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  • Vector deltas and moving in unknown areas

    - by dekz
    Hi All, I was in need of a little math help that I can't seem to find the answer to, any links to documentation would be greatly appreciated. Heres my situation, I have no idea where I am in this maze, but I need to move around and find my way back to the start. I was thinking of implementing a waypoint list of places i've been offset from my start at 0,0. This is a 2D cartesian plane. I've been given 2 properties, my translation speed from 0-1 and my rotation speed from -1 to 1. -1 is very left and +1 is very right. These are speed and not angles so thats where my problem lies. If I'm given 0 as a translation speed and 0.2 I will continually turn to my right at a slow speed. How do I figure out the offsets given these 2 variables? I can store it every time I take a 'step'. I just need to figure out the offsets in x and y terms given the translations and rotation speeds. And the rotation to get to those points. Any help is appreciated.

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  • Finding the digit root of a number

    - by Jessica M.
    Study question is to find the digit root of a already provided number. The teacher provides us with the number 2638. In order to find the digit root you have to add each digit separately 2 + 6 + 3 + 8 = 19. Then you take the result 19 and add those two digits together 1 + 9 = 10. Do the same thing again 1 + 0 = 1. The digit root is 1. My first step was to use the variable total to add up the number 2638 to find the total of 19. Then I tried to use the second while loop to separate the two digits by using the % I have to try and solve the problem by using basic integer arithmetic (+, -, *, /). 1.Is it necessary and or possible to solve the problem using nested while loops? 2.Is my math correct? 3. As I wrote it here it does not run in Eclipse. Am I using the while loops correctly? import acm.program.*; public class Ch4Q7 extends ConsoleProgram { public void run(){ println("This program attempts to find the digit root of your number: "); int n = readInt("Please enter your number: "); int total = 0; int root = total; while (n > 0 ){ total = total + (n %10); n = (n / 10); } while ( total > 0 ){ root = total; total = ((total % 10) + total / 10); } println("your root should be " + root); } }

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  • Find if a user is facing a certain Location using Digital Compass by constructing a triangle and usi

    - by Aidan
    Hi Guys, I'm constructing a geolocation based application and I'm trying to figure out a way to make my application realise when a user is facing the direction of the given location (a particular long / lat co-ord). I've done some Googling and checked the SDK but can't really find anything for such a thing. Does anyone know of a way? To clarify Android knows my location, the second location and my orientation. What I want is a way for Android to recognise when my orientation is "facing" the second location (e.g within 90 Degrees or so). We're also assuming that the user is stationary and needs updates every second or so therefore getBearing(); is useless. Alright so we get it has to be math, there appears to be no simple SDK stuff we can use. I did some searching of my own and found Barycentric Co-ords http://www.blackpawn.com/texts/pointinpoly/default.html . So what I'm trying to do now is map the camera's field of view. For Example if the person is facing a certain direction the program should construct a triangle around that field of view, e.g it should make one vertices the phone's position and then go out either side for a set distance making the 2 end points vertices constructing a triangle. If I had this I could then apply Barycentric co-ords to see if the point lay within the newly constructed triangle. Idea's anyone? Example. I could get my current orientation, add 45 to it and go up X distance one side and subtact 45 and go up X distance the other side to find my 2 other points. Though, how would I make android know which way it should go "up" I guess? It would know its baring as this stage so I need it to recognise it's bearing and go out that direction.

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  • How to parse mathematical expressions involving parentheses

    - by Rob P.
    Please forgive my title, I really don't know how to phrase it better. This isn't a school assignment or anything, but I realize it's a mostly academic question. But, what I've been struggling to do is parse 'math' text and come up with an answer. For Example - I can figure out how to parse '5 + 5' or '3 * 5' - but I fail when I try to correctly chain operations together. (5 + 5) * 3 It's mostly just bugging me that I can't figure it out. If anyone can point me in a direction, I'd really appreciate it. EDIT Thanks for all of the quick responses. I'm sorry I didn't do a better job of explaining. First - I'm not using regular expressions. I also know there are already libraries available that will take, as a string, a mathematical expression and return the correct value. So, I'm mostly looking at this because, sadly, I don't "get it". Second - What I've tried doing (is probably misguided) but I was counting '(' and ')' and evaluating the deepest items first. In simple examples, this worked; but my code is not pretty and more complicated stuff crashes. When I 'calculated' the lowest level, I was modifying the string. So... (5 + 5) * 3 Would turn into 10 * 3 Which would then evaluate to 30 But it just felt 'wrong'. I hope that helps clarify things. I'll certainly check out the links provided.

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  • Calculating distance between two X,Y coordinates

    - by Umopepisdn
    I am writing a tool for a game that involves calculating the distance between two coordinates on a spherical plane 500 units across. That is, [0,0] through [499,499] are valid coordinates, and [0,0] and [499,499] are also right next to each other. Currently, in my application, I am comparing the distance between a city with an [X,Y] location respective to the user's own [X,Y] location, which they have configured in advance. To do this, I found this algorithm, which kind of works: Math.sqrt ( dx * dx + dy * dy ); Because sorting a paged list by distance is a useful thing to be able to do, I implemented this algorithm in a MySQL query and have made it available to my application using the following part of my SELECT statement: SQRT( POW( ( ".strval($sourceX)." - cityX ) , 2 ) + POW( ( ".strval($sourceY)." - cityY ) , 2 ) ) AS distance This works fine for many calculations, but does not take into account the fact that [0,0] and [499,499] are kitty-corner to one another. Is there any way I can tweak this algorithm to generate an accurate distance, given that 0 and 499 are adjacent? Thanks, -Umo

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  • Interview question : What is the fastest way to generate prime number recursively ?

