file system – How to add strong encryption to an SD card so that I can access it both on my desktop and phone using open source non tracking offline tools?

The question says it all. So far I was able to find out

  1. Default android encrypted SD card can only be read by the phone or by a skilled hacker or their non open source tools
  2. There are tools that are clunky to use or not open source (cannot be trusted) or riddled with bugs (cannot be used with important files)
  3. I know I can zip files but that’s so impractical.
  4. Most apps use weak encryption algorhithms that are not suitable anymore for anything but a kid who finds your phone in a park

Question: Is there no solution like examples: veracrypt (for windows) or luks (for linux)? Such as mounting a volume and accessing it in your file browser like I do now when I insert an SD card in my phone, it then mounts the volume, which I then am able to access in my file browser. That is what I want with encryption ideally, it’s the simplest most practical and natural way to do it, so I assume there is a way to do it…

multi factor – What are security benefits (if any) of using multiple passwords (device, password manager, …) vs single strong password when paired with FIDO U2F?

I am evaluating security benefits of requiring employees (in enterprise context) to use multiple passwords (or passphrases) vs single strong randomly generated passphrase (minimum of 8 words) when paired with FIDO U2F (using YubiKey).

My gut feeling is that the more passwords we ask employees to create (and memorize), the weaker they tend to be, especially when enforcing regular password rotations.

I would assume many will likely write them on a piece of paper resulting is increased opsec attack surface.

Are there significant security benefits to using difference passwords (or passphrases) to unlock computer vs unlock password manager for example?

My understanding is that both are subject to same keylogger exploits.

Would love to read you guys on this!

real analysis – weak implies strong convergence in $l^1$

I want to prove Schur theorem stating that every weak convergent sequence in $l^1$ is strongly convergent. If we let $(x_n)_n=(x^i_n)in l^1$, then $(x_n)_n$ converge weakly to $x$ if for any $y in l^infty$ we have $$(x_n-x,y)_{l^1times l^infty}rightarrow 0.$$
Now, let $y=sing(x_n-x)$, we obtain $$(x_n-x,x_n-x)_{l^1times l^infty}=||x_n-x||_{l^1}rightarrow 0.$$
I this passage is correct?.
Thank you.

ag.algebraic geometry – Is the strong Suslin’s conjecture incorrect?

This post is related to this one, but I will explain everything from scratch. It seems to me that I am getting a contradiction out of the strong Suslin conjecture. So either I am making a mistake or the conjecture is incorrect. I’ll appreciate if anyone can verify my logic.

I’ll assume the reader is familiar with morphic cohomology at the level described in here. I will be using the same notation and will first describe the Suslin’s conjecture (Conjecture 5.2). Let $mathbb{Z}^{sst}(q)$ be the weight $q$ morphic cohomology sheaf on the smooth quasi-projective varieties over $mathbb{C}$. Let the $epsilon$ be the map of topologies from the classical one to the Zariski. There is a natural map $mathbb{Z}^{sst}(q)rightarrow Repsilon_*mathbb{Z}$ that factors through $tr^{leq q}Repsilon_*mathbb{Z}$ (see theorem 5.1). The strong Suslin’s conjecture claims that $mathbb{Z}^{sst}(q)rightarrow tr^{leq q}Repsilon_*mathbb{Z}$ is a quasi-isomorphism. I’ll assume that this is true and tensor it with $mathbb{C}$, so we have a quasi-isomorphism $mathbb{C}^{sst}(q)rightarrow tr^{leq q}Repsilon_*mathbb{C}$.

Let $X$ be a smooth complex projective variety. $mathbb{C}$ has a De Rahm resolution by smooth differential forms $Omega^{bullet}_{C^{infty}}$. Each element in this resolution is a fine sheaf, this especially implies that they are acyclic with respect to the functor $epsilon_*$, so $Repsilon_*mathbb{C}$ is equal to $epsilon_*(Omega^{bullet}_{C^{infty}})$. Let $Omega^{bullet}_{alg}$ be the complex of the algebraic De Rahm differential forms, there is a natural morphism $Omega^{bullet}_{alg}rightarrow Omega^{bullet}_{C^{infty}}$, which should be a quasi-isomorphism (the reason is that on every Zariski open they both calculate the singular cohomology of that open). So we can represent $Repsilon_*mathbb{C}$ by $Omega^{bullet}_{alg}$ in the derived category. By Suslin’s conjecture the natural map $mathbb{C}^{sst}(q)rightarrow Repsilon_*mathbb{C}$ which induces the cycle map from Chow groups to singular cohomology, can be replaced by $tr^{leq q}Repsilon_*mathbb{C}rightarrow Repsilon_*mathbb{C}$. Note that taking the $2q$-th cohomology of the first map gives us $CH^{q}(X)otimes mathbb{C}/alg rightarrow H^{2q}_{sing}(X, mathbb{C})$. This is the same cycle map from Chow group mod algebraic adequate equivalence relation to the singular cohomology. By Suslin’s conjecture this is the same as the following:
$$cl: H^{2q}(X,tr^{leq q}Omega_{alg}^{bullet})rightarrow H^{2q}(X, Omega_{alg}^{bullet})$$
We want to identify the image of this morphism. We have a distinguished triangle of the following form:
$$tr^{leq q}Omega_{alg}^{bullet}rightarrow Omega_{alg}^{bullet}rightarrow tr^{geq q+1}Omega_{alg}^{bullet}$$
The image of $cl$ is the kernel of the following map:
$$H^{2q}(X, Omega^{bullet}_{alg})rightarrow H^{2q}(X, tr^{geq q+1}Omega_{alg}^{bullet})$$

