Search Results

• Proving that a function f(n) belongs to a Big-Theta(g(n))

- by PLS
Its a exercise that ask to indicate the class Big-Theta(g(n)) the functions belongs to and to prove the assertion. In this case f(n) = (n^2+1)^10 By definition f(n) E Big-Theta(g(n)) <= c1*g(n) < f(n) < c2*g(n), where c1 and c2 are two constants. I know that for this specific f(n) the Big-Theta is g(n^20) but I don't know who to prove it properly. I guess I need to manipulate this inequality but I don't know how

• Programmaticaly finding the Landau notation (Big O or Theta notation) of an algorithm?

- by Julien L
I'm used to search for the Landau (Big O, Theta...) notation of my algorithms by hand to make sure they are as optimized as they can be, but when the functions are getting really big and complex, it's taking way too much time to do it by hand. it's also prone to human errors. I spent some time on Codility (coding/algo exercises), and noticed they will give you the Landau notation for your submitted solution (both in Time and Memory usage). I was wondering how they do that... How would you do it? Is there another way besides Lexical Analysis or parsing of the code? PS: This question concerns mainly PHP and or JavaScript, but I'm opened to any language and theory.

• Indicates the Big-Theta g(n) and prove it [closed]

- by PLS
Indicates the Big-Theta g(n) for the function below and prove it (n^2 + 1)^10 I don not know how to prove it.

• Big 0 theta notation

- by niggersak
Can some pls help with the solution Use big-O notation to classify the traditional grade school algorithms for addition and multiplication. That is, if asked to add two numbers each having N digits, how many individual additions must be performed? If asked to multiply two N-digit numbers, how many individual multiplications are required? Suppose f is a function that returns the result of reversing the string of symbols given as its input, and g is a function that returns the concatenation of the two strings given as its input. If x is the string hrwa, what is returned by g(f(x),x)? Explain your answer - don't just provide the result!

• Rotation Matrix calculates by column not by row

- by pinnacler
I have a class called forest and a property called fixedPositions that stores 100 points (x,y) and they are stored 250x2 (rows x columns) in MatLab. When I select 'fixedPositions', I can click scatter and it will plot the points. Now, I want to rotate the plotted points and I have a rotation matrix that will allow me to do that. The below code should work: theta = obj.heading * pi/180; apparent = [cos(theta) -sin(theta) ; sin(theta) cos(theta)] * obj.fixedPositions; But it wont. I get this error. ??? Error using == mtimes Inner matrix dimensions must agree. Error in == landmarkslandmarks.get.apparentPositions at 22 apparent = [cos(theta) -sin(theta) ; sin(theta) cos(theta)] * obj.fixedPositions; When I alter forest.fixedPositions to store the variables 2x250 instead of 250x2, the above code will work, but it wont plot. I'm going to be plotting fixedPositions constantly in a simulation, so I'd prefer to leave it as it, and make the rotation work instead. Any ideas? Also, fixed positions, is the position of the xy points as if you were looking straight ahead. i.e. heading = 0. heading is set to 45, meaning I want to rotate points clockwise 45 degrees. Here is my code: classdef landmarks properties fixedPositions %# positions in a fixed coordinate system. [x, y] heading = 45; %# direction in which the robot is facing end properties (Dependent) apparentPositions end methods function obj = landmarks(numberOfTrees) %# randomly generates numberOfTrees amount of x,y coordinates and set %the array or matrix (not sure which) to fixedPositions obj.fixedPositions = 100 * rand([numberOfTrees,2]) .* sign(rand([numberOfTrees,2]) - 0.5); end function obj = set.apparentPositions(obj,~) theta = obj.heading * pi/180; [cos(theta) -sin(theta) ; sin(theta) cos(theta)] * obj.fixedPositions; end function apparent = get.apparentPositions(obj) %# rotate obj.positions using obj.facing to generate the output theta = obj.heading * pi/180; apparent = [cos(theta) -sin(theta) ; sin(theta) cos(theta)] * obj.fixedPositions; end end end P.S. If you change one line to this: obj.fixedPositions = 100 * rand([2,numberOfTrees]) .* sign(rand([2,numberOfTrees]) - 0.5); Everything will work fine... it just wont plot.

