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  • Generating lognormally distributed random number from mean, coeff of variation

    - by Richie Cotton
    Most functions for generating lognormally distributed random numbers take the mean and standard deviation of the associated normal distribution as parameters. My problem is that I only know the mean and the coefficient of variation of the lognormal distribution. It is reasonably straight forward to derive the parameters I need for the standard functions from what I have: If mu and sigma are the mean and standard deviation of the associated normal distribution, we know that coeffOfVar^2 = variance / mean^2 = (exp(sigma^2) - 1) * exp(2*mu + sigma^2) / exp(mu + sigma^2/2)^2 = exp(sigma^2) - 1 We can rearrange this to sigma = sqrt(log(coeffOfVar^2 + 1)) We also know that mean = exp(mu + sigma^2/2) This rearranges to mu = log(mean) - sigma^2/2 Here's my R implementation rlnorm0 <- function(mean, coeffOfVar, n = 1e6) { sigma <- sqrt(log(coeffOfVar^2 + 1)) mu <- log(mean) - sigma^2 / 2 rlnorm(n, mu, sigma) } It works okay for small coefficients of variation r1 <- rlnorm0(2, 0.5) mean(r1) # 2.000095 sd(r1) / mean(r1) # 0.4998437 But not for larger values r2 <- rlnorm0(2, 50) mean(r2) # 2.048509 sd(r2) / mean(r2) # 68.55871 To check that it wasn't an R-specific issue, I reimplemented it in MATLAB. (Uses stats toolbox.) function y = lognrnd0(mean, coeffOfVar, sizeOut) if nargin < 3 || isempty(sizeOut) sizeOut = [1e6 1]; end sigma = sqrt(log(coeffOfVar.^2 + 1)); mu = log(mean) - sigma.^2 ./ 2; y = lognrnd(mu, sigma, sizeOut); end r1 = lognrnd0(2, 0.5); mean(r1) % 2.0013 std(r1) ./ mean(r1) % 0.5008 r2 = lognrnd0(2, 50); mean(r2) % 1.9611 std(r2) ./ mean(r2) % 22.61 Same problem. The question is, why is this happening? Is it just that the standard deviation is not robust when the variation is that wide? Or have a screwed up somewhere?

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  • Targeted Simplify in Mathematica

    - by Timo
    I generate very long and complex analytic expressions of the general form: (...something not so complex...)(...ditto...)(...ditto...)...lots... When I try to use Simplify, Mathematica grinds to a halt, I am assuming due to the fact that it tries to expand the brackets and or simplify across different brackets. The brackets, while containing long expressions, are easily simplified by Mathematica on their own. Is there some way I can limit the scope of Simplify to a single bracket at a time? Edit: Some additional info and progress. So using the advice from you guys I have now started using something in the vein of In[1]:= trouble = Log[(x + I y) (x - I y) + Sqrt[(a + I b) (a - I b)]]; In[2]:= Replace[trouble, form_ /; (Head[form] == Times) :> Simplify[form],{3}] Out[2]= Log[Sqrt[a^2 + b^2] + (x - I y) (x + I y)] Changing Times to an appropriate head like Plus or Power makes it possible to target the simplification quite accurately. The problem / question that remains, though, is the following: Simplify will still descend deeper than the level specified to Replace, e.g. In[3]:= Replace[trouble, form_ /; (Head[form] == Plus) :> Simplify[form], {1}] Out[3]= Log[Sqrt[a^2 + b^2] + x^2 + y^2] simplifies the square root as well. My plan was to iteratively use Replace from the bottom up one level at a time, but this clearly will result in vast amount of repeated work by Simplify and ultimately result in the exact same bogging down of Mathematica I experienced in the outset. Is there a way to restrict Simplify to a certain level(s)? I realize that this sort of restriction may not produce optimal results, but the idea here is getting something that is "good enough".

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  • Python Turtle Graphics, how to plot functions over an interval?

    - by TheDragonAce
    I need to plot a function over a specified interval. The function is f1, which is shown below in the code, and the interval is [-7, -3]; [-1, 1]; [3, 7] with a step of .01. When I execute the program, nothing is drawn. Any ideas? import turtle from math import sqrt wn = turtle.Screen() wn.bgcolor("white") wn.title("Plotting") mypen = turtle.Turtle() mypen.shape("classic") mypen.color("black") mypen.speed(10) while True: try: def f1(x): return 2 * sqrt((-abs(abs(x)-1)) * abs(3 - abs(x))/((abs(x)-1)*(3-abs(x)))) * \ (1 + abs(abs(x)-3)/(abs(x)-3))*sqrt(1-(x/7)**2)+(5+0.97*(abs(x-0.5)+abs(x+0.5))-\ 3*(abs(x-0.75)+abs(x+0.75)))*(1+abs(1-abs(x))/(1-abs(x))) mypen.penup() step=.01 startf11=-7 stopf11=-3 startf12=-1 stopf12=1 startf13=3 stopf13=7 def f11 (startf11,stopf11,step): rc=[] y = f1(startf11) while y<=stopf11: rc.append(startf11) #y+=step mypen.setpos(f1(startf11)*25,y*25) mypen.dot() def f12 (startf12,stopf12,step): rc=[] y = f1(startf12) while y<=stopf12: rc.append(startf12) #y+=step mypen.setpos(f1(startf12)*25, y*25) mypen.dot() def f13 (startf13,stopf13,step): rc=[] y = f1(startf13) while y<=stopf13: rc.append(startf13) #y+=step mypen.setpos(f1(startf13)*25, y*25) mypen.dot() f11(startf11,stopf11,step) f12(startf12,stopf12,step) f13(startf13,stopf13,step) except ZeroDivisionError: continue

