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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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  • Does Python/Scipy have a firls( ) replacement (i.e. a weighted, least squares, FIR filter design)?

    - by delicasso
    I am porting code from Matlab to Python and am having trouble finding a replacement for the firls( ) routine. It is used for, least-squares linear-phase Finite Impulse Response (FIR) filter design. I looked at scipy.signal and nothing there looked like it would do the trick. Of course I was able to replace my remez and freqz algorithsm, so that's good. On one blog I found an algorithm that implemented this filter without weighting, but I need one with weights. Thanks, David

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  • Code computing the cross-product

    - by WizardOfOdds
    This is not a dupe of my question: http://stackoverflow.com/questions/2532810/detecting-one-points-location-compared-to-two-other-points If I have the following piece of pseudo-C/Java/C# code: int a[]= { 30, 20 }; int b[] = { 40, 50 }; int c[] = {12, 12}; How do I compute the sign of the cross-product ABxAC? I'm only interested in the sign, so I have: boolean signABxAC = ? Now concretely what do I write to get the sign of the cross-product ABxAC?

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  • Can my loop be optimized any more? (C++)

    - by Sagekilla
    Below is one of my inner loops that's run several thousand times, with input sizes of 20 - 1000 or more. Is there anything I can do to help squeeze any more performance out of this? I'm not looking to move this code to something like using tree codes (Barnes-Hut), but towards optimizing the actual calculations happening inside, since the same calculations occur in the Barnes-Hut algorithm. Any help is appreciated! typedef double real; struct Particle { Vector pos, vel, acc, jerk; Vector oldPos, oldVel, oldAcc, oldJerk; real mass; }; class Vector { private: real vec[3]; public: // Operators defined here }; real Gravity::interact(Particle *p, size_t numParticles) { PROFILE_FUNC(); real tau_q = 1e300; for (size_t i = 0; i < numParticles; i++) { p[i].jerk = 0; p[i].acc = 0; } for (size_t i = 0; i < numParticles; i++) { for (size_t j = i+1; j < numParticles; j++) { Vector r = p[j].pos - p[i].pos; Vector v = p[j].vel - p[i].vel; real r2 = lengthsq(r); real v2 = lengthsq(v); // Calculate inverse of |r|^3 real r3i = Constants::G * pow(r2, -1.5); // da = r / |r|^3 // dj = (v / |r|^3 - 3 * (r . v) * r / |r|^5 Vector da = r * r3i; Vector dj = (v - r * (3 * dot(r, v) / r2)) * r3i; // Calculate new acceleration and jerk p[i].acc += da * p[j].mass; p[i].jerk += dj * p[j].mass; p[j].acc -= da * p[i].mass; p[j].jerk -= dj * p[i].mass; // Collision estimation // Metric 1) tau = |r|^2 / |a(j) - a(i)| // Metric 2) tau = |r|^4 / |v|^4 real mij = p[i].mass + p[j].mass; real tau_est_q1 = r2 / (lengthsq(da) * mij * mij); real tau_est_q2 = (r2*r2) / (v2*v2); if (tau_est_q1 < tau_q) tau_q = tau_est_q1; if (tau_est_q2 < tau_q) tau_q = tau_est_q2; } } return sqrt(sqrt(tau_q)); }

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  • posmax: like argmax but gives the position(s) of the element x for which f[x] is maximal

    - by dreeves
    Mathematica has a built-in function ArgMax for functions over infinite domains, based on the standard mathematical definition. The analog for finite domains is a handy utility function. Given a function and a list (call it the domain of the function), return the element(s) of the list that maximize the function. Here's an example of finite argmax in action: http://stackoverflow.com/questions/471029/canonicalize-nfl-team-names/472213#472213 And here's my implementation of it (along with argmin for good measure): (* argmax[f, domain] returns the element of domain for which f of that element is maximal -- breaks ties in favor of first occurrence. *) SetAttributes[{argmax, argmin}, HoldFirst]; argmax[f_, dom_List] := Fold[If[f[#1]>=f[#2], #1, #2]&, First[dom], Rest[dom]] argmin[f_, dom_List] := argmax[-f[#]&, dom] First, is that the most efficient way to implement argmax? What if you want the list of all maximal elements instead of just the first one? Second, how about the related function posmax that, instead of returning the maximal element(s), returns the position(s) of the maximal elements?

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  • Is there any sense in performing binary AND with a number where all bits are set to 1

    - by n535
    Greetings everybody. I have seen examples of such operations for so many times that I begin to think that I am getting something wrong with binary arithmetic. Is there any sense to perform the following: byte value = someAnotherByteValue & 0xFF; I don't really understand this, because it does not change anything anyway. Thanks for help. P.S. I was trying to search for information both elsewhere and here, but unsuccessfully. EDIT: Well, off course i assume that someAnotherByteValue is 8 bits long, the problem is that i don't get why so many people ( i mean professionals ) use such things in their code. For example in Jon Skeet's MiscUtil there is: uint s1 = (uint)(initial & 0xffff); where initial is int.