    - by hilal
    Generation of prime number is simple but what is the fastest way to find it and generate( prime numbers) it recursively ? Here is my solution. However, it is not the best way. I think it is O(N*sqrt(N)). Please correct me, if I am wrong. public static boolean isPrime(int n) { if (n < 2) { return false; } else if (n % 2 == 0 & n != 2) { return false; } else { return isPrime(n, (int) Math.sqrt(n)); } } private static boolean isPrime(int n, int i) { if (i < 2) { return true; } else if (n % i == 0) { return false; } else { return isPrime(n, --i); } } public static void generatePrimes(int n){ if(n < 2) { return ; } else if(isPrime(n)) { System.out.println(n); } generatePrimes(--n); } public static void main(String[] args) { generatePrimes(200); }

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  • how to reverse an angle

    - by MissHalberd
    I am no mathematician, but I somehow got into game development as a hobby. Having never studied anything beyond basic math, I have a lot of trouble figuring out how to reverse the angle of something, facing to the opposite direction, among the X axis. One image says more than 1000 words though (specially uneducated words): http://img156.imageshack.us/i/wihwin.png/ I basically want to reverse the direction of cannon objects adhered to a robot. When the robot changes from facing right to facing left, I do (180 - angle) as everyone suggested me, but it reverses the angle...literally, making the cannons aim up when they are aiming down. So, I need to do something else, but it escapes my knowledge. Anyone would be so kind to help me with this? Oh, I use regular C by the way, in case there's something built-in specific to it. To put it in other words, I work in 2D, so I want an angle that is facing right to face left. 0 being "totally to the right", 180 "left", 90 "up" and 270 "down". I want something that is aiming with an angle of 91 to turn into 89 when reversed, literally. There's no Z axis present. EDIT: Thanks for the answers! Trying them out now. I'll post which one worked in a minute!

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  • Beginner Question ; About Prime Generation in "C" - What is wrong with my code ? -

    - by alorsoncode
    I'm a third year irregular CS student and ,i just realized that i have to start coding. I passed my coding classes with lower bound grades so that i haven't a good background in coding&programming. I'm trying to write a code that generates prime numbers between given upper and lower bounds. Not knowing C well, enforce me to write a rough code then go over it to solve. I can easily set up the logic for intended function but i probably create a wrong algorithm through several different ways. Here I share my last code, i intend to calculate that when a number gives remainder Zero , it should be it self and 1 , so that count==2; What is wrong with my implementation and with my solution generating style? I hope you will warm me up to programming world, i couldn't find enough motivation and courage to get deep into programming. Thanks in Advance :) Stdio and Math.h is Included int primegen(int down,int up) { int divisor,candidate,count=0,k; for(candidate=down;candidate<=up;candidate++) { for(divisor=1;divisor<=candidate;divisor++) { k=(candidate%divisor); } if (k==0) count++; if(count==2) { printf("%d\n", candidate); count=0; } else { continue; } } } int main() { primegen(3,15); return 0; }

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  • Value types of variable size

    - by YellPika
    I'm trying to code a small math library in C#. I wanted to create a generic vector structure where the user could define the element type (int, long, float, double, etc.) and dimensions. My first attempt was something like this... public struct Vector<T> { public readonly int Dimensions; public readonly T[] Elements; // etc... } Unfortunately, Elements, being an array, is also a reference type. Thus, doing this, Vector<int> a = ...; Vector<int> b = a; a[0] = 1; b[0] = 2; would result in both a[0] and b[0] equaling 2. My second attempt was to define an interface IVector<T>, and then use Reflection.Emit to automatically generate the appropriate type at runtime. The resulting classes would look roughly like this: public struct Int32Vector3 : IVector<T> { public int Element0; public int Element1; public int Element2; public int Dimensions { get { return 3; } } // etc... } This seemed fine until I found out that interfaces seem to act like references to the underlying object. If I passed an IVector to a function, and changes to the elements in the function would be reflected in the original vector. What I think is my problem here is that I need to be able to create classes that have a user specified number of fields. I can't use arrays, and I can't use inheritance. Does anyone have a solution? EDIT: This library is going to be used in performance critical situations, so reference types are not an option.

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  • Is there an algorithm for converting quaternion rotations to Euler angle rotations?

    - by Will Baker
    Is there an existing algorithm for converting a quaternion representation of a rotation to an Euler angle representation? The rotation order for the Euler representation is known and can be any of the six permutations (i.e. xyz, xzy, yxz, yzx, zxy, zyx). I've seen algorithms for a fixed rotation order (usually the NASA heading, bank, roll convention) but not for arbitrary rotation order. Furthermore, because there are multiple Euler angle representations of a single orientation, this result is going to be ambiguous. This is acceptable (because the orientation is still valid, it just may not be the one the user is expecting to see), however it would be even better if there was an algorithm which took rotation limits (i.e. the number of degrees of freedom and the limits on each degree of freedom) into account and yielded the 'most sensible' Euler representation given those constraints. I have a feeling this problem (or something similar) may exist in the IK or rigid body dynamics domains. Solved: I just realised that it might not be clear that I solved this problem by following Ken Shoemake's algorithms from Graphics Gems. I did answer my own question at the time, but it occurs to me it may not be clear that I did so. See the answer, below, for more detail. Just to clarify - I know how to convert from a quaternion to the so-called 'Tait-Bryan' representation - what I was calling the 'NASA' convention. This is a rotation order (assuming the convention that the 'Z' axis is up) of zxy. I need an algorithm for all rotation orders. Possibly the solution, then, is to take the zxy order conversion and derive from it five other conversions for the other rotation orders. I guess I was hoping there was a more 'overarching' solution. In any case, I am surprised that I haven't been able to find existing solutions out there. In addition, and this perhaps should be a separate question altogether, any conversion (assuming a known rotation order, of course) is going to select one Euler representation, but there are in fact many. For example, given a rotation order of yxz, the two representations (0,0,180) and (180,180,0) are equivalent (and would yield the same quaternion). Is there a way to constrain the solution using limits on the degrees of freedom? Like you do in IK and rigid body dynamics? i.e. in the example above if there were only one degree of freedom about the Z axis then the second representation can be disregarded. I have tracked down one paper which could be an algorithm in this pdf but I must confess I find the logic and math a little hard to follow. Surely there are other solutions out there? Is arbitrary rotation order really so rare? Surely every major 3D package that allows skeletal animation together with quaternion interpolation (i.e. Maya, Max, Blender, etc) must have solved exactly this problem?

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  • difference equations in MATLAB - why the need to switch signs?