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  • Remove redundant SQL code

    - by Dave Jarvis
    Code The following code calculates the slope and intercept for a linear regression against a slathering of data. It then applies the equation y = mx + b against the same result set to calculate the value of the regression line for each row. Can the two separate sub-selects be joined so that the data and its slope/intercept are calculated without executing the data gathering part of the query twice? SELECT AVG(D.AMOUNT) as AMOUNT, Y.YEAR * ymxb.SLOPE + ymxb.INTERCEPT as REGRESSION_LINE, Y.YEAR as YEAR, MAKEDATE(Y.YEAR,1) as AMOUNT_DATE FROM CITY C, STATION S, YEAR_REF Y, MONTH_REF M, DAILY D, (SELECT ((avg(t.AMOUNT * t.YEAR)) - avg(t.AMOUNT) * avg(t.YEAR)) / (stddev( t.AMOUNT ) * stddev( t.YEAR )) as CORRELATION, ((sum(t.YEAR) * sum(t.AMOUNT)) - (count(1) * sum(t.YEAR * t.AMOUNT))) / (power(sum(t.YEAR), 2) - count(1) * sum(power(t.YEAR, 2))) as SLOPE, ((sum( t.YEAR ) * sum( t.YEAR * t.AMOUNT )) - (sum( t.AMOUNT ) * sum(power(t.YEAR, 2)))) / (power(sum(t.YEAR), 2) - count(1) * sum(power(t.YEAR, 2))) as INTERCEPT FROM ( SELECT AVG(D.AMOUNT) as AMOUNT, Y.YEAR as YEAR, MAKEDATE(Y.YEAR,1) as AMOUNT_DATE FROM CITY C, STATION S, YEAR_REF Y, MONTH_REF M, DAILY D WHERE $X{ IN, C.ID, CityCode } AND SQRT( POW( C.LATITUDE - S.LATITUDE, 2 ) + POW( C.LONGITUDE - S.LONGITUDE, 2 ) ) < $P{Radius} AND S.STATION_DISTRICT_ID = Y.STATION_DISTRICT_ID AND Y.YEAR BETWEEN 1900 AND 2009 AND M.YEAR_REF_ID = Y.ID AND M.CATEGORY_ID = $P{CategoryCode} AND M.ID = D.MONTH_REF_ID AND D.DAILY_FLAG_ID <> 'M' GROUP BY Y.YEAR ) t ) ymxb WHERE $X{ IN, C.ID, CityCode } AND SQRT( POW( C.LATITUDE - S.LATITUDE, 2 ) + POW( C.LONGITUDE - S.LONGITUDE, 2 ) ) < $P{Radius} AND S.STATION_DISTRICT_ID = Y.STATION_DISTRICT_ID AND Y.YEAR BETWEEN 1900 AND 2009 AND M.YEAR_REF_ID = Y.ID AND M.CATEGORY_ID = $P{CategoryCode} AND M.ID = D.MONTH_REF_ID AND D.DAILY_FLAG_ID <> 'M' GROUP BY Y.YEAR Question How do I execute the duplicate bits only once per query, instead of twice? The duplicate bit is the WHERE clause: $X{ IN, C.ID, CityCode } AND SQRT( POW( C.LATITUDE - S.LATITUDE, 2 ) + POW( C.LONGITUDE - S.LONGITUDE, 2 ) ) < $P{Radius} AND S.STATION_DISTRICT_ID = Y.STATION_DISTRICT_ID AND Y.YEAR BETWEEN 1900 AND 2009 AND M.YEAR_REF_ID = Y.ID AND M.CATEGORY_ID = $P{CategoryCode} AND M.ID = D.MONTH_REF_ID AND D.DAILY_FLAG_ID <> 'M' Related http://stackoverflow.com/questions/1595659/how-to-eliminate-duplicate-calculation-in-sql Thank you!

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  • Rewriting a for loop in pure NumPy to decrease execution time

    - by Statto
    I recently asked about trying to optimise a Python loop for a scientific application, and received an excellent, smart way of recoding it within NumPy which reduced execution time by a factor of around 100 for me! However, calculation of the B value is actually nested within a few other loops, because it is evaluated at a regular grid of positions. Is there a similarly smart NumPy rewrite to shave time off this procedure? I suspect the performance gain for this part would be less marked, and the disadvantages would presumably be that it would not be possible to report back to the user on the progress of the calculation, that the results could not be written to the output file until the end of the calculation, and possibly that doing this in one enormous step would have memory implications? Is it possible to circumvent any of these? import numpy as np import time def reshape_vector(v): b = np.empty((3,1)) for i in range(3): b[i][0] = v[i] return b def unit_vectors(r): return r / np.sqrt((r*r).sum(0)) def calculate_dipole(mu, r_i, mom_i): relative = mu - r_i r_unit = unit_vectors(relative) A = 1e-7 num = A*(3*np.sum(mom_i*r_unit, 0)*r_unit - mom_i) den = np.sqrt(np.sum(relative*relative, 0))**3 B = np.sum(num/den, 1) return B N = 20000 # number of dipoles r_i = np.random.random((3,N)) # positions of dipoles mom_i = np.random.random((3,N)) # moments of dipoles a = np.random.random((3,3)) # three basis vectors for this crystal n = [10,10,10] # points at which to evaluate sum gamma_mu = 135.5 # a constant t_start = time.clock() for i in range(n[0]): r_frac_x = np.float(i)/np.float(n[0]) r_test_x = r_frac_x * a[0] for j in range(n[1]): r_frac_y = np.float(j)/np.float(n[1]) r_test_y = r_frac_y * a[1] for k in range(n[2]): r_frac_z = np.float(k)/np.float(n[2]) r_test = r_test_x +r_test_y + r_frac_z * a[2] r_test_fast = reshape_vector(r_test) B = calculate_dipole(r_test_fast, r_i, mom_i) omega = gamma_mu*np.sqrt(np.dot(B,B)) # write r_test, B and omega to a file frac_done = np.float(i+1)/(n[0]+1) t_elapsed = (time.clock()-t_start) t_remain = (1-frac_done)*t_elapsed/frac_done print frac_done*100,'% done in',t_elapsed/60.,'minutes...approximately',t_remain/60.,'minutes remaining'

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  • What is this: main:for(...){...} doing?

