its a shame they realised that NONE filters should be used for indexed images yet they didnt provide an option to have no filters or a global filter instead of 1 per row. especially considering there are constant bytes that could have been used for such a setting like the "filter method".
Nice video; a small caveat with the normalization. 26:44 : i think the 1/n normalization is not at the w level (1/n would disappear in the power zero and we would divide for example w^{-7} by n^7, which does not match your formula for the inverse of the matrix. Instead, return y/n at the end.
06:30 why is the result (-2,9) and not (4,9) and so on? Is it because the x-coord is simply unchanged in your method, and is it important to choose set with same x's?
You clarified a lot of the math involved in JPEG compression so that now I think I understand it, or at least the most important parts. Great explanation!
how this simulation works? you can even see at 14:35 near left bottom corner that yellow and orange balls dont simply bounce but "orbits" or something for a while
The Shannon-Nyquist explanation is pretty misleading here, I think. You only need 2 points to sample a 7hz (or any other hz) wave. It's about the speed of sampling, not the number of points. The only reason you need 15 points here is specifically because of the length of the waveform shown.
That was a great explanation 👏 I'd like to add a different perspective. Suppose we have 5 disks and label our rods as 'A', 'B', and 'C'. It's important to recognize that the largest disk, disk number 5, will ultimately need to be placed at the bottom of the target rod. To achieve this, how can we remove the other 4 disks from rod 'A'? We can approach this by temporarily moving the top 4 disks (disks 1 to 4) from rod 'A' to rod 'B'. This relocation allows us to move disk 5 directly from rod 'A' to rod 'C'. Once disk 5 is placed, we then need to move the 4 disks from rod 'B' to rod 'C', starting with the smaller disks. This requires a recursive process, where each step involves moving fewer disks than the previous one until only one disk remains to be moved. Here is some Java code that illustrates this process more clearly: public class Main { public static void moveDisk(int n, char originalRod, char auxRod, char targetRod) { if (n == 1) { System.out.println("Move disk 1 from " + originalRod + " to " + targetRod); return; } moveDisk(n - 1, originalRod, targetRod, auxRod); System.out.println("Move disk " + n + " from " + originalRod + " to " + targetRod); moveDisk(n - 1, auxRod, originalRod, targetRod); } public static void main(String[] args) { int diskNumbers = 4; moveDisk(diskNumbers, 'A', 'B', 'C'); } }
