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• Project Euler 9 Understanding

  - by DMan

  This question states: A Pythagorean triplet is a set of three natural numbers, a b c, for which, a2 + b2 = c2 For example, 32 + 42 = 9 + 16 = 25 = 52. There exists exactly one Pythagorean triplet for which a + b + c = 1000. Find the product abc. I'm not sure what's it trying to ask you. Are we trying to find a^2+b^2=c^2 and then plug those numbers into a+b+c=1000?

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• Project Euler, Problem 10 java solution now working

  - by Dennis S

  Hi, I'm trying to find the sum of the prime numbers < 2'000'000. This is my solution in java but I can't seem get the correct answer. Please give some input on what could be wrong and general advice on the code is appreciated. Printing 'sum' gives: 1308111344, which is incorrect. /* The sum of the primes below 10 is 2 + 3 + 5 + 7 = 17. Find the sum of all the primes below two million. */ class Helper{ public void run(){ Integer sum = 0; for(int i = 2; i < 2000000; i++){ if(isPrime(i)) sum += i; } System.out.println(sum); } private boolean isPrime(int nr){ if(nr == 2) return true; else if(nr == 1) return false; if(nr % 2 == 0) return false; for(int i = 3; i < Math.sqrt(nr); i += 2){ if(nr % i == 0) return false; } return true; } } class Problem{ public static void main(String[] args){ Helper p = new Helper(); p.run(); } }

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• Project Euler: problem 8

  - by Marijus

  n = # some ridiculously large number, omitted N = [int(i) for i in str(n)] maxProduct = 0 for i in range(0,len(N)-4): newProduct = 1 is_cons = 0 for j in range(i,i+4): if N[j] == N[j+1] - 1: is_cons += 1 if is_cons == 5: for j in range(i,i+5): newProduct *= N[j] if newProduct > maxProduct: maxProduct = newProduct print maxProduct I've been working on this problem for hours now and I can't get this to work. I've tried doing this algorithm on paper and it works just fine.. Could you give me hints what's wrong ?

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• Project Euler #3

  - by Alex

  Question: The prime factors of 13195 are 5, 7, 13 and 29. What is the largest prime factor of the number 600851475143? I found this one pretty easy, but running the file took an extremely long time, it's been going on for a while and the highest number I've got to is 716151937. Here is my code, am I just going to have a wait or is there an error in my code? //User made class public class Three { public static boolean checkPrime(long p) { long i; boolean prime = false; for(i = 2;i<p/2;i++) { if(p%i==0) { prime = true; break; } } return prime; } } //Note: This is a separate file public class ThreeMain { public static void main(String[] args) { long comp = 600851475143L; boolean prime; long i; for(i=2;i<comp/2;i++) { if(comp%i==0) { prime = Three.checkPrime(i); if(prime==true) { System.out.println(i); } } } } }

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• Quaternion dfference + time --> angular velocity (gyroscope in physics library)