• How to write a code Newton Raphson code in R involving integration and Bessel function

- by Ahmed
I have want to estimate the parameters of the function which involves Bessel function and integration. However, when i tried to run it, i got a message that "Error in f(x, ...) : could not find function "BesselI" ". I don't know to fix it and would appreciate any related proposal. library(Bessel) library(maxLik) library(miscTools) K<-300 f <- function(theta,lambda,u) {exp(-u*theta)*BesselI(2*sqrt(t*u*theta*lambda),1)/u^0.5} F <- function(theta,lambda){integrate(f,0,K,theta=theta,lambda=lambda)\$value} tt<-function(theta,lambda){(sqrt(lambda)*exp(-t*lambda)/(2*sqrt(t*theta)))(theta(2*t*lambda-1)*F(theta,lambda)} loglik <- function(param) { theta <- param[1] lambda <- param[2] ll <-sum(log(tt(theta,lambda))) } t<-c(24,220,340,620,550,559,689,543) res <- maxNR(loglik, start=c(0.001,0.0005),print.level=1,tol = 1e-08) summary(res)

• Recurrence relation solution

- by Travis
I'm revising past midterms for a final exam this week and am trying to make sense of a solution my professor posted for one of past exams. (You can see the original pdf here, question #6). I'm given the original recurrence relation T(m)=3T(n/2) + n and am told T(1) = 1. I'm pretty sure the solution I've been given is wrong in a few places. The solution is as follows: Let n=2^m T(2^m) = 3T(2^(m-1)) + 2^m 3T(2^(m-1)) = 3^2*T(2^(m-2)) + 2^(m-1)*3 ... 3^(m-1)T(2) = T(1) + 2*3^(m-1) I'm pretty sure this last line is incorrect and they forgot to multiply T(1) by 3^m. He then (tries to) sum the expressions: T(2^m) = 1 + (2^m + 2^(m-1)*3 + ... + 2*3(m-1)) = 1 + 2^m(1 + (3/2)^1 + (3/2)^2 + ... + (3/2)^(m-1)) = 1 + 2^m((3/2)^m-1)*(1/2) = 1 + 3^m - 2^(m-1) = 1 + n^log 3 - n/2 Thus the algorithm is big Theta of (n^log 3). I'm pretty sure that he also got the summation wrong here. By my calculations this should be as follows: T(2^m) = 2^m + 3 * 2^(m-1) + 3^2 * 2^(m-2) + ... + 3^m (3^m because 3^m*T(1) = 3^m should be added, not 1) = 2^m * ((3/2)^1 + (3/2)^2 + ... + (3/2)^m) = 2^m * sum of (3/2)^i from i=0 to m = 2^m * ((3/2)^(m+1) - 1)/(3/2 - 1) = 2^m * ((3/2)^(m+1) - 1)/(1/2) = 2^(m+1) * 3^(m+1)/2^(m+1) - 2^(m+1) = 3^(m+1) - 2 * 2^m Replacing n = 2^m, and from that m = log n T(n) = 3*3^(log n) - 2*n n is O(3^log n), thus the runtime is big Theta of (3^log n) Does this seem right? Thanks for your help!

• How to analyze the efficiency of this algorithm Part 2

- by Leonardo Lopez
I found an error in the way I explained this question before, so here it goes again: FUNCTION SEEK(A,X) 1. FOUND = FALSE 2. K = 1 3. WHILE (NOT FOUND) AND (K < N) a. IF (A[K] = X THEN 1. FOUND = TRUE b. ELSE 1. K = K + 1 4. RETURN Analyzing this algorithm (pseudocode), I can count the number of steps it takes to finish, and analyze its efficiency in theta notation, T(n), a linear algorithm. OK. This following code depends on the inner formulas inside the loop in order to finish, the deal is that there is no variable N in the code, therefore the efficiency of this algorithm will always be the same since we're assigning the value of 1 to both A & B variables: 1. A = 1 2. B = 1 3. UNTIL (B > 100) a. B = 2A - 2 b. A = A + 3 Now I believe this algorithm performs in constant time, always. But how can I use Algebra in order to find out how many steps it takes to finish?

