my computer fan is running so quickly my computer's about to start flying o_O

oh wait the program finished now it's finally calming down enough that I can hear myself

don't forget to take a picture

aw...

lol :p i was worried my computer would crash to prevent itself from overheating :P but it finished in time

idek why it was doing that and not just simply being slow

@HyperNeutrino My computer has two of these. That's six GPU fans. It is insanely load at full load.

oh wow I see

mine is just a normal acer laptop though lmao

it was almost as loud as like an electric fan used in the summer o_O (slight exaggeration)

> asked 5 hours ago; viewed 3 times

rip

Between your average electric fan and my computer, I have absolutely no doubt that my computer is far louder.

lol I see. yeah I'm confident my computer is basically silent compared to yours at full load :p

@Pavel Are you sure it's not a spaceship?

Reasonably sure, yes.

@ATaco feature-request "I Am Typing" should use oxford comma for when more than two people are typing at once.

The third fan actually made the GPU not fit into my case, I had to cut a hole.

CMC: beat my time in CLI minesweeper i wrote (imgur) (github) / (or) find a bug (except the one in todo.md)

@HyperNeutrino Yeah 10/10 it avoids ambiguity perfectly.

@Mr.Xcoder I'd expect that from SO...

@betseg 0/10 no docs and does not even do anything

@HyperNeutrino github.com/betseg/minestest/blob/master/mine.c#L333 ?

#define MINE 9
#define FLAG 9

But why?

lol y ;-;

@HyperNeutrino also "does not even do anything"? doesnt compile or what?

ok got it

@betseg please add makefile

@betseg no like it just doesn't respond to anything I type into it :P but that's because i had 0 mines

@HyperNeutrino lol

@betseg what diff+size should we use to compete with you

@miles fastest or shortest?

@HyperNeutrino ./mine easy

ok

@Pavel all: gcc mine.c -o mine

i'm bad at this ;-;

rip [*]

i.stack.imgur.com/ulyv1.png (antionebox)

@betseg It's not that I can't figure out how to build it it's that make is more convenient and adding a makefile would be really easy

I must say, the UI design is pretty nice :D

@HyperNeutrino woah nice lucky seed

@HyperNeutrino thanks :D

@Pavel github.com/betseg/minestest/commit/…

+1

ikr first point was a huge 0-field

last 5 digits of the commit: cdef1 lol

@HyperNeutrino ... oh

?

just noticed it wont default to something if no arguments is given

xD

so you found any bugs else than the one in todo.md?

crowdsourcing: how far can we optimize this problem? Find a collection A of 5-element subsets of [1..21] such that every 4-element subsets of [1..21] is a subset of exactly one element of A

[serious]

Also, if we know that {1,2,3,4,5} in A, we know that other elements of A cannot intersect with {1,2,3,4,5} in 4 places, so we know that 80 5-element subsets are not in A

Is there a specific reason you have [1..21], 5, and 4? This would probably be a little easier to consider with smaller numbers.

because I saw it on a list of open problems

I'm not expecting this problem to be solved at all. I just wonder how far we can narrow down the search space.

Then {1,2,3,6,?} has to also be in A. The something doesn't matter. So A consists of {{1,2,3,4,5},{1,2,3,6,7}..}

2 down, 1195 to go, lol

One could do something similar for {1,2,3,8,9},{1,2,3,10,11}..

hmm

{1,2,3,4,5}
{1,2,3,6,7}
{1,2,3,8,9}
{1,2,3,10,11}
{1,2,3,12,13}
{1,2,3,14,15}
{1,2,3,16,17}
{1,2,3,18,19}
{1,2,3,20,21}

Then {1,2,4,6,8}?

[e.g. {1,2,3,12,?} needs to be in A, but it can't be anything before, so might as well be 13]

And {1,2,4,7,9}?

can you show me how you get {1,2,4,6,8}?

So, we need all subsets of length 4 starting with {1,2,4}. The first we don't have is {1,2,4,6}. We already have {1,2,6,7}. So {1,2,4,6,8} Is the first option

@H.PWiz but it might be wrong?

