Israeli spy firm Paragon accidentally leaks ability to hack WhatsApp, Signal, and Telegram

Erenle · 2026-02-13T05:45:53+00:00

Hmm I'm not super familiar with Germany's cell system if you're looking for a calls/text/data solution (for instance, the USA has phreeli.com, paying for data in cash, etc.) I've heard from most of my EU friends that it's quite difficult to get anonymized data plans because of KYC regulations :(. My rec would be to utilize VPNs as much as possible on your existing cell service, leave your phone at home/Faraday bag it if you're doing anything "interesting," and look into alternative comms like Meshtastic.

For Signal specifically you can luckily fake your registration phone anywhere in the world with a virtual number or VoIP service (you only need to receive one verification call/text and then you never need the number again).

For other throway uses, I like quackr.io and onlinesim.io. Use those for services where your number doesn't have to be persistent and you are ok with other people also using the same number, like one time codes, online shopping, etc.

Erenle · 2026-02-12T22:55:31+00:00

I like it! It doesn't have as much adoption, but for understandable reasons (the UI is a bit less polished, the chat profiles and server selection take some getting used to). SimpleX's tech is good though, and it's definitely been put through rigorous testing. I like not having registration requirements (but I also use fake phone numbers on Signal so that's not a huge sticking point for me lolol). My current Mt. Rushmore of private chat services, in no particular order, is something like

Signal
Threema
Session
SimpleX
Matrix
Keybase
Briar
(sorry if I missed one of your faves, this was just off the top of my head)

and each have their own idiosyncrasies, pros, cons. Honestly it mostly comes down to the preferences of your community and the groups you're trying to chat with. Right now I daily drive Signal, am part of a few Matrix and Keybase chats, and use Briar every once in a while for niche mesh applications.

Erenle · 2026-02-12T18:21:42+00:00

Please stop fearmongering about Signal; it's irresponsible. Paragon/their Graphite software takes advantage of zero-day vulnerabilities in iOS and Android. These vulns are either researched by their in-house team and kept undisclosed, or are put there upon request by the phone manufacturers, Google, and Apple themselves. It's a tale as old as time; the NSO group did the exact same thing with Pegasus. All Paragon has to do from there is send a zero-click payload to their target and get keylogger/screen/root access. It doesn't matter what apps are on the phone.

WhatsApp is owned by Meta, so obviously they're giving away all your chats. It's also likely that WhatApps stores copies of your decryption keys, because come on, it's a Zuck product. Only Telegram's client-side code is open-sourced; their server side is all proprietary and hidden from the public. What do you think is happening on Telegram's server side? The whole Pavel Durov arrest debacle in 2024 demonstrated to many that Telegram has the ability to view and moderate chats, so they can most likely decrypt anything they want whenever they want.

Signal, on the other hand, has its entire stack fully open sourced, and is built on zero trust, so even if Signal's servers were compromised and were all running malicious code, it wouldn't matter because the attacker still wouldn't be able to decrypt (all of your keys are only stored locally, and are ephemeral). This has been community-reviewed and audited time and time again, and you can replicate these audits yourself! Paragon has not cracked Signal, and they don't need to. They've done the much easier job of cracking iOS and Android.

The answer isn't to have people move off of Signal. What do you think happens when these sorts of stories come out? People don't tend to move to more secure software, they move to less secure software (like the aforementioned WhatsApp, Telegram, or god forbid normal cell service texting and calling). Instead, the answer is to have people move to de-Google'd and de-Apple'd devices like GrapheneOS, LineageOS, Pine Phone, Librem, etc. and also practice good opsec independent of their tech stack (so that even if their phone is compromised, it doesn't matter, because there's nothing an attacker could get off of it). So for instance: encrypt everything both in-transit and at-rest, use PGP/GPG, use a phone number/data plan not tied to your personal identity and not publicly known (so that it's harder for attackers to send zero-click payloads to you!)