    - by David Murdoch
    I pulled up the NWmatcher source code for some light morning reading and noticed this odd bit of code I'd never seen in javascript before: main:for(/*irrelevant loop stuff*/){/*...*/} This snippet can be found in the compileGroup method on line 441 (nwmatcher-1.1.1) return new Function('c,s,d,h', 'var k,e,r,n,C,N,T,X=0,x=0;main:for(k=0,r=[];e=N=c[k];k++){' + SKIP_COMMENTS + source + '}return r;' ); Now I figured out what main: is doing on my own. If you have a loop within a loop and want to skip to the next iteration of the outer loop (without completing the inner OR the outer loop) you can execute continue main. Example: // This is obviously not the optimal way to find primes... function getPrimes(max) { var primes = [2], //seed sqrt = Math.sqrt, i = 3, j, s; outer: for (; i <= max; s = sqrt(i += 2)) { j = 3; while (j <= s) { if (i % j === 0) { // if we get here j += 2 and primes.push(i) are // not executed for the current iteration of i continue outer; } j += 2; } primes.push(i); } return primes; } What is this called? Are there any browsers that don't support it? Are there other uses for it other than continue?

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  • parsing of mathematical expressions

    - by gcc
    (in c90) (linux) input: sqrt(2 - sin(3*A/B)^2.5) + 0.5*(C*~(D) + 3.11 +B) a b /*there are values for a,b,c,d */ c d input: cos(2 - asin(3*A/B)^2.5) +cos(0.5*(C*~(D)) + 3.11 +B) a b /*there are values for a,b,c,d */ c d input: sqrt(2 - sin(3*A/B)^2.5)/(0.5*(C*~(D)) + sin(3.11) +ln(B)) /*max lenght of formula is 250 characters*/ a b /*there are values for a,b,c,d */ c /*each variable with set of floating numbers*/ d As you can see infix formula in the input depends on user. My program will take a formula and n-tuples value. Then it calculate the results for each value of a,b,c and d. If you wonder I am saying ;outcome of program is graph. /sometimes,I think i will take input and store in string. then another idea is arise " I should store formula in the struct" but i don't know how I can construct the code on the base of structure./ really, I don't know way how to store the formula in program code so that I can do my job. can you show me? /* a,b,c,d is letters cos,sin,sqrt,ln is function*/

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  • Creating simple calculator with bison & flex in C++ (not C)

    - by ak91
    Hey, I would like to create simple C++ calculator using bison and flex. Please note I'm new to the creating parsers. I already found few examples in bison/flex but they were all written in C. My goal is to create C++ code, where classes would contain nodes of values, operations, funcs - to create AST (evaluation would be done just after creating whole AST - starting from the root and going forward). For example: my_var = sqrt(9 ** 2 - 32) + 4 - 20 / 5 my_var * 3 Would be parsed as: = / \ my_var + / \ sqrt - | / \ - 4 / / \ / \ ** 32 20 5 / \ 9 2 and the second AST would look like: * / \ my_var 3 Then following pseudocode reflects AST: ast_root = create_node('=', new_variable("my_var"), exp) where exp is: exp = create_node(OPERATOR, val1, val2) but NOT like this: $$ = $1 OPERATOR $3 because this way I directly get value of operation instead of creation the Node. I believe the Node should contain type (of operation), val1 (Node), val2 (Node). In some cases val2 would be NULL, like above mentioned sqrt which takes in the end one argument. Right? It will be nice if you can propose me C++ skeleton (without evaluation) for above described problem (including *.y file creating AST) to help me understand the way of creating/holding Nodes in AST. Code can be snipped, just to let me get the idea. I'll also be grateful if you point me to an existing (possibly simple) example if you know any. Thank you all for your time and assistance!

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  • How can I use iteration to lead targets?

    - by e100
    In my 2D game, I have stationary AI turrets firing constant speed bullets at moving targets. So far I have used a quadratic solver technique to calculate where the turret should aim in advance of the target, which works well (see Algorithm to shoot at a target in a 3d game, Predicting enemy position in order to have an object lead its target). But it occurs to me that an iterative technique might be more realistic (e.g. it should fire even when there is no exact solution), efficient and tunable - for example one could change the number of iterations to improve accuracy. I thought I could calculate the current range and thus an initial (inaccurate) bullet flight time to target, then work out where the target would actually be by that time, then recalculate a more accurate range, then recalculate flight time, etc etc. I think I am missing something obvious to do with the time term, but my aimpoint calculation does not currently converge after the significant initial correction in the first iteration: import math def aimpoint(iters, target_x, target_y, target_vel_x, target_vel_y, bullet_speed): aimpoint_x = target_x aimpoint_y = target_y range = math.sqrt(aimpoint_x**2 + aimpoint_y**2) time_to_target = range / bullet_speed time_delta = time_to_target n = 0 while n <= iters: print "iteration:", n, "target:", "(", aimpoint_x, aimpoint_y, ")", "time_delta:", time_delta aimpoint_x += target_vel_x * time_delta aimpoint_y += target_vel_y * time_delta range = math.sqrt(aimpoint_x**2 + aimpoint_y**2) new_time_to_target = range / bullet_speed time_delta = new_time_to_target - time_to_target n += 1 aimpoint(iters=5, target_x=0, target_y=100, target_vel_x=1, target_vel_y=0, bullet_speed=100)