I think it's right. I have:

I mean you don't know if the solution is that one

whereas for the sets before, they are taken WLOG

[{1,2,4,6,?} needs to be in A, can't be 3, can't be 5, can't be 7, might as well be 8 since the others are equivalent]

Oh, yeah. Continuing, I have found a problem

ok I proved it for you

{1,2,3,4,5}
{1,2,3,6,7}
{1,2,3,8,9}
{1,2,3,10,11}
{1,2,3,12,13}
{1,2,3,14,15}
{1,2,3,16,17}
{1,2,3,18,19}
{1,2,3,20,21}
{1,2,4,6,8}

{1,2,4,7,?} needs to be in A, but there are two equivalence classes: {1,2,4,7,9} and {1,2,4,7,10}

so you have two cases now

Exactly why are they different. Which would be equivalent to say: choosing 21

{1,2,4,7,9} is itself an equivalence class, and {1,2,4,7,10-21} are equivalent

case 1              case 2
{ 1, 2, 3, 4, 5}    { 1, 2, 3, 4, 5}
{ 1, 2, 3, 6, 7}    { 1, 2, 3, 6, 7}
{ 1, 2, 3, 8, 9}    { 1, 2, 3, 8, 9}
{ 1, 2, 3,10,11}    { 1, 2, 3,10,11}
{ 1, 2, 3,12,13}    { 1, 2, 3,12,13}
{ 1, 2, 3,14,15}    { 1, 2, 3,14,15}
{ 1, 2, 3,16,17}    { 1, 2, 3,16,17}
{ 1, 2, 3,18,19}    { 1, 2, 3,18,19}
{ 1, 2, 3,20,21}    { 1, 2, 3,20,21}
{ 1, 2, 4, 6, 8}    { 1, 2, 4, 6, 8}
{ 1, 2, 4, 7, 9}    { 1, 2, 4, 7,10}

Currently I've decided to continue blindly (greedily) until I hit a wall

alright

hope you succeed

(I don't really think the greedy approach works)

If it did, the problem would likely not be open

exactly

Also, you'd have to enumerate over a thousand elements.

I've plenty of time :)

Maybe write a script to brute force it and just let it sit for a while

Stuck ...

I think we've derailed

my original intention is to find how narrow we can bound our search space to

Fails at 1 2 6 13 17

so I would like to do case analysis for a while and then basically calculate our resulting search space

Ok

After case 1, what do we have?

basically doing your analysis?

but also analysing how many cases we obtain

So, {1,2,4,10,?}

right

Am I right in saying that has only one equivalence class?

case 1            case 2
{ 1, 2, 3, 4, 5}    { 1, 2, 3, 4, 5}
{ 1, 2, 3, 6, 7}    { 1, 2, 3, 6, 7}
{ 1, 2, 3, 8, 9}    { 1, 2, 3, 8, 9}
{ 1, 2, 3,10,11}    { 1, 2, 3,10,11}
{ 1, 2, 3,12,13}    { 1, 2, 3,12,13}
{ 1, 2, 3,14,15}    { 1, 2, 3,14,15}
{ 1, 2, 3,16,17}    { 1, 2, 3,16,17}
{ 1, 2, 3,18,19}    { 1, 2, 3,18,19}
{ 1, 2, 3,20,21}    { 1, 2, 3,20,21}
{ 1, 2, 4, 6, 8}    { 1, 2, 4, 6, 8}
{ 1, 2, 4, 7, 9}    { 1, 2, 4, 7,10}
{ 1, 2, 4,10,12}

to my surprise, yes

It is still not completely obvious to me why {1, 2, 4, 7, 9}, {1, 2, 4, 7,10} are different

because there is {1,2,3,8,9} connecting 8 and 9

{1,2,3,6,7} connecting 6 and 7

{1,2,4,6,8} connecting 6 and 8

so 7 and 9 are connected

7-6-8-9

Ok, so connecting 7 and 9 constrains decisions later?

right

that's what I think

And does connectedness have a direct meaning relating to the problem?

that's just my intuition

you can't freely exchange 9 and 10 without changing the situation

i.e. there's no renaming

that just exchanges 9 and 10

so now it isn't just my intuition :P

Oh, I think I grasp that

{ 1, 2, 3, 4, 5}    { 1, 2, 3, 4, 5}    { 1, 2, 3, 4, 5}    { 1, 2, 3, 4, 5}    { 1, 2, 3, 4, 5}
{ 1, 2, 3, 6, 7}    { 1, 2, 3, 6, 7}    { 1, 2, 3, 6, 7}    { 1, 2, 3, 6, 7}    { 1, 2, 3, 6, 7}
{ 1, 2, 3, 8, 9}    { 1, 2, 3, 8, 9}    { 1, 2, 3, 8, 9}    { 1, 2, 3, 8, 9}    { 1, 2, 3, 8, 9}
{ 1, 2, 3,10,11}    { 1, 2, 3,10,11}    { 1, 2, 3,10,11}    { 1, 2, 3,10,11}    { 1, 2, 3,10,11}
{ 1, 2, 3,12,13}    { 1, 2, 3,12,13}    { 1, 2, 3,12,13}    { 1, 2, 3,12,13}    { 1, 2, 3,12,13}
{ 1, 2, 3,14,15}    { 1, 2, 3,14,15}    { 1, 2, 3,14,15}    { 1, 2, 3,14,15}    { 1, 2, 3,14,15}

hmm

might as well write a program to do some initial casings