Chat minigolf

grc

12:20 AM

I've found a few more code golf missions on Check iO if anyone is interested

Sp3000

1:04 AM

@MartinBüttner Wow nice! Ahaha (?1), noted for the future :P

@grc Sounds fun! We can slowly go through half of CheckIO over time I think :)

grc

here's one: checkio.org/mission/coins-golf

top solution is 44 chars

Sp3000

:o

grc

that one could be hardcoded though

Sp3000

Not many test cases?

grc

I haven't checked

the next best is 55 chars, so 44 just seems a bit odd

Sp3000

1:13 AM

The small coin limits make it seem kind of simple

Hmm have you unlocked Mine?

grc

I think so

Sp3000

@MartinBüttner Ahaha given the spec I think this could be 39: ^(([^[-}()]|$(?1)$|\[(?1)]|{(?1)})*)$

Hmm k, I might try and look at the test cases (haven't unlocked Mine yet though)

grc

this is an interesting one: checkio.org/mission/palindromic-palindrome

Sp3000

1:28 AM

Ahaha surely we have a question like that already

27

Q: Palindromic palindrome checker

Write a program to test if a string is palindromic, with the added condition that the program be palindromic itself.

Making it a golf question though, would be interesting :P

@grc pastebin.com/Ce7D3Q6v Hmm looks like if we ever need to use range(100), we can use range(99) instead :P

34th place with

from itertools import*
golf=lambda c:min(set(range(99))-set(sum(x)for i in range(11)for x in combinations(c,i)))

XD

grc

golf=lambda l:sum(l)+1 works for 5/8 of the test cases

Sp3000

D:

Maybe that's how the 44 works

xnor

i wrote a completely different type of solution for the brackets problem

checkio=lambda s:eval("s"+311*".replace('%s','')"%(tuple('0123456789+-/*')+('()','[]','{}')*99)‌)==''

Sp3000

1:46 AM

Weird, I seem to have left out a test case @grc or golf=lambda c:[len(c)==4,4][len(c)in[3,6]]or sum(c)+1 would have worked

@xnor :o lambda-ified!

xnor

the magic of eval

letting you do loops in a lambda

i'll take a look at the coin problem

grc

[10,2,2,1] 6 is a test case

Sp3000

Added

@xnor I was wondering how to do consecutive .replace's efficiently when doing the 12 days of golfmas problem

But wow...

Hmm it seems to fail one of their test cases though, I wonder why

Oh. They have an insane amount of brackets

xnor

oops, that should be a 999

Sp3000

:P

xnor

1:55 AM

which means the 33 should be a 3*999+14

3011

Sp3000

2:08 AM

Hmmm how do you even get the first element not in a list efficiently, without min(set(range(99))-L)

Or do you think that's the wrong way to go about it

xnor

7

Q: Minimum excluded number

This is intended to be an easy, bite-size code-golf. The mex (minimal excluded number) of a finite collection of numbers is the smallest non-negative integer 0, 1, 2, 3, 4, ... that does not appear in the collection. In other words, it's the minimum of the complement. The mex operation is centra...

code-golf math

there's no a generally shorter way

Sp3000

:/ that's 21 bytes already

xnor

here's what I have so far:

def golf(l,s={0}):
 for x in l:s|={x+y for y in s}
 return min(set(range(99))-s)

yeah, that min is really the killer

we could be tracking the complement is s instead though

Sp3000

If only you could do x+s to mean "add x to every element in s" :P

xnor

what if we could ...

we could store the list of possibilities as a bit-string

def f(l,s=1):
 for x in l:s|=s<<x
 return s

this returns the bitstring of achievable sums from l

now how to find the index of the least 0 in a bitstring....

thanks for comments about x+s -- it was a great idea even if you didn't mean it!

Sp3000

2:21 AM

Speaking of << I was thinking of the lookup table idea from your golf tip

Not sure if that's short enough though

xnor

are there actually only 8 test cases?

Sp3000

At least 10 - I think only some are revealed when you solve the problem, and there's a few extras you hit when you fail

xnor

i see

this bit-string thing gives me an idea for a prime-finding golf...

Sp3000

2:35 AM

This doesn't even work properly and it's already 42:

golf=lambda c:(7247757764>>sum(c[:-3]))&7

:/

xnor

why doesn't it work?

Sp3000

I was trying to match the small cases, because as grc pointed out some of the larger ones can be done by sum(c)+1

[1,1,1,1] 5 is a test case :/

xnor

unrelatedly, here's my effort to find primes via bitstrings

b=0
N=4
exec("b|=N*N*~-N**99//~-N;N*=2;"*99)

after this is run, the bitstring b is a lookup table for composites up to 100

Sp3000

:o funky

grc

looks efficient :P

xnor

2:43 AM

i don't think it beats existing methods, unfortunately

unless maybe you actually want a bitstring as the result

Sp3000

Oh I see why I'm missing so many test cases now

There's two tabs for test cases, you can change the test case number, but also the test case set

... one of the cases is [1] 2

There goes that plan

Updated: pastebin.com/Ce7D3Q6v That looks pretty difficult to hardcode now

xnor

2:59 AM

is there a really short way to find the great power of 2 that divides a number?

like 48 -> 4

Sp3000

@xnor stackoverflow.com/questions/1551775/…

Looks like you want the log of that though

xnor

whoa, thanks

that's really cool

Sp3000

I swear we could almost have a Golf Practice: Bitwise

xnor

go write it!

