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  • solving origin of a vectors

    - by Mike
    I have two endpoints (xa,ya) and (xb,yb) of two vectors, respectively a and b, originating from a same point (xo, yo). Also, I know that |a|=|b|+s, where s is a constant. I tried to compute the origin (xo, yo) but seem to fail at some point. How to solve this?

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  • Reducing Integer Fractions Algorithm - Solution Explanation?

    - by Andrew Tomazos - Fathomling
    This is a followup to this problem: Reducing Integer Fractions Algorithm Following is a solution to the problem from a grandmaster: #include <cstdio> #include <algorithm> #include <functional> using namespace std; const int MAXN = 100100; const int MAXP = 10001000; int p[MAXP]; void init() { for (int i = 2; i < MAXP; ++i) { if (p[i] == 0) { for (int j = i; j < MAXP; j += i) { p[j] = i; } } } } void f(int n, vector<int>& a, vector<int>& x) { a.resize(n); vector<int>(MAXP, 0).swap(x); for (int i = 0; i < n; ++i) { scanf("%d", &a[i]); for (int j = a[i]; j > 1; j /= p[j]) { ++x[p[j]]; } } } void g(const vector<int>& v, vector<int> w) { for (int i: v) { for (int j = i; j > 1; j /= p[j]) { if (w[p[j]] > 0) { --w[p[j]]; i /= p[j]; } } printf("%d ", i); } puts(""); } int main() { int n, m; vector<int> a, b, x, y, z; init(); scanf("%d%d", &n, &m); f(n, a, x); f(m, b, y); printf("%d %d\n", n, m); transform(x.begin(), x.end(), y.begin(), insert_iterator<vector<int> >(z, z.end()), [](int a, int b) { return min(a, b); }); g(a, z); g(b, z); return 0; } It isn't clear to me how it works. Can anyone explain it? The equivilance is as follows: a is the numerator vector of length n b is the denominator vector of length m

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  • KD-Trees and missing values (vector comparison)

    - by labratmatt
    I have a system that stores vectors and allows a user to find the n most similar vectors to the user's query vector. That is, a user submits a vector (I call it a query vector) and my system spits out "here are the n most similar vectors." I generate the similar vectors using a KD-Tree and everything works well, but I want to do more. I want to present a list of the n most similar vectors even if the user doesn't submit a complete vector (a vector with missing values). That is, if a user submits a vector with three dimensions, I still want to find the n nearest vectors (stored vectors are of 11 dimensions) I have stored. I have a couple of obvious solutions, but I'm not sure either one seem very good: Create multiple KD-Trees each built using the most popular subset of dimensions a user will search for. That is, if a user submits a query vector of thee dimensions, x, y, z, I match that query to my already built KD-Tree which only contains vectors of three dimensions, x, y, z. Ignore KD-Trees when a user submits a query vector with missing values and compare the query vector to the vectors (stored in a table in a DB) one by one using something like a dot product. This has to be a common problem, any suggestions? Thanks for the help.

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  • How to make an equation span the whole page / line in LaTeX?

    - by Reed Richards
    I have this equation and it's quite big (basically a FDM one) but it aligns with the text and then continues out on the right side to the nothingness. I've tried stuff like \begin{center} and \hspace*{-2.5cm} but to no avail. I want it to use the whole line not just from the left-margin and out to the right. How do I do it and do I need to install some special package for it? I use the \[ instead of the displaymath like this \[ Equation arrays here \] The code \[ \left( \begin{array}{cccccc} -(2\kappa+\frac{hV\rho}{2}) & (\frac{hV\rho}{2}-\kappa) & 0 & \cdots & 0 \\ -\kappa & -(2\kappa+\frac{hV\rho}{2}) & (\frac{hV\rho}{2}-\kappa) & 0 & \cdots \\ 0 & -\kappa & -(2\kappa+\frac{hV\rho}{2}) & (\frac{hV\rho}{2}-\kappa) & 0 & \cdots \\ \vdots & 0 & \ddots & \vdots \\ \vdots & \vdots & \vdots & -\kappa & -(2\kappa+\frac{hV\rho}{2}) & (\frac{hV\rho}{2}-\kappa) \\ 0 & \vdots & \vdots & 0 & \kappa - \frac{2h\kappa_{v}}{\kappa}(\frac{hv\rho}{2} - \kappa) & -2\kappa \\ \end{array} \right) \left( \begin{array}{c} T_{1} \\ T_{2} \\ \vdots \\ T_{n} \\ \end{array} \right) = \left( \begin{array}{c} Q(0) + \kappa T_{0} \\ Q(h) \\ Q(2h) \\ \vdots \\ Q((n-1)h) \\ 2\frac{\kappa_{v}}{\kappa_{v}}T_{out} \\ \end{array} \right) \]

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  • how do I make a portable isnan/isinf function.