  - by AndrewK

  I am using Bullet Physic library to program some function, where I have difference between orientation from gyroscope given in quaternion and orientation of my object, and time between each frame in milisecond. All I want is set the orientation from my gyroscope to orientation of my object in 3D space. But all I can do is set angular velocity to my object. I have orientation difference and time, and from that I calculate vector of angular velocity [Wx,Wy,Wz] from that formula: W(t) = 2 * dq(t)/dt * conj(q(t)) My code is: btQuaternion diffQuater = gyroQuater - boxQuater; btQuaternion conjBoxQuater = gyroQuater.inverse(); btQuaternion velQuater = ((diffQuater * 2.0f) / d_time) * conjBoxQuater; And everything works well, till I get: 1 rotating around Y axis, angle about 60 degrees, then I have these values in 2 critical frames: x: -0.013220 y: -0.038050 z: -0.021979 w: -0.074250 - diffQuater x: 0.120094 y: 0.818967 z: 0.156797 w: -0.538782 - gyroQuater x: 0.133313 y: 0.857016 z: 0.178776 w: -0.464531 - boxQuater x: 0.207781 y: 0.290452 z: 0.245594 - diffQuater -> euler angles x: 3.153619 y: -66.947929 z: 175.936615 - gyroQuater -> euler angles x: 4.290697 y: -57.553043 z: 173.320053 - boxQuater -> euler angles x: 0.138128 y: 2.823307 z: 1.025552 w: 0.131360 - velQuater d_time: 0.058000 x: 0.211020 y: 1.595124 z: 0.303650 w: -1.143846 - diffQuater x: 0.089518 y: 0.771939 z: 0.144527 w: -0.612543 - gyroQuater x: -0.121502 y: -0.823185 z: -0.159123 w: 0.531303 - boxQuater x: nan y: nan z: nan - diffQuater -> euler angles x: 2.985240 y: -76.304405 z: -170.555054 - gyroQuater -> euler angles x: 3.269681 y: -65.977966 z: 175.639420 - boxQuater -> euler angles x: -0.730262 y: -2.882153 z: -1.294721 w: 63.325996 - velQuater d_time: 0.063000 2 rotating around X axis, angle about 120 degrees, then I have these values in 2 critical frames: x: -0.013045 y: -0.004186 z: -0.005667 w: -0.022482 - diffQuater x: -0.848030 y: -0.187985 z: 0.114400 w: 0.482099 - gyroQuater x: -0.834985 y: -0.183799 z: 0.120067 w: 0.504580 - boxQuater x: 0.036336 y: 0.002312 z: 0.020859 - diffQuater -> euler angles x: -113.129463 y: 0.731925 z: 25.415056 - gyroQuater -> euler angles x: -110.232368 y: 0.860897 z: 25.350458 - boxQuater -> euler angles x: -0.865820 y: -0.456086 z: 0.034084 w: 0.013184 - velQuater d_time: 0.055000 x: -1.721662 y: -0.387898 z: 0.229844 w: 0.910235 - diffQuater x: -0.874310 y: -0.200132 z: 0.115142 w: 0.426933 - gyroQuater x: 0.847352 y: 0.187766 z: -0.114703 w: -0.483302 - boxQuater x: -144.402298 y: 4.891629 z: 71.309158 - diffQuater -> euler angles x: -119.515343 y: 1.745076 z: 26.646086 - gyroQuater -> euler angles x: -112.974533 y: 0.738675 z: 25.411509 - boxQuater -> euler angles x: 2.086195 y: 0.676526 z: -0.424351 w: 70.104248 - velQuater d_time: 0.057000 2 rotating around Z axis, angle about 120 degrees, then I have these values in 2 critical frames: x: -0.000736 y: 0.002812 z: -0.004692 w: -0.008181 - diffQuater x: -0.003829 y: 0.012045 z: -0.868035 w: 0.496343 - gyroQuater x: -0.003093 y: 0.009232 z: -0.863343 w: 0.504524 - boxQuater x: -0.000822 y: -0.003032 z: 0.004162 - diffQuater -> euler angles x: -1.415189 y: 0.304210 z: -120.481873 - gyroQuater -> euler angles x: -1.091881 y: 0.227784 z: -119.399445 - boxQuater -> euler angles x: 0.159042 y: 0.169228 z: -0.754599 w: 0.003900 - velQuater d_time: 0.025000 x: -0.007598 y: 0.024074 z: -1.749412 w: 0.968588 - diffQuater x: -0.003769 y: 0.012030 z: -0.881377 w: 0.472245 - gyroQuater x: 0.003829 y: -0.012045 z: 0.868035 w: -0.496343 - boxQuater x: -5.645197 y: 1.148993 z: -146.507187 - diffQuater -> euler angles x: -1.418294 y: 0.270319 z: -123.638245 - gyroQuater -> euler angles x: -1.415183 y: 0.304208 z: -120.481873 - boxQuater -> euler angles x: 0.017498 y: -0.013332 z: 2.040073 w: 148.120056 - velQuater d_time: 0.027000 The problem is the most visible in diffQuater - euler angles vector. Can someone tell me why it is like that? and how to solve that problem? All suggestions are welcome.

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• what does AngleVectors method in quake 3 source code does

  - by kypronite

  I just downloaded quake 3 for learning purposes. I know some of some linear algebra(basic vector math ie: dot,cross product). However I can't decipher what below method does, I know what is yaw,pitch and roll. But I can't connect these with vector. Worse, I'm not sure this fall under what math 'category', so I don't really know how to google. Hence the question here. Anyone? void AngleVectors( const vec3_t angles, vec3_t forward, vec3_t right, vec3_t up) { float angle; static float sr, sp, sy, cr, cp, cy; // static to help MS compiler fp bugs angle = angles[YAW] * (M_PI*2 / 360); sy = sin(angle); cy = cos(angle); angle = angles[PITCH] * (M_PI*2 / 360); sp = sin(angle); cp = cos(angle); angle = angles[ROLL] * (M_PI*2 / 360); sr = sin(angle); cr = cos(angle); if (forward) { forward[0] = cp*cy; forward[1] = cp*sy; forward[2] = -sp; } if (right) { right[0] = (-1*sr*sp*cy+-1*cr*-sy); right[1] = (-1*sr*sp*sy+-1*cr*cy); right[2] = -1*sr*cp; } if (up) { up[0] = (cr*sp*cy+-sr*-sy); up[1] = (cr*sp*sy+-sr*cy); up[2] = cr*cp; } } ddddd

