SSL Cert for client side web application, is it needed?


I have tried to find good answer for it, but I haven’t gotten good article about this topic.

Since there are 2 types of client applications (in bigger picture) – one that runs on server and one that you download and runs in your browser.

My question comes in about the one, that runs in your machine (that you download at first visit – Blazor WebAssembly to be specific).


Do I need to enable SSL (HTTPS) for this application or web server, that hosts this application as well or is it not needed in the end?

Does only having API connection encrypted be enough?


Yes, this is a cost saving masure, since this is for my hobby project and I would like to keep running costs as minimal as possible. But since I still exchange data, that should not be seen by 3rd party, this application needs to be secure.

To enable HTTPS I would need second Static IP, which is 3$ a month (which is not much), but again, it is additional cost for me, that I would rather not have.

Algorithm to get the number of iterations needed, if possible, to get an specific 2 element array

I am fairly new to algorithms and I am dealing with a problem I cannot fully translate into mathematical language.

So, I am given the array [1,1] and I can only perform one sum between their numbers per step. Thus,

0: [1,1] 1: [2,1], [1,2] 2: [3,1], [2,3], [3,2], [1,3] 3: [4,1], [3,4], [5,3], [2,5], [5,2], [3,5], [4,3], [1,4] ...and so on. 

The goal is to know how many steps are needed in order to get a given [x,y] array.

This far, I know that

if (min(x,y)==1) --> steps =max(x,y)-1  if (x%2 ==0 and y%2==0) (both even) --> steps= not possible if (max(x,y)%min(x,y) == 0) (one is multiple of the other or x,y are the same ) --> steps= not possible if (x%3 ==0 and y%3==0) (both divisble by 3) --> steps= not possible 

Also I plotted for each pair (x,y) how many steps are needed, and I can see a pattern happening for every multiple of x or y, but I can’t write it as a mathematical function when x or y is >= 5.

Any guidance will be much appreciated.

Steps needed per (x,y)

How much time and money would be needed to make a giant magic submarine?

My group is level 19 and we are currently messing around on an island chain near the coast. Our whole team is into the idea of making an epic-level submarine. Our Wizard, Artificer and me (the Bard) are willing to sacrifice alot of time to help make it work. We also have a long standing favor to call upon with a Dwarven city state that we saved.,the%20expedition%20to%20find%20Atlantis.&text=It%20is%20big%20enough%20to,maintaining%20some%20moving%20space%20within.

We would use this as the basic layout and size but with magic cannon and harpoon cannon instead of torpedo launchers. Between us and the dwarven blacksmiths’ and tinkers’ guides helping us as a favor, How much time and GP it will take going off the handbook?

PS. In our group if we come to the DM with a fully formed idea and alot of the work done already he is more likely to clear it. So he is unaware of this plan so far.

Help needed with Tabs and Sub Tabs Creation

Hi Everyone,
I'm new to Web site Page Design Just learning I have Web site
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Help needed with Tabs and Sub Tabs Creation

Freelance PDF Creator Needed

Hi, Folks.

I’m not sure if this is the right forum, so please let me know if I need to post elsewhere.

OK, I have 4300 words of text that I want to convert into an easy-to-read PDF file, something like the attached image. I know how to convert OpenOffice and Word files, etc., into PDF, but I want something more polished. I guess that means a template of some kind with colors, boxes, a few simple graphics, and a format that flows. I also need to be able to edit the file for when I add…

Freelance PDF Creator Needed

Help needed: How to set up a campaign/project for non-English speaking regions properly?


I’ve been trying to to find my way around the Search Engine Ranker lately. I’m pretty sure that you can do some things wrong when you want to start a project for a language area outside of English.

Does anyone have experience with campaigns for other languages like German, French or Spanish. What to look for and what data to translate into the target language.

Your help is highly appreciated.

Thanks in advance

How to calculate wierd number in efficient way – optimization of algorithm needed

I am trying to print n weird numbers where n is really big number (eg: 15000).

