I made one small iOS app the other day and it's just a button / label kind of thing. When I press the button, it generates a random number. That's really all I could come up with. Can you 'learn' that? It probably comes down to creativity, but evidently, I'm not too great at being creative. Are there any sites or resources out there that provide something like a basic list of things you can program when you're just starting out? Maybe I'm focusing on too many things at once. I want to keep my HTML/CSS at a decent level, while learning PHP and CodeIgniter, while diving into Ruby on Rails and learning Objective-C and the iOS SDK at the same time. I just want to be good at something, I guess. The problem is that I can't seem to be happy with my PHP stuff. I want more, something 'harder'; that's why I decided to pick up the iOS thing. Like I said, I have the basics down of a lot of different languages. I can program something simple in Java, in C, in Objective-C as of this week.. but it ends there. Mostly because I can't come up with ideas for more complex applications, and also because I just doubt myself: 'Oh, that's too complex, I can never do that'. And then it ends there. To conclude my rant, let me basically rephrase my questions into a 'tl;dr' part. A. I want to get into iOS programming and I have basic knowledge of C/Objective-C. However, I struggle to come up with ideas of my own and implement them and I also suck at math which is something that isn't directly related to, yet often needed while programming. What can I do? B. I have an interest in a lot of different programming languages and I can't stop reading/learning. However, I don't feel like I'm good in anything. Should I perhaps focus on just one language for a year or longer, or keep taking it all in at the same time and hope I'll finally get them all down? C. Are there any resources out there that provide basic ideas of things I can program? I'm thinking about 'simple' command-line applications here to help me while studying C/Obj-C away from the whole iPhone SDK. Like I said, the examples in my book are mainly math-based (fraction calculator) and it's kinda hard. :( Thanks a lot for reading my post. I didn't plan it to be this long but oh well. Thanks in advance for any answers.

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  • Apress Deal of the day - 23/Feb/2011 - Ultra-Fast ASP.NET: Building Ultra-Fast and Ultra-Scalable Websites Using ASP.NET and SQL Server

    - by TATWORTH
    Today's $10 deal of the day at http://www.apress.com/info/dailydeal  is Ultra-Fast ASP.NET: Building Ultra-Fast and Ultra-Scalable Websites Using ASP.NET and SQL Server by Richard Kessig - ISBN 978-1-4302-2383-2 I won a copy of this book at 101 Books. Richard Kessig is an all-star member of forums.asp.net - see http://forums.asp.net/members/RickNZ.aspx - this book has been on before as deal of the day. If you did not get a copy then, I suggest getting it today. " Ultra-Fast ASP.NET provides a practical guide to building extremely fast and scalable web sites using ASP.NET and SQL Server. It strikes a balance between imparting usable advice and backing that advice up with supporting background information. $49.99 | Published Nov 2009 | Rick Kiessig"

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  • Apress Deal of the day - 6/Feb/2011 - Ultra-Fast ASP.NET: Building Ultra-Fast and Ultra-Scalable Websites Using ASP.NET and SQL Server

    - by TATWORTH
    Today's $10 deal of the day at http://www.apress.com/info/dailydeal  is Ultra-Fast ASP.NET: Building Ultra-Fast and Ultra-Scalable Websites Using ASP.NET and SQL Server by Richard Kessig - ISBN 978-1-4302-2383-2 I won a copy of this book at 101 Books. Richard Kessig is an all-star member of forums.asp.net - see http://forums.asp.net/members/RickNZ.aspx " Ultra-Fast ASP.NET provides a practical guide to building extremely fast and scalable web sites using ASP.NET and SQL Server. It strikes a balance between imparting usable advice and backing that advice up with supporting background information. $49.99 | Published Nov 2009 | Rick Kiessig"

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  • Extracting dates from html meta data in FAST-ESP