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• Alternatives to Project Euler for improving Excel ability

  - by Jonathan Deamer

  I've recently been enjoying using the mathematical problems listed at Project Euler to learn Python. My Excel ability is better than my Python, but I think I'd still benefit from the sort of inductive learning that comes with solving a series of increasingly difficult puzzles using a particular tool. I know Project Euler can be completed using Excel, but are there any other puzzle series similar to this or The Python Challenge specifically tailored for people trying to increase their knowledge of Excel and what it can do? NB. I'm not looking for a "tutorial", I know there are plenty of these. And apologies if this isn't completely appropriate for programmers.SE.com - some of the folks at SuperUser suggested it was a better fit here than there!

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• Check if an object is facing another based on angles

  - by Isaiah

  I already have something that calculates the bearing angle to get one object to face another. You give it the positions and it returns the angle to get one to face the other. Now I need to figure out how tell if on object is facing toward another object within a specified field and I can't find any information about how to do this. The objects are obj1 and obj2. Their angles are at obj1.angle and obj2.angle. Their vectors are at obj1.pos and obj2.pos. It's in the format [x,y]. The angle to have one face directly at another is found with direction(obj1.pos,obj2.pos). I want to set the function up like this: isfacing(obj1,obj2,area){...} and return true/false depending if it's in the specified field area to the angle to directly see it. I've got a base like this: var isfacing = function (obj1,obj2,area){ var toface = direction(obj1.pos,obj2.pos); if(toface+area >= obj1.angle && ob1.angle >= toface-area){ return true; } return false; } But my problem is that the angles are in 360 degrees, never above 360 and never below 0. How can I account for that in this? If the first object's angle is say at 0 and say I subtract a field area of 20 or so. It'll check if it's less than -20! If I fix the -20 it becomes 340 but x < 340 isn't what I want, I'd have to x 340 in that case. Is there someone out there with more sleep than I that can help a new dev pulling an all-nighter just to get enemies to know if they're attacking in the right direction? I hope I'm making this harder than it seems. I'd just make them always face the main char if the producer didn't want attacks from behind to work while blocking. In which case I'll need the function above anyways. I've tried to give as much info as I can think would help. Also this is in 2d.

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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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• Project Euler problem 214, How can i make it more efficient?

  - by Once

  I am becoming more and more addicted to the Project Euler problems. However since one week I am stuck with the #214. Here is a short version of the problem: PHI() is Euler's totient function, i.e. for any given integer n, PHI(n)=numbers of k<=n for which gcd(k,n)=1. We can iterate PHI() to create a chain. For example starting from 18: PHI(18)=6 = PHI(6)=2 = PHI(2)=1. So starting from 18 we get a chain of length 4 (18,6,2,1) The problem is to calculate the sum of all primes less than 40e6 which generate a chain of length 25. I built a function that calculates the chain length of any number and I tested it for small values: it works well and fast. sum of all primes<=20 which generate a chain of length 4: 12 sum of all primes<=1000 which generate a chain of length 10: 39383 Unfortunately my algorithm doesn't scale well. When I apply it to the problem, it takes several hours to calculate... so I stop it because the Project Euler problems must be solved in less than one minute. I thought that my prime detection function might be slow so I fed the program with a list of primes <40e6 to avoid the primality test... The code runs now a little bit faster, but there is still no way to get a solution in less than a few hours (and I don't want this). So is there any "magic trick" that I am missing here ? I really don't understand how to be more efficient on this one... I am not asking for the solution, because fighting with optimization is all the fun of Project Euler. However, any small hint that could put me on the right track would be welcome. Thanks !

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• Calculating up-vector to avoid gimbal lock using euler angles

  - by jessejuicer

  I wish to orbit a camera around a sphere, yet the problem is that when the camera rotates so that it is at the north pole (and pointing down) or the south pole (and pointing up) of the sphere the camera doesn't handle itself very well. It spins rapidly until arriving 180 degrees in the opposite direction. I believe this is known as gimbal lock. I understand you can avoid this problem using quaternions. But I also read in another forum that it's possible to avoid this easily using euler angles as well. Which I would prefer to do. It was said that all you need to do is "calculate a proper up-vector every frame, and that avoids the problem entirely." Well, I tried aligning the up-vector with the vertical axis of the camera whenever the camera changed orientation, but this didn't seem to work. Meaning that the up-vector followed exactly the orientation of the camera's y-axis (or it's up vector), instead of using a constant up-vector aligned to the up-vector of the world (0, 1, 0). How exactly do I go about calculating a proper up-vector as my camera orientation changes to avoid the gimbal lock problem mentioned above?