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  • Normalizing the direction to check if able to move

    - by spartan2417
    i have a a room with 4 walls along the x and z axis respectively. My player who is in first person (therefore the camera) should have collision detection with these walls. I'm relatively new to this so please bare with me. I believe the way to do this is to calculate the direction and distance to the wall from the camera and then normalize the directions. However i can only get this far before i dont know what to do. I think you should work out the angle and direction your facing? where _dx and _dz is the small buffer in front of the camera. float CalcDirection(float Cam_x, float Cam_z, float Wall_x, float Wall_z) { //Calculate direction and distance to obstacle. float ob_dirx = Cam_x + _dx - Wall_x; float ob_dirz = Cam_z + _dz - Wall_z; float ob_dist = sqrt(ob_dirx*ob_dirx + ob_dirz*ob_dirz); //Normalise directions float ob_norm = sqrt(ob_dirx*ob_dirx + ob_dirz*ob_dirz); ob_dirx = (ob_dirx)/ob_norm; ob_dirz = (ob_dirz)/ob_norm; can anyone explain in laymen's terms how i work out the angle?

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  • Increasing efficiency of N-Body gravity simulation

    - by Postman
    I'm making a space exploration type game, it will have many planets and other objects that will all have realistic gravity. I currently have a system in place that works, but if the number of planets goes above 70, the FPS decreases an practically exponential rates. I'm making it in C# and XNA. My guess is that I should be able to do gravity calculations between 100 objects without this kind of strain, so clearly my method is not as efficient as it should be. I have two files, Gravity.cs and EntityEngine.cs. Gravity manages JUST the gravity calculations, EntityEngine creates an instance of Gravity and runs it, along with other entity related methods. EntityEngine.cs public void Update() { foreach (KeyValuePair<string, Entity> e in Entities) { e.Value.Update(); } gravity.Update(); } (Only relevant piece of code from EntityEngine, self explanatory. When an instance of Gravity is made in entityEngine, it passes itself (this) into it, so that gravity can have access to entityEngine.Entities (a dictionary of all planet objects)) Gravity.cs namespace ExplorationEngine { public class Gravity { private EntityEngine entityEngine; private Vector2 Force; private Vector2 VecForce; private float distance; private float mult; public Gravity(EntityEngine e) { entityEngine = e; } public void Update() { //First loop foreach (KeyValuePair<string, Entity> e in entityEngine.Entities) { //Reset the force vector Force = new Vector2(); //Second loop foreach (KeyValuePair<string, Entity> e2 in entityEngine.Entities) { //Make sure the second value is not the current value from the first loop if (e2.Value != e.Value ) { //Find the distance between the two objects. Because Fg = G * ((M1 * M2) / r^2), using Vector2.Distance() and then squaring it //is pointless and inefficient because distance uses a sqrt, squaring the result simple cancels that sqrt. distance = Vector2.DistanceSquared(e2.Value.Position, e.Value.Position); //This makes sure that two planets do not attract eachother if they are touching, completely unnecessary when I add collision, //For now it just makes it so that the planets are not glitchy, performance is not significantly improved by removing this IF if (Math.Sqrt(distance) > (e.Value.Texture.Width / 2 + e2.Value.Texture.Width / 2)) { //Calculate the magnitude of Fg (I'm using my own gravitational constant (G) for the sake of time (I know it's 1 at the moment, but I've been changing it) mult = 1.0f * ((e.Value.Mass * e2.Value.Mass) / distance); //Calculate the direction of the force, simply subtracting the positions and normalizing works, this fixes diagonal vectors //from having a larger value, and basically makes VecForce a direction. VecForce = e2.Value.Position - e.Value.Position; VecForce.Normalize(); //Add the vector for each planet in the second loop to a force var. Force = Vector2.Add(Force, VecForce * mult); //I have tried Force += VecForce * mult, and have not noticed much of an increase in speed. } } } //Add that force to the first loop's planet's position (later on I'll instead add to acceleration, to account for inertia) e.Value.Position += Force; } } } } I have used various tips (about gravity optimizing, not threading) from THIS question (that I made yesterday). I've made this gravity method (Gravity.Update) as efficient as I know how to make it. This O(N^2) algorithm still seems to be eating up all of my CPU power though. Here is a LINK (google drive, go to File download, keep .Exe with the content folder, you will need XNA Framework 4.0 Redist. if you don't already have it) to the current version of my game. Left click makes a planet, right click removes the last planet. Mouse moves the camera, scroll wheel zooms in and out. Watch the FPS and Planet Count to see what I mean about performance issues past 70 planets. (ALL 70 planets must be moving, I've had 100 stationary planets and only 5 or so moving ones while still having 300 fps, the issue arises when 70+ are moving around) After 70 planets are made, performance tanks exponentially. With < 70 planets, I get 330 fps (I have it capped at 300). At 90 planets, the FPS is about 2, more than that and it sticks around at 0 FPS. Strangely enough, when all planets are stationary, the FPS climbs back up to around 300, but as soon as something moves, it goes right back down to what it was, I have no systems in place to make this happen, it just does. I considered multithreading, but that previous question I asked taught me a thing or two, and I see now that that's not a viable option. I've also thought maybe I could do the calculations on my GPU instead, though I don't think it should be necessary. I also do not know how to do this, it is not a simple concept and I want to avoid it unless someone knows a really noob friendly simple way to do it that will work for an n-body gravity calculation. (I have an NVidia gtx 660) Lastly I've considered using a quadtree type system. (Barnes Hut simulation) I've been told (in the previous question) that this is a good method that is commonly used, and it seems logical and straightforward, however the implementation is way over my head and I haven't found a good tutorial for C# yet that explains it in a way I can understand, or uses code I can eventually figure out. So my question is this: How can I make my gravity method more efficient, allowing me to use more than 100 objects (I can render 1000 planets with constant 300+ FPS without gravity calculations), and if I can't do much to improve performance (including some kind of quadtree system), could I use my GPU to do the calculations?