    - by Neil
    During document processing I want to extract all dates from html meta data and then identify the latest date which will be used to populate a date field (dtgeneric1). <meta name="OriginalPublicationDate" content="2010/04/21 12:06:36" /> <meta name="LastModificationDate" content="2010/04/22 14:10:16" /> + other non-date meta data Inspection using spy stages shows that our pipeline already adds meta_* attributes but the meta data names will be different across documents from different sources. #### ATTRIBUTE meta_originalpublicationdate <class 'docproc.DocumentAttributes.TextChunks'>: 2010/04/21 12:06:36 #### ATTRIBUTE meta_lastmodificationdate <class 'docproc.DocumentAttributes.TextChunks'>: 2010/04/22 14:10:16 + other non-date meta attributes Ideally we would like to pass all the meta_* attributes to a Python stage and use that to work out which are dates and which is the largest but there seems to be no way of specifying "all meta attributes" as input. Has anyone done something similar and can offer any advice on the best way to do this. Thanks Neil

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  • Is Fast Enumeration messing with my text output?

    - by Dan Ray
    Here I am iterating through an array of NSDictionary objects (inside the parsed JSON response of the EXCELLENT MapQuest directions API). I want to build up an HTML string to put into a UIWebView. My code says: for (NSDictionary *leg in legs ) { NSString *thisLeg = [NSString stringWithFormat:@"<br>%@ - %@", [leg valueForKey:@"narrative"], [leg valueForKey:@"distance"]]; NSLog(@"This leg's string is %@", thisLeg); [directionsOutput appendString:thisLeg]; } The content of directionsOutput (which is an NSMutableString) contains ALL the values for [leg valueForKey:@"narrative"], wrapped up in parens, followed by a hyphen, followed by all the parenthesized values for [leg valueForKey:@"distance"]. So I put in that NSLog call... and I get the same thing there! It appears that the for() is somehow batching up our output values as we iterate, and putting out the output only once. How do I make it not do this but instead do what I actually want, which is an iterative output as I iterate? Here's what NSLog gets. Yes, I know I need to figure out NSNumberFormatter. ;-) This leg's string is ( "Start out going NORTH on INFINITE LOOP.", "Turn LEFT to stay on INFINITE LOOP.", "Turn RIGHT onto N DE ANZA BLVD.", "Merge onto I-280 S toward SAN JOSE.", "Merge onto CA-87 S via EXIT 3A.", "Take the exit on the LEFT.", "Merge onto CA-85 S via EXIT 1A on the LEFT toward GILROY.", "Merge onto US-101 S via EXIT 1A on the LEFT toward LOS ANGELES.", "Take the CA-152 E/10TH ST exit, EXIT 356.", "Turn LEFT onto CA-152/E 10TH ST/PACHECO PASS HWY. Continue to follow CA-152/PACHECO PASS HWY.", "Turn SLIGHT RIGHT.", "Turn SLIGHT RIGHT onto PACHECO PASS HWY/CA-152 E. Continue to follow CA-152 E.", "Merge onto I-5 S toward LOS ANGELES.", "Take the CA-46 exit, EXIT 278, toward LOST HILLS/WASCO.", "Turn LEFT onto CA-46/PASO ROBLES HWY. Continue to follow CA-46.", "Merge onto CA-99 S toward BAKERSFIELD.", "Merge onto CA-58 E via EXIT 24 toward TEHACHAPI/MOJAVE.", "Merge onto I-15 N via the exit on the LEFT toward I-40/LAS VEGAS.", "Keep RIGHT to take I-40 E via EXIT 140A toward NEEDLES (Passing through ARIZONA, NEW MEXICO, TEXAS, OKLAHOMA, and ARKANSAS, then crossing into TENNESSEE).", "Merge onto I-40 E via EXIT 12C on the LEFT toward NASHVILLE (Crossing into NORTH CAROLINA).", "Merge onto I-40 BR E/US-421 S via EXIT 188 on the LEFT toward WINSTON-SALEM.", "Take the CLOVERDALE AVE exit, EXIT 4.", "Turn LEFT onto CLOVERDALE AVE SW.", "Turn SLIGHT LEFT onto N HAWTHORNE RD.", "Turn RIGHT onto W NORTHWEST BLVD.", "1047 W NORTHWEST BLVD is on the LEFT." ) - ( 0.0020000000949949026, 0.07800000160932541, 0.14000000059604645, 7.827000141143799, 5.0329999923706055, 0.15299999713897705, 5.050000190734863, 20.871000289916992, 0.3050000071525574, 2.802999973297119, 0.10199999809265137, 37.78000259399414, 124.50700378417969, 0.3970000147819519, 25.264001846313477, 20.475000381469727, 125.8580093383789, 4.538000106811523, 1693.0350341796875, 628.8970336914062, 3.7990000247955322, 0.19099999964237213, 0.4099999964237213, 0.257999986410141, 0.5170000195503235, 0 )