    - by jefflovejapan
    Perhaps this is more of a math question than a MATLAB one, not really sure. I'm using MATLAB to compute an economic model - the New Hybrid ISLM model - and there's a confusing step where the author switches the sign of the solution. First, the author declares symbolic variables and sets up a system of difference equations. Note that the suffixes "a" and "2t" both mean "time t+1", "2a" means "time t+2" and "t" means "time t": %% --------------------------[2] MODEL proc-----------------------------%% % Define endogenous vars ('a' denotes t+1 values) syms y2a pi2a ya pia va y2t pi2t yt pit vt ; % Monetary policy rule ia = q1*ya+q2*pia; % ia = q1*(ya-yt)+q2*pia; %%option speed limit policy % Model equations IS = rho*y2a+(1-rho)yt-sigma(ia-pi2a)-ya; AS = beta*pi2a+(1-beta)*pit+alpha*ya-pia+va; dum1 = ya-y2t; dum2 = pia-pi2t; MPs = phi*vt-va; optcon = [IS ; AS ; dum1 ; dum2; MPs]; He then computes the matrix A: %% ------------------ [3] Linearization proc ------------------------%% % Differentiation xx = [y2a pi2a ya pia va y2t pi2t yt pit vt] ; % define vars jopt = jacobian(optcon,xx); % Define Linear Coefficients coef = eval(jopt); B = [ -coef(:,1:5) ] ; C = [ coef(:,6:10) ] ; % B[c(t+1) l(t+1) k(t+1) z(t+1)] = C[c(t) l(t) k(t) z(t)] A = inv(C)*B ; %(Linearized reduced form ) As far as I understand, this A is the solution to the system. It's the matrix that turns time t+1 and t+2 variables into t and t+1 variables (it's a forward-looking model). My question is essentially why is it necessary to reverse the signs of all the partial derivatives in B in order to get this solution? I'm talking about this step: B = [ -coef(:,1:5) ] ; Reversing the sign here obviously reverses the sign of every component of A, but I don't have a clear understanding of why it's necessary. My apologies if the question is unclear or if this isn't the best place to ask.

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  • JavaScript + Maths: Image zoom with CSS3 Transforms, How to set Origin? (with example)

    - by Sunday Ironfoot
    My Math skills really suck! I'm trying to implement an image zoom effect, a bit like how the Zoom works with Google Maps, but with a grid of fix position images. I've uploaded an example of what I have so far here: http://www.dominicpettifer.co.uk/Files/MosaicZoom.html (uses CSS3 transforms so only works with Firefox, Opera, Chrome or Safari) Use your mouse wheel to zoom in/out. The HTML source is basically an outer div with an inner-div, and that inner-div contains 16 images arranged using absolute position. It's going to be a Photo Mosaic basically. I've got the zoom bit working using CSS3 transforms: $(this).find('div').css('-moz-transform', 'scale(' + scale + ')'); ...however, I'm relying on the mouse X/Y position on the outer div to zoom in on where the mouse cursor is, similar to how Google Maps functions. The problem is that if you zoom right in on an image, move the cursor to the bottom/left corner and zoom again, instead of zooming to the bottom/left corner of the image, it zooms to the bottom/left of the entire mosaic. This has the effect of appearing to jump about the mosaic as you zoom in closer while moving the mouse around, even slightly. That's basically the problem, I want the zoom to work exactly like Google Maps where it zooms exactly to where your mouse cursor position is, but I can't get my head around the Maths to calculate the transform-origin: X/Y values correctly. Please help, been stuck on this for 3 days now. Here is the full code listing for the mouse wheel event: var scale = 1; $("#mosaicContainer").mousewheel(function(e, delta) { if (delta > 0) { scale += 1; } else { scale -= 1; } scale = scale < 1 ? 1 : (scale > 40 ? 40 : scale); var x = e.pageX - $(this).offset().left; var y = e.pageY - $(this).offset().top; $(this).find('div').css('-moz-transform', 'scale(' + scale + ')') .css('-moz-transform-origin', x + 'px ' + y + 'px'); return false; });

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  • Collision point of 2 curves in a 3d-room

    - by Frank
    Hello, i am programming a small game for quite some time. We started coding a small FPS-Shooter inside of a project at school to get a bit experience using directX. I dont know why, but i couldnt stop the project and started programming at home aswell. At the moment i am trying to create some small AI. Of cause thats definatlly not easy, but thats my personal goal anyways. The topic could prolly fill multiple books hehe. I've got the walking part of my bots done so far. They walk along a scriped path. I am not working on the "aiming" of the bots. While programming that i hit on some math problem i couldnt solve yet. I hope of your input on this to help me get further. Concepts, ideas and everything else are highly appreciated. Problem: Calculate the position (D3DXVECTOR3) where the curve of the projectile (depends on gravity, speed), hit the curved of the enemys walking path (depends on speed). We assume that the enemy walks in a constant line. Known variables: float projectilSpeed = 2000 m/s //speed of the projectile per second float gravitation = 9.81 m/s^2 //of cause the gravity lol D3DXVECTOR3 targetPosition //position of the target stored in a vector (x,y,z) D3DXVECTOR3 projectilePosition //position of the projectile D3DXVECTOR3 targetSpeed //stores the change of the targets position in the last second Variabledefinition ProjectilePosition at time of collision = ProjectilePos_t TargetPosition at time of collision = TargetPos_t ProjectilePosition at time 0, now = ProjectilePos_0 TargetPosition at time 0, now = TargetPos_0 Time to impact = t Aim-angle = theta My try: Found a formular to calculate "drop" (Drop of the projectile based on the gravity) on Wikipedia: float drop = 0.5f * gravity * t * t The speed of the projectile has a horizontal and a vertical part.. Found a formular for that on wikipedia aswell: ProjectilVelocity.x = projectilSpeed * cos(theta) ProjectilVelocity.y = projectilSpeed * sin(theta) So i would assume this is true for the projectile curve: ProjectilePos_t.x = ProjectilePos_0.x + ProjectileSpeed * t ProjectilePos_t.y = ProjectilePos_0.y + ProjectileSpeed * t + 0.5f * gravity * t * t ProjectilePos_t.z = ProjectilePos_0.z + ProjectileSpeed * t The target walk with a constant speed, so we can determine his curve by this: TargetPos_t = TargetPos_0 + TargetSpeed * D3DXVECTOR3(t, t, t) Now i dont know how to continue. I have to solve it somehow to get a hold on the time to impact somehow. As a basic formular i could use: float time = distanz / projectileSpeed But that wouldnt be truly correct as it would assume a linear "Trajectory". We just find this behaivor when using a rocket. I hope i was able to explain the problem as much as possible. If there are questions left, feel free to ask me! Greets from germany, Frank

