Is “Solving two-variable quadratic polynomials over the Integers” is an NP-Complete Problem?

On this Wikipedia article, it claims that given $ A, B, C \geq 0, \; \in \mathbb{Z}$ , finding $ x, \,y \geq 0, \, \in \mathbb{Z}$ for $ Ax^2+Bx^2-C=0$ is NP-complete? Given by how easy I can solve some (with nothing but Wolfram), it doesn’t seem right. I’m sure it’s either written incorrectly or I’m just misunderstanding something.

Does Pathfinder 2e Playtest fix the “linear fighters, quadratic wizards” problem?

It is generally accepted that a reasonably optimized DnD 3.5e wizard will wipe the floor with even a well optimized DnD 3.5e fighter by high levels, often known as “linear fighters, quadratic wizards”. Pathfinder 1e removed or nerfed a few of the easiest tricks for this (e.g. polymorph) but probably actually makes the problem worse.

Is there any evidence from the Pathfinder 2e playtest materials currently available that there is any serious attempt to fix this issue, or the more general issue of class balance?

Is deciding solvability of systems of quadratic equations with integer coefficients over the reals in NP?

In the book ‘Computational Complexity’ by Arora and Barak the following question is posed (exercise 2.20.):

Let REALQUADEQ be the language of all satisfiable sets of quadratic equations over real variables. Show that REALQUADEQ is NP-complete.

I know how to show NP-hardness, but I’m stuck when it comes to proving that this problem is in NP, in particular how to show that we can describe a solution using a polynomial number of bits.

I did some research and found out that over the complex numbers, it remains an open question if the problem is in NP [1]. It also seems closely related to the existential theory of the reals, which again is not known to be in NP.

Thus my question: Is this problem known to be in NP? And if so, could somebody point me in the right direction regarding the proof.

[1] Pascal Koiran, Hilbert’s Nullstellensatz is in the Polynomial Hierarchy, Journal of Complexity 12 (1996), no. 4, pp. 273–286.

Is solving a quadratic equation using Turing machine impossible?

I’ve just started Algorithms at university. There’s a task to write an algorithm for a Turing machine to solve quadratic equations. The task doesn’t specify if it’s x^2+bx+c or ax^2+bx+c. I’ve searched whole bunch of information over Russian and English Internet.

I did find articles, which say it’s not possible because we’ve got real numbers A, B, C. Please confirm if that’s true. I may not get it correct.. But I think that’s impossible. I still don’t know how to prove my thoughts.

Thanks in advance!

Expected search times with linear vs quadratic probing

Why exactly does quadratic probing lead to a shorter avg. search time than linear probing?

I fully get that linear probing leads to a higher concentration of used slots in the hash table (i.e. higher “clustering” of used consecutive indices). However, it’s not immediately trivial (to me at least) why that translates to higher search times in expectation than in quadratic probing, since in both linear and quadratic probing the first value of the probing sequence determines the rest of the sequence.

I suppose this has to do more with the probability of collisions between different probing sequences. Perhaps different auxiliary hash values are less likely to lead to collisions early in the probing sequence in quadratic than in linear hashing, but I haven’t seen this result derived or formalized.

How are Fighters Linear but Wizards Quadratic?

The phrase “Linear Fighters, Quadratic Wizards” gets bandied about a lot, but I’ve found I don’t have a good way to explain it to newer players.

The tier system post has some examples of how wizards are better than fighters in specific situations, but I don’t find the examples very satisfactory: the wizards in the examples seem to mostly rely on cheesy abuses that wouldn’t happen in an actual game. For example the post says a wizard can kill a dragon using shivering touch from Frostburn, or using mindrape and love’s pain from the Book of Vile Darkness, but many games won’t allow those books.

In an actual play scenario, with no access to any expansion books, and assuming a group of characters that aren’t grossly evil: what sorts of trends make wizards (or, more generally, full spellcasters) more powerful than non-spellcasting classes? At what character level does this start to happen, and what spells available at that level are responsible for the change?

I’m interested in responses pertaining to both 3.5e and Pathfinder; if there are important differences between the two, I’d be interested in hearing about those as well.

Mathematica returns uneditable long solutions for two simple quadratic equations

I tried to get positive solution(or any solution) of the following two quadratic equations with two variables. My code is:

Solve[(1/8)(-A1+x2+α+x1(-2+β)-2 β x2^2-θ x1^2)==0 &&       (1/16)(A1+3 x2-α-2β x2+x1(-2+3β))^2 - θ x2^2==0, {x1, x2}] 

It shows that there is large output, then i clicked show fulloutput, it took 5 minutes to display …and the result is in weird format, there is only one symbol in each line in the last part, and they are very difficult to identify, i can’t even find where x2 appears