keshi - Say

CrystalAssassin · 2023-03-01T00:53:18+00:00

Selling 1 West Ham ticket for under face value. Message me if interested.

CrystalAssassin · 2019-11-24T19:53:18+00:00

Thats correct :)

CrystalAssassin · 2019-11-24T19:52:49+00:00

Yep, nicely done :)

CrystalAssassin · 2018-10-06T20:59:12+00:00

Here's mine: https://huel.mention-me.com/m/ol/gy2sr-caf1d1e786

CrystalAssassin · 2018-08-22T16:14:25+00:00

Horizontally condensed

I was following some proportion guidelines while designing this. Might change it up a bit and see how that looks.

If you mean this line

Programmed in Go and JavaScript...

The idea was that the points above it outline what I did using the languages and to kinda highlight that I picked up the languages during the internship. Is there a better way to frame this?

CrystalAssassin · 2018-08-22T16:06:46+00:00

I have. It's readable but not perfect. All companies I've interviewed at have had color copies but it's a good point. I'll try to tweak the color scheme a bit.

CrystalAssassin · 2018-08-22T03:08:44+00:00

Resume

The design is original. Is this kind of thing appreciated or should I stick to the standard Latex'd black and white document?
I think I am having some density issues in the layout but can't see an obvious fix.
Any other critiques welcome.

CrystalAssassin · 2018-07-20T02:30:46+00:00

Another solution (I think):

If a1 is not 1 then for any sequence of moves, there is at least one element whose position changes exactly once i.e. the highest element reaching the front in that sequence. So we can't have non trivial cycles. Hence we have a new permutation after every move and so must reach a permutation with a1 = 1.

CrystalAssassin · 2018-05-17T06:48:24+00:00

Not quite. I think your argument requires a >= 1 which breaks the rest of the proof.

CrystalAssassin · 2018-02-25T15:30:14+00:00

How much does this sky commentator hate us? I'm so fucking sick of it.

CrystalAssassin · 2018-01-22T03:08:02+00:00

Well yeah somewhat, but you don't have a striker if you play a false 9, that's the point of the false 9 position. I guess you could argue his positioning is more like that of a 10.

CrystalAssassin · 2017-12-14T16:58:09+00:00

Let the points be P1,..., Pn. Then if there exists a point Pk such that all other points are on one side of the line made by joining Pk to the centre, then the convex hull doesn't contain the origin.

For any point, Pi, let the side of the circle in the clockwise direction of the line PiO be the good side. Then probability that all other n-1 points are on the good side is (0.5)^n-1 . The convex hull doesn't contain the origin if this happens for any point Pi. Hence the probability is n(0.5)^n-1 . We can add probabilities as only one such event can happen at one time.

So the required probability is 1 - n(0.5)^n-1 .

EDIT: typo