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  • Checking for Repeated Strings in 2d list

    - by Zach Santiago
    i have a program where i have a list of names and classes. i have the list in alphabetical order. now im trying to check if names repeat, add the classes to one single name. im trying to write some code like go through names if name is already in list, add the class to the one name. so an example would be, instead of having 'Anita ','phys 1443', and 'Anita','IE 3312' i would just have 'Anita','PHYS 1443','IE 3312'. How would i go about doing this in a logival way, WITHOUT using any sort of built in functions? i tried comparing indexe's like if list[i][0] == list[i+1][0], append list[i+1][1] to an emptylist. while that almost worked, it would screw up at some points along the way. here is my attempt size = len(c) i = 0 c = [['Anita', 'PHYS 1443'], ['Anita', 'IE 3312'], ['Beihuang', 'PHYS 1443'], ['Chiao-Lin', 'MATH 1426'], ['Chiao-Lin', 'IE 3312'], ['Christopher', 'CSE 1310'], ['Dylan', 'CSE 1320'], ['Edmund', 'PHYS 1443'], ['Ian', 'IE 3301'], ['Ian', 'CSE 1320'], ['Ian', 'PHYS 1443'], ['Isis', 'PHYS 1443'], ['Jonathan', 'MATH 2325'], ['Krishna', 'MATH 2325'], ['Michael', 'IE 3301'], ['Nang', 'MATH 2325'], ['Ram', 'CSE 1320'], ['Taesu', 'CSE 1320'], ["Tre'Shaun", 'IE 3312'], ["Tre'Shaun", 'MATH 2325'], ["Tre'Shaun", 'CSE 1310']] ## Check if any names repeat d.append(c[0][0]) while i < size - 1 : if c[i][0] == c[i+1][0] : d.append(c[i][1]) d.append(c[i+1][1]) else : d.append(c[i+1][0]) d.append(c[i+1][1]) i = i + 1 print d output was. ['Anita', 'PHYS 1443', 'IE 3312', 'Beihuang', 'PHYS 1443', 'Chiao-Lin', 'MATH 1426', 'MATH 1426', 'IE 3312', 'Christopher', 'CSE 1310', 'Dylan', 'CSE 1320', 'Edmund', 'PHYS 1443', 'Ian', 'IE 3301', 'IE 3301', 'CSE 1320', 'CSE 1320', 'PHYS 1443', 'Isis', 'PHYS 1443', 'Jonathan', 'MATH 2325', 'Krishna', 'MATH 2325', 'Michael', 'IE 3301', 'Nang', 'MATH 2325', 'Ram', 'CSE 1320', 'Taesu', 'CSE 1320', "Tre'Shaun", 'IE 3312', 'IE 3312', 'MATH 2325', 'MATH 2325', 'CSE 1310']

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  • mapping rect in small image to larger image (in order to do a copyPixels operation)

    - by skinnyTOD
    Hi all - this is (I think) a relatively simple math question but I've spent a day banging my head against it and have only the dents and no solution... I'm coding in actionscript 3 - the functionality is: large image loaded at runtime. The bitmapData is stored and a smaller version is created to display on the available screen area (I may end up just scaling the large image since it is in memory anyway). The user can create a rectangle hotspot on the smaller image (the functionality will be more complex: multiple rects with transparency: example a donut shape with hole, etc) 3 When the user clicks on the hotspot, the rect of the hotspot is mapped to the larger image and a new bitmap "callout" is created, using the larger bitmap data. The reason for this is so the "callout" will be better quality than just scaling up the area of the hotspot. The image below shows where I am at so far- the blue rect is the clicked hotspot. In the upper left is the "callout" - copied from the larger image. I have the aspect ratio right but I am not mapping to the larger image correctly. Ugly code below... Sorry this post is so long - I just figured I ought to provide as much info as possible. Thanks for any tips! --trace of my data values *source BitmapDada 1152 864 scaled to rect 800 600 scaled BitmapData 800 600 selection BitmapData 58 56 scaled selection 83 80 ratio 1.44 before (x=544, y=237, w=58, h=56) (x=544, y=237, w=225.04, h=217.28) * Image here: http://i795.photobucket.com/albums/yy237/skinnyTOD/exampleST.jpg public function onExpandCallout(event:MouseEvent):void{ if (maskBitmapData.getPixel32(event.localX, event.localY) != 0){ var maskClone:BitmapData = maskBitmapData.clone(); //amount to scale callout - this will vary/can be changed by user var scale:Number =150 //scale percentage var normalizedScale :Number = scale/=100; var w:Number = maskBitmapData.width*normalizedScale; var h:Number = maskBitmapData.height*normalizedScale; var ratio:Number = (sourceBD.width /targetRect.width); //creat bmpd of the scaled size to copy source into var scaledBitmapData:BitmapData = new BitmapData(maskBitmapData.width * ratio, maskBitmapData.height * ratio, true, 0xFFFFFFFF); trace("source BitmapDada " + sourceBD.width, sourceBD.height); trace("scaled to rect " + targetRect.width, targetRect.height); trace("scaled BitmapData", bkgnImageSprite.width, bkgnImageSprite.height); trace("selection BitmapData", maskBitmapData.width, maskBitmapData.height); trace("scaled selection", scaledBitmapData.width, scaledBitmapData.height); trace("ratio", ratio); var scaledBitmap:Bitmap = new Bitmap(scaledBitmapData); var scaleW:Number = sourceBD.width / scaledBitmapData.width; var scaleH:Number = sourceBD.height / scaledBitmapData.height; var scaleMatrix:Matrix = new Matrix(); scaleMatrix.scale(ratio,ratio); var sRect:Rectangle = maskSprite.getBounds(bkgnImageSprite); var sR:Rectangle = sRect.clone(); var ss:Sprite = new Sprite(); ss.graphics.lineStyle(8, 0x0000FF); //ss.graphics.beginFill(0x000000, 1); ss.graphics.drawRect(sRect.x, sRect.y, sRect.width, sRect.height); //ss.graphics.endFill(); this.addChild(ss); trace("before " + sRect); w = uint(sRect.width * scaleW); h = uint(sRect.height * scaleH); sRect.inflate(maskBitmapData.width * ratio, maskBitmapData.height * ratio); sRect.offset(maskBitmapData.width * ratio, maskBitmapData.height * ratio); trace(sRect); scaledBitmapData.copyPixels(sourceBD, sRect, new Point()); addChild(scaledBitmap); scaledBitmap.x = offsetPt.x; scaledBitmap.y = offsetPt.y; } }