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  • Stochastic calculus library in python

    - by LeMiz
    Hello, I am looking for a python library that would allow me to compute stochastic calculus stuff, like the (conditional) expectation of a random process I would define the diffusion. I had a look a at simpy (simpy.sourceforge.net), but it does not seem to cover my needs. This is for quick prototyping and experimentation. In java, I used with some success the (now inactive) http://martingale.berlios.de/Martingale.html library. The problem is not difficult in itself, but there is a lot non trivial, boilerplate things to do (efficient memory use, variable reduction techniques, and so on). Ideally, I would be able to write something like this (just illustrative): def my_diffusion(t, dt, past_values, world, **kwargs): W1, W2 = world.correlated_brownians_pair(correlation=kwargs['rho']) X = past_values[-1] sigma_1 = kwargs['sigma1'] sigma_2 = kwargs['sigma2'] dX = kwargs['mu'] * X * dt + sigma_1 * W1 * X * math.sqrt(dt) + sigma_2 * W2 * X * X * math.sqrt(dt) return X + dX X = RandomProcess(diffusion=my_diffusion, x0 = 1.0) print X.expectancy(T=252, dt = 1./252., N_simul= 50000, world=World(random_generator='sobol'), sigma1 = 0.3, sigma2 = 0.01, rho=-0.1) Does someone knows of something else than reimplementing it in numpy for example ?

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  • Resizing an image using mouse dragging (C#)

    - by Gaax
    Hi all. I'm having some trouble resizing an image just by dragging the mouse. I found an average resize method and now am trying to modify it to use the mouse instead of given values. The way I'm doing it makes sense to me but maybe you guys can give me some better ideas. I'm basically using the distance between the current location of the mouse and the previous location of the mouse as the scaling factor. If the distance between the current mouse location and the center of of the image is less than the distance between previous mouse location and the center of the image then the image gets smaller, and vice-versa. With the code below I'm getting an Argument Exception (invalid parameter) when creating the new bitmap with the new height and width and I really don't understand why... any ideas? private static Image resizeImage(Image imgToResize, System.Drawing.Point prevMouseLoc, System.Drawing.Point currentMouseLoc) { int sourceWidth = imgToResize.Width; int sourceHeight = imgToResize.Height; float dCurrCent = 0; //Distance between current mouse location and the center of the image float dPrevCent = 0; //Distance between previous mouse location and the center of the image float dCurrPrev = 0; //Distance between current mouse location and the previous mouse location int sign = 1; System.Drawing.Point imgCenter = new System.Drawing.Point(); float nPercent = 0; imgCenter.X = imgToResize.Width / 2; imgCenter.Y = imgToResize.Height / 2; // Calculating the distance between the current mouse location and the center of the image dCurrCent = (float)Math.Sqrt(Math.Pow(currentMouseLoc.X - imgCenter.X, 2) + Math.Pow(currentMouseLoc.Y - imgCenter.Y, 2)); // Calculating the distance between the previous mouse location and the center of the image dPrevCent = (float)Math.Sqrt(Math.Pow(prevMouseLoc.XimgCenter.X,2) + Math.Pow(prevMouseLoc.Y - imgCenter.Y, 2)); // Calculating the sign value if (dCurrCent >= dPrevCent) { sign = 1; } else { sign = -1; } nPercent = sign * (float)Math.Sqrt(Math.Pow(currentMouseLoc.X - prevMouseLoc.X, 2) + Math.Pow(currentMouseLoc.Y - prevMouseLoc.Y, 2)); int destWidth = (int)(sourceWidth * nPercent); int destHeight = (int)(sourceHeight * nPercent); Bitmap b = new Bitmap(destWidth, destHeight); // exception thrown here Graphics g = Graphics.FromImage((Image)b); g.InterpolationMode = InterpolationMode.HighQualityBicubic; g.DrawImage(imgToResize, 0, 0, destWidth, destHeight); g.Dispose(); return (Image)b; }

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  • How to cycle through matrix blocks?

    - by luiss
    I have some matrix which I want to cycle through blocks, the matrix could be of many different sizes, but I can know the size, is there a way to fast cycle through blocks? i.e: to fast output the indexes of the blocks, suppose a matrix of 4*4 I should have: Block1: (0,0),(0,1)(1,0)(1,1) Block2: (0,2),(0,3)(1,2)(1,3) Block3: (2,0),(2,1)(3,0)(3,1) Block4: (2,2),(2,3)(3,2)(3,3) Where the indexes are (row,col). For blocks I mean a submatrix of size sqrt(matrixSize)* sqrt(matrixSize) where matrix is a matrix of matrixSize*matrixSize. For example a matrix of 4*4 has 4 blocks of 2*2, a 9*9 has 9 blocks of 3*3... I'm workdeing in C, but I think that the pseudocode is useful also, I only need the loop on the indexes... Thanks

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  • How to get ouput from expect

    - by Mallikarjunarao
    i wrote a script for spawing the bc command package require Expect proc bc {eq} { spawn e:/GnuWin32/bc/bin/bc send "$eq\r" expect -re "(.*)\r" return "$expect_out(0,string)" } set foo "9487294387234/sqrt(394872394879847293847)" puts "the valule [bc $foo]" how to get the output from this. When i am running this one i get ouput like this bc 1.06 Copyright 1991-1994, 1997, 1998, 2000 Free Software Foundation, Inc. This is free software with ABSOLUTELY NO WARRANTY. For details type `warranty'. 9487294387234/sqrt(394872394879847293847) 477 can't read "expect_out(0,string)": no such element in array while executing "return "The values is $expect_out(0,string)"" (procedure "bc" line 6) invoked from within "bc $foo" invoked from within "puts "the valule [bc $foo]"" (file "bc.tcl" line 21) how to resolve this one.

