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  • When do you use float and when do you use double

    - by Jakub Zaverka
    Frequently in my programming experience I need to make a decision whether I should use float or double for my real numbers. Sometimes I go for float, sometimes I go for double, but really this feels more subjective. If I would be confronted to defend my decision, I would probably not give sound reasons. When do you use float and when do you use double? Do you always use double, only when memory constraints are present you go for float? Or you use always float unless the precision requirement requires you to use double? Are there some substantial differences regarding computational complexity of basic arithemtics between float and double? What are the pros and cons of using float or double? And have you even used long double?

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  • Entry Level Programming Jobs with Applied Math Degree

    - by Mark
    I am about to finish my B.Sc. in Applied Math. I started out in CS a few years back had a bit of a change of heart and decided to go the math route. Now that I am looking for career options finishing up and I'm just wonder how my Applied Math degree will look when applying for programming jobs. I have taken CS courses in C++/Java/C and done 2 semester of Scientific Computing with MATLAB/Mathematica and the like, so I feel like i at least know how to program. Of course I am lacking some of the theoretical courses on the CS. I'd very much like to know how I stack up for a programming job as a math major. Thanks.

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  • Parsing basic math equations for children's educational software?

    - by Simucal
    Inspired by a recent TED talk, I want to write a small piece of educational software. The researcher created little miniature computers in the shape of blocks called "Siftables". [David Merril, inventor - with Siftables in the background.] There were many applications he used the blocks in but my favorite was when each block was a number or basic operation symbol. You could then re-arrange the blocks of numbers or operation symbols in a line, and it would display an answer on another siftable block. So, I've decided I wanted to implemented a software version of "Math Siftables" on a limited scale as my final project for a CS course I'm taking. What is the generally accepted way for parsing and interpreting a string of math expressions, and if they are valid, perform the operation? Is this a case where I should implement a full parser/lexer? I would imagine interpreting basic math expressions would be a semi-common problem in computer science so I'm looking for the right way to approach this. For example, if my Math Siftable blocks where arranged like: [1] [+] [2] This would be a valid sequence and I would perform the necessary operation to arrive at "3". However, if the child were to drag several operation blocks together such as: [2] [\] [\] [5] It would obviously be invalid. Ultimately, I want to be able to parse and interpret any number of chains of operations with the blocks that the user can drag together. Can anyone explain to me or point me to resources for parsing basic math expressions? I'd prefer as much of a language agnostic answer as possible.

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  • Return Double from Boost thread

    - by Benedikt Wutzi
    Hi I have an Boost thread which should return a double. The function looks like this: void analyser::findup(const double startwl, const double max, double &myret){ this->data.begin(); for(int i = (int)data.size() ; i >= 0;i--){ if(this->data[i].lambda > startwl){ if(this->data[i].db >= (max-30)) { myret = this->data[i+1].lambda; std::cout <<"in thread " << myret << std::endl; return; } } } } this function is called by another function: void analyser::start_find_up(const double startwl, const double max){ double tmp = -42.0; boost::thread up(&analyser::findup,*this, startwl,max,tmp); std::cout << "before join " << tmp << std::endl; up.join(); std::cout << "after join " << tmp << std::endl; } Anyway I've tried and googled almost anything but i can't get it to return a value. The output looks like this right now. before join -42 in thread 843.487 after join -42 Thanks for any help.

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  • Underbraces in Word math zones and dealing with stretchy parentheses

    - by Johannes Rössel
    Parentheses in Word usually stretch with whatever they're containing. This might be un-noticeable for things like but for stuff like it's definitely nice, especially compared to the fact that naïve LaTeX users often produce uglinesses such as There is a problem, however, when using under-/overbraces in math and putting parentheses around the complete term it becomes ugly. For simple things like shown here this can be solved by not letting the parentheses stretch which looks almost right. However, for more complex things it's certainly not an option: Both variants look horrible. So is there a way of letting the parentheses only stretch around the actual term parts, not including the under-/overbraces? Those are frequently used for annotations of individual pieces, so simply not using them is a bad idea too. In LaTeX you can get away with guesswork and using explicit sizes for the parentheses instead of relying on \left and \right but I haven't found a comparable option in Word yet. Since the underbrace is (tree-wise) a sibling of the term in parentheses it probably simply has to stretch and there probably can't be an algorithm that determines when to stretch or when not, considering that \above and \below are used for annotations as well but also for other things where perentheses have to stretch. Also, since the parenthesized expression is opaque from the outside one has to put the underbrace inside. From a markup point of view, at least. One can probably draw the rest around but that falls apart when styles change and wouldn't be a good idea either.

