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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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• 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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• Confusion about Rotation matrices from Euler Angles

  - by xEnOn

  I am trying to learn more about Euler Angles so as to help myself in understanding how I can control my camera better in the game. I came across the following formula that converts Euler Angles to rotation matrices: In the equation, I could see that the first matrix from the left is the rotation matrix about x-axis, the second is about y-axis and the third is about z-axis. From my understanding about ordinary matrix transformations, the later transformation is always applied to the right hand side. And if I'm right about this, then the above equation should have a rotation order starting from rotating about z-axis, y-axis, then finally x-axis. But, from the symbols it seems that the rotation order start rotating about x-axis, then y-axis, then finally z-axis. What should the actual order of the rotation be? Also, I am confuse about if the input vector, in this case, would be a row vector on the left, or a column vector on the right?

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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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• 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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• 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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• 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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• 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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• 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 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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• 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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• 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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• Euler Problem 13

  - by MarkPearl

  The Problem Work out the first ten digits of the sum of the following one-hundred 50-digit numbers. 37107287533902102798797998220837590246510135740250 46376937677490009712648124896970078050417018260538 74324986199524741059474233309513058123726617309629 91942213363574161572522430563301811072406154908250 23067588207539346171171980310421047513778063246676 89261670696623633820136378418383684178734361726757 28112879812849979408065481931592621691275889832738 44274228917432520321923589422876796487670272189318 47451445736001306439091167216856844588711603153276 70386486105843025439939619828917593665686757934951 62176457141856560629502157223196586755079324193331 64906352462741904929101432445813822663347944758178 92575867718337217661963751590579239728245598838407 58203565325359399008402633568948830189458628227828 80181199384826282014278194139940567587151170094390 35398664372827112653829987240784473053190104293586 86515506006295864861532075273371959191420517255829 71693888707715466499115593487603532921714970056938 54370070576826684624621495650076471787294438377604 53282654108756828443191190634694037855217779295145 36123272525000296071075082563815656710885258350721 45876576172410976447339110607218265236877223636045 17423706905851860660448207621209813287860733969412 81142660418086830619328460811191061556940512689692 51934325451728388641918047049293215058642563049483 62467221648435076201727918039944693004732956340691 15732444386908125794514089057706229429197107928209 55037687525678773091862540744969844508330393682126 18336384825330154686196124348767681297534375946515 80386287592878490201521685554828717201219257766954 78182833757993103614740356856449095527097864797581 16726320100436897842553539920931837441497806860984 48403098129077791799088218795327364475675590848030 87086987551392711854517078544161852424320693150332 59959406895756536782107074926966537676326235447210 69793950679652694742597709739166693763042633987085 41052684708299085211399427365734116182760315001271 65378607361501080857009149939512557028198746004375 35829035317434717326932123578154982629742552737307 94953759765105305946966067683156574377167401875275 88902802571733229619176668713819931811048770190271 25267680276078003013678680992525463401061632866526 36270218540497705585629946580636237993140746255962 24074486908231174977792365466257246923322810917141 91430288197103288597806669760892938638285025333403 34413065578016127815921815005561868836468420090470 23053081172816430487623791969842487255036638784583 11487696932154902810424020138335124462181441773470 63783299490636259666498587618221225225512486764533 67720186971698544312419572409913959008952310058822 95548255300263520781532296796249481641953868218774 76085327132285723110424803456124867697064507995236 37774242535411291684276865538926205024910326572967 23701913275725675285653248258265463092207058596522 29798860272258331913126375147341994889534765745501 18495701454879288984856827726077713721403798879715 38298203783031473527721580348144513491373226651381 34829543829199918180278916522431027392251122869539 40957953066405232632538044100059654939159879593635 29746152185502371307642255121183693803580388584903 41698116222072977186158236678424689157993532961922 62467957194401269043877107275048102390895523597457 23189706772547915061505504953922979530901129967519 86188088225875314529584099251203829009407770775672 11306739708304724483816533873502340845647058077308 82959174767140363198008187129011875491310547126581 97623331044818386269515456334926366572897563400500 42846280183517070527831839425882145521227251250327 55121603546981200581762165212827652751691296897789 32238195734329339946437501907836945765883352399886 75506164965184775180738168837861091527357929701337 62177842752192623401942399639168044983993173312731 32924185707147349566916674687634660915035914677504 99518671430235219628894890102423325116913619626622 73267460800591547471830798392868535206946944540724 76841822524674417161514036427982273348055556214818 97142617910342598647204516893989422179826088076852 87783646182799346313767754307809363333018982642090 10848802521674670883215120185883543223812876952786 71329612474782464538636993009049310363619763878039 62184073572399794223406235393808339651327408011116 66627891981488087797941876876144230030984490851411 