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• libGDX using Stage and Actor produces different camera angles on desktop and Android Phone

  - by Brandon

  libGDX using Stage and Actor produces different camera angles on desktop and Android Phone. Here are pictures demonstrating the problem: http://brandonyuh.minus.com/mFpdTSgN17VUq On the desktop version, the image takes up most all the screen. On the Android phone it only takes up a bit of the screen. Here's the code (not my actual project but I isolated the problem): package com.me.mygdxgame2; import com.badlogic.gdx.*; import com.badlogic.gdx.graphics.*; import com.badlogic.gdx.graphics.Texture.TextureFilter; import com.badlogic.gdx.graphics.g2d.*; import com.badlogic.gdx.scenes.scene2d.*; public class MyGdxGame2 implements ApplicationListener { private Stage stage; public void create() { stage = new Stage(); stage.addActor(new ActorHi()); } public void render() { Gdx.gl.glClearColor(0, 1, 0, 1); Gdx.gl.glClear(GL10.GL_COLOR_BUFFER_BIT); stage.draw(); } public void dispose() {} public void resize(int width, int height) {} public void pause() {} public void resume() {} public class ActorHi extends Actor { private Sprite sprite; public ActorHi() { Texture texture = new Texture(Gdx.files.internal("data/hi.png")); texture.setFilter(TextureFilter.Linear, TextureFilter.Linear); sprite = new Sprite(new TextureRegion(texture, 0, 0, 128, 128)); sprite.setBounds(0, 0, 300.0f, 300.0f); } public void draw(SpriteBatch batch, float parentAlpha) { sprite.draw(batch); } } } hi.png is included in the above link Thank you very much for answering my question. I've spent 3 days trying to figure it out.

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• algorithm for project euler problem no 18

  - by Valentino Ru

  Problem number 18 from Project Euler's site is as follows: By starting at the top of the triangle below and moving to adjacent numbers on the row below, the maximum total from top to bottom is 23. 3 7 4 2 4 6 8 5 9 3 That is, 3 + 7 + 4 + 9 = 23. Find the maximum total from top to bottom of the triangle below: 75 95 64 17 47 82 18 35 87 10 20 04 82 47 65 19 01 23 75 03 34 88 02 77 73 07 63 67 99 65 04 28 06 16 70 92 41 41 26 56 83 40 80 70 33 41 48 72 33 47 32 37 16 94 29 53 71 44 65 25 43 91 52 97 51 14 70 11 33 28 77 73 17 78 39 68 17 57 91 71 52 38 17 14 91 43 58 50 27 29 48 63 66 04 68 89 53 67 30 73 16 69 87 40 31 04 62 98 27 23 09 70 98 73 93 38 53 60 04 23 NOTE: As there are only 16384 routes, it is possible to solve this problem by trying every route. However, Problem 67, is the same challenge with a triangle containing one-hundred rows; it cannot be solved by brute force, and requires a clever method! ;o) The formulation of this problems does not make clear if the "Traversor" is greedy, meaning that he always choosed the child with be higher value the maximum of every single walkthrough is asked The NOTE says, that it is possible to solve this problem by trying every route. This means to me, that is is also possible without! This leads to my actual question: Assumed that not the greedy one is the max, is there any algorithm that finds the max walkthrough value without trying every route and that doesn't act like the greedy algorithm? I implemented an algorithm in Java, putting the values first in a node structure, then applying the greedy algorithm. The result, however, is cosidered as wrong by Project Euler. sum = 0; void findWay(Node node){ sum += node.value; if(node.nodeLeft != null && node.nodeRight != null){ if(node.nodeLeft.value > node.nodeRight.value){ findWay(node.nodeLeft); }else{ findWay(node.nodeRight); } } }

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• Viewing at Impossible Angles