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  • Underbraces in Word math zones and dealing with parentheses

    - by Johannes Rössel
    Parentheses in Word usually stretch with whatever they're containing. This might be un-noticeable for things like but for stuff like it's definitely nice, especially compared to the fact that naïve LaTeX users often produce uglinesses such as There is a problem, however, when using under-/overbraces in math and putting parentheses around the complete term it becomes ugly. For simple things like shown here this can be solved by not letting the parentheses stretch which looks almost right. However, for more complex things it's certainly not an option: Both variants look horrible. So is there a way of letting the parentheses only stretch around the actual term parts, not including the under-/overbraces? Those are frequently used for annotations of individual pieces, so simply not using them is a bad idea too. In LaTeX you can get away with guesswork and using explicit sizes for the parentheses instead of relying on \left and \right but I haven't found a comparable option in Word yet. Since the underbrace is (tree-wise) a sibling of the term in parentheses it probably simply has to stretch and there probably can't be an algorithm that determines when to stretch or when not, considering that \above and \below are used for annotations as well but also for other things where perentheses have to stretch. Also, since the parenthesized expression is opaque from the outside one has to put the underbrace inside. From a markup point of view, at least. One can probably draw the rest around but that falls apart when styles change and wouldn't be a good idea either.

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  • Opening password protected Excel 2007 documents by double clicking from My documents does not work u

    - by erik-van-gorp
    When all of the following conditions are true, excel will open (most of the time) but will not open the document itself. No error is displayed. This only occurs with Excel files, Word and powerpoint do open perfectly. Conditions : OS is "Windows 7 Professional 64-bit" office is "Office 2007 Ultimate". excel file is in .xls (2003 format) excel file is password protected excel file is in "My Documents" (or a subfolder of it) file is double-clicked from explorer under Windows 7. Following options do open the excel file as it should : right click and selecting the (bold) open action single click the file and pressing enter moving the file to the desktop and double-click it. non password protected files do open from the same directory. Actions taken not resolving the problem: - reboot - repair office installation - system restore does not work because of Antivirus application installed (message from system restore, using "Symantec Internet Security 2010") Anyone any idea ?

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  • Haskell math performance

    - by Travis Brown
    I'm in the middle of porting David Blei's original C implementation of Latent Dirichlet Allocation to Haskell, and I'm trying to decide whether to leave some of the low-level stuff in C. The following function is one example—it's an approximation of the second derivative of lgamma: double trigamma(double x) { double p; int i; x=x+6; p=1/(x*x); p=(((((0.075757575757576*p-0.033333333333333)*p+0.0238095238095238) *p-0.033333333333333)*p+0.166666666666667)*p+1)/x+0.5*p; for (i=0; i<6 ;i++) { x=x-1; p=1/(x*x)+p; } return(p); } I've translated this into more or less idiomatic Haskell as follows: trigamma :: Double -> Double trigamma x = snd $ last $ take 7 $ iterate next (x' - 1, p') where x' = x + 6 p = 1 / x' ^ 2 p' = p / 2 + c / x' c = foldr1 (\a b -> (a + b * p)) [1, 1/6, -1/30, 1/42, -1/30, 5/66] next (x, p) = (x - 1, 1 / x ^ 2 + p) The problem is that when I run both through Criterion, my Haskell version is six or seven times slower (I'm compiling with -O2 on GHC 6.12.1). Some similar functions are even worse. I know practically nothing about Haskell performance, and I'm not terribly interested in digging through Core or anything like that, since I can always just call the handful of math-intensive C functions through FFI. But I'm curious about whether there's low-hanging fruit that I'm missing—some kind of extension or library or annotation that I could use to speed up this numeric stuff without making it too ugly.