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  • Fast 4x4 matrix multiplication in Java with NIO float buffers

    - by kayahr
    I know there are LOT of questions like that but I can't find one specific to my situation. I have 4x4 matrices implemented as NIO float buffers (These matrices are used for OpenGL). Now I want to implement a multiply method which multiplies Matrix A with Matrix B and stores the result in Matrix C. So the code may look like this: class Matrix4f { private FloatBuffer buffer = FloatBuffer.allocate(16); public Matrix4f multiply(Matrix4f matrix2, Matrix4f result) { {{{result = this * matrix2}}} <-- I need this code return result; } } What is the fastest possible code to do this multiplication? Some OpenGL implementations (Like the OpenGL ES stuff in Android) provide native code for this but others doesn't. So I want to provide a generic multiplication method for these implementations.

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  • Fast ceiling of an integer division in C / C++

    - by andand
    Given integer values x and y, C and C++ returns as the quotient q = x/y the floor of the floating point valued equivalent. I'm interestd in a method of returning the ceiling instead? For example, ceil(10/5) = 2 and ceil(11/5) = 3. The obvious approach involves something like: q = x / y; if (q * y < x) ++q; This requires an extra comparison and multiplication; and other methods I've seen (used in fact) involve casting as a float or double. Is there a more direct method that avoids the additional multiplication (or a second division) and branch, and that also avoids casting as a floating point number?

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  • how to follow python polymorphism standards with math functions

    - by krishnab
    So I am reading up on python in Mark Lutz's wonderful LEARNING PYTHON book. Mark makes a big deal about how part of the python development philosophy is polymorphism and that functions and code should rely on polymorphism and not do much type checking. However, I do a lot of math type programming and so the idea of polymorphism does not really seem to apply--I don't want to try and run a regression on a string or something. So I was wondering if there is something I am missing here. What are the applications of polymorphism when I am writing functions for math--or is type checking philosophically okay in this case.

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  • If a library doesn't provide all my needs, how should I proceed?

    - by 9a3eedi
    I'm developing an application involving math and physics models, and I'd like to use a Math library for things like Matrices. I'm using C#, and so I was looking for some libraries and found Math.NET. I'm under the impression, from past experience, that for math, using a robust and industry-approved third party library is much better than writing your own code. It seems good for many purposes, but it does not provide support for Quaternions, which I need to use as a type. Also, I need some functions in Vector and Matrix that also aren't provided, such as rotation matrices and vector rotation functions, and calculating cross products. At the same time, it provides a lot of functions/classes that I simply do not need, which might mean a lot of unnecessary bloat and complexity. At this rate, should I even bother using the library? Should I write my own math library? Or is it a better idea to stick to the third party library and somehow wrap around it? Perhaps I should make a subclass of the Matrix and Vector type of the library? But isn't that considered bad style? I've also tried looking for other libraries but unfortunately I couldn't find anything suitable.

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  • What are atan and atan2 used for in games?

    - by kyrogue
    I am having some trouble understanding Math.tan() and Math.atan() and Math.atan2(). I have basic knowledge of trigonmetry but the usage of SIN, COS, and TAN etc for game development is very new to me. I am reading on some tutorials and I see that by using tangent we can get the angle in which one object needs to be rotated by how much to face another object for example my mouse. So why do we still need to use atan or atan2?

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  • How can I fall in love with Math? Again?

    - by gotts
    After reading How to not sort by average rating by Evan Miller I was really inspired to learn some more math. But after thinking about it for a while I didn't find a way I can use beyond-trivial math in my pet projects.. Or probably it is a moment like "You are not aware that you are not aware" and I should learn more math before I can start to see great examples of how I can apply it?

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