• Asymptotic runtime question

- by 2hype
If f(n) is T(g(n)), then the function 2^(f(n)) is always T(2^(g(n))). Is this statement true or false, and why?

• OpenGL texture on sphere

- by Cilenco
I want to create a rolling, textured ball in OpenGL ES 1.0 for Android. With this function I can create a sphere: public Ball(GL10 gl, float radius) { ByteBuffer bb = ByteBuffer.allocateDirect(40000); bb.order(ByteOrder.nativeOrder()); sphereVertex = bb.asFloatBuffer(); points = build(); } private int build() { double dTheta = STEP * Math.PI / 180; double dPhi = dTheta; int points = 0; for(double phi = -(Math.PI/2); phi <= Math.PI/2; phi+=dPhi) { for(double theta = 0.0; theta <= (Math.PI * 2); theta+=dTheta) { sphereVertex.put((float) (raduis * Math.sin(phi) * Math.cos(theta))); sphereVertex.put((float) (raduis * Math.sin(phi) * Math.sin(theta))); sphereVertex.put((float) (raduis * Math.cos(phi))); points++; } } sphereVertex.position(0); return points; } public void draw() { texture.bind(); gl.glEnableClientState(GL10.GL_VERTEX_ARRAY); gl.glVertexPointer(3, GL10.GL_FLOAT, 0, sphereVertex); gl.glDrawArrays(GL10.GL_TRIANGLE_FAN, 0, points); gl.glDisableClientState(GL10.GL_VERTEX_ARRAY); } My problem now is that I want to use this texture for the sphere but then only a black ball is created (of course because the top right corner s black). I use this texture coordinates because I want to use the whole texture: 0|0 0|1 1|1 1|0 That's what I learned from texturing a triangle. Is that incorrect if I want to use it with a sphere? What do I have to do to use the texture correctly?

• Arbitrary Rotation about a Sphere

- by Der
I'm coding a mechanic which allows a user to move around the surface of a sphere. The position on the sphere is currently stored as theta and phi, where theta is the angle between the z-axis and the xz projection of the current position (i.e. rotation about the y axis), and phi is the angle from the y-axis to the position. I explained that poorly, but it is essentially theta = yaw, phi = pitch Vector3 position = new Vector3(0,0,1); position.X = (float)Math.Sin(phi) * (float)Math.Sin(theta); position.Y = (float)Math.Sin(phi) * (float)Math.Cos(theta); position.Z = (float)Math.Cos(phi); position *= r; I believe this is accurate, however I could be wrong. I need to be able to move in an arbitrary pseudo two dimensional direction around the surface of a sphere at the origin of world space with radius r. For example, holding W should move around the sphere in an upwards direction relative to the orientation of the player. I believe I should be using a Quaternion to represent the position/orientation on the sphere, but I can't think of the correct way of doing it. Spherical geometry is not my strong suit. Essentially, I need to fill the following block: public void Move(Direction dir) { switch (dir) { case Direction.Left: // update quaternion to rotate left break; case Direction.Right: // update quaternion to rotate right break; case Direction.Up: // update quaternion to rotate upward break; case Direction.Down: // update quaternion to rotate downward break; } }

• Android, how important is deltaTime?