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  • optimization math computation (multiplication and summing)

    - by wiso
    Suppose you want to compute the sum of the square of the differences of items: $\sum_{i=1}^{N-1} (x_i - x_{i+1})^2$, the simplest code (the input is std::vector<double> xs, the ouput sum2) is: double sum2 = 0.; double prev = xs[0]; for (vector::const_iterator i = xs.begin() + 1; i != xs.end(); ++i) { sum2 += (prev - (*i)) * (prev - (*i)); // only 1 - with compiler optimization prev = (*i); } I hope that the compiler do the optimization in the comment above. If N is the length of xs you have N-1 multiplications and 2N-3 sums (sums means + or -). Now suppose you know this variable: sum = $x_1^2 + x_N^2 + 2 sum_{i=2}^{N-1} x_i^2$ Expanding the binomial square: $sum_i^{N-1} (x_i-x_{i+1})^2 = sum - 2\sum_{i=1}^{N-1} x_i x_{i+1}$ so the code becomes: double sum2 = 0.; double prev = xs[0]; for (vector::const_iterator i = xs.begin() + 1; i != xs.end(); ++i) { sum2 += (*i) * prev; prev = (*i); } sum2 = -sum2 * 2. + sum; Here I have N multiplications and N-1 additions. In my case N is about 100. Well, compiling with g++ -O2 I got no speed up (I try calling the inlined function 2M times), why?

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  • C++ double division by 0.0 versus DBL_MIN

    - by wonsungi
    When finding the inverse square root of a double, is it better to clamp invalid non-positive inputs at 0.0 or MIN_DBL? (In my example below double b may end up being negative due to floating point rounding errors and because the laws of physics are slightly slightly fudged in the game.) Both division by 0.0 and MIN_DBL produce the same outcome in the game because 1/0.0 and 1/DBL_MIN are effectively infinity. My intuition says MIN_DBL is the better choice, but would there be any case for using 0.0? Like perhaps sqrt(0.0), 1/0.0 and multiplication by 1.#INF000000000000 execute faster because they are special cases. double b = 1 - v.length_squared()/(c*c); #ifdef CLAMP_BY_0 if (b < 0.0) b = 0.0; #endif #ifdef CLAMP_BY_DBL_MIN if (b <= 0.0) b = DBL_MIN; #endif double lorentz_factor = 1/sqrt(b); double division in MSVC: 1/0.0 = 1.#INF000000000000 1/DBL_MIN = 4.4942328371557898e+307

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  • Swap bits in c++ for a double

    - by hidayat
    Im trying to change from big endian to little endian on a double. One way to go is to use double val, tmp = 5.55; ((unsigned int *)&val)[0] = ntohl(((unsigned int *)&tmp)[1]); ((unsigned int *)&val)[1] = ntohl(((unsigned int *)&tmp)[0]); But then I get a warning: "dereferencing type-punned pointer will break strict-aliasing rules" and I dont want to turn this warning off. Another way to go is: #define ntohll(x) ( ( (uint64_t)(ntohl( (uint32_t)((x << 32) >> 32) )) << 32) | ntohl( ((uint32_t)(x >> 32)) ) ) val = (double)bswap_64(unsigned long long(tmp)); //or val = (double)ntohll(unsigned long long(tmp)); But then a lose the decimals. Anyone know a good way to swap the bits on a double without using a for loop?

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  • Math.max and Math.min outputting highest and lowest values allowed

    - by user1696162
    so I'm trying to make a program that will output the sum, average, and smallest and largest values. I have everything basically figured out except the smallest and largest values are outputting 2147483647 and -2147483647, which I believe are the absolute smallest and largest values that Java will compute. Anyway, I want to compute the numbers that a user enters, so this obviously isn't correct. Here is my class. I assume something is going wrong in the addValue method. public class DataSet { private int sum; private int count; private int largest; private int smallest; private double average; public DataSet() { sum = 0; count = 0; largest = Integer.MAX_VALUE; smallest = Integer.MIN_VALUE; average = 0; } public void addValue(int x) { count++; sum = sum + x; largest = Math.max(x, largest); smallest = Math.min(x, smallest); } public int getSum() { return sum; } public double getAverage() { average = sum / count; return average; } public int getCount() { return count; } public int getLargest() { return largest; } public int getSmallest() { return smallest; } } And here is my tester class for this project: public class DataSetTester { public static void main(String[] arg) { DataSet ds = new DataSet(); ds.addValue(13); ds.addValue(-2); ds.addValue(3); ds.addValue(0); System.out.println("Count: " + ds.getCount()); System.out.println("Sum: " + ds.getSum()); System.out.println("Average: " + ds.getAverage()); System.out.println("Smallest: " + ds.getSmallest()); System.out.println("Largest: " + ds.getLargest()); } } Everything outputs correctly (count, sum, average) except the smallest and largest numbers. If anyone could point me in the right direction of what I'm doing wrong, that would be great. Thanks.

