Video

In this example the order of result is necessary after that I try to calculate by myself with n=4 this algorithm seems not work very well

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.

should've included what line of code the action is taking place, for better understanding i got lost

Wow man. Love you

Very nicely presented, clear and concise. You are the great teacher.

this video is so well done, thank you so much!

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?

By Far the Best Explanation Ever for Recursion with Those Tips😀😀

Teach better than my professor

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!

Why not using recursion?

I don't see why the coefficient to value representation algorithm is O(n log n)

I got lost at “We’re given 2 polynomials”

Instant confusion. Right out of the gate lol.

char* ptr0 = {'h', 'i'}; char* ptr1 = {' ', 'm'}; char* ptr2 = {'a', 'n'}; char** aptr = {*ptr0, *ptr1, *ptr2}; char** ptptptaotp = ***aptr;

part 2?

thank youuuuuu

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.

The best explanation!!This guy is a gem

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'); } }

legend

Brilliant!

im so lost lmao Im just hoping that this wont come out in interviews because it's so popular lmao

legend

i got confused at @5:50 because what if the element of at LIS[4] is 1 therefore the formula would not be valid, right?

Why other = 6 - (start + end) ?

1:50 I thought of this by accident; Interestingly, if you assign zero as the start of your acceptable range there is an extra sequence that could be valid: [2, 6, 6, 9] --> 4 I assume if you continue to add or decrease your starting position the Goldilocks zone changes. I remember from somewhere that the resulting numbers of a functions subset, remains constant in any set. Assuming of course that each node remains constant, without the set changing mid step.

Very helpful, thank you.

your last example is totally unclear. When you clearly said m goes from 0 up to m, why didn't you don't consider m = 0 when doing partition for (4,3) and (5, 3)? Also you considered m=0 for partition for (3, 3), SMH. At least be consistent in your examples.

Can you please provide a solution to the Box Stack problem in case if you are allowed to rotate the boxes? The original Box Stack problem states that you can rotate the boxes however you want and any side of the box may serve as a base of that box. I think it complicates the problem a bit. Thank you!

Great video but I still find it hard to wrap my mind around, by no fault of yours lol. Got a test tomorrow too.

Very clear explanation！

B - Boss F - Fighting S - Stages

An interesting twist on count_partitions would be to return the actual partitions instead of the count.

Step 1: collect underpants Step 2: …… Step 3: profit

Are you going to add move videos or is this channel dead?

around 6:30, you said (row, col): (2,2). but it seems like (3,3). what am I missing here?

Best explanation to Hanoi tower and its basis in discrete mathematics and algorithms, This is multi target video!!

Looking at that paper for the polynomial approximation of field strength, I think I have found a potential improvement. Specifically, I found a visually indistinguishable polynomial which can be computed in fewer arithmetic operations. If we let s = r²/R², the paper recommends using p = 1 - (22/9)s + (17/9)s² - (4/9)s³, which they say can be computed in 5 multiplications and 3 additions. However, if we let u = (1 - s)² then we can use q = (3 + u)u/4 as our polynomial. This takes only 2 multiplications (1 each for u and q) and 3 additions if we count division-by-4 as an addition. If we count it instead as a multiplication then we’re at 3-and-2, but either way it’s fewer total operations and fewer multiplications than the method from the paper. And in fact, we can completely eliminate the division-by-4 if we’re happy with a different overall scale for the field strength. That would bring the operation counts down to 2-and-2. Note that my polynomial q satisfies all the same constraints on its values and slopes that the paper requires of p. Here is a Desmos graph showing the two curves on top of each other: www.desmos.com/calculator/dde9rpjtj2 (Okay, technically my curve crosses y=1/2 at a slightly different x-value, but that was not one of their original constraints. They added it when they had an extra degree of freedom. My approach can still match it though, if we use q_2 = (47/16 + u)u(16/63), where again the final multiplication by 16/63 can be eliminated if we don’t care about the overall scaling of the field.) And of course, if we we just want a vaguely-similar shape, and don’t care too much about the specifics, then we can simply use u itself. That only takes 1 multiplication and 1 addition.

I have a test tomorrow thank you ❤

Can you do count_partitions(n-1,m) + count_partitions(n,m-1)?

What an awesome video, I was struggling to understand recursion until I saw this video. keept them comming

I feel bad for our teachers who had to teach us this 😂

Now I just gotta do it in assembly...

What a clean video anyone can understand from images itself Thanks a lot for this

I have never been good in math, even simple math. I have an app called IMPULSE. One of the games is Tower of Hanoi. It started out relatively simple but I was taking so long to finish each level and was ending up at 2% of the number of moves and time used. So i searched out a video to help me understand how this game works. I never thought it was a mathematical problem. One of my issues is my ADHD & ASD brain. Trying to keep organized in my thinking is not easy. But now I hope to finish my next level in much fewer moves. I will never reach a faster time, but improving in fewer moves is now my biggest goal. Thank you for this video

love it... awesome explanation of recursion.

You're the only cs channel using manim at that level. Respect bro, keep going