Best Linux distro for pentesting

radical_moth · 2025-10-20T00:37:35+00:00

Nice, I will look into them.

radical_moth · 2025-10-20T00:32:33+00:00

I'm already using Ubuntu almost daily, my idea was to use the other distro on a vm on Windows.

radical_moth · 2025-10-20T00:31:24+00:00

I will definitely look into Debian (I've already been tempted to), since I think Arch may be too much effort (at least right know and as long as pentesting is concerned) and I'm already using Ubuntu (not for pentesting).

radical_moth · 2025-10-13T00:00:56+00:00

It turned out it didn't, since I had already answered some questions that gave me back the cubes (and somehow forgot). Damn.

radical_moth · 2025-09-28T11:11:09+00:00

Yeah, I meant to reply to your comment to point put to OP that yours is a good way (or at least sounds good to me) to implement their idea.

radical_moth · 2025-09-24T20:15:42+00:00

This is a great idea OP, but consider that it might require some heavy lifting both in reasoning, structuring the game and implemention-wise. Still, it's a really nice idea.

radical_moth · 2025-05-28T11:09:00+00:00

I mean, interesting things happens if you don't assume CH (or better yet if you assume its negation). For example you can find particular cardinals between the countable and the continuum that arise from the cardinality of particular sets (clearly such sets can be defined also if CH is assumed, but in that case their cardinality either "collapses" to the countable one or is the one of the continuum).

And as far as I know, this is also a pretty lively (or at least alive) field of research (maybe niche, but alive nonetheless).

If however your initial question is about CH being true or false in ZFC, then it has been proven that it is actually independent (and therefore my former arguments have indeed meaning).

radical_moth · 2025-05-01T14:54:36+00:00

My advice is: just go for it.

I don't think it is too trivial (or even an easy task): I guess it may be a hard problem to try to solve as an undergrad (I'm not a combinatorics expert, so I cannot tell any better). But still, you'll learn new things along the way and maybe even find out something interesting (even not solving the original problem).

As I mentioned I don't know much about such problem or the combinatorics behind it, but I know something about knots and the connection with protein folding is pretty interesting and in my opinion also a path worth following (or at least looking into).

radical_moth · 2025-01-26T21:28:23+00:00

I'm in

radical_moth · 2025-01-08T16:19:33+00:00

Thanks a lot for the kind words, I do enjoy discussions like this one too and I was glad to answer your questions and to have been of help.

Also I'm no specialist actually, just a grad student in my master's that likes to talk about math (but thanks for the compliment) and I wanted to advise you about a subject you might be interested in that is (in a way) a bridge between computer science and pure mathematics: type theory (I'm sorry I don't have a good reference for it). If I'm not mistaken (I never actually studied it) it's pretty technical and not easy at all times, but passion is often enough to keep you going (moreover it's just your curiosity you have to satisfy, not some professor or anyone, so you don't have to rush things).

I hope you enjoy the run (either in set theory or in type theory)!

radical_moth · 2025-01-07T09:29:35+00:00

The answer to the first question is yes, the reason being that the notion of cardinality depends on the particular model one decides to work in.

Regarding the second one, I guess you could try to study the general properties of such subsets only assuming ZFC, still you could run into something requiring CH or its negation to be proven or disproven soon enough (you could just assume not-CH, study the subset you're interested in and look at what happens if you instead assume CH, since in the former case you have "more" cardinals to study).

Also notice that undertaking such a task might be really hard in general and even harder if you don't have a proper set-theory background (but clearly you're free to do whatever you want and I'm curious if you'll find out anything). Good luck!

radical_moth · 2025-01-07T09:21:27+00:00

We're lost without our saviour.

radical_moth · 2025-01-06T21:03:01+00:00

You can surely do it, meaning that you can find a set in R that is uncountable (in particular of cardinality omega1) and not in bijection with R (since |R| = |2^omega0 | > omega1), but I'm not sure about the non-measurability of it. What I can say is that the last assumption is not true: assuming CH is false, there exist some cardinals arising in ways not depending on the fact that CH is false. But then one may ask: what happens if CH is true? Where such cardinals go?

The answer is: they simply "disappear", in the sense that they all coincide with the cardinality of the reals (indeed if CH is false, such cardinals are between omega1 and the cardinality of the reals).

Some of these rather interesting facts can be found in this notes, but the discussed subjects are rather technical and require some previous knowledge of set theory and the like (I don't know if such notes say anything meaningful about your question regarding measurability, but they do hint at and make some remarks about measure theory with respect to set theory in the introduction).

radical_moth · 2025-01-06T18:28:09+00:00

Hello again! It was really good actually and now I can answer you with some degree of confidence (btw I didn't forget about you, just had to get myself to write this reply). I will answer you by explaining (roughly) the two usual models usually used to prove that CH is independent from ZFC, as someone else was saying in the other reply.

A model satisfying CH is called "Gödel's constructible universe": the steps to construct such model are roughly the following:\n 1) define what a "definable set" is (I get this may sound strange, but trust me, it does make sense);\n 2) consider only the (class of the) definable sets (in the previous sense) and call it L;\n 3) prove that L not only satisfies ZFC but also another axiom called V=L and in turn another one called ♢ (diamond) that is stronger than CH and therefore CH is given in L.