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  • Android-Java: Constructing a triangle based on Coordinates on a map and your bearing

    - by Aidan
    Hi Guys, I'm constructing a geolocation based application and I'm trying to figure out a way to make my application realise when a user is facing the direction of the given location (a particular long / lat co-ord). I've got the math figured, I just have the triangle to construct. //UPDATE So I've figured out a good bit of this... Below is a method which takes in a long / lat value and attempts to compute a triangle finding a point 700 meters away and one to its left + right. It'd then use these to construct the triangle. It computes the correct longitude but the latitude ends up somewhere off the coast of east Africa. (I'm in Ireland!). public void drawtri(double currlng,double currlat, double bearing){ bearing = (bearing < 0 ? -bearing : bearing); System.out.println("RUNNING THE DRAW TRIANGLE METHOD!!!!!"); System.out.println("CURRENT LNG" + currlng); System.out.println("CURRENT LAT" + currlat); System.out.println("CURRENT BEARING" + bearing); //Find point X(x,y) double distance = 0.7; //700 meters. double R = 6371.0; //The radius of the earth. //Finding X's y value. Math.toRadians(currlng); Math.toRadians(currlat); Math.toRadians(bearing); distance = distance/R; Global.Alat = Math.asin(Math.sin(currlat)*Math.cos(distance)+ Math.cos(currlat)*Math.sin(distance)*Math.cos(bearing)); System.out.println("CURRENT ALAT!!: " + Global.Alat); //Finding X's x value. Global.Alng = currlng + Math.atan2(Math.sin(bearing)*Math.sin(distance) *Math.cos(currlat), Math.cos(distance)-Math.sin(currlat)*Math.sin(Global.Alat)); Math.toDegrees(Global.Alat); Math.toDegrees(Global.Alng); //Co-ord of Point B(x,y) // Note: Lng = X axis, Lat = Y axis. Global.Blat = Global.Alat+ 00.007931; Global.Blng = Global.Alng; //Co-ord of Point C(x,y) Global.Clat = Global.Alat - 00.007931; Global.Clng = Global.Alng; } From debugging I've determined the problem lies with the computation of the latitude done here.. Global.Alat = Math.asin(Math.sin(currlat)*Math.cos(distance)+ Math.cos(currlat)*Math.sin(distance)*Math.cos(bearing)); I have no idea why though and don't know how to fix it. I got the formula from this site.. http://www.movable-type.co.uk/scripts/latlong.html It appears correct and I've tested multiple things... I've tried converting to Radians then post computations back to degrees, etc. etc. Anyone got any ideas how to fix this method so that it will map the triangle ONLY 700 meters in from my current location in the direction that I am facing? Thanks, EDIT/// Converting the outcome to radians gives me a lat of 5.6xxxxxxxxxxxxxx .I have a feeling this bug has something to do with conversions but its not THAT simple. The equation is correct, it just.. outputs wrong..