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  • New grad; To overcome complete lack of experience, should I ditch a creative pet project in lieu of one that would demonstrate more applicable skills?

    - by Hart Simha
    I am currently working on a project on github that I think would be a good demonstration of my initiative, creativity and enthusiasm. It is an educational game I am developing in pygame that enables the user to learn to improve their development productivity by using vim, specifically with python, though learning to code faster with vim should be transferable to any language. I think this is something that might have a mass appeal and benefit to a lot of people in a measurable way. -However- I am graduating from college in a month (my degree is computer science with a minor in English), with no experience that is relevant to helping me get any kind of job in the field, and a gpa that doesn't tout my merits. I could pursue a career in game development, but it's not necessarily what I'm most interested in, and see myself applying to startups around the country. To the places I am looking at applying, showing that I have experience with pygame is going to be largely irrelevant, except in demonstration of my ability to code, period. A lot of skills that ARE more marketable, such a data modeling, GIS, mobile application, development, javascript, .net framework, and various web development technologies, are not going to be showcased by this project (on the upside, employers do like to see familiarity with git and python). I'm wondering if I should sink all my free time in the next couple of months into this project, since I'm motivated and interested in it, and if the value of being able to demonstrate ambition and 'good ideas' (for lack of a better term, and in my own opinion) will compensate for the absence of demonstrating more sought-after skills. I am probably at a point where I should either commit fully to this project now, or put it on the backburner in favor of something else, and I am leaning towards continuing with what I am already working on, because I think it's a great idea, and something achievable to me with enough dedication over the next couple months. But the most important thing to me is being able to get a job out of college, which I am exceedingly concerned about as the professional landscape which I am navigating for the first time is a lot more intimidating than I could have anticipated, with almost every job (even short-term contract positions) requiring years of experience which I lack. So in brief, the common denominator to answering the question "How can I overcome experience requirements for a job" seems to be "Show off your own project." I want to know WHICH project I should work on to best increase my chances of getting a job out of college, keeping in mind that I have no experience. I believe this question is applicable to any new grad that lacks demonstrable experience.

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  • Have you been stuck with the math in a Flash project?

    - by VideoDnd
    Have you been stuck with the math in a Flash project? It's a loose design pattern my director formulated. My goal is to keep the project object oriented, and get 'non Flash obstacles' off my plate. XML values going to AS3, updating a changing acceleration formula. I don't hate math, but it just doesn't seem OOP or good project planning to have the math stuck in Flash. Your comments are welcome.

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  • Stuck with the math in a Flash project, would parsing engine help?

    - by VideoDnd
    I've been stuck with the math in a Flash project? It's a loose design pattern my director formulated. My goal is to keep the project object oriented, and get 'non Flash obstacles' off my plate. Do you recommend using parsing engines for processing math? XML values going to AS3, updating a changing acceleration formula. I don't hate math, but it just doesn't seem OOP or good project planning to have the math stuck in Flash. Your comments are welcome.

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  • Logic error for Gauss elimination