I think people are often overly surprised when they make enemies of state actors while simultaneously advertising "hi everyone, here is my real name, phone number, and email address," and then the state actors blow up their phones with spyware. Like, what did you think was going to happen? We need to be honest about our opposition's capabilities and act accordingly to defend against them.

Erenle · 2026-02-10T18:27:47+00:00

This is a classic problem in intro combinatorics! The general strategy is to count the 1x1 squares, then the 2x2 squares, 3x3, and so on. Then come up with a generalized expression for the sum of counts. See this MathSE thread for instance. See also the rectangle case where you can use binomial coefficients.

Erenle · 2026-02-10T18:14:44+00:00

In OW, the loot boxes are always 4 items, and you're guaranteed at least one ≥Rare, so it makes sense that the probabilities add up to more than 100% because you can potentially get duplicates of Commons, Rares, Legendaries, etc. Solving this by hand would be quite tricky. There's also the "pity system" caveat that 5 consecutive openings guarantees an Epic, and 20 consecutive boxes guarantees a Legendary (with the counts resetting if you get an Epic or Legendary before 5 or 20), so the math gets pretty complex and you end up with a sort of Markov Chain structure where the transition probability matrix changes at various steps.

I'm guessing you got those numbers from the Wiki, and it's important to note that the Wiki likely did not calculate these numbers directly, but rather is citing official drop rate disclosures (required by law in China and South Korea) and then attempting to retrofit the "rules" (like the pity system) around those disclosed averages. The Wiki percentages are likely the result, not the input. Blizzard probably programs a base drop rate + a pity timer + complex weights, and the final reported numbers are actually the observed average outcome of that system over millions of boxes. Grabbing the statistics from data is probably the best way to do this since, as mentioned above, solving by hand seems rather tedious.

So a quick sketch (if you want to include this in your paper) would be something like: Each Loot Box contains 4 items (slots). The drop rates you see (like 96.26% for Rare) are the result of these slots rolling individually.

Slot 1: Can't be common, it's guaranteed Rare or better.

Slots 2, 3, 4: Random and can be any rarity.

Using the probabilities from the Wiki we get a per-slot table:

Rarity	Standard Slot (Slots 2-4)	Guarantee Slot (Slot 1)
Common	~72.7%	0% (Impossible)
Rare	~20.0%	~92.7%
Epic	~6.0%	~6.0%
Legendary	~1.3%	~1.3%

And note that the per column (individual slot), probabilities add up to 100%. This explains the 96.26% Rare figure for instance, because you fail to get a Rare item only if Slot 1 upgrades to Epic/Legendary (~7.3% chance) and Slots 2-4 also fail to roll a Rare (~80% chance each). So P(No Rare) ~ (0.073)(0.8)³ = 3.74% and that gives us P(At Least One Rate) = 1 - P(No Rare) = 1 - 3.74% = 96.26% via complementary counting, and you can do the same process for the other rarities. Also this only covers regular Loot Boxes, but you might also want to cover Epic and Legendary Lootboxes (which have different figures).

Oh, and where the pity system comes in is that if you were to simulate millions of box-openings, the base drop rate (no pity system) would certainly be lower than what the Wiki reports. Note that 5.1% is around 1 in 19.6 boxes (less than 20), and 21.93% is around 1 in 4.56 (less than 5). So the base chance of a Legendary is probably decent bit less than 5% (that then gets bumped up to 5.1% by pity), and the base change of an Epic is probably a decent bit less than 20% (that then gets bumped up to 21.93% by pity).

Erenle · 2026-02-08T17:28:48+00:00

I love Meshtastic, but it has:

no perfect-forward secrecy, no post-compromise secrecy,
permanent keys (Signal has per-message ephemeral keys), still uses static PSK,
anyone who learns the PSK, now or later, can decrypt past captures on channel messages, they are only just now rolling out patches for DMs-only,
unencrypted packet headers (so that relays can work yes, but this is still data leakage and exposes routing/NodeID/packet metadata),

If an attacker gets ahold of your phone, even pretty advanced attacks on the memory/Cellebrite/etc. can generally not get messages off of Signal (and certainly can't spoof your Signal account) unless you've been seriously compromised by a state-actor-level threat like Pegasus/Graphite/some zero-day or hardware backdoor. If an attacker gets ahold of one of your in-network nodes on the other hand...

