The Nineteenth Byte

4:00 PM

Hello

@flawr If you wanted to speed it up to allow higher resolution in a reasonable time, bear in mind that the standard Mandelbrot set tells you where the non-escaping paths are, so you can avoid calculating most of these. Also, the points of the exterior of the Mandelbrot set that are far from the set boundary have very short escape paths (few steps) so will not contribute much to the final image. You could omit or reduce these to give a much faster program.

Martin Büttner

@TimmyD would a (as opposed to a!) yield an error?

because I can make it if(($a=(gc $PSCommandPath).Length)-ne67){a}

TimmyD

Yes, as a is still not a valid cmdlet name

Martin Büttner

then why do you need ! in the first place?

TimmyD

>.>

Umm...

I mean ... Thanks for golfing 2 bytes! :)

Martin Büttner

4:10 PM

How about this: if(($a=(gc $PSCommandPath).Length)-ne65-ne65){a}

(5-ne6-ne6 gives False in my PS)

Optimizer

this should be converted to cops and robber XD

TimmyD

@MartinBüttner Yep. Solved by surrounding the second $a-ne65 with parens and swapping 65 to 67 ... there went those two bytes

trichoplax

At the moment it's a shoot out. More than half the answers are dead already :(

TimmyD

This is one of the trickiest challenges I've seen on here.

Martin Büttner

@TimmyD if(($a)){a}

looks like you'll need another pair of parentheses

Agawa001

4:16 PM

hello i have a question about taylor's challenge

trichoplax

link?

Agawa001

why isnt {3, 4} a free-sum set ?

trichoplax

13

Q: Finding sum-free partitions

Executive summary Given input k, find a partition of integers 1 to n into k sum-free subsets for the largest n you can within 10 minutes. Background: Schur numbers A set A is sum-free if its self-sum A + A = { x + y | x, y in A} has no elements in common with it. For every positive integer k ...

Agawa001

yes ^

3+4=7 not included in {1:4}

Martin Büttner

@Agawa001 it is

(sum-free)

where does it say that it isn't?

Agawa001

4:18 PM

quote: "We can partition {1, 2, 3, 4} as {1, 4}, {2, 3}"

trichoplax

The problem would be {1,2}, not {3,4}

Agawa001

a missing partition

?

Martin Büttner

and in {1,3},{2,4} the problem is {2,4}

@Agawa001 partition != subset. A partition is a list of sets (technically a set of sets) which share no elements and whose union forms the initial list

trichoplax

@Agawa001 1+1=2 (the numbers being added do not need to be distinct)

Agawa001

so any missing partition is excluding that choice

?

Martin Büttner

4:20 PM

I'm afraid I don't understand the question

Agawa001

@MartinBüttner ok get it

so

the partition {1,3},{2,4} is excluded because of {2,4}

Martin Büttner

yes

for reference, these are all possible partitions of {1,2,3,4} into two sets:

{1}, {2,3,4}
{2}, {1,3,4}
{3}, {1,2,4}
{4}, {1,2,3}
{1,2}, {3,4}
{1,3}, {2,4}
{1,4}, {2,3}

but except for the last one, each partition includes one set which isn't sum-free

Agawa001

and the empty set ?

Martin Büttner

well then the other set would be {1,2,3,4} which isn't sum-free either, but I think we're generally considering partitions into non-empty sets.

Agawa001

of course

Martin Büttner

4:23 PM

no mathjax in chat

Agawa001

why not link it ?

Martin Büttner

it's not sum-free because both all of 1+1=2, 1+2=3, 1+3=4 and 2+2=4 are in the set

Agawa001

it appears like a charm in my screen

Martin Büttner

right, then I guess you're using some browser extension

Agawa001

yes ,js

yes yes i understood

ll see what gonna do about it

New Main Posts

4:27 PM

0

Q: Are these trees isomorphic?

Introduction In this challenge, your task is to write a program that decides whether two given trees are isomorphic. A tree means a directed acyclic graph where every node has exactly one outgoing edge, except the root, which has none. Two trees are isomorphic if one can be transformed into the ...