@Riker random and urandom use the same entropy pool. The difference is that random blocks when the entropy is exhausted, while urandom uses a PRNG to stretch the available entropy. As long as the the PRNG is cryptographically secure and the implementation isn't vulnerable to side-channel attacks, urandom is as secure as random, but it cannot possible be more random.
@Riker That doesn't mean you should prefer random over urandom though. random's blocking will stall your entire application if the entropy pool is exhausted, and a malicious application (no special privileges needed) can essentially DoS random by simply cat'ing it.
In general, you have to specially request a manual transmission here in the States when purchasing a new vehicle. Automatic transmission is technically an option (and thus an upcharge), but it's practically common.
I have basic cable only because my Internet is cheaper with it (as a bundle) than without it (by about $5 / month). I never watch it and it's not even hooked up.
@Riker I've actually had really good experiences with Comcast so far. They set me up in like 20 minutes and were really friendly and easygoing. Nothing shitty yet
@DJMcMayhem See? They're nice because you have the option of going to CenturyLink. Many areas don't have options so Comcast can be as shitty as they want without losing business.
I have been designing Carrot since November of 2015, but I submitted it anyway since it was most likely not TC, too many interpreter bugs made the language very unusable for big things (but to be fair, it still isn't TC yet)
> When they hit eleven, give them a plaintext file with ten-thousand WPA2 keys and tell them that the real one is in there somewhere. See how quickly they discover Python or Bash then.
I've puzzled over this not usual task for a days to figure out an appropriate algorithm. I have a 2D Matrix, of dimension MxN. And I should rotate the matrix n times.
Let's say I have the following 2D matrix:
int[][] jagged = new int[4][];
jagged[0] = new int[4] { 1, 2, 3, 4 };
jagged[1] = new ...
So from what I have found, the full adjacency matrix for trackety track has a size of 331776*n^2 if you account for all possible combinations of rails.
That clashes with our general don't polish turds philosophy. It doesn't matter for a deleted post, but visible ones may get bumped to the front page by edits, which is not at all desirable.
if anything, I'd expect [1, 2, 3] to rotate to [3, 2, 1] (and then back to itself when doing two steps). But it's probably more sensible to exclude such cases from the challenge altogether.
suggestion for a new sort of code-golf challenge. You have to write code that compiles with godbolt.org and you are judged on the length of the assembly
@LliwTelracs yes I suppose so.. but in a way all challenges are
@MartinEnder I suppose you could write assembly inline in C?
oh that's what you said :)
well actually I am not sure it defeats the point.. the reasons to do it this way are a) most people don't like to code in assembly so it widens the audience and b) it's nice to have something online where you can test your code
"It is guaranteed that the minimum of M and N will be even." You still have a 3x3 example in your spec, as well as Riker's test cases which can now be removed. — Martin Ender ♦12 mins ago
Given an non-negative integer n >= 0, output forever the sequence of integers x_i >= 3 that are palindromes in n different bases b, where the base can be 2 <= b <= x_i-2.
So for n=0, you get the output of this challenge, because you get numbers that are palindromes in n=0 bases.
This is basical...
I've puzzled over this not usual task for a days to figure out an appropriate algorithm. I have a 2D Matrix, of dimension MxN. And I should rotate the matrix n times.
Let's say I have the following 2D matrix:
int[][] jagged = new int[4][];
jagged[0] = new int[4] { 1, 2, 3, 4 };
jagged[1] = new ...
Given an non-negative integer n >= 0, output forever the sequence of integers x_i >= 3 that are palindromes in n different bases b, where the base can be 2 <= b <= x_i-2.
So for n=0, you get the output of this challenge, because you get numbers that are palindromes in n=0 bases.
This is basical...