    - by monkeyking
    I've been using isinf,isnan functions on linux platforms which worked perfectly. But this didn't work on osx, so I decided to use std::isinf std::isnan which works on both linux and osx. But the intel compiler doesn't recognize it, and I guess its a bug in the intel compiler according to http://software.intel.com/en-us/forums/showthread.php?t=64188 So now I just want to avoid the hassle and define my own isinf,isnan implementation. Does anyone know how this could be done Thanks edit: I ended up doing this in my sourcecode for making isinf/isnan working #include <iostream> #include <cmath> #ifdef __INTEL_COMPILER #include <mathimf.h> #endif int isnan_local(double x) { #ifdef __INTEL_COMPILER return isnan(x); #else return std::isnan(x); #endif } int isinf_local(double x) { #ifdef __INTEL_COMPILER return isinf(x); #else return std::isinf(x); #endif } int myChk(double a){ std::cerr<<"val is: "<<a <<"\t"; if(isnan_local(a)) std::cerr<<"program says isnan"; if(isinf_local(a)) std::cerr<<"program says isinf"; std::cerr<<"\n"; return 0; } int main(){ double a = 0; myChk(a); myChk(log(a)); myChk(-log(a)); myChk(0/log(a)); myChk(log(a)/log(a)); return 0; }

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  • Choosing circle radius to fully fill a rectangle

    - by Andy
    Hi, the pixman image library can draw radial color gradients between two circles. I'd like the radial gradient to fill a rectangular area defined by "width" and "height" completely. Now my question, how should I choose the radius of the outer circle? My current parameters are the following: A) inner circle (start of gradient) center pointer of inner circle: (width*0.5|height*0.5) radius of inner circle: 1 color: black B) outer circle (end of gradient) center pointer of outer circle: (width*0.5|height*0.5) radius of outer circle: ??? color: white How should I choose the radius of the outer circle to make sure that the outer circle will entirely fill my bounding rectangle defined by width*height. There shall be no empty areas in the corners, the area shall be completely covered by the circle. In other words, the bounding rectangle width,height must fit entirely into the outer circle. Choosing outer_radius = max(width, height) * 0.5 as the radius for the outer circle is obviously not enough. It must be bigger, but how much bigger? Thanks!

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  • How to solve such system with given parts of it? (maple)

    - by Kabumbus
    So I had a system #for given koefs k:=3; n:=3; #let us solve system: koefSolution:= solve({ sum(a[i], i = 0 .. k) = 0, sum(a[i], i = 0 .. k)-(sum(b[i], i = 0 .. k)) = 0, sum(i^n*a[i], i = 0 .. k)-(sum(i^(n-1)*b[i], i = 0 .. k)) = 0 }); So I have a vector like koefSolution := { a[0] = 7*a[2]+26*a[3]-b[1]-4*b[2]-9*b[3], a[1] = -8*a[2]-27*a[3]+b[1]+4*b[2]+9*b[3], a[2] = a[2], a[3] = a[3], b[0] = -b[1]-b[2]-b[3], b[1] = b[1], b[2] = b[2], b[3] = b[3]} I have a[0] so I try solve({koefSolution, a[0] = 1}); why it does not solve my system for given a[0]? ( main point here is to fill koefSolution with given a[] and b[] and optimize.)

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  • draw an arc given 3 points in SWT

    - by Ahmed Kotb
    iam using the swt java library and iam having a problem. the gc draw arc method takes the following arguments GC.drawArc(int x, int y, int width, int height, int startAngle, int endAngle); but i want to be able to draw the arc using 3 arguments : the source ,destination and control points. is there any formula to convert between those parameters ? QuadCurve2D class do exactly what i want but it is not AWT not swt ...and i tried to use java2d under swt but it was very slow .... any solutions ?