Sp3000

XD too hard, I don't know many tricks :P

Maybe a golf tips: bitwise might be better

grc

3:10 AM

54 chars for coin golf :)

Sp3000

:o how the!?

grc

studying the test data ;)

Sp3000

Ahaha figures :P

A legit solution doesn't seem to be able to break 55...

xnor

i have 63 for a legit sol'n

grc

nice

xnor

3:12 AM

is there a fast way to take the log base 2 of a power of 2?

Sp3000

Have you unlocked the challenge? You should post it :P

xnor

i haven't unlocked it

len(bin(n))-3 seems really silly...

Sp3000

:P k well I don't think it takes long

45

A: Tips for golfing in C

Use bitwise XOR to check for inequality between integers: if(a^b) instead of if(a!=b) saves 1 character.

D: damn

Never thought about that before

Oh there's already a "Tips for golfing in <all languages>"

xnor

3:29 AM

oh, where?

that sounds useful

Sp3000

18

Q: Tips for golfing in <all languages>

The aim of this post is to gather all the golfing tips that can be easily applied to <all languages> rather than a specific one. Only post answers that its logic can be applied to the majority of the languages Please, one post per answer

code-golf tips

... they're mostly quite general though

xnor

@grc, do you have a list of the test cases?

Sp3000

pastebin.com/Ce7D3Q6v

xnor

thanks!

Sp3000

That should be all of them

grc

3:40 AM

hint: the answer is always the sum of reachable numbers plus one

Sp3000

... that makes sense, right? Or by "reachable" do you mean something other than values that can be formed by summing any subset?

Sub-challenge: Can we beat the current leader (Python only, 44 bytes) on checkio.org/mission/coins-golf knowing the test cases are pastebin.com/Ce7D3Q6v ?

grc

you can use the test cases to simplify the definition of reachable

Sp3000

sum(c[i:j])? :o

grc

I was thinking of the gap between reachable and unreachable numbers (e.g. 1, 2, 3, 8)

Sp3000

Oh, my one fails on [1,2,3,4,5] 16 anyway

xnor

3:51 AM

oh, are they all one plus the sum of numbers you can reach without crossing a gap of 2?

grc

yeah 2 or 3

xnor

4:05 AM

i can tie 44 chars if you give me a few hundred thousand tries: golf=lambda c:1+sum(x*(x<4&id(0))for x in c)

Sp3000

XD

I'm still not sure about id(0) though

xnor

what about it?

Sp3000

I don't think it gives a random number, if that's what you're doing

xnor

do you mind giving it a try?

i know it gives different numbers each time

Sp3000

... oh I see

xnor

4:07 AM

if it can get through 3 test cases, that would be promising

Sp3000

I need to do it in different sessions

xnor

oh, no

each test case has a separate one

it seems to reset id(0) each time

Sp3000

Yeah I meant I was just trying id(0) consecutively in the shell :P

xnor

oh, i see

grc

id(x) always seems to be divisible by 8 on my computer

Sp3000

4:12 AM

Seems to be 4 mod 8 on mine

Maybe id(0)>>3?

Too many chars though :P

Hmm only 5 of the test cases aren't sum(c)+1

xnor

this works golf=lambda c:1+sum(x*(x<4 or 4 in c or 5 in c)for x in c)

unfortunately, both 4 and 5 are relevant

grc

that's basically mine:
golf=lambda l:sum(x for x in l if{4,5}&set(l)or x<6)+1

Sp3000

{4,5}&set(l), nice

That should be a golf tip for a in L or b in L or c in L :P

Sp3000

4:39 AM

Just unlocked Hubspot ... it's not a place of good golfing challenges D:

Sp3000

5:34 AM

Hah, this problem: checkio.org/mission/solution-for-anything

Sp3000

9:03 AM

@MartinBüttner Or maybe even [*-9]

Martin Büttner

9:38 AM

@Sp3000 yup :)

Sp3000

10:17 AM

Hmm I wonder if I should put up what people have come up with so far somewhere. Not sure how to do that well though (preferably so that everyone can add their own solutions)

Github?

Martin Büttner

github then

Sp3000

10:51 AM

A Github has been made for keeping track of solutions! github.com/Sp3000/Minigolf I'll try and keep it updated

(I've only ever used Github for my own purposes so I've never had to deal with any fancy problems like edit conflicts :P)

@xnor for square roots - nice, I didn't think of that :o

Sp3000

12:41 PM

Hmm I have a question picked for the next one but I think it might just work as a PPCG question itself...

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