    - by iwanttoprogram
    Logic error problem with the Gaussian Elimination code...This code was from my Numerical Methods text in 1990's. The code is typed in from the book- not producing correct output... Sample Run: SOLUTION OF SIMULTANEOUS LINEAR EQUATIONS USING GAUSSIAN ELIMINATION This program uses Gaussian Elimination to solve the system Ax = B, where A is the matrix of known coefficients, B is the vector of known constants and x is the column matrix of the unknowns. Number of equations: 3 Enter elements of matrix [A] A(1,1) = 0 A(1,2) = -6 A(1,3) = 9 A(2,1) = 7 A(2,2) = 0 A(2,3) = -5 A(3,1) = 5 A(3,2) = -8 A(3,3) = 6 Enter elements of [b] vector B(1) = -3 B(2) = 3 B(3) = -4 SOLUTION OF SIMULTANEOUS LINEAR EQUATIONS The solution is x(1) = 0.000000 x(2) = -1.#IND00 x(3) = -1.#IND00 Determinant = -1.#IND00 Press any key to continue . . . The code as copied from the text... //Modified Code from C Numerical Methods Text- June 2009 #include <stdio.h> #include <math.h> #define MAXSIZE 20 //function prototype int gauss (double a[][MAXSIZE], double b[], int n, double *det); int main(void) { double a[MAXSIZE][MAXSIZE], b[MAXSIZE], det; int i, j, n, retval; printf("\n \t SOLUTION OF SIMULTANEOUS LINEAR EQUATIONS"); printf("\n \t USING GAUSSIAN ELIMINATION \n"); printf("\n This program uses Gaussian Elimination to solve the"); printf("\n system Ax = B, where A is the matrix of known"); printf("\n coefficients, B is the vector of known constants"); printf("\n and x is the column matrix of the unknowns."); //get number of equations n = 0; while(n <= 0 || n > MAXSIZE) { printf("\n Number of equations: "); scanf ("%d", &n); } //read matrix A printf("\n Enter elements of matrix [A]\n"); for (i = 0; i < n; i++) for (j = 0; j < n; j++) { printf(" A(%d,%d) = ", i + 1, j + 1); scanf("%lf", &a[i][j]); } //read {B} vector printf("\n Enter elements of [b] vector\n"); for (i = 0; i < n; i++) { printf(" B(%d) = ", i + 1); scanf("%lf", &b[i]); } //call Gauss elimination function retval = gauss(a, b, n, &det); //print results if (retval == 0) { printf("\n\t SOLUTION OF SIMULTANEOUS LINEAR EQUATIONS\n"); printf("\n\t The solution is"); for (i = 0; i < n; i++) printf("\n \t x(%d) = %lf", i + 1, b[i]); printf("\n \t Determinant = %lf \n", det); } else printf("\n \t SINGULAR MATRIX \n"); return 0; } /* Solves the system of equations [A]{x} = {B} using */ /* the Gaussian elimination method with partial pivoting. */ /* Parameters: */ /* n - number of equations */ /* a[n][n] - coefficient matrix */ /* b[n] - right-hand side vector */ /* *det - determinant of [A] */ int gauss (double a[][MAXSIZE], double b[], int n, double *det) { double tol, temp, mult; int npivot, i, j, l, k, flag; //initialization *det = 1.0; tol = 1e-30; //initial tolerance value npivot = 0; //mult = 0; //forward elimination for (k = 0; k < n; k++) { //search for max coefficient in pivot row- a[k][k] pivot element for (i = k + 1; i < n; i++) { if (fabs(a[i][k]) > fabs(a[k][k])) { //interchange row with maxium element with pivot row npivot++; for (l = 0; l < n; l++) { temp = a[i][l]; a[i][l] = a[k][l]; a[k][l] = temp; } temp = b[i]; b[i] = b[k]; b[k] = temp; } } //test for singularity if (fabs(a[k][k]) < tol) { //matrix is singular- terminate flag = 1; return flag; } //compute determinant- the product of the pivot elements *det = *det * a[k][k]; //eliminate the coefficients of X(I) for (i = k; i < n; i++) { mult = a[i][k] / a[k][k]; b[i] = b[i] - b[k] * mult; //compute constants for (j = k; j < n; j++) //compute coefficients a[i][j] = a[i][j] - a[k][j] * mult; } } //adjust the sign of the determinant if(npivot % 2 == 1) *det = *det * (-1.0); //backsubstitution b[n] = b[n] / a[n][n]; for(i = n - 1; i > 1; i--) { for(j = n; j > i + 1; j--) b[i] = b[i] - a[i][j] * b[j]; b[i] = b[i] / a[i - 1][i]; } flag = 0; return flag; } The solution should be: 1.058824, 1.823529, 0.882353 with det as -102.000000 Any insight is appreciated...

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  • How to make efficient code emerge through unit testing

    - by Jean
    Hi, I participate in a TDD Coding Dojo, where we try to practice pure TDD on simple problems. It occured to me however that the code which emerges from the unit tests isn't the most efficient. Now this is fine most of the time, but what if the code usage grows so that efficiency becomes a problem. I love the way the code emerges from unit testing, but is it possible to make the efficiency property emerge through further tests ? Here is a trivial example in ruby: prime factorization. I followed a pure TDD approach making the tests pass one after the other validating my original acceptance test (commented at the bottom). What further steps could I take, if I wanted to make one of the generic prime factorization algorithms emerge ? To reduce the problem domain, let's say I want to get a quadratic sieve implementation ... Now in this precise case I know the "optimal algorithm, but in most cases, the client will simply add a requirement that the feature runs in less than "x" time for a given environment. require 'shoulda' require 'lib/prime' class MathTest < Test::Unit::TestCase context "The math module" do should "have a method to get primes" do assert Math.respond_to? 'primes' end end context "The primes method of Math" do should "return [] for 0" do assert_equal [], Math.primes(0) end should "return [1] for 1 " do assert_equal [1], Math.primes(1) end should "return [1,2] for 2" do assert_equal [1,2], Math.primes(2) end should "return [1,3] for 3" do assert_equal [1,3], Math.primes(3) end should "return [1,2] for 4" do assert_equal [1,2,2], Math.primes(4) end should "return [1,5] for 5" do assert_equal [1,5], Math.primes(5) end should "return [1,2,3] for 6" do assert_equal [1,2,3], Math.primes(6) end should "return [1,3] for 9" do assert_equal [1,3,3], Math.primes(9) end should "return [1,2,5] for 10" do assert_equal [1,2,5], Math.primes(10) end end # context "Functionnal Acceptance test 1" do # context "the prime factors of 14101980 are 1,2,2,3,5,61,3853"do # should "return [1,2,3,5,61,3853] for ${14101980*14101980}" do # assert_equal [1,2,2,3,5,61,3853], Math.primes(14101980*14101980) # end # end # end end and the naive algorithm I created by this approach module Math def self.primes(n) if n==0 return [] else primes=[1] for i in 2..n do if n%i==0 while(n%i==0) primes<<i n=n/i end end end primes end end end

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  • what's best language to mate with Adobe Flex-based GUI for math crunching?

    - by gkdsp
    Hi, I'm not a software expert but need to outsource a web-based scientific GUI application, and I'm considering Adobe Flex. My math routines are currently in Javascript and C/C+. Having no experience with Flex, was hoping someone could help me understand what options are available for performing (preferably fast and efficient) CLIENT-side calculations. That is, can Flex interact with Javascript and/or C easily? If not, is actionscript or other language preferred? Downsides/tradeoffs? Need functions like LOG10, LN, SQRT, and would be nice to also have the error function (ERF) and complementary error function (ERFC), although I may be able to derive these last two from more basic functions if necessary. Thanks!

