# Boards

## Russian Mathlete solves conjectures, wins $1m, tells Americans to go f*ck themselves

http://www.guardian.co.uk/world/2010/mar/23/grigory-perelman-rejects-1m-dollars

What a guy!!!

(Apologies for the Guardian link)

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http://www.guardian.co.uk/world/2010/mar/23/grigory-perelman-rejects-1m-dollars

What a guy!!!

(Apologies for the Guardian link)

## Yes!

He looks exactly as I'd hoped.

Hairy IS awesome.

## I think he's moulding himself on this ficitonal character

http://images.paxholley.net/blog/geeks/lazlo_hollyfeld.jpg

## The Clay Questions are great

http://www.claymath.org/millennium/

but I'm sure that P vs. NP looks solvable.

## Sheeeeeeeeettttttt

## what a guy

"He severed all contact with the community, and wanted to find a job unrelated to maths," Kisliakov says. "I don't know whether he succeeded in that."

## He must've been every careers advisor's nightmare.

"What you good at?"

"Maths"

"You should become a mathlete"

"I fooking 'ate mathletes"

## Just been reading a bit about the old Poincaré Conjecture......

having read through i can confirm that there is a pretty decent chance that I wouldn't have been able to work it out.

## or to put it another way......

what.the.fuck?

## Hmm

I wonder how his mother and sister feel about him turning down $1m, if he is living with them in a small flat and is unemployed.

"WHO NEEDS A MILLION DOLLARS? I BROUGHT YOU SOME MUSHROOMS!"

## Did anyone else just read through the other 6 problems

in the hope that you knew the answer intuitively (and your lack of knowledge of advanced mathematics actually allowed you to approach the problem in a way that experience mathematicians hadn't thought of)?

## I got a headache just reading the first line of the poincare conjecture.

## is this the russian troll?

## well I think he is a brilliant example to the world

Its great that there are humans who have real values rather than artificial societorial ones

## OK most of those are totally beyond me

BUT the 'P vs NP' looks suprisingly solvable to me.......Im not quite sure what they want proving here........but given a list of the 400 students and the list of pairs to not include, I know how I would go about doing it with a computer

## OK I see the dificulty with this now

It depends on the size and connections of the pairs of mismatches list, if this list is relatively small like say 400 - 10000 or 100000 or even a million pairs of mismatches then it is relatively simples......to come up with a list that is OK.

the issue is when it gets to the other extreme (supposing the list of mismatched pairs excludes all combinations of 100 students from the 400....except for 1) so really I just have to come up with a formula/algorithm for being able to keep going the same way consistantly......hmmmmmmm

## creaky are you good at maths then?

if i was to put on a 10 fold accumulator on games with 3 possible outcomes (lose,win,draw) how much more likely would it be for me to back all ten losing bets than back all ten winners? please forgive the wording, i'm knackered and can't get this to read correctly.

## Im not sure what a 10 fold acumulator is (I've never bet) but the likelyhood would be

exactly the same unless you have inside or some knowledge of the outcomes because If an accumulator is what I think it is, then you will probably bet mostly according to what is most likely, so you would mostly be able to predict some games....

However if the choices were random and you are not cursed or blessed by gypsies, then the likleyhood is the same

## it's when you pick ten results and the odds are multiplied togther

i'm willing to accept that but just making sure you read right -for each bet i'm choosing one of three outcomes which would leave two outcomes that would lead to me losing.

argh i'm so tired. hate maths.

## oh no....you said all ten losing

did you mean not winning rather than losing? i.e. including draws in with losing?

## 59049

(1/3)to the 10th

## yeah sorry

i meant the bet losing as opposed to the outcome of the bet being a loss

http://pluckyoutoo.files.wordpress.com/2008/10/scanners4.jpg

cheers

## Im not particularly good at maths

but I am good with big volumes of numbers and sorting them out. I use SQL a lot and look for patterns in big numbers, sometimes in hex, and have done in binary and octal (I once did it as a joke as we knew there was a pattern to a problem but couldnt see it, properly, and it made it more apparent)

## actually the p vs np is a lot worse than I thought

forget I said it seemed easy

## my initial assumption of ease was based upon the list of mismatched pairs being a list that a dean would conceivably ever compile

whereas of course this is a theoretical dean, who could (in theory come up with a list of.....oh no.....hang on there are only about 5000 unique pairings , that actually looks a bit more promising,

## no there isnt...theres only about 5000 pairings for 100

why do i never read the question properly

## 79800 unique pairings for 400 students

1+399 = 400 2+ 398 = 400 3+397 = 400 etc 100 which we can do 199 times plus 200.

so a theoretical dean could come up with a list 79800 long, obviously there would not be any 100 students that could fit in then.

OK its about making lists and then lists of deltas between lists

so list 1 - generate a list of the 400 students that has zero occurances on the mismatched pairs list

if this is < 100 then make

list 2a - generate a list of the 400 students that only occur once in the mismatch list, simultaniously make a list of the their mismatch partners when their partners only occur once in the list - list2b

generate a list where students occurs once in the list but their mismatch partner occurs twice in the list

etc etc

keep creating lists up until when the sum of the lists = 100, it isnt quite as simple as that as the lists this way might not generate 100 if there are a lot of mismatched pairs.....so each list will then need to start to transfer students from one list to another to see if this generates enough there will be lots of movements of students between lists but at least the number of lists should be finite (I cant imagine there would be more than a 5 figure number of lists, at a guess, although there might need to be more movements between lists than this. biut I do think there can be a formula for this

## shame im not very good at algebra and formulas :(

## I miss gowman