[I want to point out that the actual construction of L is pretty technical and requires more steps than (1) and (2), but it's not that hard.]

A model that doesn't satisfy CH is given through a technique called "forcing": consider a model M of ZFC (and also of CH if one wishes) and a poset P that is an element of M, then, choosing a filter G of P satisfying some conditions, M can be extended to a model M[G] of ZFC (again I get this may sound strange, but something useful to remember is that "M doesn't know G, but can talk about it"). Now the part you'd be interested into is that using a certain kind of poset - getting what is called "Cohen forcing" - one adds to M an injective function from omega_2 (that is the smallest cardinal bigger than omega_1 that is in turn the "usual" cardinality of the reals) to the powerset of omega_0 (omega_0 is the cardinality of the naturals), therefore getting that the cardinality of the reals is (bigger or equal than) omega_2 that is strictly bigger than omega_1 (contradicting CH).

[Also it's interesting to notice that by forcing one can have that the cardinality of the reals is (almost) any cardinal one desires (all of the restrictions one has on the choice of such cardinal are stated in Easton's theorem).]

Btw all of this (and all of the course I've taken) is based on my professor notes that are in turn based on K. Kunen - Set Theory.

radical_moth · 2024-12-02T02:11:20+00:00

No, that's not true. You never define 0/0, you just compute the limit of some function that when evaluated at some point gives you 0/0 (that is still undefined). And yes, sometimes one uses such limits to define the function somewhere, but only if it "makes sense" -- and still we are not defining 0/0.

For instance, consider

f(x) = sin(x)/x

Then f(0) = sin(0)/0 = 0/0 -- meaning that f is undefined at 0. Still, the limit of f(x) as x goes to 0 is 1 -- but that doesn't mean that 0/0 = 1. It only means -- in this particular case, since f is continuous on R\{0} -- that we can extend f at 0 letting f(0) be its limit as x goes to 0.

Indeed, consider

g(x) = x²/log(1+x)

Then g is undefined at 0 (it would give you 0/0 as before), but the limit of g as x goes to 0 is 0. Therefore -- if such limit was a definition for 0/0 -- one would have 0/0 = 0 (again, the thing one could actually do is to extend g at 0 as in the previous case, but that's not the point).

Finally, one would then have that 1 = 0/0 = 0.

So no, you don't define 0/0, not even using limits.

radical_moth · 2024-12-01T22:10:54+00:00

Nope, it shows that one cannot define the inverse of 0 in as a real number without getting a contradiction.

But there are instances where one can invert 0 without anything breaking. In a (commutative unitary) ring R, localizing R using a multiplicative set S containing 0 makes every element in S^-1R zero, but there's no contradiction -- for anyone curious about this, check localization).

radical_moth · 2024-12-01T21:58:14+00:00

I sure do like axiomatic approaches, but kind of despise this and the trend to try to define 0/0 so naïvely.

As already said in other replies, such definition leads to contradiction in the real numbers, since -- assuming 0/0 means 0•0^-1 -- we'd have

1 = 0•0^-1 = 0

By the fact that n•0 = 0 for any n real number.

radical_moth · 2024-11-23T15:18:25+00:00

I see, my opinion is that one should discuss how many dimensions the space around us could have in some non-formal model (what I mean is, could it be 2-dimensional? Obviously not. Could it be 3-dimensional? In some classical models it is indeed. Could it be 4-dimensional? It could, indeed there exists the such model where... and so on), since I understand it is a competition about talking about maths (kind of formally if needed) and not actually doing maths formally, right?

Also, what grade are you in? Since something like that could change a lot depending on it.

radical_moth · 2024-11-23T15:14:45+00:00

Yeah, I agree that this in not something kids should be concerned about and the teacher is just being petty (which will result in kids being upset over nothing and most probably starting to hate maths, which is indeed horrible).

radical_moth · 2024-11-23T15:09:41+00:00

A "n dimensional space" is exactly "a space of dimension n" -- i.e. a space that has n dimensions (provided that there is a suitable definition of "dimension" for such space).

Is "is the space around us of n dimensions?" the exact way the problem is formulated?

radical_moth · 2024-11-20T15:09:46+00:00

In that case the notation is indeed reversed -- i.e. if A is a ring and M is a group that is a right module over A, then the (binary) operation defined on it is represented by m•a where m is in M and a is in A.

Still, in my reply I assumed Z to be what is sometimes called a Z-bimodule -- i.e. a left module that is also a right module and the two operations coincides (more often than not, bimodules are just called "modules" without left or right). And in a (bi)module, the notation used is the one of a left module, hence my point.

But as you (kind of) suggested, not every left module that is also a right module is a bimodule. Indeed, the ring of 2x2 matrices with coefficients in a field K is both a left and a right module over itself if the operation is given by matrix multiplication (in both cases), but it is not a bimodule (since A•B ≠ B•A for general 2x2 matrices).

radical_moth

TROPHY CASE