  - by kemer

  The picture of the little screwdriver with the Allen wrench head to the right is bound to invoke a little nostalgia for those readers who were Sun customers in the late 80s. This tool was a very popular give-away: it was essential for installing and removing Multibus (you youngsters will have to look that up on Wikipedia…) cards in our systems. Back then our mid-sized systems were gargantuan: it was routine for us to schlep around a 200 lb. desk side box and 90 lb. monitor to demo a piece of software your smart phone will run better today. We were very close to the hardware, and the first thing a new field sales systems engineer had to learn was how put together a system. If you were lucky, a grizzled service engineer might run you through the process once, then threaten your health and existence should you ever screw it up so that he had to fix it. Nowadays we make it much easier to learn the ins and outs of our hardware with simulations–3D animations–that take you through the process of putting together or replacing pieces of a system. Most recently, we have posted three sophisticated PDFs that take advantage of Acrobat 9 features to provide a really intelligent approach to documenting hardware installation and repair: Sun Fire X4800/X4800 M2 Animations for Chassis Components Sun Fire X4800/X4800 M2 Animations for Sub Assembly Module (SAM) Sun Fire X4800/X4800 M2 Animations for CMOD Download one of these documents and take a close look at it. You can view the hardware from any angle, including impossible ones. Each document has a number of procedures, that break down into steps. Click on a procedure, then a step and you will see it animated in the drawing. Of course hardware design has generally eliminated the need for things like our old giveaway tools: components snap and lock in. Often you can replace redundant units while the system is hot, but for heaven’s sake, you’ll want to verify that you can do that before you try it! Meanwhile, we can all look forward to a growing portfolio of these intelligent documents. We would love to hear what you think about them. –Kemer

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• Translate extrinsic rotations to intrinsic rotations ( Euler angles )

  - by MineMan287

  The problem I have is very frustrating: I am using the Jitter Physics library which gives Quaternion rotations, you can extract the extrinsic rotations but I need intrinsic rotations to rotate in OpenTK (There are other reasons as well so I don't want to make OpenTK use a Matrix) GL.Rotate(xr, 1, 0, 0) GL.Rotate(yr, 0, 1, 0) GL.Rotate(zr, 0, 0, 1) EDIT : Response to the first answer Like This? GL.Rotate(zr, 0, 0, 1) GL.Rotate(yr, 0, 1, 0) GL.Rotate(xr, 1, 0, 0) Or This? GL.Rotate(xr, 1, 0, 0) GL.Rotate(yr, 0, 1, 0) GL.Rotate(zr, 0, 0, 1) GL.Rotate(zr, 0, 0, 1) GL.Rotate(yr, 0, 1, 0) GL.Rotate(xr, 1, 0, 0) GL.Rotate(xr, 1, 0, 0) GL.Rotate(yr, 0, 1, 0) GL.Rotate(zr, 0, 0, 1) I'm confused, please give an example

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• Euler angles to Cartesian Coordinates for use with gluLookAt

  - by notrodash

  I have searched all of the internet but just couldn't find the answer. I am using LibGDX and this is part of my code that loops over and over: public void render() { GL11 gl = Gdx.gl11; float centerX = (float)Math.cos(yaw) * (float)Math.cos(pitch); float centerY = (float)Math.sin(yaw) * (float)Math.cos(pitch); float centerZ = (float)Math.sin(pitch); System.out.println(centerX+" "+centerY+" "+centerZ+" ~ "+GDXRacing.camera.position.x+" "+GDXRacing.camera.position.y+" "+GDXRacing.camera.position.z); Gdx.glu.gluLookAt(gl, GDXRacing.camera.position.x, GDXRacing.camera.position.y, GDXRacing.camera.position.z, centerX, centerY, centerZ, 0, 1, 0); if(Gdx.input.isKeyPressed(Keys.A)) { yaw--; } if(Gdx.input.isKeyPressed(Keys.D)) { yaw++; } } I might just be bad at the math, but I dont get it. Does someone have a good explanation and an idea about how to deal with this? I am trying to make a first person camera. By the way, the camera is translated by +10 on the Z axis. Currently when I run the application, this is what I get: Watch video in browser | Download video (for those who cant download the video, everything shakes in a clockwise/anticlockwise action, depending on if I increase or decrease the Yaw value) -Thank you. [edit] and with this code: public void render() { GL11 gl = Gdx.gl11; float centerX = (float)(MathUtils.cosDeg(yaw)*4); float centerY = 0; float centerZ = (float)(MathUtils.sinDeg(yaw)*4); System.out.println(centerX+" "+centerY+" "+centerZ+" ~ "+GDXRacing.camera.position.x+" "+GDXRacing.camera.position.y+" "+GDXRacing.camera.position.z); Gdx.glu.gluLookAt(gl, GDXRacing.camera.position.x, GDXRacing.camera.position.y, GDXRacing.camera.position.z, centerX, centerY, centerZ, 0, 1, 0); if(Gdx.input.isKeyPressed(Keys.A)) { yaw--; } if(Gdx.input.isKeyPressed(Keys.D)) { yaw++; } } it slowly swings from the left to the right. This approach worked for turning left and right for 2d games though. What am I doing wrong?

