@WheatWizard admittedly, the Add a language to a polyglot challenge doesn't have over 250 non-deleted answers, which is probably more of a problem than simply how long it's been going
This is a riddle I learned growing up, but it was phrased like this:
> You're in a hallway with 100 lightbulbs, each one connected to a single switch. They are all off. On the first pass through, you turn all of them on, then you go back to the start. Then, you toggle every other switch. (2, 4, 6, etc.), and go back to the beginning. Then, you toggle every third switch. Then every fourth switch, every fifth, etc. up until the hundredth. After 100 passes, how many light bulbs are on?
@DJMcMayhem not looking at your answer, but seeing as you're talking depending on the value of n I'm guessing my answer of "only the first bulb" is incorrect
I thought, "every light toggled an odd number of times stays on", which means it must have an odd number of divisors to stay on, so only the first bulb will stay on. Am I wrong? @DJMcMayhem
You're right about the riddle, it seemed daunting at first, but once I realised it's just looking for numbers with an odd number of divisors it doesn't seem like a massive task
Most non-mathematicians/programmers I've asked take a longer time to connect the odd number of divisors == on and then squares == odd number of divisors
Our standard policy regarding command-line flags states that one should count the space before the dash (-) too, when there aren't any "free" options, since it doesn't exist in the shortest possible invocation. For example this:
language -option program.p
is 8 bytes longer than the shortest po...
Haskell, 98 bytes, score=fε0−1(n)
_#[]=0
k#(0:a)=k#a
k#(a:b)=1+(k#(([1..k]>>fst(span(>=a)b)++[a-1])++b))
f n=[k|k<-[0..],k#[k]>n]!!0
How it works
This computes the inverse of a very fast growing function related to Beklemishev’s worm game. Its growth rate is comparable to fε0, where fα is th...
Could anyone explain this answer to me? I'm not good with Haskell
A Fine-Grained Mesh
code-golf
If you've used Matlab before, it's highly likely that you've heard of meshgrid. It's a function that has since mostly been obsoleted by broadcasting, but it still has its uses sometimes.
The function itself is relatively simple. Given two vectors x and y of le...
k#(a:b)=1+(k#..) is the last case of #. It means that x has first element a and tail b (so it's not empty), and a is not 0 (otherwise we'd be in the second case).
The number of steps from x is the number of steps from .. which is the next iteration, plus one.
.. is this list: ([1..k]>>fst(span(>=a)b)++[a-1])++b
What happens here is that we take the longest prefix of b whose elements are all at least a (this is fst(span(>=a)b)), tack a-1 to its end (this is ++[a-1]), and repeat the result k times (this is [1..k]>>). We concatenate this with b, which means that we essentially replace a with this long list.
For example, if k = 3 and x = [2,3,3,2,1,6], then a = 2 and b = [3,3,2,1,6]. The prefix is [3,3,2,1,6]. We tack a-1 to it and get [3,3,2,1,6,1]. Repeat k times to get [3,3,2,1,6,1,3,3,2,1,6,1,3,3,2,1,6,1]. Concatenate with b to get [3,3,2,1,6,1,3,3,2,1,6,1,3,3,2,1,6,1,3,3,2,1,6].
Our standard policy regarding command-line flags states that one should count the space before the dash (-) too, when there aren't any "free" options, since it doesn't exist in the shortest possible invocation. For example this:
language -option program.p
is 8 bytes longer than the shortest po...
Haskell, 98 bytes, score=fε0−1(n)
_#[]=0
k#(0:a)=k#a
k#(a:b)=1+(k#(([1..k]>>fst(span(>=a)b)++[a-1])++b))
f n=[k|k<-[0..],k#[k]>n]!!0
How it works
This computes the inverse of a very fast growing function related to Beklemishev’s worm game. Its growth rate is comparable to fε0, where fα is th...
A Fine-Grained Mesh
code-golf
If you've used Matlab before, it's highly likely that you've heard of meshgrid. It's a function that has since mostly been obsoleted by broadcasting, but it still has its uses sometimes.
The function itself is relatively simple. Given two vectors x and y of le....png)
