Box2d: High screen resolution / frequency causes high friction?

I’m using Cocos Creator with (built-in) box2d for physics.

Recently our game behaves weirdly on our new device Galaxy S20 Ultra 5G – which has screen size = 1440 x 3200 – frequency = 120Hz.

After stop pushing, all our physical bodies almost stop immediately like they has very high friction. No other device react that way.

Anyone experienced this issue can give me an advice?

How does each element in the output array of a DFT correspond to a specific frequency?

I have a basic understanding of the Fourier Transform, though I’m trying to use it in a program and I’m confused on the specifics. Based on source code I can find online, the DFT takes a set of samples/numbers, performs a summation for each term, and returns a set of these summations which is the same size as the input set. Suppose I have a periodic function. As I understand it, the output array should contain the amplitudes/weights of each frequency which sum to that function. What I can’t figure out is how each frequency is encoded in the array as just an index. In each example I read, we just assign a summation at each iteration of the inner loop to the next consecutive index in the output array. How are these indices indicative of which frequency they correspond to?

I’m attaching the source code I’m referencing to the bottom of this in case the website I linked to ever goes down.

import cmath def compute_dft_complex(input):     n = len(input)     output = []     for k in range(n):  # For each output element         s = complex(0)         for t in range(n):  # For each input element             angle = 2j * cmath.pi * t * k / n             s += input[t] * cmath.exp(-angle)         output.append(s)     return output 

Frequency of cantrips outside of battle

Despite being “at will”, how often can a cantrip be cast outside of battle? For example, how long must a player wait before shooting 2 fire bolts, and is there a limit to how many could be cast or can they just go at it continuously for a day without incurring penalties?

Some Background

I have recently started DM’ing and there’s a few concepts that still elude me, or at least I am finding some trouble reconciling my interpretation of the rules and practical balance.

As I understand it, cantrips can be cast “at-will”. That’s fine inside of battle, but outside of it, things begin to get extremely broken. I have a player whose solution to most challenges is to “fire bolt it”. Is there a reinforced door? A stone wall? A shielded altar? He will just fire bolt it for 8 consecutive hours… Now, on most occasions I have been able to come up with counter-measures, like, bouncing the fire bolts back at him when they hit the altar’s shield, but still, it sounds ridiculous to me that you could just cast continuously for 2/3 of a day if you wished it so, without incurring any penalties… it’s simply absurd on so many levels! And yes, I know you technically could say “I hit it with my club for 8 hours”, and it would be just as ridiculous in my mind that there would be no fatigue, no weapon deterioration, etc…

I have told my player he can cast once every 6 seconds (since that’s the duration of a turn in battle), but I have no idea if that is correct, and it still feels unbalanced.

Given node and value find frequency of given value from node to root in a tree

A tree with N vertices and N-1 edges is present.

A value x can be inserted in any of the nodes.

A single node may contain multiple values.

How do I answer queries of the given type?

Given a node n and value v find frequency of v in the path from node n to the root of the tree.

I require an algorithm that answers such queries in log(N) time.

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Information Theory: Comparing surprisal of words with varying count frequency

This is a very broad question, I’m not sure if cstheory is the better place.

How can I compare the conditional surprisal of words that vary in frequency?

$ S(w|context)=−log(p(w|context))=−log(\frac{count(w,context)}{count(context)})$

The $ count(w,context)$ depends on the frequency of the word w because this can be further broken down to $ p(w|context)count(w)$ . This means a word that is more frequent will have a lower surprisal.

Is there a way to compare the surprisal of words with varying count/frequency, i.e control for frequency? Do I just divide $ count(w,context)$ by $ count(w)$ to normalize by count?

Disable CPU frequency scaling?

I need to run ATLAS on my ubuntu laptop, but the software won’t run unless you dissable CPU frequency scaling. I’ve tried the instructions from here but they aren’t working on my Toshiba laptop. With freq-scaling on the software becomes useless.

Can anyone help?