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• F# Project Euler Problem 1

  - by MarkPearl

  Every now and then I give project Euler a quick browse. Since I have been playing with F# I have found it a great way to learn the basics of the language. Today I thought I would give problem 1 an attempt… Problem 1 If we list all the natural numbers below 10 that are multiples of 3 or 5, we get 3, 5, 6 and 9. The sum of these multiples is 23. Find the sum of all the multiples of 3 or 5 below 1000. My F# Solution I broke this problem into two functions… 1) be able to generate a collection of numbers that are multiples of a number but but are smaller than another number. let GenerateMultiplesOfXbelowY X Y = X |> Seq.unfold (fun i -> if (i<Y) then Some(i, i+X) else None) I then needed something that generated collections for multiples of 3 & 5 and then removed any duplicates. Once this was done I would need to sum these all together to get a result. I found the Seq object to be extremely useful to achieve this… let Multiples = Seq.append (GenerateMultiplesOfXbelowY 3 1000) (GenerateMultiplesOfXbelowY 5 1000) |> Seq.distinct |> Seq.fold(fun acc a -> acc + a) 0 |> Console.WriteLine |> Console.ReadLine My complete solution was … open System let GenerateMultiplesOfXbelowY X Y = X |> Seq.unfold (fun i -> if (i<Y) then Some(i, i+X) else None) let Multiples = Seq.append (GenerateMultiplesOfXbelowY 3 1000) (GenerateMultiplesOfXbelowY 5 1000) |> Seq.distinct |> Seq.fold(fun acc a -> acc + a) 0 |> Console.WriteLine |> Console.ReadLine   Which seemed to generate the correct result in a relatively short period of time although I am sure I will get some comments from the experts who know of some intrinsic method to achieve all of this in one method call.

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• Modular building technique with angles? (A roof)

  - by Mungoid

  Ive been spending a bit of time lately studying the modular buildings of many games and reading/viewing several tutorials about it as well, but almost every example I see uses a plain square building that does not have any angled roof or similar. In all my applications (CS6, Blender/Max, UDK) I adhere to the same grid spacing and I get pretty good results, but trying to make modular angled pieces is confusing me as I'm not sure the best way to approach it. Below is some shots of my template sheet and workflow I have been doing. Should I do the roof separately or is it possible for me to keep it in the same texture sheet? The main issue is below. I have made a couple modular roof pieces but when i try to use them, i end up needing to model multiple other parts to fill gaps based on what roof shape i want. I then model those 'filler' pieces and now i have that much less space left in my texture sheet and those pieces are usually not that reusable for anything else. This is where im not sure how to proceed. If anyone has any links to documents or papers talking about this or advice, I would greatly appreciate it! =-) My main roof pieces with the gaps My power of 2 texture sheet, with 16x16 grid squares. The texture sheet loaded into blender on a 16x16 plane and starting to separate and extrude.

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• Solving Euler Project Problem Number 1 with Microsoft Axum

