## Use a Cell Phone Number Listing to Get the Truth Behind Your Spouse’s Strange Calls

Google will inform you that many people are currently the usage of the online reverse cellular phone appearance up directory today and  Phone Number List . There can be several reasons for utilizing the carrier among the ones the maximum common are finding out whom a prank caller is or if a spouse is cheating. Reasons range from all of us, but either manner what their motive, the manner they get entry to the online opposite cell cellphone research is the same manner.

An on line reverse cellular Bahamas phone number list  lookup provider is said to be the very nice manner due to its comfort and in cost as compared to the numerous other methods that could price you a fortune and are difficult to apply.

Among services provided to, you that provide the same facts’s as a opposite cell smartphone look up are as follows:

Detective – They will follow your accomplice and supply the desired facts however are you able to truly have the funds for what they’re going to charge for this? In addition, possibilities are your spouse will discover.

Therefore, with the available options the handiest one that makes experience is the opposite cell telephone online lookup. Just honestly pay for a one-time search. Input the suspect number click on on seek now and within a few seconds, you may recognise who is at the other give up of that mysterious telephone call. You can ease your thoughts and both confront your partner or let dozing puppies lie.

## Can you attack a creature behind a Wall of Sand?

Wall of Sand’s description reads as follows (emphasis mine):

You conjure up a wall of swirling sand on the ground at a point you can see within range. You can make the wall up to 30 feet long, 10 feet high, and 10 feet thick, and it vanishes when the spell ends. It blocks line of sight but not movement. A creature is blinded while in the wall’s space and must spend 3 feet of movement for every 1 foot it moves there.

To me, it’s unclear whether or not the bolded clause provides total cover to a creature behind the wall, or if the creature is simply heavily obscured. If you and another creature are on the opposite sides of a Wall of Sand, can you perform a ranged weapon or spell attack against the creature?

## Intuition behind Max Flow Algorithm

Given :

1. A flow network G whose edges have capacity of 1
2. G’s maximum flow |F|
3. A positive integer K

Delete exactly K edges so that the flow of the network is minimized.

So I was asked to develop an algorithm for this but that’s not the issue.I would like to know if my thoughts are in the right direction because the Max Flow – Min Cut Theorem is new to me.

My thoughts :

1. If K is greater than or equal to |F| delete ALL of the edges that cross G’s Minimum cut and if K is still greater than zero delete random edges and the new max flow is zero

2. If K is equal to |F| delete ALL of the edges that cross G’s Minimum cut and the new max flow is zero

3. If K is less than |F| delete K of the edges that cross G’s Minimum cut and the new max flow is |F|-K

Lastly the way that I understand the Max Flow/Min Cut Theorem is that the Min (Source,Sink) cut works as a "bottleneck" to the maximum flow we can push in the network,right?

## What is the lore behind why soul coins have the symbol of Bhaal on them?

I am struggling to find an image of both sides of a Soul Coin (Baator’s currency), but everywhere I’ve looked, I’ve only found only one side bearing a letter and the other side depicting a skull with blood droplets flying around it.

But it is the symbol of Bhaal. Why does it appear on Baator’s currency?

Am I correct about how soul coins look? What images appear on the sides of a soul coin, and why?

## Can the Police trace an internal user which is behind a shared IP?

I’m not sure if this is the right place to ask-

Can the Police or any other lawful body obtain information about a criminal from the ISP knowing such things as the shared IP (the NAT, I believe?), timestamps, visited websites etc.?

## Hiding in a crowd / Hiding behind multiple creatures

When it comes to stealth, 5e is very vague on what can or cannot be done.

If we refer to the rules as written:

You can’t hide from a creature that can see you,… (PH. p177)

This means that being in partial cover doesn’t grant you hiding potential since others can still “see” you.

So following this thinking, you couldn’t hide behind a target since you would get half cover and could still be seen. What about hiding behind two consecutive targets? Could a DM rule that it would provide an improvement toward 3/4 cover? But then again, one could still be seen since it’s partial cover… But what if you are behind multiple (3+) consecutive targets? At what point can we say that you are behind enough covering targets to provide a total cover?

Imagine an Assassin’s creed type of scene where the hero is concealed and breaks line of sight by hiding in a massive crowd. It would definitely be an interesting scenario but the rules on hiding as they are make it difficult.

So my question is in two parts:

• Do you think by some DM ruling, it would be appropriate to allow multiple creatures to provide enough cover to completely hide someone?
• What would be a balanced ruling to permit such stealth play?

## I am unable to understand the logic behind the code (I’ve added exact queries as comments in the code)

Our local ninja Naruto is learning to make shadow-clones of himself and is facing a dilemma. He only has a limited amount of energy (e) to spare that he must entirely distribute among all of his clones. Moreover, each clone requires at least a certain amount of energy to function (m) . Your job is to count the number of different ways he can create shadow clones. Example:

e=7;m=2

ans = 4

The following possibilities occur: Make 1 clone with 7 energy

Make 2 clones with 2, 5 energy

Make 2 clones with 3, 4 energy

Make 3 clones with 2, 2, 3 energy.