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  • LWJGL SlickUtil Texture Binding

    - by Matthew Dockerty
    I am making a 3D game using LWJGL and I have a texture class with static variables so that I only need to load textures once, even if I need to use them more than once. I am using Slick Util for this. When I bind a texture it works fine, but then when I try to render something else after I have rendered the model with the texture, the texture is still being bound. How do I unbind the texture and set the rendermode to the one that was in use before any textures were bound? Some of my code is below. The problem I am having is the player texture is being used in the box drawn around the player after it the model has been rendered. Model.java public class Model { public List<Vector3f> vertices = new ArrayList<Vector3f>(); public List<Vector3f> normals = new ArrayList<Vector3f>(); public ArrayList<Vector2f> textureCoords = new ArrayList<Vector2f>(); public List<Face> faces = new ArrayList<Face>(); public static Model TREE; public static Model PLAYER; public static void loadModels() { try { TREE = OBJLoader.loadModel(new File("assets/model/tree_pine_0.obj")); PLAYER = OBJLoader.loadModel(new File("assets/model/player.obj")); } catch (Exception e) { e.printStackTrace(); } } public void render(Vector3f position, Vector3f scale, Vector3f rotation, Texture texture, float shinyness) { glPushMatrix(); { texture.bind(); glColor3f(1, 1, 1); glTranslatef(position.x, position.y, position.z); glScalef(scale.x, scale.y, scale.z); glRotatef(rotation.x, 1, 0, 0); glRotatef(rotation.y, 0, 1, 0); glRotatef(rotation.z, 0, 0, 1); glMaterialf(GL_FRONT, GL_SHININESS, shinyness); glBegin(GL_TRIANGLES); { for (Face face : faces) { Vector2f t1 = textureCoords.get((int) face.textureCoords.x - 1); glTexCoord2f(t1.x, t1.y); Vector3f n1 = normals.get((int) face.normal.x - 1); glNormal3f(n1.x, n1.y, n1.z); Vector3f v1 = vertices.get((int) face.vertex.x - 1); glVertex3f(v1.x, v1.y, v1.z); Vector2f t2 = textureCoords.get((int) face.textureCoords.y - 1); glTexCoord2f(t2.x, t2.y); Vector3f n2 = normals.get((int) face.normal.y - 1); glNormal3f(n2.x, n2.y, n2.z); Vector3f v2 = vertices.get((int) face.vertex.y - 1); glVertex3f(v2.x, v2.y, v2.z); Vector2f t3 = textureCoords.get((int) face.textureCoords.z - 1); glTexCoord2f(t3.x, t3.y); Vector3f n3 = normals.get((int) face.normal.z - 1); glNormal3f(n3.x, n3.y, n3.z); Vector3f v3 = vertices.get((int) face.vertex.z - 1); glVertex3f(v3.x, v3.y, v3.z); } texture.release(); } glEnd(); } glPopMatrix(); } } Textures.java public class Textures { public static Texture FLOOR; public static Texture PLAYER; public static Texture SKYBOX_TOP; public static Texture SKYBOX_BOTTOM; public static Texture SKYBOX_FRONT; public static Texture SKYBOX_BACK; public static Texture SKYBOX_LEFT; public static Texture SKYBOX_RIGHT; public static void loadTextures() { try { FLOOR = TextureLoader.getTexture("PNG", new FileInputStream(new File("assets/model/floor.png"))); FLOOR.setTextureFilter(GL11.GL_NEAREST); PLAYER = TextureLoader.getTexture("PNG", new FileInputStream(new File("assets/model/tree_pine_0.png"))); PLAYER.setTextureFilter(GL11.GL_NEAREST); SKYBOX_TOP = TextureLoader.getTexture("PNG", new FileInputStream(new File("assets/textures/skybox_top.png"))); SKYBOX_TOP.setTextureFilter(GL11.GL_NEAREST); SKYBOX_BOTTOM = TextureLoader.getTexture("PNG", new FileInputStream(new File("assets/textures/skybox_bottom.png"))); SKYBOX_BOTTOM.setTextureFilter(GL11.GL_NEAREST); SKYBOX_FRONT = TextureLoader.getTexture("PNG", new FileInputStream(new File("assets/textures/skybox_front.png"))); SKYBOX_FRONT.setTextureFilter(GL11.GL_NEAREST); SKYBOX_BACK = TextureLoader.getTexture("PNG", new FileInputStream(new File("assets/textures/skybox_back.png"))); SKYBOX_BACK.setTextureFilter(GL11.GL_NEAREST); SKYBOX_LEFT = TextureLoader.getTexture("PNG", new FileInputStream(new File("assets/textures/skybox_left.png"))); SKYBOX_LEFT.setTextureFilter(GL11.GL_NEAREST); SKYBOX_RIGHT = TextureLoader.getTexture("PNG", new FileInputStream(new File("assets/textures/skybox_right.png"))); SKYBOX_RIGHT.setTextureFilter(GL11.GL_NEAREST); } catch (Exception e) { e.printStackTrace(); } } } Player.java public class Player { private Vector3f position; private float yaw; private float moveSpeed; public Player(float x, float y, float z, float yaw, float moveSpeed) { this.position = new Vector3f(x, y, z); this.yaw = yaw; this.moveSpeed = moveSpeed; } public void update() { if (Keyboard.isKeyDown(Keyboard.KEY_W)) walkForward(moveSpeed); if (Keyboard.isKeyDown(Keyboard.KEY_S)) walkBackwards(moveSpeed); if (Keyboard.isKeyDown(Keyboard.KEY_A)) strafeLeft(moveSpeed); if (Keyboard.isKeyDown(Keyboard.KEY_D)) strafeRight(moveSpeed); if (Mouse.isButtonDown(0)) yaw += Mouse.getDX(); LowPolyRPG.getInstance().getCamera().setPosition(-position.x, -position.y, -position.z); LowPolyRPG.getInstance().getCamera().setYaw(yaw); } public void walkForward(float distance) { position.setX(position.getX() + distance * (float) Math.sin(Math.toRadians(yaw))); position.setZ(position.getZ() - distance * (float) Math.cos(Math.toRadians(yaw))); } public void walkBackwards(float distance) { position.setX(position.getX() - distance * (float) Math.sin(Math.toRadians(yaw))); position.setZ(position.getZ() + distance * (float) Math.cos(Math.toRadians(yaw))); } public void strafeLeft(float distance) { position.setX(position.getX() + distance * (float) Math.sin(Math.toRadians(yaw - 90))); position.setZ(position.getZ() - distance * (float) Math.cos(Math.toRadians(yaw - 90))); } public void strafeRight(float distance) { position.setX(position.getX() + distance * (float) Math.sin(Math.toRadians(yaw + 90))); position.setZ(position.getZ() - distance * (float) Math.cos(Math.toRadians(yaw + 90))); } public void render() { Model.PLAYER.render(new Vector3f(position.x, position.y + 12, position.z), new Vector3f(3, 3, 3), new Vector3f(0, -yaw + 90, 0), Textures.PLAYER, 128); GL11.glPushMatrix(); GL11.glTranslatef(position.getX(), position.getY(), position.getZ()); GL11.glRotatef(-yaw, 0, 1, 0); GL11.glScalef(5.8f, 21, 2.2f); GL11.glDisable(GL11.GL_LIGHTING); GL11.glLineWidth(3); GL11.glBegin(GL11.GL_LINE_STRIP); GL11.glColor3f(1, 1, 1); glVertex3f(1f, 0f, -1f); glVertex3f(-1f, 0f, -1f); glVertex3f(-1f, 1f, -1f); glVertex3f(1f, 1f, -1f); glVertex3f(-1f, 0f, 1f); glVertex3f(1f, 0f, 1f); glVertex3f(1f, 1f, 1f); glVertex3f(-1f, 1f, 1f); glVertex3f(1f, 1f, -1f); glVertex3f(-1f, 1f, -1f); glVertex3f(-1f, 1f, 1f); glVertex3f(1f, 1f, 1f); glVertex3f(1f, 0f, 1f); glVertex3f(-1f, 0f, 1f); glVertex3f(-1f, 0f, -1f); glVertex3f(1f, 0f, -1f); glVertex3f(1f, 0f, 1f); glVertex3f(1f, 0f, -1f); glVertex3f(1f, 1f, -1f); glVertex3f(1f, 1f, 1f); glVertex3f(-1f, 0f, -1f); glVertex3f(-1f, 0f, 1f); glVertex3f(-1f, 1f, 1f); glVertex3f(-1f, 1f, -1f); GL11.glEnd(); GL11.glEnable(GL11.GL_LIGHTING); GL11.glPopMatrix(); } public Vector3f getPosition() { return new Vector3f(-position.x, -position.y, -position.z); } public float getX() { return position.getX(); } public float getY() { return position.getY(); } public float getZ() { return position.getZ(); } public void setPosition(Vector3f position) { this.position = position; } public void setPosition(float x, float y, float z) { this.position.setX(x); this.position.setY(y); this.position.setZ(z); } } Thanks for the help.