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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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  • Mathematica Plot3D does not produce a plot when graphing a user-defined function?

    - by pythonscript
    I'm writing a simple Mathematica implementation of the black-scholes model and using Plot3D to plot the pricing surface. However, when I run this code, no plot is produced. My call and put functions to produce correct values when run separately, but no plot is produced. Code: Clear[d1, d2, call, put, stockPrice, strikePrice, riskFreeRate, timeToExp, volatility] d1[stockPrice_, strikePrice_, riskFreeRate_, timeToExp_, volatility_] := (Log[stockPrice / strikePrice] + (riskFreeRate + 0.5*volatility^2)*timeToExp) / (volatility * Sqrt[timeToExp]) d2[stockPrice_, strikePrice_, riskFreeRate_, timeToExp_, volatility_] := d1[stockPrice, strikePrice, riskFreeRate, timeToExp, volatility] - volatility*Sqrt[timeToExp] call[stockPrice_, strikePrice_, riskFreeRate_, timeToExp_, volatility_] := stockPrice * CDF[NormalDistribution[0, 1], d1[stockPrice, strikePrice, riskFreeRate, timeToExp, volatility]] - strikePrice * Exp[-riskFreeRate*timeToExp] *CDF[NormalDistribution[0, 1], d2[stockPrice, strikePrice, riskFreeRate, timeToExp, volatility]] Plot3D[call[stockPrice, 500, 0.0030, timeToExp, 0.39], {stockPrice, 10, 1000}, {timetoExp, 0.0833333, 5}] Other plots, like this sample from the reference, do work. Plot3D[{x^2 + y^2, -x^2 - y^2}, {x, -2, 2}, {y, -2, 2}, RegionFunction -> Function[{x, y, z}, x^2 + y^2 <= 4], BoxRatios -> Automatic]

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  • How to get Distance Kilometer in android?

    - by user1787493
    i am very new to Google maps i want calculate the distance between two places in android .for that i get the two places lat and lag positions for that i write the following code: private double getDistance(double lat1, double lat2, double lon1, double lon2) { double dLat = Math.toRadians(lat2 - lat1); double dLon = Math.toRadians(lon2 - lon1); double a = Math.sin(dLat / 2) * Math.sin(dLat / 2) + Math.cos(Math.toRadians(lat1)) * Math.cos(Math.toRadians(lat2)) * Math.sin(dLon / 2) * Math.sin(dLon / 2); double c = 2 * Math.atan2(Math.sqrt(a), Math.sqrt(1 - a)); double temp = 6371 * c; temp=temp*0.621; return temp; } the above code cant give the accurate distance between two places .what is the another way to find distance please give me any suggestions thanks in advance....

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  • Python beginner, strange output problem

    - by Protean
    I'm having a weird problem with the following piece of code. from math import sqrt def Permute(array): result1 = [] result2 = [] if len(array) <= 1: return array for subarray in Permute(array[1:]): for i in range(len(array)): temp1 = subarray[:i]+array[0]+subarray[i:] temp2 = [0] for num in range(len(array)-1): temp2[0] += (sqrt(pow((temp1[num+1][1][0]-temp1[num][1][0]),2) + pow((temp1[num+1][1][1]-temp1[num][1][1]),2))) result1.append(temp1+temp2) return result1 a = [['A',[50,1]]] b = [['B',[1,1]]] c = [['C',[100,1]]] array = [a,b,c] result1 = Permute(array) for i in range(len(result1)): print (result1[i]) print (len(result1)) What it does is find all the permutations of the points abc and then returns them along with the sum of the distances between each ordered point. It does this; however, it also seems to report a strange additional value, 99. I figure that the 99 is coming from the computation of the distance between point a and c but I don't understand why it is appearing in the final output as it does.

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  • How to make this C++ code more DRY?

    - by Macha
    I have these two methods on a class that differ only in one method call. Obviously, this is very un-DRY, especially as both use the same formula. int PlayerCharacter::getAttack() { attack = 1 + this.level; for(int i = 0; i <= current_equipment; i++) { attack += this.equipment[i].getAttack(); } attack *= sqrt(this.level); return attack; } int PlayerCharacter::getDefense() { defense = 1 + this.level; for(int i = 0; i <= current_equipment; i++) { defense += this.equipment[i].getDefense(); } defense *= sqrt(this.level); return defense; } How can I tidy this up in C++?

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  • Simple Physics Simulation in java not working.