60661826293682836764744779239180335110989069790714 85786944089552990653640447425576083659976645795096 66024396409905389607120198219976047599490197230297 64913982680032973156037120041377903785566085089252 16730939319872750275468906903707539413042652315011 94809377245048795150954100921645863754710598436791 78639167021187492431995700641917969777599028300699 15368713711936614952811305876380278410754449733078 40789923115535562561142322423255033685442488917353 44889911501440648020369068063960672322193204149535 41503128880339536053299340368006977710650566631954 81234880673210146739058568557934581403627822703280 82616570773948327592232845941706525094512325230608 22918802058777319719839450180888072429661980811197 77158542502016545090413245809786882778948721859617 72107838435069186155435662884062257473692284509516 20849603980134001723930671666823555245252804609722 53503534226472524250874054075591789781264330331690   The Solution private static List<string> table = new List<string> { "37107287533902102798797998220837590246510135740250", "46376937677490009712648124896970078050417018260538", "74324986199524741059474233309513058123726617309629", "91942213363574161572522430563301811072406154908250", "23067588207539346171171980310421047513778063246676", "89261670696623633820136378418383684178734361726757", "28112879812849979408065481931592621691275889832738", "44274228917432520321923589422876796487670272189318", "47451445736001306439091167216856844588711603153276", "70386486105843025439939619828917593665686757934951", "62176457141856560629502157223196586755079324193331", "64906352462741904929101432445813822663347944758178", "92575867718337217661963751590579239728245598838407", "58203565325359399008402633568948830189458628227828", "80181199384826282014278194139940567587151170094390", "35398664372827112653829987240784473053190104293586", "86515506006295864861532075273371959191420517255829", "71693888707715466499115593487603532921714970056938", "54370070576826684624621495650076471787294438377604", "53282654108756828443191190634694037855217779295145", "36123272525000296071075082563815656710885258350721", "45876576172410976447339110607218265236877223636045", "17423706905851860660448207621209813287860733969412", "81142660418086830619328460811191061556940512689692", "51934325451728388641918047049293215058642563049483", "62467221648435076201727918039944693004732956340691", "15732444386908125794514089057706229429197107928209", "55037687525678773091862540744969844508330393682126", "18336384825330154686196124348767681297534375946515", "80386287592878490201521685554828717201219257766954", "78182833757993103614740356856449095527097864797581", "16726320100436897842553539920931837441497806860984", "48403098129077791799088218795327364475675590848030", "87086987551392711854517078544161852424320693150332", "59959406895756536782107074926966537676326235447210", "69793950679652694742597709739166693763042633987085", "41052684708299085211399427365734116182760315001271", "65378607361501080857009149939512557028198746004375", "35829035317434717326932123578154982629742552737307", "94953759765105305946966067683156574377167401875275", "88902802571733229619176668713819931811048770190271", "25267680276078003013678680992525463401061632866526", "36270218540497705585629946580636237993140746255962", "24074486908231174977792365466257246923322810917141", "91430288197103288597806669760892938638285025333403", "34413065578016127815921815005561868836468420090470", "23053081172816430487623791969842487255036638784583", "11487696932154902810424020138335124462181441773470", "63783299490636259666498587618221225225512486764533", "67720186971698544312419572409913959008952310058822", "95548255300263520781532296796249481641953868218774", "76085327132285723110424803456124867697064507995236", "37774242535411291684276865538926205024910326572967", "23701913275725675285653248258265463092207058596522", "29798860272258331913126375147341994889534765745501", "18495701454879288984856827726077713721403798879715", "38298203783031473527721580348144513491373226651381", "34829543829199918180278916522431027392251122869539", "40957953066405232632538044100059654939159879593635", "29746152185502371307642255121183693803580388584903", "41698116222072977186158236678424689157993532961922", "62467957194401269043877107275048102390895523597457", "23189706772547915061505504953922979530901129967519", "86188088225875314529584099251203829009407770775672", "11306739708304724483816533873502340845647058077308", "82959174767140363198008187129011875491310547126581", "97623331044818386269515456334926366572897563400500", "42846280183517070527831839425882145521227251250327", "55121603546981200581762165212827652751691296897789", "32238195734329339946437501907836945765883352399886", "75506164965184775180738168837861091527357929701337", "62177842752192623401942399639168044983993173312731", "32924185707147349566916674687634660915035914677504", "99518671430235219628894890102423325116913619626622", "73267460800591547471830798392868535206946944540724", "76841822524674417161514036427982273348055556214818", "97142617910342598647204516893989422179826088076852", "87783646182799346313767754307809363333018982642090", "10848802521674670883215120185883543223812876952786", "71329612474782464538636993009049310363619763878039", "62184073572399794223406235393808339651327408011116", "66627891981488087797941876876144230030984490851411", "60661826293682836764744779239180335110989069790714", "85786944089552990653640447425576083659976645795096", "66024396409905389607120198219976047599490197230297", "64913982680032973156037120041377903785566085089252", "16730939319872750275468906903707539413042652315011", "94809377245048795150954100921645863754710598436791", "78639167021187492431995700641917969777599028300699", "15368713711936614952811305876380278410754449733078", "40789923115535562561142322423255033685442488917353", "44889911501440648020369068063960672322193204149535", "41503128880339536053299340368006977710650566631954", "81234880673210146739058568557934581403627822703280", "82616570773948327592232845941706525094512325230608", "22918802058777319719839450180888072429661980811197", "77158542502016545090413245809786882778948721859617", "72107838435069186155435662884062257473692284509516", "20849603980134001723930671666823555245252804609722", "53503534226472524250874054075591789781264330331690"}; static void Main(string[] args) { BigInteger result = 0; table.ForEach(x => result += BigInteger.Parse(x)); Console.WriteLine(result.ToString().Substring(0,10)); Console.ReadLine(); }