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  • Animation with Initial Velocity

    - by abustin
    I've been trying to solve this problem for a number of days now but I must be missing something. Known Variables: vi = Initial Velocity t = Animation Duration d = Distance The function I'm trying to create: D(t) = the current distance for a given time Using this information I want to be able to create a smooth animation curve with varying velocity (ease-in/ease-out). The animation must be able ease-in from an initial velocity. The animation must be exactly t seconds and must be travel exactly d units. The curve should lean towards the average velocity with acceleration occurring at the beginning and the end portions of the curve. I'm open to extra configuration variables. The best I've been able to come up with is something that doesn't factor in the initial velocity. I'm hoping someone smarter can help me out. ;) Thank you! p.s. I'm working with an ECMAScript variant

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  • How can I transform latitude and longitude to x,y in Java?

    - by hory.incpp
    Hello, I am working on a geographic project in Java. The input are coordinates : 24.4444 N etc Output: a PLAIN map (not round) showing the point of the coordinates. I don't know the algorithm to transform from coordinates to x,y on a JComponent, can somebody help me? The map looks like this: http://upload.wikimedia.org/wikipedia/commons/7/74/Mercator-projection.jpg Thank you

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  • Linear regression confidence intervals in SQL

    - by Matt Howells
    I'm using some fairly straight-forward SQL code to calculate the coefficients of regression (intercept and slope) of some (x,y) data points, using least-squares. This gives me a nice best-fit line through the data. However we would like to be able to see the 95% and 5% confidence intervals for the line of best-fit (the curves below). What these mean is that the true line has 95% probability of being below the upper curve and 95% probability of being above the lower curve. How can I calculate these curves? I have already read wikipedia etc. and done some googling but I haven't found understandable mathematical equations to be able to calculate this. Edit: here is the essence of what I have right now. --sample data create table #lr (x real not null, y real not null) insert into #lr values (0,1) insert into #lr values (4,9) insert into #lr values (2,5) insert into #lr values (3,7) declare @slope real declare @intercept real --calculate slope and intercept select @slope = ((count(*) * sum(x*y)) - (sum(x)*sum(y)))/ ((count(*) * sum(Power(x,2)))-Power(Sum(x),2)), @intercept = avg(y) - ((count(*) * sum(x*y)) - (sum(x)*sum(y)))/ ((count(*) * sum(Power(x,2)))-Power(Sum(x),2)) * avg(x) from #lr Thank you in advance.

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  • Fastest primality test

    - by Grigory Javadyan
    Hi. Could you suggest a fast, deterministic method that is usable in practice, for testing if a large number is prime or not? Also, I would like to know how to use non-deterministic primality tests correctly. For example, if I'm using such a method, I can be sure that a number is not prime if the output is "no", but what about the other case, when the output is "probably"? Do I have to test for primality manually in this case? Thanks in advance.

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  • PHP Code to Generate Simple Flowchart

    - by revbackup
    I am making a flowchart out the subjects in the curriculum of our school. a flowchart is generated through its preRequisite.... for example FIRST YEAR FIRST SEMESTER SUBJECTS ---- PREREQUISITE MATH 1 ---- NONE MATH 2 ---- NONE ENGL 1 ---- NONE SOCIO 1 ----- NONE POLSCI 1 ----- NONE FIRSTE YEAR SECOND SEMESTER SUBJECTS ---- PREREQUISITE MATH 3 ----- MATH 1 MATH 4 ----- MATH 2, MATH 1 ENGL 2 ----- ENGL 1 POLSCI 2 ----- POLSCI 1 So, I must print it this way, just using simple PHP but difficult Logic.: MATH1 -----> MATH3 -----> MATH4 MATH 2 ----->MATH 4 ENGL1 -----> ENGL 2 SOCIO 1 POLSCI 1 -----> POLSCI 2 Can anyone give me a good algorithm for this, because this is really difficult. I am planning to echo the results in an HTML table, and it makes it more complicated. Do you have suggestions how to solve this problem properly and display the results also properly???? Thank you in advance!

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  • Mapping Hilbert values to 3D points

    - by Alexander Gladysh
    I have a set of Hilbert values (length from the start of the Hilbert curve to the given point). What is the best way to convert these values to 3D points? Original Hilbert curve was not in 3D, so I guess I have to pick by myself the Hilbert curve rank I need. I do have total curve length though (that is, the maximum value in the set). Perhaps there is an existing implementation? Some library that would allow me to work with Hilbert curve / values? Language does not matter much.

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  • calculate intersection between two segments in a symmetric way

    - by Elazar Leibovich
    When using the usual formulas to calculate intersection between two 2D segments, ie here, if you round the result to an integer, you get non-symmetric results. That is, sometimes, due to rounding errors, I get that intersection(A,B)!=intersection(B,A). The best solution is to keep working with floats, and compare the results up to a certain precision. However, I must round the results to integers after calculating the intersection, I cannot keep working with floats. My best solution so far was to use some full order on the segments in the plane, and have intersection to always compare the smaller segment to the larger segment. Is there a better method? Am I missing something?