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  • Double type returns -1.#IND/NaN error when calculating pi iteratively

    - by Draak
    I am working through a problem for my MCTS certification. The program has to calculate pi until the user presses a key, at which point the thread is aborted, the result returned to the main thread and printed in the console. Simple enough, right? This exercise is really meant to be about threading, but I'm running into another problem. The procedure that calculates pi returns -1.#IND. I've read some of the material on the web about this error, but I'm still not sure how to fix it. When I change double to Decimal type, I unsurprisingly get Overflow Exception very quickly. So, the question is how do I store the numbers correctly? Do I have to create a class to somehow store parts of the number when it gets too big to be contained in a Decimal? Class PiCalculator Dim a As Double = 1 Dim b As Double = 1 / Math.Sqrt(2) Dim t As Double = 1 / 4 Dim p As Double = 1 Dim pi As Double Dim callback As DelegateResult Sub New(ByVal _callback As DelegateResult) callback = _callback End Sub Sub Calculate() Try Do While True Dim a1 = (a + b) / 2 Dim b1 = Math.Sqrt(a * b) Dim t1 = t - p * (a - a1) ^ 2 Dim p1 = 2 * p a = a1 b = b1 t = t1 p = p1 pi = ((a + b) ^ 2) / (4 * t) Loop Catch ex As ThreadAbortException Finally callback(pi) End Try End Sub End Class

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  • Java string to double conversion.

    - by wretrOvian
    Hi, I've been reading up on the net about the issues with handling float and double types in java. Unfortunately, the image is still not clear. Hence, i'm asking here direct. :( My MySQL table has various DECIMAL(m,d) columns. The m may range from 5 to 30. d stays a constant at 2. Question 1. What equivalent data-type should i be using in Java to work (i.e store, retrieve, and process) with the size of the values in my table? (I've settled with double - hence this post). Question 2. While trying to parse a double from a string, i'm getting errors Double dpu = new Double(dpuField.getText()); for example - "1" -> java.lang.NumberFormatException: empty String "10" -> 1.0 "101" -> 10.0 "101." -> 101.0 "101.1" -> 101.0 "101.19" -> 101.1 What am i doing wrong? What is the correct way to convert a string to a double value? And what measures should i take to perform operations on such values?

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  • [C#] Convert string to double with 2 digit after decimal separator

    - by st.stoqnov
    All began with these simple lines of code: string s = "16.9"; double d = Convert.ToDouble(s); d*=100; The result should be 1690.0, but it's not. d is equal to 1689.9999999999998. All I want to do is to round a double to value with 2 digit after decimal separator. Here is my function. private double RoundFloat(double Value) { float sign = (Value < 0) ? -0.01f : 0.01f; if (Math.Abs(Value) < 0.00001) Value = 0; string SVal = Value.ToString(); string DecimalSeparator = System.Globalization.CultureInfo.CurrentCulture.NumberFormat.CurrencyDecimalSeparator; int i = SVal.IndexOf(DecimalSeparator); if (i > 0) { int SRnd; try { // ????? ??????? ????? ???? ?????????? ?????????? SRnd = Convert.ToInt32(SVal.Substring(i + 3, 1)); } catch { SRnd = 0; } if (SVal.Length > i + 3) SVal = SVal.Substring(0, i + 3); //SVal += "00001"; try { double result = (SRnd >= 5) ? Convert.ToDouble(SVal) + sign : Convert.ToDouble(SVal); //result = Math.Round(result, 2); return result; } catch { return 0; } } else { return Value; } But again the same problem, converting from string to double is not working as I want. A workaround to this problem is to concatenate "00001" to the string and then use the Math.Round function (commented in the example above). This double value multiplied to 100 (as integer) is send to a device (cash register) and this values must be correct. I am using VS2005 + .NET CF 2.0 Is there another more "elegant" solution, I am not happy with this one.

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  • Is Programming == Math?

    - by moffdub
    I've heard many times that all programming is really a subset of math. Some suggest that OO, at its roots, is mathematically based. I don't get the connection. Aside from some obvious examples: using induction to prove a recursive algorithm formal correctness proofs functional languages lambda calculus asymptotic complexity DFAs, NFAs, Turing Machines, and theoretical computation in general the fact that everything on the box is binary In what ways is programming really a subset of math? I'm looking for an explanation that might have relevance to enterprise/OO development (if there is a strong enough connection, that is). Thanks in advance. Edit: as I stated in a comment to an answer, math is uber important to programming, but what I struggle with is the "subset" argument.

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  • Math Problem With Percentages

    - by TheDarkIn1978
    i'm terrible at math. trust me, you math experts will see why after reading my question. i have an object that is 300px in height. i need to calculate the percentage of that height where 90% = 300px (or the full height), 45% = 150px, 0% = 0px. so essentially, if i ask for 45% of the object's height, it will return 150px, or if i ask for 32% of the object's height, it will return __? i believe this is really basic math, so i apologize in advance.