- by iQue
Im making a game that is getting pretty big and sometimes my thread has to skip a frame, so far I'm not using deltaTime for setting the speed of my different objects in the game because it's still not a big enough game for it to matter imo. But its getting bigger then I planned, so my question is, how important is delta Time? If I should use delta time there is a problem, since speedX and speedY are integers(they have to be for eclipse to let you make a rectangle of them), I cant add delta time very functionally as far as I understand, but might be wrong? Ive tried adding deltaTime to the code below, and sometimes my enemies just not move after spawn, they just stand there and run in the same place Will add an some code for how I set / use speed: public void update(int dx, int dy) { double theta = 180.0 / Math.PI * Math.atan2(-(y - controls.pointerPosition.y), controls.pointerPosition.x - x); x +=dx * Math.cos(Math.toRadians(theta)); y +=dy * Math.sin(Math.toRadians(theta)); currentFrame = ++currentFrame % BMP_COLUMNS; } public void draw(Canvas canvas) { int srcX = currentFrame * width; int srcY = 1 * height; Rect src = new Rect(srcX, srcY, srcX + width, srcY + height); Rect dst = new Rect(x, y, x + width, y + height); canvas.drawBitmap(bitmap, src, dst, null); } So if someone with some experience with this has any thoughts, please share. Thank you! Changed code: public void update(int dx, int dy, float delta) { double theta = 180.0 / Math.PI * Math.atan2(-(y - controls.pointerPosition.y), controls.pointerPosition.x - x); double speedX = delta * dx * Math.cos(Math.toRadians(theta)); double speedY = delta * dy * Math.sin(Math.toRadians(theta)); x += speedX; y += speedY; currentFrame = ++currentFrame % BMP_COLUMNS; } public void draw(Canvas canvas) { int srcX = currentFrame * width; int srcY = 1 * height; Rect src = new Rect(srcX, srcY, srcX + width, srcY + height); Rect dst = new Rect(x, y, x + width, y + height); canvas.drawBitmap(bitmap, src, dst, null); } with this code my enemies move like before, except they wont move to the right (wont increment x), all other directions work.

• What does this piece of code in C++ mean? [closed]

- by user1838418
const double pi = 3.141592653589793; const angle rightangle = pi/2; inline angle deg2rad(angle degree) { return degree * rightangle / 90.; } angle function1() { return rightangle * ( ((double) rand()) / ((double) RAND_MAX) - .5 ); } bool setmargin(angle theta, angle phi, angle margin) { return (theta > phi-margin && theta < phi+margin); } Please help me. I'm new to C++

• algorithm analysis - orders of growth question

- by cchampion
I'm studing orders of growth "big oh", "big omega", and "big theta". Since I can't type the little symbols for these I will denote them as follows: ORDER = big oh OMEGA = big omega THETA = big theta For example I'll say n = ORDER(n^2) to mean that the function n is in the order of n^2 (n grows at most as fast n^2). Ok for the most part I understand these: n = ORDER(n^2) //n grows at most as fast as n^2 n^2 = OMEGA(n) //n^2 grows atleast as fast as n 8n^2 + 1000 = THETA(n^2) //same order of growth Ok here comes the example that confuses me: what is n(n+1) vs n^2 I realize that n(n+1) = n^2 + n; I would say it has the same order of growth as n^2; therefore I would say n(n+1) = THETA(n^2) but my question is, would it also be correct to say: n(n+1) = ORDER(n^2) please help because this is confusing to me. thanks.

• how to build an accumulator array in matlab

- by schwiz
I'm very new to matlab so sorry if this is a dumb question. I have to following matrices: im = imread('image.jpg'); %<370x366 double> [y,x] = find(im); %x & y both <1280x1 double> theta; %<370x366 double> computed from gradient of image I can currently plot points one at a time like this: plot(x(502) + 120*cos(theta(y(502),x(502))),y(502) + 120*sin(theta(y(502),x(502))),'*b'); But what I want to do is some how increment an accumulator array, something like this: acc = zeros(size(im)); acc(y,x) = acc(x + 120*cos(theta(y,x)),y + 120*sin(theta(y,x)),'*b')) + 1; It would be nice if the 120 could actually be another matrix containing different radius values as well.

