## Turing machine where the next step is determined by the state and the symbols up to the read/write head

Given a modified type of turning machine where the next step is determined as followed:

$$\delta = Q\times \Gamma^* \implies Q\times \Gamma \times \{L,R\}$$

where the next step of the machine is determined by the current state and whatever written on the tape up to the current point.

For example: if the tape content is $$a\_ab\_ a$$, and the head is on the left $, then the next step is determined by the state and the word $$a\_ab\_$$. 1. How can I prove that unrecognized languages can be recognized with these types of machines? 2. Do those machines contradict the Church–Turing thesis? Informative answer would be really appreciated as I’m finding it hard to understand this subject specifically. Thanks in advance ## How to show that the product of two binary numbers can be determined in AC1? I was working on a proof to determine that a product cannot be done in AC0, how can a proof that can be done in AC1? ## How to show that the product of two binary numbers cannot be determined in$AC^{0}$? Input $$x = x_{0} … x_{n-1}$$ for determine the xor over n-bits xi it is suffcient to multiply the following two n2-bits binary numbers: $$a = 0^{n-1} \hspace{2mm} x_{n-1} \hspace{2mm} 0^{n-1} \hspace{2mm} x_{n-2} \hspace{2mm} 0^{n-1} \hspace{2mm} … \hspace{2mm} 0^{n-1} \hspace{2mm} x_{1} \hspace{2mm} 0^{n-1} \hspace{2mm} x_{0}$$ $$b = 0^{n-1} \hspace{3mm} 1 \hspace{5mm} 0^{n-1} \hspace{5mm} 1 \hspace{5mm} 0^{n-1} \hspace{2mm} … \hspace{2mm} 0^{n-1} \hspace{5mm} 1 \hspace{5mm} 0^{n-1} \hspace{5mm} 1$$ The product of two binary numbers can be determined in $$AC^{1}$$? ## Why sum of two binary numbers cannot be determined in$NC^0$but in$AC^0\$?

Why sum of two binary numbers cannot be determined in $$NC^0$$ but it can be determined in $$AC^0$$?

## How to complete ESTA application when “THE PAYMENT STATUS COULD NOT BE DETERMINED AT THIS TIME”?

I received a payment error during my ESTA application and still need to pay. The current application status for the last few hours shows:

THE PAYMENT STATUS COULD NOT BE DETERMINED AT THIS TIME.
PLEASE CHECK BACK LATER FOR THE PAYMENT STATUS.

Your application is not complete and will not be processed by CBP until the processing fee has been paid.
You must complete payment within 7 days.

The help says to click the payment button if have not paid, however there is no payment button, I guess because already started the payment process.

I tried creating a new application however get an error about an application already existing for my passport. The application will expire automatically in 7 days, however I leave in 5 days.

I tried calling their support number but couldn’t get through, and their email support has a 2-3 week turnaround time.

Any ideas how to get the visa-waiver in time? Is there perhaps a way to cancel an ESTA application so can start a new one?

## How is the strength of a key/Password determined

I was reading about the Principle of Key Sizes and came across this table:

The deduction I made off this is that after a certain key size, increase in a single bit causes Time required for search to get doubled from the previous one.

Then, I came across this video How to crack Passwords in which Prof. Pound tell’s that your key strength is somewhat related to the character set of your password/Key.

For Instance if my password is of 8 length and its character set is lowercase characters, then total attempts required to crack my password will be 26^8 = 208827064576 (character_set ^ keysize) (on the worst case).

So I was wondering, whether the increase in Key_Size by bits, and this character set theory is somewhat related or not. If yes, then why most people measure strength of key by the number of bits it has, rather then the character set it uses?

And what is the most common character set used in password/key cracking, as most character tables (ASCII/UTF8) contains a lot of NPC’s and/or Emoji’s etc. so are they excluded during the key search process or not?

## When is a square matrix uniquely determined when the row sums and the column sums are given?

Define a $$A-matrix$$ as a matrix consisted of only the elements of set $$A$$. Let $$r_1,…,r_n$$be the row sums and $$c_1,…,c_n$$be the column sums. For what $$2n$$-tuple $$(r_1,…,r_n,c_1,…,c_n)$$ is there a unique $$A-matrix$$ consisted of only satisfying those sums?

## Find a vector that lies in the plane determined by a line and a point and is perpendicular to that line

I need to use the triple cross product to find a vector that lies in the plane determined by

the point (1, 0, 2) and the line $$\frac {x}{2}=\frac {y+1}{3}=\frac {z-2}{-1}$$ , and is perpendicular to the line

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## Error encountered with using rbindlist: column 25 of result is determined to be integer64 but maxType == ‘Character’ !=REALSXP

I used the following function to merge all .csv files in my directory into one dataframe:

multmerge = function(mypath){ filenames = list.files(path = mypath, full.names = TRUE) rbindlist(lapply(filenames,fread),fill = TRUE) }  dataframe = multmerge(path)

This code produces this error: “Error in rbindlist(lapply(filenames, fread), fill = TRUE) : Internal error: column 25 of result is determined to be integer64 but maxType==’character’ != REALSXP”

The code has worked on the same csv files before…I’m not sure what’s changed and what the error message means.