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  • Is there a 128 or 256 bit double class in .net?

    - by AKRamkumar
    I have an application that I want to be able to use large numbers and very precise numbers. For this, I needed a precision interpretation and IntX only works for integers. Is there a class in .net framework or even third party(preferably free) that would do this? Is there another way to do this?

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  • What Math topics & resources to consider as beginner to indulge the book - Introduction to Algorithm

    - by sector7
    I'm a programmer who's beginning to appreciate the knowledge & usability of Algorithms in my work as I move forward with my skill-set. I don't want to take the short path by learning how to apply algorithms "as-is" but would rather like to know the foundation and fundamentals behind them. For that I need Math, at which I'm pretty "basic". I'm considering getting tuition's for that. What I would like is to have a concise syllabus/set of topics/book which I could hand over to my math tutor to get started. HIGHLY DESIRED: one book. the silver bullet. (fingers crossed!) PS: I've got some leads but want to hear you guys/gurus out: Discrete Math, Combinatorics, Graph theory, Calculus, Linear Algebra, and Number Theory. Looking forward to your answers. Thanks!

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  • Finding the normal of OBB face with an OBB penetrating

    - by Milo
    Below is an illustration: I have an OBB in an OBB (see below for OBB2D code if needed). What I need to determine is, what face it is in, and what direction do I point the normal? The goal is to get the OBB out of the OBB so the normal needs to face outward of the OBB. How could I go about: Finding what face the line is penetrating given the 4 corners of the OBB and the class below: if we define dx=x2-x1 and dy=y2-y1, then the normals are (-dy, dx) and (dy, -dx). Which normal points outward of the OBB? Thanks public class OBB2D { // 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(Vector2D center, float w, float h, float angle) { set(center,w,h,angle); } public OBB2D(float left, float top, float width, float height) { set(new Vector2D(left + (width / 2), top + (height / 2)),width,height,0.0f); } public void set(Vector2D center,float w, float h,float angle) { Vector2D X = new Vector2D( (float)Math.cos(angle), (float)Math.sin(angle)); Vector2D Y = new Vector2D((float)-Math.sin(angle), (float)Math.cos(angle)); X = X.multiply( w / 2); Y = Y.multiply( h / 2); corner[0] = center.subtract(X).subtract(Y); corner[1] = center.add(X).subtract(Y); corner[2] = center.add(X).add(Y); corner[3] = center.subtract(X).add(Y); computeAxes(); extents.x = w / 2; extents.y = h / 2; computeDimensions(center,angle); } private void computeDimensions(Vector2D center,float angle) { this.center.x = center.x; this.center.y = center.y; this.angle = angle; boundingRect.left = Math.min(Math.min(corner[0].x, corner[3].x), Math.min(corner[1].x, corner[2].x)); boundingRect.top = Math.min(Math.min(corner[0].y, corner[1].y),Math.min(corner[2].y, corner[3].y)); boundingRect.right = Math.max(Math.max(corner[1].x, corner[2].x), Math.max(corner[0].x, corner[3].x)); boundingRect.bottom = Math.max(Math.max(corner[2].y, corner[3].y),Math.max(corner[0].y, corner[1].y)); } public void set(RectF rect) { set(new Vector2D(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; } //Updates the axes after the corners move. Assumes the //corners actually form a rectangle. private void computeAxes() { axis[0] = corner[1].subtract(corner[0]); axis[1] = corner[3].subtract(corner[0]); // 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) { axis[a] = axis[a].divide((axis[a].length() * axis[a].length())); origin[a] = corner[0].dot(axis[a]); } } public void moveTo(Vector2D center) { Vector2D centroid = (corner[0].add(corner[1]).add(corner[2]).add(corner[3])).divide(4.0f); Vector2D translation = center.subtract(centroid); for (int c = 0; c < 4; ++c) { corner[c] = corner[c].add(translation); } computeAxes(); computeDimensions(center,angle); } // 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,getWidth(),getHeight(),angle); } public float getAngle() { return angle; } public void setSize(float w,float h) { set(center,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; } };

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