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Dennis
Dennis
1:16 AM
Any idea what's going on there?
 
Alex A.
Alex A.
hahaha I love that picture
@Dennis Where?
 
Dennis
Dennis
0
A: Subsequence Substitution

DennisJulia, 62 59 58 bytes f(s,p,r)=(try s[[i=findnext(s,c,i+1)for c=p]'],i=r,0end;s) I/O is in form of character arrays. Verification julia> f(s,p,r)=(try s[[i=findnext(s,c,i+1)for c=p]'],i=r,0end;s) f (generic function with 2 methods) julia> F(s,p,r)=join(f([s...],[p...],[r...])) # string/char...

Why do I need '? Why does the tupled assignment work?
 
Alex A.
Alex A.
Why does the tupled assignment work? Because tuple assignments are a thing in Julia...
As for the transpose, I have absolutely no idea
 
Dennis
Dennis
OK, but I'm basically doing x,x[i],i=[1,2,3],4,1, so I use i before I actually define it.
 
Alex A.
Alex A.
That's... bizarre.
I honestly don't know why that would work
It looks like the left hand side of the tuple assignment all gets parsed and evaluated together rather than individually
because it knows what x is and it knows what i is and it correctly replaces x[i]
 
Dennis
Dennis
2:29 AM
@AlexA. But x[i],i,x=4,1,[1,2,3] does not work...
 
Alex A.
Alex A.
O_O
I have no idea then
What's the error?
 
Dennis
Dennis
ERROR: UndefVarError: x not defined
 
Alex A.
Alex A.
?:U
 
Dennis
Dennis
Btw, X->det(X'X)>.5 works.
 
Alex A.
Alex A.
Even for empty input?
 
Dennis
Dennis
2:32 AM
Yup.
 
Alex A.
Alex A.
X = []; X'X errors for me
 
Dennis
Dennis
You need a 0x0 matrix, not Any[].
 
Alex A.
Alex A.
Oh
In that case what I do now might work
 
Dennis
Dennis
Nope.
 
Alex A.
Alex A.
Never mind then
 
Dennis
Dennis
2:35 AM
julia> (X->det(X'X)>.5)(reshape(Int64[],(0,0)))
true
 
Alex A.
Alex A.
It's golfier to do ones(Int,0,0) :P
But that's nuts!
Sure you don't want to post that?
 
Dennis
Dennis
Quite frankly, I'm a little fuzzy on the details. Why is X'X invertible iff the columns of X are linearly independent?
 
Alex A.
Alex A.
@AlexA. The claim is true. The rows of M being dependent with coefficients given by vector v means there that v*M=0. This implies that v*M*(M^T)=0, so M*M^T is singular, since it maps some nonzero vector to zero. Conversely, if M*(M^T) is singular, then there exists a vector v so that v*M*(M^T)=0, in which case (v*M)*(v*M)^T=0. But, only the zero vector has norm zero, so v*M=0, which means the rows of M have a linear dependence. — xnor 4 hours ago
It's the same approach as Anna's Matlab answer. This is xnor's proof of the approach.
 
Dennis
Dennis
Oh, I hadn't read the MATLAB answer for some reason...
 
Alex A.
Alex A.
Nice first answer by a new user.
 
Dennis
Dennis
2:41 AM
Definitely.
Well, alrighty then. If you don't want to post the determinant answer, I will.
 
Alex A.
Alex A.
Well selfishly I want to because it's 14 bytes which will be the smallest so far, but technically you came up with it.
 
Dennis
Dennis
Completely up to you.
 
Alex A.
Alex A.
2:57 AM
Well, you came up with the approach and it happens to be identical to Anna's and I already have two other answers on that question, so you go for it. :)
 
Dennis
Dennis
3:08 AM
POSTE
D
 
Alex A.
Alex A.
hahahahaha
 
 
16 hours later…
Dennis
Dennis
7:19 PM
@AlexA. A filter would be shorter.
 
Alex A.
Alex A.
Shorter than what?
oh
 
Dennis
Dennis
Also, the Boolean logic can be replaced with ∪.
 
Alex A.
Alex A.
Not sure which logic you mean but I cut out the ternary
 
Dennis
Dennis
!=[]&&
 
Alex A.
Alex A.
Wouldn't it be an IndexError since I'm doing z[1]?
 
Dennis
Dennis
7:23 PM
Ah, nevermind, that returns a singleton array of the result...
Why on Earth is nothing not falsy?
 
Alex A.
Alex A.
Because it's not a boolean, it's nothing
 
Dennis
Dennis
I came to terms with not casting numbers to Booleans because of the loss of information, but it doesn't get falsier than nothing.
 
Alex A.
Alex A.
Well, the thing about nothing is that it means "no information." So you don't know whether something is false because you don't have any information about it at all. It's like NA in R.
Except NA is a logical in R and nothing is of type Void in Julia
 
Dennis
Dennis
Anywho: 
x->(y=sort(x))∈[circshift(x,i)for i=1:endof(x)]&&y
 
Alex A.
Alex A.
Oh nice, that's even shorter than my old solution that didn't work!
Thanks for your help :)
 
Dennis
Dennis
7:37 PM
:)
 
Dennis
Dennis
7:49 PM
@AlexA. That sum squares challenge needs a Julia answer btw.
 
Alex A.
Alex A.
I'm on it!
 
Dennis
Dennis
I'm currently at 26 bytes.
 
Alex A.
Alex A.
8:00 PM
Uh, those test cases...?
I'm at 21 based on my understanding of the challenge but I must be misunderstanding because those test cases make absolutely no sense to me
 
Dennis
Dennis
What about 'em?
Your function should sum all odd k² for which k² < n, not all odd k² for which k < n. — Dennis ♦ 14 mins ago
 
Alex A.
Alex A.
Ohhhhhhhhhh
 
Alex A.
Alex A.
8:29 PM
@Dennis 29
 
Dennis
Dennis
I'm also at 35 for driftsort.
 
Alex A.
Alex A.
._.
 

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