    - by Static Void Main
    Dear experts, I wanted to implement ball physics and as i m newbie, i adapt the code in tutorial http://adam21.web.officelive.com/Documents/JavaPhysicsTutorial.pdf . i try to follow that as i much as i can, but i m not able to apply all physical phenomenon in code, can somebody please tell me, where i m mistaken or i m still doing some silly programming mistake. The balls are moving when i m not calling bounce method and i m unable to avail the bounce method and ball are moving towards left side instead of falling/ending on floor**, Can some body recommend me some better way or similar easy compact way to accomplish this task of applying physics on two ball or more balls with interactivity. here is code ; import java.awt.*; public class AdobeBall { protected int radius = 20; protected Color color; // ... Constants final static int DIAMETER = 40; // ... Instance variables private int m_x; // x and y coordinates upper left private int m_y; private double dx = 3.0; // delta x and y private double dy = 6.0; private double m_velocityX; // Pixels to move each time move() is called. private double m_velocityY; private int m_rightBound; // Maximum permissible x, y values. private int m_bottomBound; public AdobeBall(int x, int y, double velocityX, double velocityY, Color color1) { super(); m_x = x; m_y = y; m_velocityX = velocityX; m_velocityY = velocityY; color = color1; } public double getSpeed() { return Math.sqrt((m_x + m_velocityX - m_x) * (m_x + m_velocityX - m_x) + (m_y + m_velocityY - m_y) * (m_y + m_velocityY - m_y)); } public void setSpeed(double speed) { double currentSpeed = Math.sqrt(dx * dx + dy * dy); dx = dx * speed / currentSpeed; dy = dy * speed / currentSpeed; } public void setDirection(double direction) { m_velocityX = (int) (Math.cos(direction) * getSpeed()); m_velocityY = (int) (Math.sin(direction) * getSpeed()); } public double getDirection() { double h = ((m_x + dx - m_x) * (m_x + dx - m_x)) + ((m_y + dy - m_y) * (m_y + dy - m_y)); double a = (m_x + dx - m_x) / h; return a; } // ======================================================== setBounds public void setBounds(int width, int height) { m_rightBound = width - DIAMETER; m_bottomBound = height - DIAMETER; } // ============================================================== move public void move() { double gravAmount = 0.02; double gravDir = 90; // The direction for the gravity to be in. // ... Move the ball at the give velocity. m_x += m_velocityX; m_y += m_velocityY; // ... Bounce the ball off the walls if necessary. if (m_x < 0) { // If at or beyond left side m_x = 0; // Place against edge and m_velocityX = -m_velocityX; } else if (m_x > m_rightBound) { // If at or beyond right side m_x = m_rightBound; // Place against right edge. m_velocityX = -m_velocityX; } if (m_y < 0) { // if we're at top m_y = 0; m_velocityY = -m_velocityY; } else if (m_y > m_bottomBound) { // if we're at bottom m_y = m_bottomBound; m_velocityY = -m_velocityY; } // double speed = Math.sqrt((m_velocityX * m_velocityX) // + (m_velocityY * m_velocityY)); // ...Friction stuff double fricMax = 0.02; // You can use any number, preferably less than 1 double friction = getSpeed(); if (friction > fricMax) friction = fricMax; if (m_velocityX >= 0) { m_velocityX -= friction; } if (m_velocityX <= 0) { m_velocityX += friction; } if (m_velocityY >= 0) { m_velocityY -= friction; } if (m_velocityY <= 0) { m_velocityY += friction; } // ...Gravity stuff m_velocityX += Math.cos(gravDir) * gravAmount; m_velocityY += Math.sin(gravDir) * gravAmount; } public Color getColor() { return color; } public void setColor(Color newColor) { color = newColor; } // ============================================= getDiameter, getX, getY public int getDiameter() { return DIAMETER; } public double getRadius() { return radius; // radius should be a local variable in Ball. } public int getX() { return m_x; } public int getY() { return m_y; } } using adobeBall: import java.awt.*; import java.awt.event.*; import javax.swing.*; public class AdobeBallImplementation implements Runnable { private static final long serialVersionUID = 1L; private volatile boolean Play; private long mFrameDelay; private JFrame frame; private MyKeyListener pit; /** true means mouse was pressed in ball and still in panel. */ private boolean _canDrag = false; private static final int MAX_BALLS = 50; // max number allowed private int currentNumBalls = 2; // number currently active private AdobeBall[] ball = new AdobeBall[MAX_BALLS]; public AdobeBallImplementation(Color ballColor) { frame = new JFrame("simple gaming loop in java"); frame.setSize(400, 400); frame.setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE); pit = new MyKeyListener(); pit.setPreferredSize(new Dimension(400, 400)); frame.setContentPane(pit); ball[0] = new AdobeBall(34, 150, 7, 2, Color.YELLOW); ball[1] = new AdobeBall(50, 50, 5, 3, Color.BLUE); frame.pack(); frame.setVisible(true); frame.setBackground(Color.white); start(); frame.addMouseListener(pit); frame.addMouseMotionListener(pit); } public void start() { Play = true; Thread t = new Thread(this); t.start(); } public void stop() { Play = false; } public void run() { while (Play == true) { // bounce(ball[0],ball[1]); runball(); pit.repaint(); try { Thread.sleep(mFrameDelay); } catch (InterruptedException ie) { stop(); } } } public void drawworld(Graphics g) { for (int i = 0; i < currentNumBalls; i++) { g.setColor(ball[i].getColor()); g.fillOval(ball[i].getX(), ball[i].getY(), 40, 40); } } public double pointDistance (double x1, double y1, double x2, double y2) { return Math.sqrt((x2 - x1) * (x2 - x1) + (y2 - y1) * (y2 - y1)); } public void runball() { while (Play == true) { try { for (int i = 0; i < currentNumBalls; i++) { for (int j = 0; j < currentNumBalls; j++) { if (pointDistance(ball[i].getX(), ball[i].getY(), ball[j].getX(), ball[j].getY()) < ball[i] .getRadius() + ball[j].getRadius() + 2) { // bounce(ball[i],ball[j]); ball[i].setBounds(pit.getWidth(), pit.getHeight()); ball[i].move(); pit.repaint(); } } } try { Thread.sleep(50); } catch (Exception e) { System.exit(0); } } catch (Exception e) { e.printStackTrace(); } } } public static double pointDirection(int x1, int y1, int x2, int y2) { double H = Math.sqrt((x2 - x1) * (x2 - x1) + (y2 - y1) * (y2 - y1)); // The // hypotenuse double x = x2 - x1; // The opposite double y = y2 - y1; // The adjacent double angle = Math.acos(x / H); angle = angle * 57.2960285258; if (y < 0) { angle = 360 - angle; } return angle; } public static void bounce(AdobeBall b1, AdobeBall b2) { if (b2.getSpeed() == 0 && b1.getSpeed() == 0) { // Both balls are stopped. b1.setDirection(pointDirection(b1.getX(), b1.getY(), b2.getX(), b2 .getY())); b2.setDirection(pointDirection(b2.getX(), b2.getY(), b1.getX(), b1 .getY())); b1.setSpeed(1); b2.setSpeed(1); } else if (b2.getSpeed() == 0 && b1.getSpeed() != 0) { // B1 is moving. B2 is stationary. double angle = pointDirection(b1.getX(), b1.getY(), b2.getX(), b2 .getY()); b2.setSpeed(b1.getSpeed()); b2.setDirection(angle); b1.setDirection(angle - 90); } else if (b1.getSpeed() == 0 && b2.getSpeed() != 0) { // B1 is moving. B2 is stationary. double angle = pointDirection(b2.getX(), b2.getY(), b1.getX(), b1 .getY()); b1.setSpeed(b2.getSpeed()); b1.setDirection(angle); b2.setDirection(angle - 90); } else { // Both balls are moving. AdobeBall tmp = b1; double angle = pointDirection(b2.getX(), b2.getY(), b1.getX(), b1 .getY()); double origangle = b1.getDirection(); b1.setDirection(angle + origangle); angle = pointDirection(tmp.getX(), tmp.getY(), b2.getX(), b2.getY()); origangle = b2.getDirection(); b2.setDirection(angle + origangle); } } public static void main(String[] args) { javax.swing.SwingUtilities.invokeLater(new Runnable() { public void run() { new AdobeBallImplementation(Color.red); } }); } } *EDIT:*ok splitting the code using new approach for gravity from this forum: this code also not working the ball is not coming on floor: public void mymove() { m_x += m_velocityX; m_y += m_velocityY; if (m_y + m_bottomBound > 400) { m_velocityY *= -0.981; // setY(400 - m_bottomBound); m_y = 400 - m_bottomBound; } // ... Bounce the ball off the walls if necessary. if (m_x < 0) { // If at or beyond left side m_x = 0; // Place against edge and m_velocityX = -m_velocityX; } else if (m_x > m_rightBound) { // If at or beyond right side m_x = m_rightBound - 20; // Place against right edge. m_velocityX = -m_velocityX; } if (m_y < 0) { // if we're at top m_y = 1; m_velocityY = -m_velocityY; } else if (m_y > m_bottomBound) { // if we're at bottom m_y = m_bottomBound - 20; m_velocityY = -m_velocityY; } } thanks a lot for any correction and help. jibby

