We’ve known about coronaviruses since the 1960s

Reverse image search says The Triplets of Belleville

Knowing the author’s nationality and previous cultural references, I think she’s inspired by Slavic Rusalka myths. Better read up on that archetype before considering a relationship…

This program in 35 kB of LISP-like code beat GPT 3.5 a Turing test, and some people would keep using it despite having seen the code. But I think parasocial relationships with machines date back to early cogs and pulleys…

That’s where the calculus comes in, otherwise it’s a basic Pythagorean theorem. I’ve compiled a list of possible latitudes (with questionable assumptions), knock yourself out

About “самиздат”

Except now the walking speed of 1 m/s (3.6 km/h) and running speed of 5 m/s (18 km/h) are realistic

Instructions unclear, let’s assume he started 1,312,593,975 feet further back (see my other comment)

CASIO calculators say 1, and I think it’s more intuitive with “÷2π” being equivalent to “÷(2×π)” rather than “÷2×π”. It took me a while to figure out why my results were almost but not quite one order of magnitude wrong after I was forced to switch to TI. Obviously nobody in high school or uni wrote ÷ (or Czech :) on paper, it was all fractions, but even on “natural mode” calculators I’d use the ÷ key for simple denominators to save vertical space.

You actually only need to know the latitude for that… except the local terrain will play a larger role anyway, unless they started very close to a pole and follow rhumb lines (in this case ♂︎ a meridian and ♀︎ a circle of latitude) as opposed to great circles, so better just ask for full coordinates.

What? The teacher does not want to talk about it? Let’s find out anyway, to the best of my abilities. For now, we’ll be assuming Earth is a fully walkable ellipsoid.

We don’t have many data points in the question so let’s extrapolate their movement into the past. There is the hint that they met 8 years earlier at the same spot, during which he’d have gone 1 262 304 000 ft or 384 750.2592 km, completing 9.617 polar circumferences of the Earth (40 007.863 km each).

Huh, that’s not a whole number. In some languages, “eight long years” might mean “a little over 8 years” so let’s assume he finished exactly 10 polar circumnavigations, which took 8 years and about 116 days. Her walked distance over that time is 5x smaller, 2 polar circumnavigations’ worth or 80 015.726 km. This is only exactly 2 great circles (ellipses, really) if they are polar, but we know that it’s impossible to go due east from either pole. Therefore, we’ll use the other option you pointed out, of her having gone at a constant bearing of 090, her path being a circle of latitude (aka a “parallel”). To end up in the same spot, she must have not-quite-circumnavigated-but-enough-for-Phileas-Fogg the Earth (aka crossed every meridian but not the equator) an integer number of times. After a simple conversion, we can construct a table of the options.

To calcuate latitude from circle-of-latitude circumference (colc), we’ll be using geodetic ↔ ECEF conversion equations (except those with the perverse prime vertical radius of curvature 𝑁 of course) and their notation (simplified with 𝑦 = 0, 𝜆 = 0, ℎ = 0 to ignore longitude and elevation) with values of the WGS-84 ellipsoid. The relationship we’re seeking is between colc/2𝛑 = 𝑝, circle-of-latitude radius, which is at zero longitude equal to ECEF 𝑋, and 𝜙 (latitude). See also Wikipedia on Earth radius by location but remember to skip anything with 𝑁, we’re not doing that.

The geocentric radius (𝑅) is related to 𝜙 (latitude) like this but we only need the distance to axis of rotation 𝑝.

(𝑍/𝑝)(cot 𝜙) = (1 − 𝑒²) → (𝑏²/𝑎²)(𝑍/𝑝) = 1/(cot 𝜙) = tan 𝜙 → 𝜙 = atan((𝑏²/𝑎²)(𝑍/𝑝))
(using 𝑒² = 1 −  𝑏²/𝑎²)

Since sin² 𝛂 = 1 − cos² 𝛂 and we can normalize 𝑍 and 𝑝 to the unit circle with ellipsoid radii 𝑏 and 𝑎 respectively:
𝑍²/𝑏² = (𝑍/𝑏)² = 1 − (𝑝/𝑎)² = 1 − 𝑝²/𝑎², therefore 𝑍 = √(𝑏²(1 − 𝑝²/𝑎²)).