I found this site to check the algorithm for n 600 if I have some errors:

However, my algorithm is really slow in bigger numbers:

import java.util.ArrayList; import java.util.List;  public class Test {      public static void main(String[] args) {         int n = 2;          for ( int count = 1 ; count <= 15000 ; n += 2 ) {             if (n % 6 == 0) {                 continue;             }              List<Integer> properDivisors = getProperDivisors(n);             int divisorSum = -> i.intValue()).sum();              if ( isDeficient(divisorSum, n) ) {                 continue;             }              if ( isWeird(n, properDivisors, divisorSum) ) {                 System.out.printf("w(%d) = %d%n", count, n);                 count++;             }         }     }      private static boolean isWeird(int n, List<Integer> divisors, int divisorSum) {         return isAbundant(divisorSum, n) && ! isSemiPerfect(divisors, n);     }      private static boolean isDeficient(int divisorSum, int n) {         return divisorSum < n;     }      private static boolean isAbundant(int divisorSum, int n) {         return divisorSum > n;     }      private static boolean isSemiPerfect(List<Integer> divisors, int sum) {         int size = divisors.size();          //  The value of subset[i][j] will be true if there is a subset of divisors[0..j-1] with sum equal to i          boolean subset[][] = new boolean[sum+1][size+1];          // If sum is 0, then answer is true          for (int i = 0; i <= size; i++) {             subset[0][i] = true;          }          //  If sum is not 0 and set is empty, then answer is false          for (int i = 1; i <= sum; i++) {             subset[i][0] = false;          }          // Fill the subset table in bottom up manner          for ( int i = 1 ; i <= sum ; i++ ) {             for ( int j = 1 ; j <= size ; j++ ) {                 subset[i][j] = subset[i][j-1];                 int test = divisors.get(j-1);                 if ( i >= test ) {                     subset[i][j] = subset[i][j] || subset[i - test][j-1];                  }             }          }           return subset[sum][size];     }      private static final List<Integer> getProperDivisors(int number) {         List<Integer> divisors = new ArrayList<Integer>();         long sqrt = (long) Math.sqrt(number);         for ( int i = 1 ; i <= sqrt ; i++ ) {             if ( number % i == 0 ) {                 divisors.add(i);                 int div = number / i;                 if ( div != i && div != number ) {                     divisors.add(div);                 }             }         }         return divisors;     }  } 

I have three easy breakouts:

  1. If a number is divisable by 6 it is semiperfect which means it cannot be weird

  2. If a number is deficient this means it cannot be weird

The above points are based on

  1. If a a number is odd it cannot be weird at least for 10^21 numbers (which is good for the numbers I am trying to obtain).

The other optimization that I used is the optimization for finding all the dividers of a number. Instead of looping to n, we loop to SQRT(n).

However, I still need to optimize: 1. isSemiPerfect because it is really slow 2. If I can optimize further getProperDivisors it will be good too.

Any suggestions are welcome, since I cannot find any more optimizations to find 10000 weird numbers in reasonable time.

PS: Any code in Java, C#, PHP and JavaScript are OK for me.

EDIT: I found this topic and modified isSemiPerfect to look like this. However, it looks like it does not optimize but slow down the calculations:

private static boolean isSemiPerfect(List<Integer> divisors, int n) {         BigInteger combinations = BigInteger.valueOf(2).pow(divisors.size());         for (BigInteger i = BigInteger.ZERO; i.compareTo(combinations) < 0; i = i.add(BigInteger.ONE)) {           int sum = 0;           for (int j = 0; j < i.bitLength(); j++) {             sum += i.testBit(j) ? divisors.get(j) : 0;           }            if (sum == n) {             return true;           }         }          return false;       } 

The script is still running from 11 hours and I am only at 4800th number.

Architecture of smartphone security: why FBI needed apple’s help?

I want to focus on technical aspects, not on the fact that they wanted to make a precedence.

i assume the smartphone security architecture is following:

  1. cryptography chip. it’s read only and stateless. it contains physical cryptography key. it offers some transformations of user input. it doesn’t expose the key. it doesn’t remember number of retries

  2. NAND disk. contains encrypted data

  3. OS. get input from user, talks to the chip, changes the content of the NAND

  4. retries counter. no idea where is it? is it stored on NAND disk or some other dedicated long term memory?

from what i know the FBI wanted apple to make for them less secure iOS version that doesn’t erase the disk after a few failed retries. but why do the need it? can’t they just:

  • make a copy the NAND disk (in case it has some killswitch)
  • get the chip’s spec and just send to it a few millions decrypt request (testing every possible user pin / password)
  • if the chip stores retires counter in some dedicated memory, they can always plug in a tweaked memory that always replies with the same value when read

why do they even need an OS? it’s just a simple program that can communicate with a chip. what am i missing?