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  • How to fetch managed objects sorted by calculated value

    - by Marcin Zbijowski
    Hello, I'm working on the app that uses CoreData. There is location entity that holds latitude and longitude values. I'd like to fetch those entities sorted by distance to the user's location. I tried to set sort descriptor to distance formula sqrt ((x1 - x2)^2 + (y1 - y2)^2) but it fails with exception "... keypath ... not found in entity". NSString *distanceFormula = [NSString stringWithFormat:@"sqrt(((latitude - %f) * (latitude - %f)) + ((longitude - %f) * (longitude - %f)))", location.coordinate.latitude, location.coordinate.latitude, location.coordinate.longitude, location.coordinate.longitude]; NSSortDescriptor *sortDescriptor = [[NSSortDescriptor alloc] initWithKey:distanceFormula ascending:YES]; [fetchRequest setSortDescriptors:[NSArray arrayWithObject:sortDescriptor]]; NSError *error; NSArray *result = [[self managedObjectContext] executeFetchRequest:fetchRequest error:&error]; I'd like to fetch already sorted objects rather then fetch them all and then sort in the code. Any tips appreciated.

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  • Why the valid looking statement gives error in MATLAB?

    - by user198729
    It's from this question? Why the two solutions doesn't work, though it looks very valid for me: >> t = -pi:0.1:pi; >> r = ((sin(t)*sqrt(cos(t)))*(sin(t) + (7/5))^(-1)) - 2*sin(t) + 2 ; ??? Error using ==> mtimes Inner matrix dimensions must agree. >> t = -pi:0.1:pi; >> r = ((sin(t).*sqrt(cos(t))).*(sin(t) + (7/5)).^(-1)) - 2*sin(t) + 2 ; >> plot(r,t) ??? Error using ==> plot Vectors must be the same lengths. What's wrong with the above?

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  • Please explain how Trial Division works for Primality Test

    - by mister_dani
    I came across this algorithm for testing primality through trial division I fully understand this algorithm static boolean isPrime(int N) { if (N < 2) return false; for (int i = 2; i <= Math.sqrt(N); i++) if (N % i == 0) return false; return true; } It works just fine. But then I came across this other one which works just as good but I do not fully understand the logic behind it. static boolean isPrime(int N) { if (N < 2) return false; for (int i = 2; i * i<N; i++) if (N % i == 0) return false; return true; } It seems like i *i < N behaves like i <= Math.sqrt(N). If so, why?

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