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  • Need some help understanding this problem about maximizing graph connectivity

    - by Legend
    I was wondering if someone could help me understand this problem. I prepared a small diagram because it is much easier to explain it visually. Problem I am trying to solve: 1. Constructing the dependency graph Given the connectivity of the graph and a metric that determines how well a node depends on the other, order the dependencies. For instance, I could put in a few rules saying that node 3 depends on node 4 node 2 depends on node 3 node 3 depends on node 5 But because the final rule is not "valuable" (again based on the same metric), I will not add the rule to my system. 2. Execute the request order Once I built a dependency graph, execute the list in an order that maximizes the final connectivity. I am not sure if this is a really a problem but I somehow have a feeling that there might exist more than one order in which case, it is required to choose the best order. First and foremost, I am wondering if I constructed the problem correctly and if I should be aware of any corner cases. Secondly, is there a closely related algorithm that I can look at? Currently, I am thinking of something like Feedback Arc Set or the Secretary Problem but I am a little confused at the moment. Any suggestions? PS: I am a little confused about the problem myself so please don't flame on me for that. If any clarifications are needed, I will try to update the question.

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  • reflection paths between points in2d

    - by Chris H
    Just wondering if there was a nice (already implemented/documented) algorithm to do the following Given any shape (without crossing edges) and two points inside that shape, compute all the paths between the two points such that all reflections are perfect reflections. The path lengths should be limited to a certain length otherwise there are infinite solutions. I'm not interested in just shooting out rays to try to guess how close I can get, I'm interested in algorithms that can do it perfectly. Search based, not guess/improvement based.

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  • Interview question: f(f(n)) == -n

    - by Hrvoje Prgeša
    A question I got on my last interview: Design a function f, such that: f(f(n)) == -n Where n is a 32 bit signed integer; you can't use complex numbers arithmetic. If you can't design such a function for the whole range of numbers, design it for the largest range possible. Any ideas?

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  • how to subtract circle from an arbitrary polygon

    - by George
    Given an arbitary polygon with vertices stored in either clockwise/counterclockwise fashion (depicted as a black rectangle in the diagram), I need to be able to subtract an arbitrary number of circles (in red on the diagram) from that polygon. Removing a circle could possibly split the polygon into two seperate polygons (as depicted by the second line in the diagram). I'm not sure where to start. http://www.freeimagehosting.net/image.php?89a0276d9d.jpg

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  • Merge overlapping triangles into a polygon

    - by nornagon
    I've got a bunch of overlapping triangles from a 3D model projected into a 2D plane. I need to merge each island of touching triangles into a closed, non-convex polygon. The resultant polygons shouldn't have any holes in them (since the source data doesn't). Many of the source triangles share (floating point identical) edges with other triangles in the source data. What's the easiest way to do this? Performance isn't particularly important, since this will be done at design time.

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  • Need a formula for calculating the tax portion of a total amount.

    - by pawz
    In Australia we have to advertise products with tax already added, so rather than say a product is $10 + $1 GST = $11, we normally work backwards and say "ok, total is $10, how much of that is GST ?" For example, for a $10 total, you do 10 * (1 /11) = 0.91, which is the tax component of the $10 total. My problem is I need calculate a formula for working out the taxable component when the tax rate is a variable. So far I've made this calculation although I'm not sure how correct an assertion it is: 10 * (1 / x) = 0.09 * (1 / y) where y = 10, x = 11 Basically i want to work out x on the left hand side when I know that the tax rate is 0.05 for example, which will give me a formula that I can use to calculate the taxable component of an total figure. I want a function into which I can plug in the total price and the tax rate, and get back the taxable component of the total price. I'd really appreciate the help with this as it really makes my head hurt ! :")

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  • Sparse quadratic program solver

    - by Jacob
    This great SO answer points to a good sparse solver, but I've got constraints on x (for Ax = b) such that each element in x is >=0 an <=N. The first thing which comes to mind is an QP solver for large sparse matrices. Also, A is huge (around 2e6x2e6) but very sparse with <=4 elements per row. Any ideas/recommendations?