• MATLAB intersection of 2 surfaces

- by caglarozdag
Hi everyone, I consider myself a beginner in MATLAB so bear with me if the answer to my question is an obvious one. Phi=0:pi/100:2*pi; Theta=0:pi/100:2*pi; [PHI,THETA]=meshgrid(Phi,Theta); R=(1 + cos(PHI).*cos(PHI)).*(1 + cos(THETA).*cos(THETA)); [X,Y,Z]=sph2cart(THETA,PHI,R); surf(X,Y,Z); %display hold on; x1=-4:.1:4; [X1,Y1] = meshgrid(x1); a=1.8; b=0; c=3; d=0; Z1=(d- a * X1 - b * Y1)/c; shading flat; surf(X1,Y1,Z1); I have written this code which plots a 3d cartesian plot of a plane intersecting a peanut shaped object at an angle. I need to get the intersection of these on 2D (going to be the outline of a peanut, but a bit skewed since the intersection happens at an angle), but don't know how. Thanks

• Taking fixed direction on hemisphere and project to normal (openGL)

- by Maik Xhani
I am trying to perform sampling using hemisphere around a surface normal. I want to experiment with fixed directions (and maybe jitter slightly between frames). So I have those directions: vec3 sampleDirections[6] = {vec3(0.0f, 1.0f, 0.0f), vec3(0.0f, 0.5f, 0.866025f), vec3(0.823639f, 0.5f, 0.267617f), vec3(0.509037f, 0.5f, -0.700629f), vec3(-0.509037f, 0.5f, -0.700629), vec3(-0.823639f, 0.5f, 0.267617f)}; now I want the first direction to be projected on the normal and the others accordingly. I tried these 2 codes, both failing. This is what I used for random sampling (it doesn't seem to work well, the samples seem to be biased towards a certain direction) and I just used one of the fixed directions instead of s (here is the code of the random sample, when i used it with the fixed direction i didn't use theta and phi). vec3 CosWeightedRandomHemisphereDirection( vec3 n, float rand1, float rand2 ) float theta = acos(sqrt(1.0f-rand1)); float phi = 6.283185f * rand2; vec3 s = vec3(sin(theta) * cos(phi), sin(theta) * sin(phi), cos(theta)); vec3 v = normalize(cross(n,vec3(0.0072, 1.0, 0.0034))); vec3 u = cross(v, n); u = s.x*u; v = s.y*v; vec3 w = s.z*n; vec3 direction = u+v+w; return normalize(direction); } ** EDIT ** This is the new code vec3 FixedHemisphereDirection( vec3 n, vec3 sampleDir) { vec3 x; vec3 z; if(abs(n.x) < abs(n.y)){ if(abs(n.x) < abs(n.z)){ x = vec3(1.0f,0.0f,0.0f); }else{ x = vec3(0.0f,0.0f,1.0f); } }else{ if(abs(n.y) < abs(n.z)){ x = vec3(0.0f,1.0f,0.0f); }else{ x = vec3(0.0f,0.0f,1.0f); } } z = normalize(cross(x,n)); x = cross(n,z); mat3 M = mat3( x.x, n.x, z.x, x.y, n.y, z.y, x.z, n.z, z.z); return M*sampleDir; } So if my n = (0,0,1); and my sampleDir = (0,1,0); shouldn't the M*sampleDir be (0,0,1)? Cause that is what I was expecting.