Note: <2, 5> is the same as <5, 2>. Make sure the ways are not counted multiple times because of different ordering.

``int count(int n, int k){     if((n<k)||(k<1)) return 0;     else if ((n==k)||(k==1)) return 1;     else return count(n-1,k-1)+ count(n-k,k);   // logic behind this? }  int main() {     int e,m;            // e is total energy and m is min energy per clone     scanf("%d %d", &e, &m);     int max_clones= e/m;     int i,ans=0;     for(i=1;i<=max_clones;i++){         int available = e - ((m-1)*i);   // why is it (m-1)*i instead of m*i         ans += count(available, i);     }     return 0; } ``

## The reason behind IPv6 adoption rate dramatical drop in China according to Google measurements?

Google has an IPv6 measurement page that reports that their numbers report on the percentage of users that access Google over IPv6.

According to the report by Jan 2020 0.3% of users in China used IPv6 to access Google

However, looking at this metric in dynamic we see the substantial drop starting from June 2019.

I failed to find any solid news that may cause such behavior. I have two hypotheses in mind.

1. Also as it is a percentage metric, they can adjust their calculation on the total internet penetration rate in China.
2. Previously open discussions between netizens took place on Google Plus groups. In April 2019, Google shut down Google Plus. Technical discussions continue on Chinese-language blogs, forums, and groups. For obvious reasons, discussions must be hosted outside China, and posters must register under pseudonyms. So probably that caused the shift from Google services but I hardly believe that it may cause such plummet.

## Logic behind a single-tape NTM solving the TSP in \$O({n}^4)\$ time at most

I was going through the classic text “Introduction to Automata Theory, Languages, and Computation” by Hofcroft, Ullman, Motwani where I came across a claim that a “single-tape NTM can solving the TSP in $$O({n}^4)$$ time at most” where $$n$$ is the length of the input given to the turing machine (an instance of the TSP). The authors have assumed the encoding scheme as follows:

The TSP’s problem could be couched as: “Given this graph $$G$$ and limit $$W$$ , does $$G$$ have a hamiltonian circuit of weight $$W$$ or less?”

Let us consider a possible code for the graphs and weight limits that could be the input. The code has five symbols, $$0$$, $$1$$, the left and right parentheses, and the comma.

1. Assign integers $$1$$ through $$m$$ to the nodes.

2. Begin the code with the value of $$m$$ in binary and the weight limit $$W$$ in binary, separated by a comma.

3. If there is an edge between nodes $$i$$ and $$j$$ with weight $$w$$, place $$(i, j, w)$$ in the code. The integers $$i$$, $$j$$ , and $$w$$ are coded in binary. The order of $$i$$ and $$j$$ within an edge, and the order of the edges within the code are immaterial. Thus, one of the possible codes for the graph of Fig. with limit $$W = 40$$ is

$$100, 101000(1, 10, 1111)(1, 11, 1010)(10, 11, 1100)(10, 100, 10100)(11, 100, 10010)$$

The authors move to the claim as follows:

It appears that all ways to solve the TSP involve trying essentially all cycles and computing their total weight. By being clever, we can eliminate some obviously bad choices. But it seems that no matter what we do, we must examine an exponential number of cycles before we can conclude that there is none with the desired weight limit $$W$$ , or to find one if we are unlucky in the order in which we consider the cycles.

On the other hand, if we had a nondeterministic computer, we could guess a permutation of the nodes, and compute the total weight for the cycle of nodes in that order. If there were a real computer that was nondeterministic, no branch would use more than $$O(n)$$ steps if the input was of length $$n$$. On a multitape $$NTM$$, we can guess a permutation in $$O({n}^2)$$ steps and check its total weight in a similar amount of time. Thus, a single-tape $$NTM$$ can solve the TSP in $$O({n}^4)$$ time at most.

I cannot understand the part of the above paragraph in bold. Let me focus on the points of my problem.

1. What do they mean that the branch would take no more than $$O(n)$$ ?
2. On a multitape $$NTM$$, how can we guess a permutation in $$O({n}^2)$$ steps ?
3. How are we getting the final $$O({n}^4)$$ ?

I neither get the logic nor the computation of the complexity as very little is written in the text.

## Need help to understand the math behind the logic for scheduling problem using Reinforcement Learning

I am working on a problem for scheduling VMs considering efficient resource and energy utilisation and I came across this paper. I understand RL and how Q-Learning works which they are trying to use in paper. However, I am not able to achieve an intuitive understanding of the algorithm suggested (page 3).

I understand that equal importance has been given to utilisation and power consumption but with reverse, let’s say signs. But Step-3 is not intuitive. Can someone help me get a better understanding of the same algorithm?