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  • mod,prime -> inverse possible

    - by Piet
    Hi all. I was wondering if one can do the following: We have: X is a product of N-primes, thus I assume unique. C is a constant. We can assure that C is a number that is part of the N-primes or not. Whichever will work best. Thus: X mod C = Z We have Z and C and we know that X was a product of N-primes, where N is restricted lets say first 100 primes. Is there anyway we can get back X?

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  • Fibonacci sequence subroutine returning one digit too high...PERL

    - by beProactive
    #!/usr/bin/perl -w use strict; sub fib { my($num) = @_; #give $num to input array return(1) if ($num<=1); #termination condition return($num = &fib($num-1) + &fib($num-2)); #should return sum of first "n" terms in the fibonacci sequence } print &fib(7)."\n"; #should output 20 This subroutine should be outputting a summation of the first "x" amount of terms, as specified by the argument to the sub. However, it's one too high. Does this have something to do with the recursion? Thanks.

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  • BigInteger.pow(BigInteger) ?

    - by PeterW
    I'm playing with numbers in Java, and want to see how big a number I can make. It is my understanding that BigInteger can hold a number of infinite size, so long as my computer has enough Memory to hold such a number, correct? My problem is that BigInteger.pow accepts only an int, not another BigInteger, which means I can only use a number up to 2,147,483,647 as the exponent. Is it possible to use the BigInteger class as such? BigInteger.pow(BigInteger) Thanks.

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  • Divide and conquer method to compute roots [SOLVED]

    - by hellsoul153
    Hello, Knowing that we can use Divide-and-Conquer algorithm to compute large exponents, for exemple 2 exp 100 = 2 exp(50) * 2 exp(50), which is quite more efficient, is this method efficient using roots ? For exemple 2 exp (1/100) = (2 exp(1/50)) exp(1/50) ? In other words, I'm wondering if (n exp(1/x)) is more efficient to (n exp(1/y)) for x < y and where x and y are integers.

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  • Runge-Kutta Method with adaptive step

    - by infoholic_anonymous
    I am implementing Runge-Kutta method with adaptive step in matlab. I get different results as compared to matlab's own ode45 and my own implementation of Runge-Kutta method with fixed step. What am I doing wrong in my code? Is it possible? function [ result ] = rk4_modh( f, int, init, h, h_min ) % % f - function handle % int - interval - pair (x_min, x_max) % init - initial conditions - pair (y1(0),y2(0)) % h_min - lower limit for h (step length) % h - initial step length % x - independent variable ( for example time ) % y - dependent variable - vertical vector - in our case ( y1, y2 ) function [ k1, k2, k3, k4, ka, y ] = iteration( f, h, x, y ) % core functionality performed within loop k1 = h * f(x,y); k2 = h * f(x+h/2, y+k1/2); k3 = h * f(x+h/2, y+k2/2); k4 = h * f(x+h, y+k3); ka = (k1 + 2*k2 + 2*k3 + k4)/6; y = y + ka; end % constants % relative error eW = 1e-10; % absolute error eB = 1e-10; s = 0.9; b = 5; % initialization i = 1; x = int(1); y = init; while true hy = y; hx = x; %algorithm [ k1, k2, k3, k4, ka, y ] = iteration( f, h, x, y ); % error estimation for j=1:2 [ hk1, hk2, hk3, hk4, hka, hy ] = iteration( f, h/2, hx, hy ); hx = hx + h/2; end err(:,i) = abs(hy - y); % step adjustment e = abs( hy ) * eW + eB; a = min( e ./ err(:,i) )^(0.2); mul = a * s; if mul >= 1 % step length admitted keepH(i) = h; k(:,:,i) = [ k1, k2, k3, k4, ka ]; previous(i,:) = [ x+h, y' ]; %' i = i + 1; if floor( x + h + eB ) == int(2) break; else h = min( [mul*h, b*h, int(2)-x] ); x = x + keepH(i-1); end else % step length requires further adjustments h = mul * h; if ( h < h_min ) error('Computation with given precision impossible'); end end end result = struct( 'val', previous, 'k', k, 'err', err, 'h', keepH ); end The function in question is: function [ res ] = fun( x, y ) % res(1) = y(2) + y(1) * ( 0.9 - y(1)^2 - y(2)^2 ); res(2) = -y(1) + y(2) * ( 0.9 - y(1)^2 - y(2)^2 ); res = res'; %' end The call is: res = rk4( @fun, [0,20], [0.001; 0.001], 0.008 ); The resulting plot for x1 : The result of ode45( @fun, [0, 20], [0.001, 0.001] ) is:

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  • Counting problem: possible sudoko tables?

    - by Sorush Rabiee
    Hi, I'm working on a sudoko solver. my method is using a game tree and explore possible permutations for each set of digits by DFS Algorithm. in order to analyzing problem, i want to know what is the count of possible valid and invalid sudoko tables? - a 9*9 table that have 9 one, 9 two, ... , 9 nine. (this isn't exact duplicate by this question) my solution is: 1- First select 9 cells for 1s: (*) 2- and like (1) for other digits (each time, 9 cells will be deleted from remaining available cells): C(81-9,9) , C(81-9*2,9) .... = 3- finally multiply the result by 9! (permutation of 123456789 in (*)) this is not equal to accepted answer of this question but problems are equivalent. what did i do wrong?

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