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  • Calculate new position of player

    - by user1439111
    Edit: I will summerize my question since it is very long (Thanks Len for pointing it out) What I'm trying to find out is to get a new position of a player after an X amount of time. The following variables are known: - Speed - Length between the 2 points - Source position (X, Y) - Destination position (X, Y) How can I calculate a position between the source and destion with these variables given? For example: source: 0, 0 destination: 10, 0 speed: 1 so after 1 second the players position would be 1, 0 The code below works but it's quite long so I'm looking for something shorter/more logical ====================================================================== I'm having a hard time figuring out how to calculate a new position of a player ingame. This code is server sided used to track a player(It's a emulator so I don't have access to the clients code). The collision detection of the server works fine I'm using bresenham's line algorithm and a raycast to determine at which point a collision happens. Once I deteremined the collision I calculate the length of the path the player is about to walk and also the total time. I would like to know the new position of a player each second. This is the code I'm currently using. It's in C++ but I am porting the server to C# and I haven't written the code in C# yet. // Difference between the source X - destination X //and source y - destionation Y float xDiff, yDiff; xDiff = xDes - xSrc; yDiff = yDes - ySrc; float walkingLength = 0.00F; float NewX = xDiff * xDiff; float NewY = yDiff * yDiff; walkingLength = NewX + NewY; walkingLength = sqrt(walkingLength); const float PI = 3.14159265F; float Angle = 0.00F; if(xDes >= xSrc && yDes >= ySrc) { Angle = atanf((yDiff / xDiff)); Angle = Angle * 180 / PI; } else if(xDes < xSrc && yDes >= ySrc) { Angle = atanf((-xDiff / yDiff)); Angle = Angle * 180 / PI; Angle += 90.00F; } else if(xDes < xSrc && yDes < ySrc) { Angle = atanf((yDiff / xDiff)); Angle = Angle * 180 / PI; Angle += 180.00F; } else if(xDes >= xSrc && yDes < ySrc) { Angle = atanf((xDiff / -yDiff)); Angle = Angle * 180 / PI; Angle += 270.00F; } float WalkingTime = (float)walkingLength / (float)speed; bool Done = false; float i = 0; while(i < walkingLength) { if(Done == true) { break; } if(WalkingTime >= 1000) { Sleep(1000); i += speed; WalkTime -= 1000; } else { Sleep(WalkTime); i += speed * WalkTime; WalkTime -= 1000; Done = true; } if(Angle >= 0 && Angle < 90) { float xNew = cosf(Angle * PI / 180) * i; float yNew = sinf(Angle * PI / 180) * i; float NewCharacterX = xSrc + xNew; float NewCharacterY = ySrc + yNew; } I have cut the last part of the loop since it's just 3 other else if statements with 3 other angle conditions and the only change is the sin and cos. The given speed parameter is the speed/second. The code above works but as you can see it's quite long so I'm looking for a new way to calculate this. btw, don't mind the while loop to calculate each new position I'm going to use a timer in C# Thank you very much

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  • Finding the intersection of two vector equations.

    - by Matthew Mitchell
    I've been trying to solve this and I found an equation that gives the possibility of zero division errors. Not the best thing: v1 = (a,b) v2 = (c,d) d1 = (e,f) d2 = (h,i) l1: v1 + ?d1 l2: v2 + µd2 Equation to find vector intersection of l1 and l2 programatically by re-arranging for lambda. (a,b) + ?(e,f) = (c,d) + µ(h,i) a + ?e = c + µh b +?f = d + µi µh = a + ?e - c µi = b +?f - d µ = (a + ?e - c)/h µ = (b +?f - d)/i (a + ?e - c)/h = (b +?f - d)/i a/h + ?e/h - c/h = b/i +?f/i - d/i ?e/h - ?f/i = (b/i - d/i) - (a/h - c/h) ?(e/h - f/i) = (b - d)/i - (a - c)/h ? = ((b - d)/i - (a - c)/h)/(e/h - f/i) Intersection vector = (a + ?e,b + ?f) Not sure if it would even work in some cases. I haven't tested it. I need to know how to do this for values as in that example a-i. Thank you.

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  • Area of a irregular shape

    - by Naveen
    I have set of points which lies on the image. These set of points form a irregular closed shape. I need to find the area of this shape. Does any body which is the normal algorithm used for calculating the area ? Or is there any support available in libraries such as boost? I am using C++.

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  • Diagonal of polygon is inside or outside?

    - by Himadri
    I have three consecutive points of polygon, say p1,p2,p3. Now I wanted to know whether the orthogonal between p1 and p3 is inside the polygon or outside the polygon. I am doing it by taking three vectors v1,v2 and v3. And the point before the point p1 in polygon say p0. v1 = (p0 - p1) v2 = (p2 - p1) v3 = (p3 - p1) With reference to this question, I am using the method shown in the accepted answer of that question. It is only for counterclockwise. What if my points are clockwise. I am also knowing my whole polygon is clockwise or counterclockwise. And accordingly I select the vectors v1 and v2. But still I am getting some problem. I am showing one case where I am getting problem. This polygon is counterclockwise. and It is starting from the origin of v1 and v2.