  - by Jeff Ferguson

  Note: The code below applies to version 0.3 of Microsoft Axum. If you are not using this version of Axum, then your code may differ from that shown here. I have just solved Problem 1 of Project Euler using Microsoft Axum. The problem statement is as follows: If we list all the natural numbers below 10 that are multiples of 3 or 5, we get 3, 5, 6 and 9. The sum of these multiples is 23. Find the sum of all the multiples of 3 or 5 below 1000. My Axum-based solution is as follows: namespace EulerProjectProblem1{ // http://projecteuler.net/index.php?section=problems&id=1 // // If we list all the natural numbers below 10 that are multiples of 3 or 5, we get 3, 5, 6 and 9. // The sum of these multiples is 23. // Find the sum of all the multiples of 3 or 5 below 1000. channel SumOfMultiples { input int Multiple1; input int Multiple2; input int UpperBound; output int Sum; } agent SumOfMultiplesAgent : channel SumOfMultiples { public SumOfMultiplesAgent() { int Multiple1 = receive(PrimaryChannel::Multiple1); int Multiple2 = receive(PrimaryChannel::Multiple2); int UpperBound = receive(PrimaryChannel::UpperBound); int Sum = 0; for(int Index = 1; Index < UpperBound; Index++) { if((Index % Multiple1 == 0) || (Index % Multiple2 == 0)) Sum += Index; } PrimaryChannel::Sum <-- Sum; } } agent MainAgent : channel Microsoft.Axum.Application { public MainAgent() { var SumOfMultiples = SumOfMultiplesAgent.CreateInNewDomain(); SumOfMultiples::Multiple1 <-- 3; SumOfMultiples::Multiple2 <-- 5; SumOfMultiples::UpperBound <-- 1000; var Sum = receive(SumOfMultiples::Sum); System.Console.WriteLine(Sum); System.Console.ReadLine(); PrimaryChannel::ExitCode <-- 0; } }} Let’s take a look at the various parts of the code. I begin by setting up a channel called SumOfMultiples that accepts three inputs and one output. The first two of the three inputs will represent the two possible multiples, which are three and five in this case. The third input will represent the upper bound of the problem scope, which is 1000 in this case. The lone output of the channel represents the sum of all of the matching multiples: channel SumOfMultiples{ input int Multiple1; input int Multiple2; input int UpperBound; output int Sum;} I then set up an agent that uses the channel. The agent, called SumOfMultiplesAgent, received the three inputs from the channel sent to the agent, stores the results in local variables, and performs the for loop that iterates from 1 to the received upper bound. The agent keeps track of the sum in a local variable and stores the sum in the output portion of the channel: agent SumOfMultiplesAgent : channel SumOfMultiples{ public SumOfMultiplesAgent() { int Multiple1 = receive(PrimaryChannel::Multiple1); int Multiple2 = receive(PrimaryChannel::Multiple2); int UpperBound = receive(PrimaryChannel::UpperBound); int Sum = 0; for(int Index = 1; Index < UpperBound; Index++) { if((Index % Multiple1 == 0) || (Index % Multiple2 == 0)) Sum += Index; } PrimaryChannel::Sum <-- Sum; }} The application’s main agent, therefore, simply creates a new SumOfMultiplesAgent in a new domain, prepares the channel with the inputs that we need, and then receives the Sum from the output portion of the channel: agent MainAgent : channel Microsoft.Axum.Application{ public MainAgent() { var SumOfMultiples = SumOfMultiplesAgent.CreateInNewDomain(); SumOfMultiples::Multiple1 <-- 3; SumOfMultiples::Multiple2 <-- 5; SumOfMultiples::UpperBound <-- 1000; var Sum = receive(SumOfMultiples::Sum); System.Console.WriteLine(Sum); System.Console.ReadLine(); PrimaryChannel::ExitCode <-- 0; }} The result of the calculation (which, by the way, is 233,168) is sent to the console using good ol’ Console.WriteLine().

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• (Libgdx) Move Vector2 along angle?

  - by gemurdock

  I have seen several answers on here about moving along angle, but I can't seem to get this to work properly for me and I am new to LibGDX... just trying to learn. These are my Vector2's that I am using for this function. public Vector2 position = new Vector2(); public Vector2 velocity = new Vector2(); public Vector2 movement = new Vector2(); public Vector2 direction = new Vector2(); Here is the function that I use to move the position vector along an angle. setLocation() just sets the new location of the image. public void move(float delta, float degrees) { position.set(image.getX() + image.getWidth() / 2, image.getY() + image.getHeight() / 2); direction.set((float) Math.cos(degrees), (float) Math.sin(degrees)).nor(); velocity.set(direction).scl(speed); movement.set(velocity).scl(delta); position.add(movement); setLocation(position.x, position.y); // Sets location of image } I get a lot of different angles with this, just not the correct angles. How should I change this function to move a Vector2 along an angle using the Vector2 class from com.badlogic.gdx.math.Vector2 within the LibGDX library? I found this answer, but not sure how to implement it. Update: I figured out part of the issue. Should convert degrees to radians. However, the angle of 0 degrees is towards the right. Is there any way to fix this? As I shouldn't have to add 90 to degrees in order to have correct heading. New code is below public void move(float delta, float degrees) { degrees += 90; // Set degrees to correct heading, shouldn't have to do this position.set(image.getX() + image.getWidth() / 2, image.getY() + image.getHeight() / 2); direction.set(MathUtils.cos(degrees * MathUtils.degreesToRadians), MathUtils.sin(degrees * MathUtils.degreesToRadians)).nor(); velocity.set(direction).scl(speed); movement.set(velocity).scl(delta); position.add(movement); setLocation(position.x, position.y); }

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• Euler Problem 1 : Code Optimization / Alternatives [on hold]