• Matlab: Optimization by perturbing variable

- by S_H
My main script contains following code: %# Grid and model parameters nModel=50; nModel_want=1; nI_grid1=5; Nth=1; nRow.Scale1=5; nCol.Scale1=5; nRow.Scale2=5^2; nCol.Scale2=5^2; theta = 90; % degrees a_minor = 2; % range along minor direction a_major = 5; % range along major direction sill = var(reshape(Deff_matrix_NthModel,nCell.Scale1,1)); % variance of the coarse data matrix of size nRow.Scale1 X nCol.Scale1 %# Covariance computation % Scale 1 for ihRow = 1:nRow.Scale1 for ihCol = 1:nCol.Scale1 [cov.Scale1(ihRow,ihCol),heff.Scale1(ihRow,ihCol)] = general_CovModel(theta, ihCol, ihRow, a_minor, a_major, sill, 'Exp'); end end % Scale 2 for ihRow = 1:nRow.Scale2 for ihCol = 1:nCol.Scale2 [cov.Scale2(ihRow,ihCol),heff.Scale2(ihRow,ihCol)] = general_CovModel(theta, ihCol/(nCol.Scale2/nCol.Scale1), ihRow/(nRow.Scale2/nRow.Scale1), a_minor, a_major, sill/(nRow.Scale2*nCol.Scale2), 'Exp'); end end %# Scale-up of fine scale values by averaging [covAvg.Scale2,var_covAvg.Scale2,varNorm_covAvg.Scale2] = general_AverageProperty(nRow.Scale2/nRow.Scale1,nCol.Scale2/nCol.Scale1,1,nRow.Scale1,nCol.Scale1,1,cov.Scale2,1); I am using two functions, general_CovModel() and general_AverageProperty(), in my main script which are given as following: function [cov,h_eff] = general_CovModel(theta, hx, hy, a_minor, a_major, sill, mod_type) % mod_type should be in strings angle_rad = theta*(pi/180); % theta in degrees, angle_rad in radians R_theta = [sin(angle_rad) cos(angle_rad); -cos(angle_rad) sin(angle_rad)]; h = [hx; hy]; lambda = a_minor/a_major; D_lambda = [lambda 0; 0 1]; h_2prime = D_lambda*R_theta*h; h_eff = sqrt((h_2prime(1)^2)+(h_2prime(2)^2)); if strcmp(mod_type,'Sph')==1 || strcmp(mod_type,'sph') ==1 if h_eff<=a cov = sill - sill.*(1.5*(h_eff/a_minor)-0.5*((h_eff/a_minor)^3)); else cov = sill; end elseif strcmp(mod_type,'Exp')==1 || strcmp(mod_type,'exp') ==1 cov = sill-(sill.*(1-exp(-(3*h_eff)/a_minor))); elseif strcmp(mod_type,'Gauss')==1 || strcmp(mod_type,'gauss') ==1 cov = sill-(sill.*(1-exp(-((3*h_eff)^2/(a_minor^2))))); end and function [PropertyAvg,variance_PropertyAvg,NormVariance_PropertyAvg]=... general_AverageProperty(blocksize_row,blocksize_col,blocksize_t,... nUpscaledRow,nUpscaledCol,nUpscaledT,PropertyArray,omega) % This function computes average of a property and variance of that averaged % property using power averaging PropertyAvg=zeros(nUpscaledRow,nUpscaledCol,nUpscaledT); %# Average of property for k=1:nUpscaledT, for j=1:nUpscaledCol, for i=1:nUpscaledRow, sum=0; for a=1:blocksize_row, for b=1:blocksize_col, for c=1:blocksize_t, sum=sum+(PropertyArray((i-1)*blocksize_row+a,(j-1)*blocksize_col+b,(k-1)*blocksize_t+c).^omega); % add all the property values in 'blocksize_x','blocksize_y','blocksize_t' to one variable end end end PropertyAvg(i,j,k)=(sum/(blocksize_row*blocksize_col*blocksize_t)).^(1/omega); % take average of the summed property end end end %# Variance of averageed property variance_PropertyAvg=var(reshape(PropertyAvg,... nUpscaledRow*nUpscaledCol*nUpscaledT,1),1,1); %# Normalized variance of averageed property NormVariance_PropertyAvg=variance_PropertyAvg./(var(reshape(... PropertyArray,numel(PropertyArray),1),1,1)); Question: Using Matlab, I would like to optimize covAvg.Scale2 such that it matches closely with cov.Scale1 by perturbing/varying any (or all) of the following variables 1) a_minor 2) a_major 3) theta Thanks.

• Does this code follow the definition of recursion?

