In my last rant (https://devrant.com/rants/5523458/...) I regaled you lovely folks of how I had to diplomatically yet firmly defend my work/life boundaries during off-work hours for non-life threatening affairs (a frustratingly common occurrence), and concluded the thread by mentioning that I still had a job, but would make a note of my frustration of that for whatever exit interview happens.

Well, no need for those notes any longer.

I and half of the engineering force, along with several senior managers were laid off this morning in the form of a "mandatory on-site all hands".

I live and work in NYC. Several people took trains and booked rooms from as far away as Boston to be here (or at least I know of specifically two people who commuted up here on Sunday to be here for the "all hands"). I presume those people used their travel benefits to get here and back.

We were dismissed before the meeting even took place, and according to a coworker I became friends with (yes, despite my snarky comments in other threads, I *do* actually have coworkers I became friends with lol) who survived at least this round of layoffs, once the actual all-hands commenced, the company first disclosed the layoffs, then announced being awarded a major contract with the very client the entire org had been working on overdrive to win for the last nine months. He had already been looking for a new job and got an offer last Friday, had been mulling it over, but told me once we were off the phone he was calling them up and accepting. He had three people reporting to him, and lost two. Even he had no idea it was coming until one of his now-former subordinates asked him to come outside and told him they'd just been let go.

I knew going in to this startup that "it's a startup, anything can happen, just mind the gap". That's why I asked on numerous occasions and tried to get time with our CFO to ask about revenue and earnings; things that in my years at this place were never disclosed to the rank and file, I'm not a professional accountant or CPA by any means, but I did take a pair of corporate accounting classes in community college because I like the numbers (see my other rants about leaving the field and becoming a math teacher), and I was really curious to know how the financial health of the business was.

It wasn't so much a red flag as it was an orangish-yellow that no one ever answered those questions, or that the CFO was distant but not necessarily cagey about my requests for his time; other indicators were good while interviewing--they had multiple fully integrated, paying customers (one of which being a former employer from years ago, which aided me in having strong product familiarity during the job interview), but I guess not enough to be sustainable.

Anyway. I'm gonna use the rest of the week to be a bum, might get out of the city and go hang with friends Pittsburgh, eat some hoagies and just vibe for a while. I've got assets and money stashed up to float pretty easily for a while, plus a bit of fun money so losing the job isn't world ending. Generalized anxiety because everything is going to shit worldwide, but that quickly faded into the backdrop of the generalized anxiety I always have because existentialism or something like that.

Thanks for reading. Pay the teachers.

https://en.wikipedia.org/wiki/...

Is an interesting read.
May have applications of measuring the randomness of RNGs, as deviation from Viswanath's constant

Heres a fairly useless but interesting tidbit:

if i = n
then
r = (abs(((((p)-(9**i)-9)+1))-((((9**i)-(p)-9)-2)))-p+1+1)

then r%a will (almost*) always return 0. when n = floor(a/2) for the lowest non-trivial factor of a two factor product.

Thats not really the interesting bit though. The interesting bit is the result of r will always be some product with a *larger* factor tree that includes the factor A of p, but not p's other larger factor, B.

So, useless from what I can see. But its an interesting function on its own, simply because of what it does.

I wrote a script to test it. For all two-factor products of the first 1000 primes, (with no repeating combinations, so if we calculated say, 23*31, we skip 31*23), only 3262 products failed this little formula, out of half a million.

All others reliably returned 0 for the following..
~~~
i = floor(a/2)

r = (abs(((((p)-(9**i)-9)+1))-((((9**i)-(p)-9)-2)))-p+1+1)

r%a
~~~

The distribution of failures was *very* early on in the set of factors, and once fixed at the value of 3262, stopped increasing for the rest of the run.

I didn't calculate if some primes were more likely to cause a product to fail or not. Nor the factor trees, nor if the factor trees had any factors in common between products, or anything of that nature.

All in all I count this as a worthwhile experiment.

If you want to run the code yourself, I posted it to pastebin here:

https://pastebin.com/Q4LFKBjB

edit:

Tried wolfram alpha just to see what it says, but apparently not much. Wish it could tell me more.

https://i.postimg.cc/4ycRFNZf/...

The factorization shit I'm always ranting about? I decided for once to explain it visually in this handy dandy little infographic.

We're essentially transforming the product from an unsmooth set of potential factors in its factor tree, to a factorization tree that guarantees first that the set of potential factors are all 2, 3, 5, and a or b of p, and second, that all the factors are *smooth integers* of a or b.

This is basically what Adi Shamir was trying to do with TWINKLE and TWIRL, despite checking a hundred thousand+ potential primes.

I did it in four.

So I made a couple slight modifications to the formula in the previous post and got some pretty cool results.
The original post is here:
https://devrant.com/rants/5632235/...

The default transformation from p, to the new product (call it p2) leads to *very* large products (even for products of the first 100 primes).

Take for example
a = 6229, b = 10477, p = a*b = 65261233

While the new product the formula generates, has a factor tree that contains our factor (a), the product is huge.

How huge?
6489397687944607231601420206388875594346703505936926682969449167115933666916914363806993605...
and

So huge I put the whole number in a pastebin here:
https://pastebin.com/1bC5kqGH

Now, that number DOES contain our example factor 6229. I demonstrated that in the prior post.
But first, it's huge, 2972 digits long, and second, many of its factors are huge too.

Right from the get go I had hunch, and did (p2 mod p) and the result was surprisingly small, much closer to the original product. Then just to see what happens I subtracted this result from the original product.

The modification looks like this:

(p-(((abs(((((p)-(9**i)-9)+1))-((((9**i)-(p)-9)-2)))-p+1)-p)%p))
The result is '49856916'
Thats within the ballpark of our original product.

And then I factored it.
1, 2, 3, 4, 6, 12, 23, 29, 46, 58, 69, 87, 92, 116, 138, 174, 276, 348, 667, 1334, 2001, 2668, 4002, 6229, 8004, 12458, 18687, 24916, 37374, 74748, 143267, 180641, 286534, 361282, 429801, 541923, 573068, 722564, 859602, 1083846, 1719204, 2167692, 4154743, 8309486, 12464229, 16618972, 24928458, 49856916

Well damn. It's not a-smooth or b-smooth (where 'smoothness' is defined as 'all factors are beneath some number n')
but this is far more approachable than just factoring the original product.

It still requires a value of i equal to
i = floor(a/2)

But the results are actually factorable now if this works for other products.

I rewrote the script and tested on a couple million products and added decimal support, and I'm happy to report it works.
Script is posted here if you want to test it yourself:
https://pastebin.com/RNu1iiQ8

What I'll do next is probably add some basic factorization of trivial primes
(say the first 100), and then figure out the average number of factors in each derived product.

I'm also still working on getting to values of i < a/2, but only having sporadic success.

It also means *very* large numbers (either a subset of them or universally) with *lots* of factors may be reducible to unique products with just two non-trivial factors, but thats a big question mark for now.

@scor if you want to take a look.