  - by Sudhakar

  I am new bee into the world of Datastructures and algorithms from ground up. This is my attempt to learn. If the question is very plain/simple . Please bear with me. Problem: Find the sum of all the multiples of 3 or 5 below 1000. Code i worte: package problem1; public class Problem1 { public static void main(String[] args) { //******************Approach 1**************** long start = System.currentTimeMillis(); int total = 0; int toSubtract = 0; //Complexity N/3 int limit = 10000; for(int i=3 ; i<limit ;i=i+3){ total = total +i; } //Complexity N/5 for(int i=5 ; i<limit ;i=i+5){ total = total +i; } //Complexity N/15 for(int i=15 ; i<limit ;i=i+15){ toSubtract = toSubtract +i; } //9N/15 = 0.6 N System.out.println(total-toSubtract); System.out.println("Completed in "+(System.currentTimeMillis() - start)); //******************Approach 2**************** for(int i=3 ; i<limit ;i=i+3){ total = total +i; } for(int i=5 ; i<limit ;i=i+5){ if ( 0 != (i%3)) total = total +i; } } } Question 1 - Which best approach from the above code and why ? 2 - Are there any better alternatives ?

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• Project Euler Problem 14

  - by MarkPearl

  The Problem The following iterative sequence is defined for the set of positive integers: n n/2 (n is even) n 3n + 1 (n is odd) Using the rule above and starting with 13, we generate the following sequence: 13 40 20 10 5 16 8 4 2 1 It can be seen that this sequence (starting at 13 and finishing at 1) contains 10 terms. Although it has not been proved yet (Collatz Problem), it is thought that all starting numbers finish at 1. Which starting number, under one million, produces the longest chain? NOTE: Once the chain starts the terms are allowed to go above one million. The Solution   public static long NextResultOdd(long n) { return (3 * n) + 1; } public static long NextResultEven(long n) { return n / 2; } public static long TraverseSequence(long n) { long x = n; long count = 1; while (x > 1) { if (x % 2 == 0) x = NextResultEven(x); else x = NextResultOdd(x); count++; } return count; } static void Main(string[] args) { long largest = 0; long pos = 0; for (long i = 1000000; i > 1; i--) { long temp = TraverseSequence(i); if (temp > largest) { largest = temp; pos = i; } } Console.WriteLine("{0} - {1}", pos, largest); Console.ReadLine(); }

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• Euler Problem 20

  - by MarkPearl

  This was probably one of the easiest ones to complete – a quick bash got me the following… The Problem n! means n (n 1) ... 3 2 1 For example, 10! = 10 9 ... 3 2 1 = 3628800, and the sum of the digits in the number 10! is 3 + 6 + 2 + 8 + 8 + 0 + 0 = 27. Find the sum of the digits in the number 100! The Solution   private static BigInteger Factorial(int num) { if (num > 1) return (BigInteger)num * Factorial(num - 1); else return 1; } private static BigInteger SumDigits(string digits) { BigInteger result = 0; foreach (char number in digits) { result += Convert.ToInt32(number)-48; } return result; } static void Main(string[] args) { Console.WriteLine(SumDigits(Factorial(100).ToString())); Console.ReadLine(); }

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• Non-mathematical Project Euler (or similar)?

  - by Juha Untinen

  I checked the post (Where can I find programming puzzles and challenges?) where there's a lot of programming challenges and such, but after checking several of them, they all seem to be about algorithms and mathematics. Is there a similar site for purely logic/functionality-based challenges? For example: - Retrieve data using a web service - Generate output X from a CSV file - Protect this code against SQL injection - Make this code more secure - What is wrong with this code (where the error is in logic, not syntax) - Make this loop more efficient Does a challenge site like that exist? Especially one that provides hints and/or correct solutions. That would be a very helpful learning site.

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• Better way for calculating project euler's 2nd problem Fibonacci sequence)

  - by firephil

  object Problem_2 extends App { def fibLoop():Long = { var x = 1L var y = 2L var sum = 0L var swap = 0L while(x < 4000000) { if(x % 2 ==0) sum +=x swap = x x = y y = swap + x } sum } def fib:Int = { lazy val fs: Stream[Int] = 0 #:: 1 #:: fs.zip(fs.tail).map(p => p._1 + p._2) fs.view.takeWhile(_ <= 4000000).filter(_ % 2 == 0).sum } val t1 = System.nanoTime() val res = fibLoop val t2 = (System.nanoTime() - t1 )/1000 println(s"The result is: $res time taken $t2 ms ") } Is there a better functional way for calculating the fibonaci sequence and taking the sum of the the even values below 4million ? (projecteuler.net - problem 2) The imperative method is 1000x faster ?

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