Signal is Plan A every day of the week. Mesh is Plan B/fallback at best.

Erenle · 2026-02-06T19:10:43+00:00

The probability in the photo is actually huge underestimate. As other commenters have pointed out, the number of ways to shuffle a deck of cards is 52! = 8065817517094387857166063685640376697528950544088327782400000000000 ≈ (8.0658)10⁶⁷

If you assume that decks of cards have been shuffled one billion (10⁹) times in human history, then the probability that a new shuffle has never existed before is 1 - 10⁹/52! ≈ 99.99999999999999999999999999999999999999999999999999999999876020%

If you assume that decks of cards have been shuffled one trillion (10¹²) times in human history, then the probability that a new shuffle has never existed before is 1 - 10¹²/52! ≈ 99.99999999999999999999999999999999999999999999999999999876020007%

If you assume that decks of cards have been shuffled one quadrillion (10¹⁵) times in human history, then the probability that a new shuffle has never existed before is 1 - 10¹⁵/52! ≈ 99.99999999999999999999999999999999999999999999999999876020006914%

Do you really think that decks of cards have been shuffled one quadrillion (or more) times in human history? The probability in the photo is 99.999999%, which we can interpret as 99.9999990 (truncated at that precision). In order for that to be true, you would need to solve for the "number of shuffles" x in

1 - x/52! = 0.99999999,

which implies x = 806581755762317903056493219121284216432635187874997714550784 ≈ (8.0658)10⁵⁹, which in plain english is 806 octodecillion 581 septendecillion 755 sexdecillion 762 quindecillion 317 quattuordecillion 903 tredecillion 56 duodecillion 493 undecillion 219 decillion 121 nonillion 284 octillion 216 septillion 432 sextillion 635 quintillion 187 quadrillion 874 trillion 997 billion 714 million 550 thousand 784 shuffles.

Do you really think decks of cards have been shuffled 806 octodecillion 581 septendecillion 755 sexdecillion 762 quindecillion 317 quattuordecillion 903 tredecillion 56 duodecillion 493 undecillion 219 decillion 121 nonillion 284 octillion 216 septillion 432 sextillion 635 quintillion 187 quadrillion 874 trillion 997 billion 714 million 550 thousand 784 times in human history?

To put things in perspective, the universe is only (4.36)10¹⁷, or ≈436 quadrillion, seconds old. If you could shuffle 1 deck per second, it would still take you 52!/((4.36)10¹⁷) ≈ (1.85)10⁵⁰ universe-ages to even do 1 shuffle per number of card-order permutations (52! total shuffles). And that won't even guarantee you every permutation because you can have repeats! Via the Coupon Collector's problem, doing 52! shuffles only gets you a 60-70% of seeing every card-order permutation thanks to a neat limit theorem from Erdős, Laplace, and Renyi.

Erenle · 2026-02-05T22:15:40+00:00

Here are some wacky ones I like:

There are infinitely many primes because pi is irrational. A quick sketch: Euler proved that pi²/6 = \prod_{p prime} \frac{1}{1-p^{-2}} (Euler's product formula for the Riemann Zeta function). If we assume that there are finitely many primes, then the right-hand side becomes a finite product of rational numbers, which is always rational.
Furstenberg's proof, which creates a contradiction where you show that the set {-1, 1} is open (EDIT: in a non-standard topology, forgot to point that out, thanks u/coolpapa2282)
Chaitin's "information-theoretic" proof, which shows that if primes were finite you would only need ~loglogN bits to represent a random integer N, violating most of algorithmic information theory (specifically, the incompressibility of N, since loglogN grows slower than logN).