Clearly the natural morphism $Omega^q_{alg}(-q)rightarrow Omega^{bullet}_{alg}$ becomes zero when composed with $Omega^{bullet}_{alg}rightarrow tr^{geq q+1}Omega_{alg}^{bullet}$. This implies that the image of $cl$ contains the image of $H^{2q}(X, Omega^q_{alg}(-q))rightarrow H^{2q}(X,Omega^{bullet}_{alg})$. Note that the image of $H^{2q}(X, Omega^q_{alg}(-q))rightarrow H^{2q}(X,Omega^{bullet}_{alg})$ is precisely $H^{q,q}(X, mathbb{C})$ (You might need to use GAGA to go from alg to hol and it becomes clear).

This whole thing implies that the cycle map from Chow group to the complex singular cohomology covers all $H^{q,q}(X,mathbb{C})$ cohomologies classes. This is stronger than the Hodge’s conjecture, not only this implies the Hodge’s conjecture but it also implies that $Hdg^q(X,mathbb{Q})otimes mathbb{C}=H^{q,q}(X, mathbb{C})$ (here $Hdg^q(X, mathbb{Q}):=H^{q,q}(X, mathbb{C})cap H_{sing}^{2q}(X, mathbb{Q})$). This statement I believe is not true. (I was able to find the following example see Remark 1)

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bitcoin vs

chia

The thirstily hoped-for Chia digital currency (XCH), charged as associate eco-accommodating possibility in distinction to Bitcoin, has endured a lofty call in esteem since dispatch on Mon.

According to CoinMarketCap data, XCH appeared at $1,600 per unit and climbed momentarily to a high higher than $1,800, but forthwith shed the larger a part of its price within the hours that followed.

At the hour of composing, the new cash is drifting at a value of $690 per coin, down sixty one on its greatest value, proposing “ranchers” smelled an opportunity to require advantage of altcoin craze.

Chia cryptographical cash

The Chia network was planned by Bram Cohen, author of BitTorrent, as a counteractant to the increasing centralization of the digital cash mining business, during which individuals are crushed out by immense mining organizations.

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The money is likewise supposed to deal with one in every of the elemental reactions of Bitcoin, that must do with the natural value of mining. What separates Chia from alternative vital cryptographical types of cash in such manner is that the system wont to get the organization and boost cooperation.

Bitcoin, as an example, utilizes a proof-of-work (PoW) agreement system, that sets excavators con to every alternative and is surprisingly energy-escalated. associate examination from the University of Cambridge proposes Bitcoin goes through a lot of energy on a yearly premise than the state of Sverige.

The planners of the Chia network selected associate alternate framework all, known as verification of area, that depends on capability limit as against registering power. Here, acknowledged ranchers (note the aware distinction in wording) place away further area to carry cryptographical numbers, known as plots.

“When the blockchain communicates a take a look at for the subsequent sq., ranchers will filter their plots to envision whether or not they have the hash that’s nearest to the take a look at. A sodbuster’s probability of winning a sq. is that the level of the whole house that a rancher has contrasted with the total organization,” the location clarifies.

While this framework gets obviate the necessity for energy-concentrated mining, completely different problems have arisen. within the increase to dispatch, as an example, Chia content prompted deficiencies of high-limit warehousing in varied areas, pushing prices through the top side. This more disturbed existing half deficiencies, achieved by the worldwide chip lack associated combined by an growth in excitement for customary mining.

Notwithstanding, if Chia cannot recuperate from its initial slide, the capability deficiency is probably inconceivable to stay going for long.

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authentication – Do I need TFA if I have a strong password?

This is the inverse of this question here.

Google keeps prompting me to enable Two-Factor Authentication for my Google account. I don’t like the inconvenience of TFA, such as mandating that I will always have my mobile device handy on me. Assuming that I have a secure (i.e. reasonably long and complex) password that isn’t likely to cracked by the most high-end consumer hardware for at least the next decade, can I get away with keeping TFA off?

algorithms – Find strong connected components of a Graph in O(V^2 + V*E) time complexity

Take a look at this explanation. It is pretty well known that we can find all strongly connected components in $O(|V|+|E|)$.

The basic “gist” is to use DFS from a node $v$, and then transpose the graph, and apply DFS again. The nodes that have been explored in both DFS instances, must be the strongly connected component that $v$ resides in. Apply this a few times, and you will end up finding all strongly connected components (but you need to be a bit more careful in order to ensure the time complexity stays $O(|V|+|E|)$).