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  • 3D Camera Problem

    - by Chris
    I allow the user to look around the scene by holding down the left mouse button and moving the mouse. The problem that I have is I can be facing one direction, I move the mouse up and the view tilts up, I move down and the view titles down. If I spin around 180 my left and right still works fine, but when I move the mouse up the view tilts down, and when I move the mouse down the view titles up. This is the code I am using, can anyone see what the problem with the logic is? var viewDir = g_math.subVector(target, g_eye); var rotatedViewDir = []; rotatedViewDir[0] = (Math.cos(g_mouseXDelta * g_rotationDelta) * viewDir[0]) - (Math.sin(g_mouseXDelta * g_rotationDelta) * viewDir[2]); rotatedViewDir[1] = viewDir[1]; rotatedViewDir[2] = (Math.cos(g_mouseXDelta * g_rotationDelta) * viewDir[2]) + (Math.sin(g_mouseXDelta * g_rotationDelta) * viewDir[0]); viewDir = rotatedViewDir; rotatedViewDir[0] = viewDir[0]; rotatedViewDir[1] = (Math.cos(g_mouseYDelta * g_rotationDelta * -1) * viewDir[1]) - (Math.sin(g_mouseYDelta * g_rotationDelta * -1) * viewDir[2]); rotatedViewDir[2] = (Math.cos(g_mouseYDelta * g_rotationDelta * -1) * viewDir[2]) + (Math.sin(g_mouseYDelta * g_rotationDelta * -1) * viewDir[1]); g_lookingDir = rotatedViewDir; var newtarget = g_math.addVector(rotatedViewDir, g_eye);

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  • Automatically triggering standard spaceship controls to stop its motion

    - by Garan
    I have been working on a 2D top-down space strategy/shooting game. Right now it is only in the prototyping stage (I have gotten basic movement) but now I am trying to write a function that will stop the ship based on it's velocity. This is being written in Lua, using the Love2D engine. My code is as follows (note- object.dx is the x-velocity, object.dy is the y-velocity, object.acc is the acceleration, and object.r is the rotation in radians): function stopMoving(object, dt) local targetr = math.atan2(object.dy, object.dx) if targetr == object.r + math.pi then local currentspeed = math.sqrt(object.dx*object.dx+object.dy*object.dy) if currentspeed ~= 0 then object.dx = object.dx + object.acc*dt*math.cos(object.r) object.dy = object.dy + object.acc*dt*math.sin(object.r) end else if (targetr - object.r) >= math.pi then object.r = object.r - object.turnspeed*dt else object.r = object.r + object.turnspeed*dt end end end It is implemented in the update function as: if love.keyboard.isDown("backspace") then stopMoving(player, dt) end The problem is that when I am holding down backspace, it spins the player clockwise (though I am trying to have it go the direction that would be the most efficient at getting to the angle it would have to be) and then it never starts to accelerate the player in the direction opposite to it's velocity. What should I change in this code to get that to work? EDIT : I'm not trying to just stop the player in place, I'm trying to get it to use it's normal commands to neutralize it's existing velocity. I also changed math.atan to math.atan2, apparently it's better. I noticed no difference when running it, though.

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  • Hexagonal Grid Coordinates To Pixel Coordinates

    - by CaptnCraig
    I am working with a hexagonal grid. I have chosen to use this coordinate system because it is quite elegant. This question talks about generating the coordinates themselves, and is quite useful. My issue now is in converting these coordinates to and from actual pixel coordinates. I am looking for a simple way to find the center of a hexagon with coordinates x,y,z. Assume (0,0) in pixel coordinates is at (0,0,0) in hex coords, and that each hexagon has an edge of length s. It seems to me like x,y, and z should each move my coordinate a certain distance along an axis, but they are interrelated in an odd way I can't quite wrap my head around it. Bonus points if you can go the other direction and convert any (x,y) point in pixel coordinates to the hex that point belongs in.

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