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• Why do I not get the correct answer for Euler 56 in J?

  - by Gregory Higley

  I've solved 84 of the Project Euler problems, mostly in Haskell. I am now going back and trying to solve in J some of those I already solved in Haskell, as an exercise in learning J. Currently, I am trying to solve Problem 56. Let me stress that I already know what the right answer is, since I've already solved it in Haskell. It's a very easy, trivial problem. I will not give the answer here. Here is my solution in J: digits =: ("."0)":"0 eachDigit =: adverb : 'u@:digits"0' NB. I use this so often I made it an adverb. cartesian =: adverb : '((#~ #) u ($~ ([:*~#)))' >./ +/ eachDigit x: ^ cartesian : i. 99 This produces a number less than the desired result. In other words, it's wrong somehow. Any J-ers out there know why? I'm baffled, since it's pretty straightforward and totally brute force.

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• How to derive euler angles from matrix or quaternion?

  - by KlashnikovKid

  Currently working on steering behavior for my AI and just hit a little mathematical bump. I'm in the process of writing an align function, which basically tries to match the agent's orientation with a target orientation. I've got a good source material for implementing this behavior but it uses euler angles to calculate the rotational delta, acceleration, and so on. This is nice, however I store orientation as a quaternion and the math library I'm using doesn't provide any functionality for deriving the euler angles. But if it helps I also have rotational matrices at my disposal too. What would be the best way to decompose the quaternion or rotational matrix to get the euler information? I found one source for decomposing the matrix, but I'm not quite getting the correct results. I'm thinking it may be a difference of column/row ordering of my matrices but then again, math isn't my strong point. http://nghiaho.com/?page_id=846

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• JavaScript 3D space ship rotation

