pull down to refresh
169 sats \ 0 replies \ @Scroogey 23 Jan \ parent \ on: Stacker Saloon
I've seen the Letterman interview with Malala and respect her more than a Twitter troll ranting about nano GPS tracking in bank notes.
This also works with SN, as imgprxy.stacker.news seems to use CloudFront.
You can roughly locate another user by commenting where you expect only him to read it (first), including a unique image. You wait until the user has loaded the image, then query all CloudFront caches to see which one served it, giving a rough location as per the article.
If you didn't use Tor or VPN to hide your location (because you trusted @k00b to not divulge it), now maybe you should.
There seem to be two different ways to chat with the peer (either HTTP API or websockets). Turtle mode switches between them. If you have trouble chatting with the peer, try switching. The default should be more reliable (for the average user), and it shouldn't depend on the peer (only on your setup) which is more reliable for you. If you have no trouble chatting with peers, you can safely ignore the feature. If you're curious, try it, it's harmless.
At least that's my understanding.
The answer depends on whether you are familiar with decision trees.
It might help to assemble four quarter circles into a full one:
Let's say the doubly shaded right triangles have shorter sides
The outer square's side length is
We have to subtract the four white rectangles with area
Further add the doubly shaded triangle areas
Then the answer is obviously a fourth of that.
That's how far I got, but I couldn't get
30 sats \ 0 replies \ @Scroogey OP 9 Jan \ parent \ on: [Math puzzle] Fastest route from A to B science
Just now reading the HN comments. It didn't occur to me to make the browser window smaller to accelerate the DVD bounces.
Played Stimulation Clicker, bet on Bitcoin, won.
Or it was the hot baths.
I got strange results with Python's built-in floats, as if the precision wasn't good enough.
Here's my sloppy code.
Closest match so far:
u=97366344686.000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000 z=305885393174.00000000000140601378546118371359432952443993406280164926158704474801197648048400878906250000000000000000000000000000 r=305885393174.00000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000 y=29052179.885292215619356553510385828347735846057466489994871801023624482995165949265643254134563120164232130092252951481137523825 haosz.vvc6yx5pfugb5u1knqd1
The base36 part is probably the easiest, all you need is
We can try various
But this seems to require arbitrary precision floating points, and with a naive search (m=1, 2, 3, ...) I haven't found anything yet.