google seems not to be the real google

I have a problem with google searches that occasionally point me to sites containing only nonsense. The nasty thing is, it happens only occasionally, like every hour or so of browsing which made me believe several times I solved the issue. A redirection of the link always happens but this is not the problem. The link displayed is already doggy and if I type in the found url directly in the browser, I always get "No Permission". If the response is an URL that i happen to know, then I am always redirected to the correct site.

The problem started during the weekend on my computer and that of my wife and for every browser, Firefox, Chrome, Edge. Mobiles are not affected though. I run Linux (Tumbleweed) and my wife Windows 10.

So I thought it must be my router, ASUS ac68u. Hence I factory reseted it, changed the password and installed the newest firmware (well, it did it automatically). But the problem persisted.

I ran a virus scanner on my wife’s Windows 10 and it found nothing. I also set another DNS server on all machines, 8.8.8.8, the only DNS address I remembered from ancient times. But also no luck.

My conclusion is that only google queries are affected (no problem with duckduckgo). But that is not plausible, because the SSH keys would also ad to be faked by the attacker. My browser says it’s google when right clicking on the lock.

I really do not know where to look further. Next thing I will probably do is install a fresh OS, which is easy enough on Linux, but on Windows this will be a lot more work.

BTW, everything I install in Linux is from the package manager. I have compiled a few projects from github though.

The first dodgy site I saw was the infamous "You won a price" scam. But all security sites say it is a browser hack. I cannot imagine five browsers have been hacked at the same time on two different operating systems.

Any idea?

[ Renting & Real Estate ] Open Question : Can I get a VA home loan with a credit score of 600?

I’m in the national guard I have been for 5 years I graduated college 6 months ago and got a job where I make 60,000 a year I have been renting an apartment by myself for 700$ / months and probably pay 200 I’m utilities It looks like owning a house maybe cheaper or just much as I’m paying plus I’ll own it one day PS I don’t wanna hear about my credit score I’m working on it 

[ Renting & Real Estate ] Open Question : Mortgage refinance question – Loan and Deed?

Hi, in my current situation, my deed has both mine and my mother’s name. If I want to re-finance my loan where the loan would have solely my name, do I need to change the deed? Is it possible where the loan is on 1 person who lives in the household name, where as the deed has 2 people’s name?

Encoding huge number of tape-symbols of a turing machine in the simulation of the turing-machine using real computer

I was going through the classic text “Introduction to Automata Theory, Languages and Computation” by Hopcroft,Ullman,Motwani where I came across the simulation of a turing machine using a real computer. There the author argued that it shall be almost impossible to carryout the above simulation (not considering universal turing machine) if the number of tape symbols are quite huge. In such a situation it might happen that the code of a tape symbol might not fit in the single hard-disk of a computer.

Then the author makes a claim as:

There would have to be very many tape symbols indeed, since a 30 gigabyte disk, for instance, can represent any of $ 2^{240000000000}$ symbols.

Now I can’t figure out the specific mathematics that the author does…

I assume the usual encoding as we do in digital logic say to encode $ 8$ symbols we need atleast $ 3$ bits of code. Then to represent $ n$ symbols if $ k$ bits are required then we should have the following relation,

$ 2^{k} = n$

$ \implies$ $ k=log_2(n)$

Now,

$ 30$ $ gigabytes$ $ = 30 \times 2^{33}$ $ bits$ $ = \beta (say)$

Now if our disk can hold all the $ n$ symbols then we shall have the following relation true:

$ Disk$ $ size$ $ = (Bits$ $ required$ $ for$ $ each$ $ symbol$ ) $ \times$ ($ number$ $ of$ $ symbols)$

$ \implies$ $ \beta = k \times n = (log_2(n)\times n)$

Solving graphically I have:

SOLUTION

$ n= 7.8 \times 10^{9}$ , which is no where close to the number $ 2^{240000000000}$

Where am I making the mistake?