Why is it impossible to iterate over all TMs with $n$ states and $k$ symbols that halt after $m$ steps on $\epsilon$?

Define $ \{\sigma(n,k,m,i)\}_{i=1}^{l_m}$ an ordered set of all TMs with $ n$ states and $ k$ symbols that halt after $ m$ steps on $ \epsilon$

There are $ (2kn)^{kn}$ TMs with $ n$ states and $ k$ symbols, so $ l_m$ is always finite and so is the range of $ m$ .

If we could iterate over $ \sigma$ given $ n,k$ for all $ m,i$ , then we could decide the halting problem:

  1. Get $ n,k$ from input machine
  2. Iterate over $ \sigma$ for all $ m,i$ and check for each one if $ \sigma(n,k,m,i)$ equals the input machine, if they are equal accept.
  3. Reject.

This is odd because It seems very feasible to constract $ \{\sigma(n,k,m,i)\}_{i=1}^{l_m}$ using elementary combinatorics. Even settling on a (computable) combinatorial formula for $ l_m$ for all $ m$ given $ n,k$ will be enough to solve the haling problem since we will know how many machines are there that halt for each $ m$ , and we can simulate one step at a time for all the $ (2kn)^{kn}$ machines and compare each machine that halts with the input machine until reaching $ l_m$ for all $ m$ without a correct comparison and reject.

Is there a fact (independent of it contradicting the Halting problem) that makes it clear why there is no computable combinatorial formula for $ l_m(n,k)$ ? or am I missing some detail here?

If a decision problem’s solution is exponentially large, then is it impossible for it to be in $NP$?

Decision Problem: Is $ 2^k$ + $ M$ a prime?

The inputs for both $ K$ and $ M$ are integers only. The solution is the sum of $ 2^k$ +$ M$ . (Use AKS to decide prime)

The powers of 2 have approximately $ 2^n$ digits. Consider $ 2^k$ where $ K$ = 100000. Compare the amount of digits in $ K$ to the amount of digits in it’s solution!

Question

Could the solution be verified in polynomial time even though the solution is exponentially large?

If no, can I say it is not in $ NP$ ?

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!

Is longest-path with a specific source and destination impossible in polynomial time?

The problem of finding the longest path in a graph is known to be not be possible in polynomial time, that I am aware of. I am also aware that using DFS or BFS can give the shortest distance between a given origin and destination in a graph. Is it possible to find a longest path from a source vertex to a destination vertex in polynomial time?

Are tiny point-symmetric grids of two Pentominos impossible to solve?

The game of Pentomino is a tiling puzzle game played on a grid. A Pentomino piece is a two dimensional shape of five non-diagonally connected tiles. There are exactly 12 unique pieces (ignoring rotation/reflection). The goal of the game is to fill the board with the available pieces without overlapping.

Pentomino pieces and names

This question concerns point-symmetric boards (not necessarily rectangular) of 10 spaces, i.e. with room for exactly two pieces.

Filling such a board is, of course, trivial when using the same piece twice.

Using the Y-piece twice on a point-symmetric hourglass board: ####  O# OOOO 

However, it seems impossible to do when using two different pieces. Is this true? How could this be proved?

This is not the case for larger boards or larger pieces (more than 5 tiles). Of course, the standard 6×10 board using 12 Pentomino pieces is also a counter example when using more than two pieces. Small counter examples:

Three 5-tile Pentomino pieces (V, P and L pieces) on a 3x5 grid: vvvpp vLppp vLLLL  The 5-tile pieces P, T and F on an oval-shaped point-symmetric grid.  ppTF pppTFFF   TTTF  Two 8-tile pieces on a 4x4 grid: #### #OOO #OOO ##OO 

It feels impossible to play non-D&D RPG’s

I dislike D&D, but it feels like the only choice I have when I want to play an RPG with friends. D&D is so popular and so many people play it exclusively, that it takes way too much effort to even introduce other RPG’s to D&D players, as they generally are too scared to try anything besides it. It feels like D&D is popular sure, but all other RPG’s are still incredibly niche.

How to prevent 2FA (or MFA) being bypassed 100%, impossible to penetrate?

what can I do in order to make 2FA impenetrable, provided that the service has befitting safety measures?

For example, if I have a bank app I can see my bank balance with or I can make purchases or money transfers with, and it has a 2FA having single-usage codes sent to a phone number, would it be safe enough to put a secondary phone unknown to everybody (except the bank of course)?

I mentioned a secondary phone number because obviously potential perpetrator knowing my number could potentially drain out my (hypothetical) bank balance if I put my original phone number as a 2FA step, by using the sim swap technique

And also if there are banks providing a physical code generator, then obviously the problem doesn’t stand in and of itself.

I’d gladly accept any answer mentioning other scenarios or from which I can take additional advices unknown to me.