Searching for an image tagging and ocr apk

S

Sartaj Ansari

I have tons of Screenshots, Pictures of people and docs. I want to tag them so that I can search it sort of how files by Google does by recognising text from screenshots. So i was trying to find if there is an app for Android. it should have following requirements requirements

  1. it should detect and recognise photos by face
  2. it should do it on device and should not upload on any server
  3. It should embed it's face recognition database into image itself preferably xmp tagging
  4. It should read text from screenshots and make them searchable preferably by other tagging apks also

Preferences

  1. it would be very nice if that app can recognize nature scenes, objects, animals, birds, etc but face recognition is my priority
 

Unreplied Threads

What the garlic is that pt.2? (Plant identification help needed)

  • Derptastic
  • Social
  • Replies: 0
I have sown some garlic cloves in a pot (I only have a balcony, don't judge). And... something came out of the pot, but it's not garlic, and I'm very curious what it might be: Mystery plant Mystery plant, image two

I even mistook it for garlic initially, but it's completely inedible, and in fact, can barely be chewed through in any manner - the leaves are pretty tough. As far as I could tell by rummaging in the ground, there are no cloves to this beastie, and the roots seem to go quite far down. Google Lens identifies that as Chives, but I can guarantee you it's not (I actually do have chives, in a pot beside this one). In this case, you can see some actual garlic that caught on, chilling in the background, wondering what kind of neighbor is that. It definitely feels like a weed, but I still wanted to find out what it could be before it goes out.

Any educated guesses?

What would happen if you use warm base on a compressor ice cream machine?

  • Lenny Markus
  • Social
  • Replies: 0
Note: This question is for an ice cream machine WITH a compressor

Pretty much all recipes for ice cream/gelato call for completely chilling the mixture before starting to churn.

I get that for "passively" cooled machines (compressorless) it's important, so that you don't loose any of the pre-cooling you've done to the hardware.

Does it actually matter for a compressor-based machine? The machine can certainly cool it down without a hitch, but would it impact the final texture?

Attempting to simplify Fiege-Fiat-Shamir Identification

  • deeplyconfused
  • Technology
  • Replies: 0
My understanding of the Fiege-Fiat-Shamir protocol, sourced from "RSA and Public-Key Cryptography" by Richard A. Mollin is as below:

  1. A third party chooses two large primes $p$ and $q$, sharing $n = pq$.
  2. Alice has a secret value $s_a$ and shares a number $t_a = s_a^2 \mod n$.
  3. Alice picks a random $m \in \mathbb{N}$ and sends $w = m^2 \mod n$ to Bob.
  4. Bob picks a challenge $c \in \{0, 1\}$ and sends that to Alice.
  5. Alice computes $r = ms_a^c \mod m$ and sends that to Bob.
  6. Bob computes $r^2 = m^2s_a^{2c}$, and he can compare that to the publicly known $wt_a^2$, which should be identical if Alice does indeed know $s_a$. The security of this protocol is based on the fact that it is difficult to solve quadratic modular congruences with unfactored moduli.

My text elaborates that there is a method that an attacker can use to impersonate Alice with a 50% success rate, so this is done $a$ times and the probability that this works becomes $2^{-a}$.


If my understanding of this is correct, then one could use a similar protocol, but without the 50% chance of failure:

  1. As before, a third party selects $p$ and $q$, sharing $n = pq$.
  2. Alice picks a random $m$ and sends $w = m^2 \mod n$ to Bob.
  3. Alice sends $r = ms_a \mod n$ to Bob.
  4. Bob computes $r^2 \mod n$ and confirms that it's equal to $mt_a$.

Again, because neither $w$ nor $t_a$ are easy to compute, this seems secure.

I assume that this doesn't work. Why not? Why does Fiege-Fiat-Shamir require that Bob send a challenge?

Zencash uses your passphrase to identify AND unlock a wallet - how is this secure?

When looking into Zencash I stumbled into myzenwallet.io (by the Zencash creators), which gives you the option to enter a passphrase to generate a wallet (seems normal), but then after creating your wallet you can come back later and open your wallet by simply re-entering your passphrase again... this was unexpected and had me not only wondering about how it worked, but also wondering whether whatever means were used to achieve this would remain secure in the future.

Given that I don't need to identify the resource I'm unlocking, but rather just hand over the passphrase, I'm left to assume that their system generates a token by hashing my passphrase a number of times to generate my "master address" or "account ID" (I just made those terms up, because Zencash gives you many addresses, so I also assume you have some sort of master address / account ID), and then also using that same passphrase to lock / authenticate (not to encrypt though, right?) the wallet.

What common proverb does this string refer to?

What common proverb is referenced by this three-character string?

ᚷᛖᚨ

Chromatic number of distance graphs on the integers with restricted prime divisors

  • MistressMadeline
  • Mathematics
  • Replies: 0
Let $P$ be a finite set of prime numbers. Define the distance graph $G(P, n)$ on the vertex set $V = \{1, 2, ..., n\}$ where two vertices $a$ and $b$ are connected by an edge if and only if $|a-b|$ is a product of primes in $P$. Denote the chromatic number of $G(P, n)$ by $\chi(G(P, n))$.

Can we find a function $f(k)$ such that $$ \chi(G(P, n)) \leq f(k) $$ for all $n$ and all sets $P$ with $|P| = k$?

Minimal bipartite graph density in the log-edge-density version

A graphon is a bounded measurable symmetric function $W:[0,1]^2\to [0,1]$. Let $\mathcal{W}$ denote the set of all graphons. For any $p\in [0,1]$, let $\mathcal{W}_p$ denote the set of all graphons $W$ such that $\int W=p$. For any graphon $W$ and any bipartite graph $H$ with bipartition $(A,B)$ where $|A|=v_1,\ |B|=v_2$, the density of $H$ in $W$ is defined as: $$t(H,W)=\int_{[0,1]^{v(H)}}\prod_{ij\in E(H)}W(x_i,y_j)\prod_{i\in A} dx_i \prod_{i\in B} dy_j .$$ Define $h(H,W)=\log_{\int W}t(H,W).$ For any $p\in (0,1)$ does the following hold? $$\max_{W\in\mathcal{W}}h(H,W)=\max_{W\in\mathcal{W}_p}h(H,W).$$
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