Newbe Dev Stack

Answer : Note the example given is a left-handed, row major matrix . So the operation is: Translate to the origin first (move by - eye ), then rotate so that the vector from eye to At lines up with +z: Basically you get the same result if you pre-multiply the rotation matrix by a translation - eye : [ 1 0 0 0 ] [ xaxis.x yaxis.x zaxis.x 0 ] [ 0 1 0 0 ] * [ xaxis.y yaxis.y zaxis.y 0 ] [ 0 0 1 0 ] [ xaxis.z yaxis.z zaxis.z 0 ] [ -eye.x -eye.y -eye.z 1 ] [ 0 0 0 1 ] [ xaxis.x yaxis.x zaxis.x 0 ] = [ xaxis.y yaxis.y zaxis.y 0 ] [ xaxis.z yaxis.z zaxis.z 0 ] [ dot(xaxis,-eye) dot(yaxis,-eye) dot(zaxis,-eye) 1 ] Additional notes: Note that a viewing transformation is (intentionally) inverted : you multiply every vertex by this matrix to "move the world" so that the portion you want to s...

Answer : Implementation 1 returns the magnitude of the vector that would result from a regular 3D cross product of the input vectors, taking their Z values implicitly as 0 (i.e. treating the 2D space as a plane in the 3D space). The 3D cross product will be perpendicular to that plane, and thus have 0 X & Y components (thus the scalar returned is the Z value of the 3D cross product vector). Note that the magnitude of the vector resulting from 3D cross product is also equal to the area of the parallelogram between the two vectors, which gives Implementation 1 another purpose. In addition, this area is signed and can be used to determine whether rotating from V1 to V2 moves in an counter clockwise or clockwise direction. It should also be noted that implementation 1 is the determinant of the 2x2 matrix built from these two vectors. Implementation 2 returns a vector perpendicular to the input vector still in the same 2D plane. Not a cross product in the classical sense but c...

Answer : This link might be helpful to you, as it details the use of the Haversine formula to calculate the distance. Excerpt: This script [in Javascript] calculates great-circle distances between the two points – that is, the shortest distance over the earth’s surface – using the ‘Haversine’ formula. function getDistanceFromLatLonInKm(lat1,lon1,lat2,lon2) { var R = 6371; // Radius of the earth in km var dLat = deg2rad(lat2-lat1); // deg2rad below var dLon = deg2rad(lon2-lon1); var a = Math.sin(dLat/2) * Math.sin(dLat/2) + Math.cos(deg2rad(lat1)) * Math.cos(deg2rad(lat2)) * Math.sin(dLon/2) * Math.sin(dLon/2) ; var c = 2 * Math.atan2(Math.sqrt(a), Math.sqrt(1-a)); var d = R * c; // Distance in km return d; } function deg2rad(deg) { return deg * (Math.PI/180) } I needed to calculate a lot of distances between the points for my project, so I went ahead and tried to optimize the code, I have found here. On average in different bro...

Answer : The Formula It's actually very easy to compute N choose K without even computing factorials. We know that the formula for (N choose K) is: N! -------- (N-K)!K! Therefore, the formula for (N choose K+1) is: N! N! N! N! (N-K) ---------------- = --------------- = -------------------- = -------- x ----- (N-(K+1))!(K+1)! (N-K-1)! (K+1)! (N-K)!/(N-K) K!(K+1) (N-K)!K! (K+1) That is: (N choose K+1) = (N choose K) * (N-K)/(K+1) We also know that (N choose 0) is: N! ---- = 1 N!0! So this gives us an easy starting point, and using the formula above, we can find (N choose K) for any K > 0 with K multiplications and K divisions. Easy Pascal's Triangle Putting the above together, we can easily generate Pascal's triangle as follows: for (int n = 0; n < 10; n++) { int nCk = 1; for (int k = 0; k <= n; k++) { System.out.print(nCk +...