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Cornstalks

Member Since 25 Sep 2004
Offline Last Active Oct 28 2014 10:11 AM

Posts I've Made

In Topic: **Passwords are sent over clear-text on GameDev.net**

08 October 2014 - 04:05 PM

Anyway, I imagine the reason the URLs are hardcoded is because without the protocol browsers can decide that the address is relative and not absolute and thereby break the whole thing (in other words, the protocol is required, pretty much).

The protocol is never really required. You can start with "/" and it's taken as the root of the current domain (for example, "/foobar" is <current-protocol>://<current-domain>/foobar, so here on this site it would be http://www.gamedev.net/foobar). Alternatively, you can use "//" to inherit the protocol of the current page (that is, "//foobar" is <current-protocol>://foobar, which would be http://foobar here on GameDev.net).

Either way, HTTPS should be hard coded for a login POST with a password being sent.

(ugh, the editor ate my post, and I don't have time to retype it)

In Topic: **Passwords are sent over clear-text on GameDev.net**

08 October 2014 - 11:42 AM

Yeah, it's been this way for ages. Definitely avoid reusing login credentials on any site that doesn't use good authentication.

Certainly, but I'll bet lots of people have (even though they shouldn't). I think secure should be the default for this site, rather than insecure.
 

You can use twitter or google to login, so logging in happens via Google's https login page.

That's a good alternative, but given that GameDev.net already has a valid SSL/TLS cert, they might as well use it...

In Topic: Why is infinite technically not a number.

03 June 2014 - 09:25 AM

I don't know what 1+2+3+4+...=-1/12 has to do with p-adic numbers. Care to explain?

Because using the "normal"/simple definitions of summation and integers, that summation does diverge. The natural numbers are closed over addition, and yet -1/12 is not a natural number, which breaks that closure of natural numbers. So in order to make sense of this contradiction, alternative/fancier definitions of summations and numbers must be used. Specifically, p-adic numbers, which converge for large values rather than diverge. Once you're using p-adic numbers, you're not using the natural numbers and aren't restricted to the closure of natural numbers, and so can achieve 1 + 2 + 3 + ... = -1/12.
 

Isn't 2-1 equivalent to 1/2, so 1/2-1 would be 1/(1/2), which would be equivalent to 2? So 1/1-1 + 1/2-1 + 1/3-1 + 1/4-1 ... = 1+2+3+4...?
 
Like I said, I'm not good at the math, so it's interesting why two seemingly equivalent statements aren't equivalent.
i.e. sum n where n=1..∞ != sum 1/(1/n) where n=1..∞

As others have said, you can't treat infinity like a variable and do algebra with it. You can do some things (which actually involve evaluating a limit), but there are several things one might be tempted to do with infinity that would seem valid, but in reality aren't.

But maybe I'm not understanding, as I can't see the contradiction/inequality in sum n where n=1..∞ != sum 1/(1/n) where n=1..∞.

In Topic: Why is infinite technically not a number.

02 June 2014 - 09:44 PM

This is in fact, not true. If you evaluate the sum of all the natural numbers (1 + 2 + 3 + 4...) it is infinitely large. To evaluate this sum to have a value of -1/12th is not really correct. =/
If you want to learn the mathematical reasoning behind why the answer is infinity, but why you could evaluate a similar looking sum to have a value of -1/12th, I suggest looking up the zeta function.If you have an interest in maths, I really recommend it, there are some surprising results and really beautiful mathematics to be found there. If you don't want to learn the maths, just take it as 1 + 2 + 3 + 4... = infinity

I'm not a mathematician, but according to wikipedia and wolframalpha, ζ(−1) = -1/12

Mats1 is actually kind of correct. The thing is that the sum of the natural numbers is indeed infinite. In order to get -1/12 you have to use a different concept of numbers, called p-adic numbers.

For the curious, this question and answer give good a good introduction to the subject.

Anyway, this is further complicated by the fact that we aren't actually talking about 1 + 2 + 3 + ... = -1/12. What we're really talking about are limits and convergence, which isn't necessarily the same (or as strict) as equality. Because it's a limit we're computing, there are more ways to show 1 + 2 + 3 + ... = -1/12 than just the zeta function. So you might say 1 + 2 + 3 + ... is infinity just as much as you might say it's -1/12.

In Topic: Function that takes unknown amount of strings, how do i vector?

02 June 2014 - 01:18 PM

Possible i don't have latest c++11?

I'm guessing you're using Visual Studio. VS <2013 doesn't have initializer lists.

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