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Suever
Suever
18:34
@LuisMendo There has to be a better way, right? codegolf.stackexchange.com/a/85836/51939
SEems like we need a third output from sort to indicate the location in the final array
Luis Mendo
Luis Mendo
19:25
@Suever I think &S&Sq (on the phone now, but it works for the example)
Luis Mendo
Luis Mendo
19:42
Yep, it seems to work for other cases (I haven't tested them all)
 
2 hours later…
Suever
Suever
21:19
Hmm I had tried that but it didn't seem to work, I'll try again
Luis Mendo
Luis Mendo
21:50
Putting the finally statement to good use!
0
A: Calculate the Super-Logarithm

Luis MendoMATL, 15 12 bytes 0`ZetG>~}x@q Try it online! Or verify all test cases (slightly modified version to handle several inputs). How it works Starting with 0, apply iterated exponentiation until exceeding the input. The output is the number of iterations minus 1. 0 % Push 0 ` % Do.....

beaker
beaker
22:31
evening, @Luis
Luis Mendo
Luis Mendo
@beaker Hello!
beaker
beaker
reading up on the super-logarithm
Luis Mendo
Luis Mendo
Nothing terribly fancy. I just noticed that the result is essentially how many iterated exponentials are needed to exceed the input
beaker
beaker
i was playing around with the finally statement before... can't remember which challenge it was for
Luis Mendo
Luis Mendo
The only use I've found for it (but an important one) is to be able to use the index of the last iteration before it's lost when exiting the loop
beaker
beaker
22:39
yeah, that saves you having to keep a counter
Luis Mendo
Luis Mendo
On a different subject: that J language looks cool. It has verbs, adverbs, conjunctions and the like :-) I might take a look at it when I have some time
beaker
beaker
actually taking the logarithm was longer?
Luis Mendo
Luis Mendo
@beaker It was 15 bytes (same as the initial version of the exponential) because 0 and 1 needed to be special-cased
beaker
beaker
:D
Luis Mendo
Luis Mendo
Perhaps special casing could be avoided somehow, but I didn't see how
Thanks for the upvote! :-)
beaker
beaker
22:44
log(0) gives -Inf :/
:D
it's always those edge cases that get you
that dijkstra challenge would have been easy if it weren't for null inputs
Luis Mendo
Luis Mendo
@beaker Yes, totally. I initially asked that (but then deleted the comment when the challenge was edited)
3-output find does most of the work
I initially had a neater (but longer!) version with 2D arrays
Here, under "old version"
beaker
beaker
I did one kinda like that that didn't account for null 
dj=@(m)nnz(m==[1:([a,b]=size(m))]'+[1:b])
Luis Mendo
Luis Mendo
Good use of that Octave "in-between assignment" feature!
Does it have a name?
The fact that you assign to a variable and use the value at the same time
beaker
beaker
i don't think so
Luis Mendo
Luis Mendo
Does this work? dj=@(m)nnz(m==[1:(a=size(m))]'+[1:a(2)])
It would be shorter
beaker
beaker
22:53
it's just a side-effect that if you use a vector where it expects a scalar, it will only take the first element of the vector
Luis Mendo
Luis Mendo
@beaker Yes, that's nice too
But I meant the fact that you assign to a variable for later use
beaker
beaker
i don't know... i had problems with the operator precedence
oh
um
don't know about that either
Luis Mendo
Luis Mendo
I think you need two sum's rather than nnz?
beaker
beaker
I can't find anything relevant... I guess that the assignment just evaluates to its lhs
Luis Mendo
Luis Mendo 
>> dj=@(m)sum(sum((m.*(m==[1:(a=size(m))]'+[1:a(2)]))));
>> dj([3 -1 3 3; 3 -1 3 1])
ans =  3
beaker
beaker
22:58
sum(...)(:)
this is octave ;)
Luis Mendo
Luis Mendo
Oh, true :-)
Also, you can remove some parens 
 dj=@(m)sum(sum((m.*(m==[1:a=size(m)]'+[1:a(2)])))
Or dj=@(m)sum((m.*(m==(1:a=size(m)])+(1:a(2)))(:))), that is
You should post it!
beaker
beaker
that doesn't get around the null problem though
Luis Mendo
Luis Mendo
Oh, you mean empty input?
beaker
beaker
yeah, and i'm getting errors trying to define the function
Luis Mendo
Luis Mendo
Yes. Error with empty input
beaker
beaker
23:01
who knew a 1x0 array would mismatch a 0x1 array ;)
Luis Mendo
Luis Mendo
Yeah :-(
This would work except that the result is logical 
dj=@(m)~~numel(m)&&sum((m.*(m==(1:a=size(m)])+(1:a(2)))(:)))
We need a short-circuiting * operator :-)
This seems to solve the issue 
 dj=@(m)sum(((m=blkdiag(m,0)).*(m==[1:(a=size(m))]'+[1:a(2)]))(:))
It makes sure m is not empty by diagonally including a harmless 0
But blkdiag is long
I'm going to bed. See you! Good evening for you! @beaker
beaker
beaker
hehehe
g'nite @LuisMendo

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