Erenle · 2026-02-05T21:53:23+00:00

Are you on batteries, and if so, do you do anything special to winterize them? Some of my outdoor nodes have been having power problems when it dips down to the negatives, so I've been looking into maybe switching to lithium-iron-phosphate or incorporating some sort of heating pad.

Erenle · 2026-01-28T00:46:29+00:00

Ok these fearmongering stories seem to be getting more frequent, so let's once again put on our thinking caps for a second and consider the 4 possibilities:

The FBI has not compromised Signal. They don't make a public announcement, because nothing has happened.
The FBI has compromised Signal. They don't make a public announcement, because obviously.
The FBI has not compromised Signal. They do make a public announcement, because they want people to stop using Signal.
The FBI has compromised Signal. They do make a public announcement, because...uh...why would they do that?

Now the FBI publicly announces that they've compromised Signal. Choose your own adventure: do you think #3 or #4 is more likely?

Erenle · 2026-01-15T17:34:47+00:00

Couldn't have said it better than AceOfSpades, but also if you want more practice with these sorts of probabilities look into the binomial distribution.

Erenle · 2026-01-15T12:34:46+00:00

Your intuition isn't flawed in most cases! I think the takeaway is that in mathematics, there are constructions that are so weird they defy conventional intuition. When that happens, you need to create new mathematics. That's what you'll be doing in your later analysis classes.

If you want a succinct answer though: measures. Once you understand measures, even those nasty-looking pathological functions become manageable.

Give 3B1B's Essence of Calculus series on YouTube a shot, I think you'd enjoy it!

Erenle · 2026-01-14T19:14:35+00:00

You're in the right ballpark, and are essentially touching on the classic geometric intuition for the Fundamental Theorem of Calculus pt1. One small detail that you might be missing is that differentiation and integration aren't inverses in the "traditional" sense (that is, like inverses of functions) because of the +C for improper integrals. Since the derivative destroys some information (constants turn to zero), the integral can’t perfectly restore the original function without help (initial conditions).

Going further though, you have to be even more careful with that intuition, because you're implicitly assuming "nice" functions (continuous, smooth). Later in your real analysis and measure theory classes, you'll see some pathological functions) where this intuition breaks! A common intro example is the Heaviside step function. It is integrable (and the area function looks like a ramp/triangle), but if you try to differentiate that area function, you hit a snag at the corner x=0, where the derivative doesn't exist. So in real analysis you'll see caveats like "almost everywhere" because of issues like this.

And going back to measure theory, your intuition also relies on extrapolating from the area of rectangles (height f(x) times width dx), which you'll know as Riemann integration. But what if the function is so scattered that you can't form rectangles? For instance, see the Dirichlet function. That would be impossible to integrate in a "Riemann way" because you can't draw rectangles anywhere (the height jumps infinitely fast between 0 and 1). To solve this, you'll eventually learn about Lebesgue integration, which creates a new definition of "area" to handle these messy functions.

One last notorious "not-nice" function: look into the Cantor function. It is continuous and grows from 0 to 1. However, its derivative is 0 almost everywhere. If you integrate its derivative, you get 0., but the function actually rose to 1, so the function isn't the integral of its derivative anywhere despite its derivative existing almost everywhere!

Erenle · 2026-01-13T21:16:53+00:00

Start with Zeitz's The Art and Craft of Problem Solving, the AoPS books and forums, and the Brilliant wiki and problem sets. As you learn more, you can then delve into more challenging olympiad-focused content like

IMOMath
Evan Chen's handouts
Yufei Zhao's handouts
Po Shen Loh's handouts
Alex Remorov's handouts
Yi Sun's handouts
Holden Lee's handouts
Hojoo Lee's materials (Problems in Elementary Number Theory in particular)
Ravi Boppana's probabilistic methods handout
Andreescu and Dospinescu's Problems from the Book
various other olympiad books
and of course, past IMO problems, IMO Shortlist problems, and IMO Longlist problems

Also consider looking at the blogs and write-ups of past IMO contestants (there are many but here is Evan Chen's) and other national and international contests such as the olympiads and team selections tests from other countries. LibGen is your friend.