- by dekz
Hi All, I have a piece of code which I am doubting it as a implementation of recursion by its definition. My understanding is that the code must call itself, the exact same function. I also question whether writing the code this way adds additional overhead which can be seen with the use of recursion. What are your thoughts? class dhObject { public: dhObject** children; int numChildren; GLdouble linkLength; //ai GLdouble theta; //angle of rot about the z axis GLdouble twist; //about the x axis GLdouble displacement; // displacement from the end point of prev along z GLdouble thetaMax; GLdouble thetaMin; GLdouble thetaInc; GLdouble direction; dhObject(ifstream &fin) { fin >> numChildren >> linkLength >> theta >> twist >> displacement >> thetaMax >> thetaMin; //std::cout << numChildren << std::endl; direction = 1; thetaInc = 1.0; if (numChildren > 0) { children = new dhObject*[numChildren]; for(int i = 0; i < numChildren; ++i) { children[i] = new dhObject(fin); } } } void traverse(void) { glPushMatrix(); //draw move initial and draw transform(); draw(); //draw children for(int i = 0; i < numChildren; ++i) { children[i]->traverse(); } glPopMatrix(); } void update(void) { //Update the animation, if it has finished all animation go backwards if (theta <= thetaMin) { thetaInc = 1.0; } else if (theta >= thetaMax) { thetaInc = -1.0; } theta += thetaInc; //std::cout << thetaMin << " " << theta << " " << thetaMax << std::endl; for(int i = 0; i < numChildren; ++i) { children[i]->update(); } } void draw(void) { glPushMatrix(); glColor3f (0.0f,0.0f,1.0f); glutSolidCube(0.1); glPopMatrix(); } void transform(void) { //Move in the correct way, R, T, T, R glRotatef(theta, 0, 0, 1.0); glTranslatef(0,0,displacement); glTranslatef(linkLength, 0,0); glRotatef(twist, 1.0,0.0,0.0); } };

hi i m finding cosine similarity between documents ..i did like dis D1=(8,0,0,1) where 8,0,0,1 are the tf-idf scores of the terms t1, t2, t3 , t4 D2=(7,0,0,1) cos(theta) = (56 + 0 + 0 + 1) / sqrt(64 + 49) sqrt(1 +1 ) which comes out to be cos(theta)= 5 now what do i evaluate from this value...i dont get it wat does cos(theta)=5 signify about the similarity between them...pls reply ..Am i doing things right ??????????..pls do reply guys.. will be thank ful to you..

• Draw LINE_STRIP with Unity

- by Boozzz
For a new project I am thinking about whether to use OpenGL or Unity3d. I have a bit of experience with OpenGL, but I am completely new to Unity. I already read through the Unity documentation and tutorials on the Unity Website. However, I could not find a way to draw a simple Line-Strip with Unity. In the following example (C#, OpenGL/SharpGL) I draw a round trajectory from a predifined point to an obstacle, which can be imagined as a divided circle with midpoint [cx,cy] and radius r. The position (x-y coordinates) of the obstacle is given by obst_x and obst_y. Question 1: How could I do the same with Unity? Question 2: In my new project, I will have to draw quite a lot of such geometric primitives. Does it make any sense to use Unity for those things? void drawCircle(float cx, float cy, float r, const float obst_x, const float obst_y) { float theta = 0.0f, pos_x, pos_y, dist; const float delta = 0.1; glBegin(GL_LINE_STRIP); while (theta < 180) { theta += delta; //get the current angle float x = r * cosf(theta); //calculate the x component float y = r * sinf(theta); //calculate the y component pos_x = x + cx; //calculate current x position pos_y = y + cy; //calculate current y position //calculate distance from current vertex to obstacle dist = sqrt(pow(pos_x - obst_x) + pow(pos_y - obst_y)); //check if current vertex intersects with obstacle if dist <= 0 { break; //stop drawing circle } else { glVertex2f(pos_x, pos_y); //draw vertex } } glEnd(); }

• Calculating angle a segment forms with a ray

- by kr1zz
I am given a point C and a ray r starting there. I know the coordinates (xc, yc) of the point C and the angle theta the ray r forms with the horizontal, theta in (-pi, pi]. I am also given another point P of which I know the coordinates (xp, yp): how do I calculate the angle alpha that the segment CP forms with the ray r, alpha in (-pi, pi]? Some examples follow: I can use the the atan2 function.