All in all, 𝑝 → 𝜙 conversion is:
𝜙 = atan((𝑏²/𝑎²)(√(𝑏²(1 − 𝑝²/𝑎²))/𝑝))

(Presumably, this could be simpified further but I can just put this into a calculator so idc)
Per WGS-84:
𝑎 = 6378.137 km
𝑏 = 6356.752 km

Here are the results. Finding appropriate meeting locations at some of the 25+ possible latitudes on either hemisphere is left as an exercise to the reader. Also note that “rainy days” don’t occur in some places, which is why I didn’t bother adding more rows after I got within 500 km of the pole.

Rows where the number of not-quite-circumnavigations is divisible by 2, 5, or 10 are especially interesting because then the couple would meet 3, 6 and 11 times over the 8.32-year relationship, respectively, rather than just twice.

(Fun fact: another set of latitude circles whose sizes are an inverse-integer sequence is a lesser-known solution to the famous cheeky bear color problem (the well-known solution is a 1-radian arc around the North Pole), although obtaining the ultimate answer then fails due to the lack of bears in Antarctica)

There won’t be a date number 2, she will propose on date number 1

wheel spinning

That’s the point, innit?

Or you can do this and find out how squeaky the house frame is!

Yup. I don’t like the original T-shirt’s custom, green Tux. And WSL (even with its oddly reverse name) is better than the other penguin vector graphics that ship with Windows…

I’m pretty sure the designer was told “Nothing like Tux”. The first emoji shows a corporate, decidedly uncute emperor penguin, the other one is reminiscent of Club Penguin and screams “how do you do, fellow kids”. Still, it could - and has been - made worse with the pivot to AI and gradients.

That’s the WSL penguin…

Recovered image, attempt 1:

I used exact 7x7 box blur to remove high-frequency contents, then median blur and contrast increase… Very basic but pretty good for a FOSS Android app.

Now that I’m on my computer and got GIMP, a single filter can almost recover the original image from behind the 90%-opacity checkerboard, which turns out is actually 1bpp (unfortunately, the JPEG noise, especially around tile edges, remains). The filter in question is a basic lightness curve, with steps at 13 (5%), 128 (50%) and 243 (95%):
_|¯¯¯¯¯¯¯¯¯|_________|¯
The result is:

Here’s the original dithered image BTW, and the source photograph:

Source & author’s notes

Here’s a comparison of my recovery, limited by JPEG noise, as opposed to what could be recovered if the author used a PNG (could have reached <100 kB losslessly, as opposed to lossy >200 kB!):

🟪 = false white; 🟨 = false black; as a result, the original image is in the green channel and the noisy one in the blue channel, and reddish tint highlights error

Edit: text below is from before I knew there is detail within the squares, I’d thought they were flat color, the space saving is not that substantial with that in mind (a 16x16 lossless as 4-color indexed PNG saves only cca 67 % (shown below), or about 50 % as a reasonably lossy JPEG):

There’s so much JPEG artifacting in the original. All of it, and I mean all of it, could have been avoided, at much smaller file size, by using tile size of 16 instead of 15 (ImageMagick’s default pattern:checkerboard).

(Yes, the sides are off but that’s because I’m on the phone and don’t have my usual software. Still, the tiles align with JPEG’s 8x8 blocks (usually 16x16 for chroma but that’s blank here) so they’re all clean, at ⅛ the file size.)

How does that help? There is definitely a tiny cat but can you see a big one after Gaussian blur?

https://www.mobygames.com/game/29361/millie-meters-nutrition-adventure/

A megaphone contains 10¹² (a trillion) microphones. A total surveillance state does not sound that impossible now, right?

By leveraging teamwork and unparalleled knowledge of the local ecosystem, we saved $1.2 million and cut through 7+ years’ worth of red tape

https://english.radio.cz/beavers-build-planned-dams-protected-landscape-area-while-local-officials-still-8841536