  - by user36202

  I am working with a fairly low-level JavaScript 3D API (not Three.js) which uses euler angles for rotation. In most cases, euler angles work quite well for doing things like aligning buildings, operating a hovercraft, or looking around in the first-person. However, in space there is no up or down. I want to control the ship's roll, pitch, and yaw. To do that, some people would use a local coordinate system or a permenant matrix or quaternion or whatever to represent the ship's angle. However, since I am not most people and am using a library that deals exclusively in euler angles, I will be using relative angles to represent how to rotate the ship in space and getting the resulting non-relative euler angles. For you math nerds, that means I need to convert 3 euler angles (with Y being the vertical axis, X representing the pitch, and Z representing a roll which is unaffected by the other angles, left-handed system) into a 3x3 orientation matrix, do something fancy with the matrix, and convert it back into the 3 euler angles. Euler to matrix to euler. Somebody has posted something similar to this on SO (http://stackoverflow.com/questions/1217775/rotating-a-spaceship-model-for-a-space-simulator-game) but he ended up just working with a matrix. This will not do for me. Also, I am using JavaScript, not C++. What I want essentially are the functions do_roll, do_pitch, and do_yaw which only take in and put out euler angles. Many thanks.

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• Calculate lookat vector from position and Euler angles

  - by Jaap

  I've implemented an FPS style camera, with the camera consisting of a position vector, and Euler angles pitch and yaw (x and y rotations). After setting up the projection matrix, I then translate to camera coordinates by rotating, then translating to the inverse of the camera position: // Load projection matrix glMatrixMode(GL_PROJECTION); glLoadIdentity(); // Set perspective gluPerspective(m_fFOV, m_fWidth/m_fHeight, m_fNear, m_fFar); // Load modelview matrix glMatrixMode(GL_MODELVIEW); glLoadIdentity(); // Position camera glRotatef(m_fRotateX, 1.0, 0.0, 0.0); glRotatef(m_fRotateY, 0.0, 1.0, 0.0); glTranslatef(-m_vPosition.x, -m_vPosition.y, -m_vPosition.z); Now I've got a few viewports set up, each with its own camera, and from every camera I render the position of the other cameras (as a simple box). I'd like to also draw the view vector for these cameras, except I haven't a clue how to calculate the lookat vector from the position and Euler angles. I've tried to multiply the original camera vector (0, 0, -1) by a matrix representing the camera rotations then adding the camera position to the transformed vector, but that doesn't work at all (most probably because I'm way off base): vector v1(0, 0, -1); matrix m1 = matrix::IDENTITY; m1.rotate(m_fRotateX, 0, 0); m1.rotate(0, m_fRotateY, 0); vector v2 = v1 * m1; v2 = v2 + m_vPosition; // add camera position vector glBegin(GL_LINES); glVertex3fv(m_vPosition); glVertex3fv(v2); glEnd(); What I'd like is to draw a line segment from the camera towards the lookat direction. I've looked all over the place for examples of this, but can't seem to find anything. Thanks a lot!

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• [Physics]Why Verlet better then Euler ?

  - by GameProgrammer89

  Hello! Can someone explain me why verlet integration is better then euler ? And why rk4 better then verlet ? I don't understand why it is better method :/

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• what is the Euler rotation convention in truevision3d?

  - by Fredriku73

  In what axis order are Euler angles rotated in the graphics API true vision 3d (tv3d)? I haven't been able to find it in any documentation.

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• Encrypt text using a number

  - by Adam Matan

  Project Euler I have recently begun to solve some of the Project Euler riddles. I found the discussion forum in the site a bit frustrating (most of the discussions are closed and poorly-threaded), So I have decided to publish my Python solutions on launchpad for discussion. The problem is that it seems quite unethical to publish these solutions, as it would let other people gain reputation without doing the programming work, which the site deeply discourages. My Encryption problem I want to encrypt my answers so that only those who have already solved the riddles can see my code. The logical key would be the answer to the riddle, which is always numeric. In order to prevent brute-force attacks on my answers, I want to find an encryption algorithm that takes a significantly long time (few seconds) to run. Do you know any such algorithm? I would fancy a Python package, which I can attach to the code, over an external program that might have portability issues. Thanks, Adam

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• More ruby-like solution to this problem?

  - by RaouL

  I am learning ruby and practicing it by solving problems from Project Euler. This is my solution for problem 12. # Project Euler problem: 12 # What is the value of the first triangle number to have over five hundred divisors? require 'prime' triangle_number = ->(num){ (num *(num + 1)) / 2 } factor_count = ->(num) do prime_fac = Prime.prime_division(num) exponents = prime_fac.collect { |item| item.last + 1 } fac_count = exponents.inject(:*) end n = 2 loop do tn = triangle_number.(n) if factor_count.(tn) >= 500 puts tn break end n += 1 end Any improvements that can be made to this piece of code?

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