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20:00
> Simplify a fraction of the form
> Find the average of the average and median
> Find the median
(Leaky told me to ignore the median for now)
@Mr.Xcoder when did he tell you?
@cairdcoinheringaahing 2 days ago
@Mr.Xcoder Which one of those do you want to try?
@cairdcoinheringaahing Wait
2 days ago, by Leaky Nun
@Mr.Xcoder I suggest you skip that :p
@Mr.Xcoder ^^^
20:02
@cairdcoinheringaahing fraction
Don't spoiler!
@Mr.Xcoder do you want a bytecount?
@cairdcoinheringaahing No
ooo what are you doing right now?
2 or 3 bytes
@HyperNeutrino Hypertraining?
@HyperNeutrino simplify a fraction [4, 6] => [2, 3]
20:03
... I meant what specific problem :P
@cairdcoinheringaahing oh I remember this one! I spent quite a bit of time on it :P
@HyperNeutrino don't post your solution
@HyperNeutrino I found it weirdly easy, for some reason.
No wait I'm recalling the wrong thing. This one is a bit easier than the one I was recalling. I can give you the other problem later if you would like. Is Mr. Xcoder working on the fraction one right now?
@HyperNeutrino Yes
20:05
@HyperNeutrino Do you have one for me?
Let me find the exact wording.
@HyperNeutrino actually, don't give me anything, I want to finish the easy challenges
@cairdcoinheringaahing I have 3 bytes: ,÷g
@Mr.Xcoder that works for the extension. Now try it as a monad
20:09
@cairdcoinheringaahing Sure
For later:
in The Nineteenth Byte, 55 secs ago, by HyperNeutrino
CMC: Given a positive integer n, output the representation of sqrt(n) as a * sqrt(b) with maximized a.
@HyperNeutrino you are 145 rep away from seeing little blue circles in chat.
@cairdcoinheringaahing one-byte monad? (been off a bit)
If I get 15 rep on Math.SE, I'll get 10k network rep (math.SE currently doesn't count because I have <200 rep)
20:16
@Mr.Xcoder for the fraction CMC?
@cairdcoinheringaahing yes
@Mr.Xcoder all dyads
@cairdcoinheringaahing I'm stuck with the [] fraction
@Mr.Xcoder hint: it uses a quick
2 mins ago, by caird coinheringaahing
@Mr.Xcoder all dyads
:/
20:19
what's the exact task? is this still simplify a fraction?
@Mr.Xcoder yeah I was wrong (bad wording)
@HyperNeutrino yeah, but argument must be taken as a list
@HyperNeutrino Yes, the one with [6, 4], not two separate arg
so how is [] a fraction?
@HyperNeutrino You take 6/4 as [6, 4]
@HyperNeutrino he meant that the program takes it as a list
20:20
oh that's what you meant by [] fraction lol
OK got it 3 bytes (no spoiler link)
@cairdcoinheringaahing 3 bytes: ÷g/
@Mr.Xcoder that's it
nice job
20:21
Can I give another problem or did you have one in mind already?
I will leave in ~10', so let's pick the last one.
@HyperNeutrino Give me one, relatively easy
@HyperNeutrino up to you
Oh um
Well the one I had in mind was to simplify a square root, but that's a bit more complicated (Leaky had a bit of trouble with it at first so it'll probably take a bit longer)
Nearly hit repcap today :)
lemme see if I can think of an easier one
20:22
@Mr.Xcoder I'm at 100
@cairdcoinheringaahing I'm at 188
Come on Hyper, I have to go soon
CMC: Given a list nested to any depth, return the max depth of the list (most layers of nesting)
@HyperNeutrino I just saw Eriks solution to the square root problem :/ rep
a bit long I'd say :P it's just a starting point
for my cmc, test cases:
@HyperNeutrino [[[1]],[2,3]] -> 2/3?
20:24
[1, 2, 3] -> 1
[1, 2, [3, 4]] -> 2
[[[1]], [2, 3]] -> 3
@Mr.Xcoder added test case
PFFF
I don't even know where to start
@HyperNeutrino that's not a 10 minute one
Product Flatten Flatten Flatten?
ok to be honest I don't see an immediate starting point lol
sorry that was harder than I thought it would be
Here's a 5-minute one (since I already wasted 5 minutes rip)
Given a list, replace each element in it with that list
[1, 2, 3] -> [[1, 2, 3], [1, 2, 3], [1, 2, 3]]
@HyperNeutrino Nice one
[1, 2, [3, 4]] -> [[1, 2, [3, 4]], [1, 2, [3, 4]], [[1, 2, [3, 4]], [1, 2, [3, 4]]]]
@HyperNeutrino thanks, but stolen from Leaky lol
I have 2 bytes btw
20:28
ẋL?
No, not really
no that would only work for depth-1 lists
@HyperNeutrino 3 bytes: WẋL
Er, still only works for flat lists.
@Mr.Xcoder no doesn't work for the second test case
20:31
... no check again
@Mr.Xcoder look again
I don't get it, just tell me what is wrong
Oh NVM
@HyperNeutrino for all the string ones, I'm getting 1 byte over what they say is the min. Do you know how I can remove them? It seems that its with the ”# that a byte can be removed
Yeah don't use a character :P Use a number
Just give me the solution, I am afraid I have to go
20:35
Hint: You need to get the lists into the right shape. There's a dyad that shapes a list to match the "form" of another list. That is one of the two atoms you need. The other is a monad. I'm afraid I won't give you the solution, but I can give you hints. You can come back to this later.
yay nice
@HyperNeutrino I always wondered what the point of <that dyad> was
it can be quite helpful sometimes :P
@HyperNeutrino why does this work with 1 or 2 inputs?
20:42
x×”#Y  Main link
x      Repeat (left) (right) times
 ×     Multiply
  ”#   (Generates a list of lists)
    Y  Join by newline
No matter, though. I've finished all 30 of the easy challenges!
Is that every easy challenge solution put end-to-end xD
@HyperNeutrino yep. Doesn't segfault though! (errors)
20:46
lol
14 bytes over the minimum they give us!
:o
I should try this sometime
like now
 
2 hours later…
22:40
sure
hi \o/
I have a challenge
Leaky Nun posted it originally, so you might remember it
ok gib :3
Given an arbitrarily nested list, replace every leaf node/non-list element with the original list itself.
Examples:
[1, 2, 3] -> [[1, 2, 3], [1, 2, 3], [1, 2, 3]]
[1, [2, 3]] -> [[1, [2, 3]], [[1, [2, 3]], [1, [2, 3]]]]
*nesting intensifies*
hmm
*aggressive parenthesizing with things that aren't actually parentheses*
22:43
oh hey we got a new atom today apparently
ooo \o/ what does it do?
distinct sieve
def distinct_sieve(array):
	array = iterable(array, make_digits = True)
	result = []
	for (i, x) in enumerate(array):
		result.append(1 if i == array.index(x) else 0)
	return result
Distinct Sieve... Hold wait so how does this differ from Q exactly?
I can tell it's very similar to Q from the symbol of the atom.
Oh wait never mind I'm stupid
For each element, map to whether or not it is the first of its kind to show up in the array
huh
that sounds useful
So [1, 2, 3, 1, 5, 5] -> [1, 1, 1, 0, 1, 0]
22:46
why wasn't it there before lol
That seems like it could be useful
I wonder: What did they use to do before this atom was created?
ok now on to your cmc
22:49
i have no idea where to start >.>
hrm
@totallyhuman any progress?
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