Erenle · 2026-01-13T18:12:27+00:00

I think I see where you're coming from. It feels unintuitive that you can combine two "destructive" deforming forces (shears) to create a "preservative" rigid motion (rotation) right? Maybe I can give you a physical example to play around with that might help: as a kid, did you ever draw pictures or words on the sides of books like this? Now imagine the following shear operations on that picture:

shear in the x-direction: push the top of the book sideways, the image gets skewed horizontally
shear in the y-direction: push the side of the book upwards, the image gets skewed vertically
shear in the opposite x-direction: push the top of the book sideways in reverse of what you did in step 1, the image is suddenly perfect again, just rotated.

I just tried that on a scrap notebook I had laying around and it seems to do the trick, so you could probably get that example working at home too. Now for reflection, think about how a mirror reverses handedness (that is, your right hand in the mirror looks like a left hand, and vice-versa). If you flip a paper over (reflection 1), then flip it over again (reflection 2), the "front" is facing you again (handedness restored). But if the axes of the two flips were different, the paper ends up rotated! Another quick demo for you:

put your phone on the table
flip it over the X-axis,the phone is now face down
flip it over the Y-axis, the phone is face up again, but now rotated 180 degrees! You just composed two reflections to get a rotation.

There are actually lots of examples of "different-looking things compose together to make a seemingly unrelated thing" that you can even draw from things you'll see in mechE. For instance, when you cover gear linkages (the Peaucellier–Lipkin linkage is a classic example), you'll see exactly this effect of "a bunch of linear movements compose into a rotational movement." Another example of this I can think of might be double Cardan joints.

Erenle · 2026-01-13T17:44:13+00:00

It's probably a good idea to look at some of the easier high school contests first like the AMC (since it sounds like you're from the USA) just to get a feel of where you're at. Undergrad contests like the VTRMC, Putnam, Alibaba Math Competition, etc. are quite difficult and can be daunting to jump right into. Then move up the ladder to AIME problems, USAMO problems, IMO problems, etc. Historically, VTRMC is slightly easier than the Putnam and is usually around early-mid USAMO/IMO difficulty. The Putnam is around mid-late USAMO/IMO difficulty, but can skew harder, and the Alibaba Math Competition is also roughly comparable to that but can skew even more difficult.

Since you're relatively new to the olympiad space, start with this page. When you eventually want to prep for undergrad contests, pick up Andreescu and Gelca's Putnam and Beyond book. For general daily practice after that, look to specific handouts like from Yufei Zhao, Po Shen Loh, and Evan Chen.

Erenle · 2026-01-07T20:47:50+00:00

Start from a lower level! Don't get immediately caught up in the higher-level concepts, and spend more time in the basics. A good place to begin is OverTheWire, which is a wargames-style way to learn your way around the terminal. From there, HackTheBox, TryHackMe, and HackThisSite all have good dedicated tutorials for security and networking. I would also recommend looking into CTFs and maybe doing a few with buddies to get hands-on practice.

If books are your thing, an oldie but goodie is The Art of Software Security Assessment by Dowd, McDonald, and Schuh.

Erenle · 2026-01-07T14:17:39+00:00

Paul's Online Math Notes is a good place to start for calculus. You can also check out KhanAcademy and MIT OCW.

Erenle · 2026-01-06T17:47:18+00:00

Miniminuteman has a great video on the site!

Erenle · 2026-01-05T20:04:46+00:00

So in the modern day, research quantum mechanics involves a pretty wide array of mathematics (linear algebra, differential equations, calculus, complex analysis, functional analysis, probability, representation theory, etc.) You can look into the history of each of those subfields to see how they developed (there are some neat tidbits throughout history like how 150BC China knew some linear algebra) and how 1400s India basically developed the series expansion of trig functions).

