Proof of Riemann hypothesis [on hold]

This looks like a really interesting proof for the Riemann hypothesis. What do you think? Is it any good?

We first find a Hamiltonian H that has the Hurwitz zeta functions ζ(s, x) as eigenfunctions. Then we continue constructing an operator G that is self-adjoint, with appropriate boundary conditions. We will find that the ζ(s, x)-functions do not meet these boundary conditions, except for the ones where s is a nontrivial zero of the Riemann zeta, with the real part of s being greater than 1/2. Finally, we find that these exceptional functions cannot exist, proving the Riemann hypothesis, that all nontrivial zeros have real part equal to 1/2.

Proof of lemma from Hong’s article about multi-threaded max flow algorithm

I’m struggling to prove Lemma 3 and Lemma 4 from an article about parallel version of push-relabel algorithm: A lock-free multi-threaded algorithm for the maximum flow problem.

Lemma 3. Any trace of two push and/or lift operations is equivalent to either a stage-clean trace or a stage-stepping trace.

And

Lemma 4. For any trace of three or more push and/or lift operations, there exists an equivalent trace consisting of a sequence of non-overlapping traces, each of which is either stage-clean or stage-stepping.

Pdf version of the article can be found here

Absorption Law Proof by Algebra

I’m struggling to understand the absorption law proof and I hope maybe you could help me out.

The absorption law states that: $ X + XY = X$
Which is equivalent to $ (X \cdot 1) + (XY) = X$

No problem yet, it’s this next step that stumps me. How can I apply the distributive law when there are two “brackets”?

How can I manipulate $ (X \cdot 1) + (XY) = X$ to give me $ X \cdot (1+Y)$ ?

An image to clarify

I understand that the absorption law works. I would just like to see how the algebra proof works.

Thank you!

Reduction to proof undecidability of the problem: machine M and N accept infinitely many words

I am struggling with the following problem: Decide whether this problem is decidable or not: For two given Turing Machines M and N, there exists infinitely many words accepted by both machine M and machine N. In other words, is language { encodedMachine(M)#encodedMachine(N) | intersection of language of M and language of N is infinite } decidable?

Intuitively it feels like this is undecidable problem and halting reduction might be used to proof this, but I have no idea how to start this reduction.

Is it normal for a company to ask me for a picture of my ID for proof of identity?

I bought a game off a third party website and they declined my card and said to send a picture of the cardholders ID to prove that the debt card is mine. i don’t feel safe sending a picture of my ID to this website. They charged my card and i didn’t receive my game so i don’t know what to do. the website is https://www.instant-gaming.com/en/

TCP Sack panic proof of concept?

For the vulnerability called TCP SACK panic([1], [2], [3], and many more): is there a proof of concept code out there that can be used to test vulnerability status and effectiveness of remedies?

[1] https://github.com/Netflix/security-bulletins/blob/master/advisories/third-party/2019-001.md

[2] https://access.redhat.com/security/vulnerabilities/tcpsack

[3] https://isc.sans.edu/diary/What+You+Need+To+Know+About+TCP+%22SACK+Panic%22/25046

Is this proof of Godel’s incompleteness wrong, or am I misunderstading?

L is the set of all finite length strings.

Some subset of L are legitimate computer programs, set named F.

Some subset of F are functions that turn a Natural Number into a 1 or 0, call this set A.

Here the proof states that A contains all computable functions that fit its requirements.

If we define Function I, Fi, such that it takes the nth function from A and returns it with the nth output of that function flipped, then we can say that we now have a function which is not in A, therefore we have a function which is not computable.

The trouble I have with this proof centers on either a mistake in the proof, or my misunderstanding of what computable means, being: The program which creates L is a finite string, and with a finite modification to L, we could determine which outputs are part of A, order them, and change the nth values just as Fi does.

Since this can all be written in a finite program, despite the arguments above, I am not convinced that Fi is not a member of L, and admittedly, not certain what the term computability really means.

This is the source of this proof: https://youtu.be/9JeIG_CsgvI?t=1272 Timestamped hopefully so those informed can skip to the meat of it.

Mistake in a proof of termination phase of Simplex algorithm in CLRS?

There is a pseudo-code for Simplex algorithm in CLRS:

enter image description here

The proof consists from three-part loop invariant:

Proof We use the following three-part loop invariant:

At the start of each iteration of the while loop of lines 3–12,

  1. the slack form is equivalent to the slack form returned by the call of INITIALIZE-SIMPLEX,
  2. foreach $ i \in B$ , we have $ b_i$ >= 0, and
  3. the basic solution associated with the slack form is feasible.

The initialisation and maintenance phases are clear for me. But, the termination case seems(not sure) wrong to me:

enter image description here

So, my question is:

It is unclear for me that $ x_e$ is equal to $ \infty$ if $ i = e$ . But, if we look at the pseudo-code, it is obvious that such thing happens only if $ x_e \leq 0$ . What am I missing ?

Proof Of The Poincare Conjecture: An Unofficial Erratum [on hold]

We read and checked the detailed proof of the Poincare conjecture. One can find the article (Ricci Flow And The Poincare Conjecture by Morgan and Tian) on arXiv. Since the proof contains some gaps and errors/mistakes, it took us 6 months to get through the book. The detailed proof is over 500 pages long, so many things can go wrong in the process of writing. We filled all gaps and fixed all errors/mistakes that we found. That took us over hundred hours. At the end we were tired and checked only 99,99% of the book, but we are sure Perelman solved the problem. The Clay Institute, ArXiv and the authors gave us no response (maybe we are not famous enough), but an incorrect proof stays an incorrect proof, so we decided to publish our notes (109 comments) here. We did this for everybody who wants to understand the proof. This will save them time. Our notes can be found in our dropbox https://www.dropbox.com/s/73i5wz5390o1lx3/Final_Version_2.pdf?dl=0

If you think we understood something wrong, please inform us. Of course, some comments are only accessible to readers of the book/article, but we tried to make the errors/mistakes clear to any graduate student. The hardest error/mistake to fix was 97. You are welcome to check and comment our notes after reading them!!! Of course, some people will not take us for real. We only show that the article/book is not perfect. Note that the page numbers refer to the book, but our work can also be used for the arXiv version.

Wembley and friends

We are not interested in self-promotion. We just want to help the readers. We all have a PhD.

Here is the question: Can you find more errors? Did we make some mistakes?

How to proof problem L $\notin \texttt{DSPACE}(f)$

I want to proof a language $ L$ is not in $ \texttt{DSPACE}(f(n))$ the languages, that a deterministic Turing machine can decide with fixed tape length of $ f(n)$ (wiki). Show:

$ $ \begin{align*} L \notin \texttt{DSPACE}(f(n)) \end{align*}$ $

How would I do this?

My first attempt was to reduce a language $ L’$ from $ \texttt{DSPACE}$ to the language $ L$ , but I don’t know of any language in $ \texttt{DSPACE}$ and I don’t know if I can “simply” reduce it. Does anyone have a reduction that I see and try to understand?

edit:

as david-richerby mentioned: $ f(n) \in \Omega(\log{n})$

The exact exercise is: $ L = \{ w\mid w \in T(M_w),M_w \text{ needs space |w|}\}$