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  • Calculus? Need help solving for a time-dependent variable given some other variables.

    - by user451527
    Long story short, I'm making a platform game. I'm not old enough to have taken Calculus yet, so I know not of derivatives or integrals, but I know of them. The desired behavior is for my character to automagically jump when there is a block to either side of him that is above the one he's standing on; for instance, stairs. This way the player can just hold left / right to climb stairs, instead of having to spam the jump key too. The issue is with the way I've implemented jumping; I've decided to go mario-style, and allow the player to hold 'jump' longer to jump higher. To do so, I have a 'jump' variable which is added to the player's Y velocity. The jump variable increases to a set value when the 'jump' key is pressed, and decreases very quickly once the 'jump' key is released, but decreases less quickly so long as you hold the 'jump' key down, thus providing continuous acceleration up as long as you hold 'jump.' This also makes for a nice, flowing jump, rather than a visually jarring, abrupt acceleration. So, in order to account for variable stair height, I want to be able to calculate exactly what value the 'jump' variable should get in order to jump exactly to the height of the stair; preferably no more, no less, though slightly more is permissible. This way the character can jump up steep or shallow flights of stairs without it looking weird or being slow. There are essentially 5 variables in play: h -the height the character needs to jump to reach the stair top<br> j -the jump acceleration variable<br> v -the vertical velocity of the character<br> p -the vertical position of the character<br> d -initial vertical position of the player minus final position<br> Each timestep:<br> j -= 1.5; //the jump variable's deceleration<br> v -= j; //the jump value's influence on vertical speed<br> v *= 0.95; //friction on the vertical speed<br> v += 1; //gravity<br> p += v; //add the vertical speed to the vertical position<br> v-initial is known to be zero<br> v-final is known to be zero<br> p-initial is known<br> p-final is known<br> d is known to be p-initial minus p-final<br> j-final is known to be zero<br> j-initial is unknown<br> Given all of these facts, how can I make an equation that will solve for j? tl;dr How do I Calculus? Much thanks to anyone who's made it this far and decides to plow through this problem.

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  • Drawing Directed Acyclic Graphs: Minimizing edge crossing?

    - by Robert Fraser
    Laying out the verticies in a DAG in a tree form (i.e. verticies with no in-edges on top, verticies dependent only on those on the next level, etc.) is rather simple without graph drawing algorithms such as Efficient Sugimiya. However, is there a simple algorithm to do this that minimizes edge crossing? (For some graphs, it may be impossible to completely eliminate edge crossing.) A picture says a thousand words, so is there an algorithm that would suggest: instead of: EDIT: As the picture suggests, a vertex's inputs are always on top and outputs are always below, which is another barrier to just pasting in an existing layout algorithm.

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  • Using bitwise operators on > 32 bit integers

    - by dqhendricks
    I am using bitwise operations in order to represent many access control flags within one integer. ADMIN_ACCESS = 1; EDIT_ACCOUNT_ACCESS = 2; EDIT_ORDER_ACCESS = 4; var myAccess = 3; // ie: ( ADMIN_ACCESS | EDIT_ACCOUNT_ACCESS ) if ( myAccess & EDIT_ACCOUNT_ACCESS ) { // check for correct access // allow for editing of account } Most of this is occurring on the PHP side of my project. There is one piece however where Javascript is used to join several access flags using | when saving someone's access level. This works fine to a point. I have found that once an integer (flag) gets too large ( 32bit), it no longer works correctly with bitwise operators in Javascript. For instance: alert( 4294967296 | 1 ); // equals 1, but should equal 4294967297 I am trying to find a workaround for this so that I do not have to limit my number of access control flags to 32. Each access control flag is two times the previous control flag so that each control flag will not interfere with other control flags. dec(4) = bin(100) dec(8) = bin(1000) dec(16) = bin(10000) I have noticed that when adding two of these flags together with a simple +, it seems to come out with the same answer as a bitwise or operation, but am having trouble wrapping my head around whether this is a simple substitution, or if there might be problems with doing this. Can anyone comment on the validity of this workaround? Example: (4294967296 | 262144 | 524288) == (4294967296 + 262144 + 524288)

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