How is the Longest Path Problem NP complete?

From the following link:

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So basically, in our iff proof, we have to show two directions:

Forward: If Hamiltonian Path has a yes-instance, so does longest path. This makes sense because we can just let “k” = |V| – 1 if hamiltonian path is yes. Then clearly there is a longest simple path with |V| – 1 edges.

I’m having trouble with the backward part

Backward: If Longest Path has a yes instance, so does longest path. Let’s assume that there is a longest path from s to t of length k (this can be a different k than the one we defined above?). How does that guarantee that there is a hamiltonian path from s-t? If it is the same k, where k = |V| – 1, then I agree there is a hamiltonian path, but what if this k is something different?

What is the complete lifecycle of souls in the Nine Hells?

The Descent into Avernus adventure provides more information about what happens to souls in the Nine Hells/Baator than was previously available (soul coins and their uses, in particular).

It’s also made it clear that I really don’t understand the paths that souls in the Hells can follow, and thus don’t feel able to convey the moral/logistical implications of any particular action relating to souls to players.

I’m looking for a complete picture of the soul lifecycle in the Hells (and I’m not expecting it to be linear). I could ask a bunch of smaller questions, but that wouldn’t actually help me understand the lifecycle as a whole so I can reason about it in unusual situations.

Answers should probably include:

  • States such as: being a living lawful evil mortal who hasn’t made a devilish pact, being a living mortal who has sold their soul, being a lemure, being in the River Styx, being tortured to draw out soul energy, being a higher rank devil, being trapped in a soul coin
  • Transitions such as: dying, being forged into a soul coin, being utterly annihilated, being freed from a soul coin, being promoted, being demoted
  • How the states and transitions relate to each other

5e materials & designer statements should take precedence, but materials from earlier editions are welcome in answers.

How to use Scrapebox as a complete beginner without using an API?

I have purchased as much as I can in the starting up of backlinking and I have followed advice from the best of them. I have bought Scrapebox, GSA, and The Best Spinner and Now I am out of money until next week.

Can anybody here guide me to using Scrapebox the best or only way possible without needing an api (just until next week) in conjunction with GSA Search Engine Ranker?

I have gsa set up and im just trying to get some lists because apparently my campaign has stopped because of that (it needs more urls to post to).

Can anyone tell me, am I able to extract Urls from Scrapebox and move forward with GSA just for the time being and if so can you tell me how?



Special Monotone SAT problem: NP complete?

Say we have the set $ X=\{ x_1, x_2, \dots \}$ of variables. Then we consider the following problem: Is the formula $ $ \bigwedge_{(a,b,c) \in A}(a \vee b \vee c) \wedge \bigwedge_{(a,b,c) \in B}(\neg a \vee \neg b \vee \neg c)$ $ for $ A,B \subset X^3$ satisfiable.

This problem has been called Montone-3-Sat before and it is known to be NP-complete.

My question is: If we assume that $ A=B$ has to hold, is the problem still NP-complete? I.e. we ask if we can color vertices in such a way that always specific three of them do not have the same color.

The results I have found on the internet confuse me a bit, because I believe the definitions of Monotone-Sat, Not-All-Equal-Sat are not always the same.

Proof that an almost complete binary tree with n nodes has at least $\frac{n}{2}$ leaf nodes

I’m having some trouble proving what my title states. Some textbooks refer to almost complete binary trees as complete, so to make myself clear, when I say almost complete binary tree I mean a binary tree whose levels are all full except for the last one in which all the nodes are as far left as possible.

I thought of proving it using induction but I’m not really sure how to do so. Any ideas?

How to find a lambda term to complete a function?

I tried to complete this exercise but i stopped… Defining a $ \lambda $ -term M such that: $ $ (<M,u>)<M,v> \: \simeq_{\beta} \: <M,u>$ $

I chose $ M=\lambda m \lambda a \lambda b \lambda p \,((p)m)b \:$ then i have to find a representation T of a function using M that value true if the sequence is empty and false if it’s not. A sequence is defined as: $ $ []=\lambda x_0\lambda x_1 \lambda z z \ [b]=\lambda x_0 \lambda x_1 \lambda z (z) x_b\ [b_1 b_2]=\lambda x_0 \lambda x_1 \lambda z ((z)x_{b_1})x_{b_2} \ .\. \ . \ [b_1 .. b_n]= \lambda x_0 \lambda x_1 \lambda z (…((z) x_{b_1})x_{b_2}…)x_{b_n} $ $ so the sequence of exercise is : $ $ [01101]= \lambda x_0 \lambda x_1 \lambda z (((((z)x_0)x_1)x_1)x_0)x_1 $ $ For example T need to be: $ (T)[01101] \simeq_{\beta}$ false while $ (T) []\simeq_{\beta}$ true. I really find that difficult. How i can do that?

RSolve produces a solution which is not complete?

Consider the following solution of a recurrence relation in two variables

RSolve[f[x + 1, y] == f[x, y + 1], f[x, y], {x, y}] 

{{f[x, y] -> C[1][x + y]}}

It suggests that the recurrence relation is in general satisfied by any function that depends on the sum of x and y only. However, it is trivial to see that we also have

f[x + 1, y] == f[x, y + 1] /. f[x_, y_] -> (-1)^(x - y) C[2][x + y] // Simplify 


which uses a function different from the type of solution returned by RSolve, but still properly solves the recurrence relation.

Does this mean that RSolve does not return the most general solution? Is this a bug, or am I misunderstanding how RSolve should work?