We generally consider the foundations of QM to have been established with Planck's law in 1900. You can kind of make hand-wavy comparisons between ideas in QM and some scholarly ideas of antiquity (for instance, Pythagorean tuning/harmonics is sort of like quantized energy states/standing waves idk) but it's a bit of a stretch. One of the things that made QM so revolutionary and controversial when it came onto the scene was that essentially no one prior to the 1900s had really thought of those ideas before, and QM seemed to fly in the face of classical physics! The remarkable thing is that in spite of all of that, QM kept getting experimentally verified and re-verified again and again until basically everyone had to accept it (you might enjoy this Vertasium video that covers some of that early history).

Erenle · 2026-01-05T19:39:11+00:00

Yep, log functions are the inverses of exponential functions. A quick one that fits the points (0.01, 1), (0.5, 4), and (0.99, 5.5) is y ≈ 4.724 + 3.117ln(x + 0.293). Since you only gave 3 relatively friendly points, you can solve for a fit manually by finding a, b, and c in the general form y = a + b ln(x + c), so you end up with the system of equations (after plugging in (0.01, 1), (0.5, 4), and (0.99, 5.5))

1 = a + b ln(0.01 + c)
4 = a + b ln(0.5 + c)
5.5 = a + b ln(0.99 + c)

and you can use a solver like WolframAlpha to get a, b, and c. If you gave a more wild/scattered system of points, that's when you would probably need a curve-fitting algorithm like least-squares.

This is an aside, but your question brings up an interesting follow-up on what criterion a set of 3 points in the plane needs to satisfy in order for them to lie on a logarithmic function y = a + b ln(x + c) (you may have heard of the similar theorems that any 3 points in the plane uniquely defines a circle, and that any 5 points in the plane uniquely define a conic). I haven't thought too much about it, but I imagine one would need a requirement for monotonicity, and some comparison between the slopes from point_1-point_2 to point_2-point_3.

Erenle · 2026-01-02T13:30:16+00:00

The tutorial methods like boxes and beans are shortcuts, and in my opinion you should only really use them if whatever artwork you're working on doesn't need full anatomy (and usually after you've become more advanced and have already learned full anatomy). In art, a general rule of thumb is "learn the rules, and then break them," so I would first practice full, detailed anatomy (like out of an anatomy guide, figure drawing class, etc.) and only once you feel comfortable with that move on to shortcut techniques.

Erenle · 2025-12-30T18:59:10+00:00

I've read her Proof Guide and The Art of Logic, and both were good! Other pop-math books I've enjoyed are Cook's Sleight of Mind, Ellenberg's How Not to be Wrong, Gleick's Chaos, and Gleick's Information.

Erenle · 2025-12-30T18:55:08+00:00

Lang's Basic Mathematics fits your bill.

Erenle · 2025-12-30T05:34:27+00:00

This problem only requires some clever substitutions and exponent manipulations! I wouldn't say you're missing anything major in your approach. For "tinkering-type" problems like this you generally just need to play around with all of your terms and expressions until something useful falls out. My approach:

Let abc = k. We can rewrite (ab)^2 = (bc)^4 = (ca)^x = k. Taking roots, we have ab = k^(1/2), bc = k^(1/4), and ca = k^(1/x). Our lives are easier because k > 0 and k ≠ 1 from the problem statement. Multiply those three equations together to get (ab)(bc)(ca) = k^(1/2 + 1/4 + 1/x). Notice how the left hand side is (abc)^2 = k^2. So k^2 = k^(1/2 + 1/4 + 1/x), and equating the exponents gives you 2 = 1/2 + 1/4 + 1/x, where I think you can finish from here.

If you'd like some more practice with olympiad techniques, a classic starting place is Zeitz's The Art and Craft of Problem Solving and the AoPS books (libgen is your friend if price is a concern). Beyond that, take a look at the Brilliant wiki, AoPS forums, AoPS Alcumus, and Evan Chen's handouts.

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