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This is not true. Green cards are distributed across countries respecting the 7% limit, but the remaining green cards are distributed to those in the backlog.


This is specifically not happening. There have been multiple attempts to pass a bill to recapture the unused greencards - none of the bills have been passed.

I am talking about the employment based greencard backlog. I’m not familiar with the situation with other categories.


This depends on the state, e.g. in California:

"Strategic default works in California because this is a "non-recourse" state. Unlike in many states, a mortgage lender in California has only one type of legal recourse in the event that a borrower defaults on a loan: to foreclose on the property. All that the bank can do is repossess your house; they cannot sue you in court to recover payment for the deficiency, which is the difference between the outstanding loan balance and the amount that the bank recovers by selling the home in a foreclosure auction. Strategic default has the advantage that in many cases it allows the homeowner to remain in the property for an extended period because the process of foreclosure often takes a considerable amount of time. During this period, you can save the money you would normally be spending on your mortgage and use it to pay down other debts."

https://www.sandiegolegalpros.com/other-practice-areas/forec....


Oh, sorry, I didn't know that. I stand corrected!


The numbers at the bottom of the check are all that someone needs to initiate an ACH transfer out of your account!


The company becomes equally less valuable after a buyback.

Consider a company with value of $1000, with 100 shares outstanding. Each share is $10. Buying back 10 shares, the company spent $100, so the company is now worth $900 and has 90 shares outstanding. Each share is still $10.

This is the basic model that shows share price should be unaffected by buybacks, but there are other effects. The buyback could signal to investors that the company is unlikely to be inefficient with capital, so investors would value the company at $910 instead of $900.

Alternatively, in a demand/supply model of shares, the buybacks could have exhausted some of the supply of shares, so the valuation for the company settled on by the rest of the market is higher.

There's no clear answer here, but reality is probably somewhere between these models.


> Consider a company with value of $1000, with 100 shares outstanding. Each share is $10. Buying back 10 shares, the company spent $100

How would you buy 10% of the outstanding stock of a real company and pay exactly the market price for the entire block? On a public exchange, you'd need to bid higher than the market price for someone to sell you a block that big (otherwise it wouldn't be worth their while to sell).

If you're buying in the public market, other sellers will see that some party is willing to pay above the market price for this stock. They'll raise their own ask price as a result. In private, off-exchange sales, you'll be dealing with seasoned investors (family offices, hedge funds etc) where they're likely to know your situation, and you'll almost definitely pay a premium to acquire that stock.


I think you may be confusing market cap with share price. Market cap does not generally increase with buybacks, only share price does.


It's a little more complicated than that.

Say a company is worth $1000, and has 100 shares outstanding, for a share price of $10.

They can pay a dividend of $100, reducing their value to $900 (share price of $9) and paying out $1 per share to shareholders. In the end, the shareholders still have $10 of value per share, but some was forcibly liquidated!

Alternatively, the company can do a stock buyback of 10% of their shares. They spend $100 destroying 10 shares. The company is now worth $900 and there are 90 shares outstanding - investors still have $10 worth of value, whether they sold their shares or not.

In the end, no value is created in either scenario - capital is just returned to shareholders.

This is still a simplification because: 1. Stock prices often rise on the announcement of new dividends/share repurchases for similar reasons. It's a positive signal to investors that either of these things will happen, and affects their valuation of the firm. 2. Investors like capital-efficient businesses, and either of these methods of returning capital can actually create value for investors.


Yes!

Funnily enough, the fact that equity grants are "options" to purchase stock at a strike price less than the real share price is much less valuable than the optionality you describe of continuing working to earn the rest of a stock grant after more information is known.

Unfortunately I think it's very hard to pin a value on the optionality to continue working, and I haven't seen anyone mention it when considering joining a startup over a larger company.


Interestingly enough, you don't need Apple's support to do that: https://github.com/st3fan/dovecot-xaps-daemon.

Apple didn't have to build an API for them, they simply provided documentation for how to use the existing API.


Ah, that's nice! Though their blog post at the time seemed to indicate that Apple helped them a bit:

https://blog.fastmail.com/2015/07/17/push-email-now-availabl...


I was about to correct you until I realized you used big-O instead of big-theta notation.

Informally we tend to use big-O to mean big-theta which only adds to the confusion.


Actually we don't use it informally as big-Theta, because big-Theta assumes that the lower and upper bound are asymptotically the same.

For example Quicksort is O(n^2) but Omega(nlogn) it is neither Theta(nlogn) nor Theta(n^2).

You probably meant that informally we just assume that the stated bound is as tight as possible.


> For example Quicksort is O(n^2) but Omega(nlogn) it is neither Theta(nlogn) nor Theta(n^2).

No, this is flat out wrong. Big O is an upper bound on an asymptotic growth rate. Big Omega is a lower bound on the asymptotic growth rate. Big Theta is a tight bound on the asymptotic growth rate. These are independent to the average case, best case, or worst case run time of a given algorithm.

Quicksort has an average run time of Theta(n lg n). Equivalently, its average run time is O(n lg n) and Omega(n lg n). It has a worst case run time of Theta(n^2). Equivalently, its worst case run time is O(n^2) and Omega(n^2).

> Actually we don't use it informally as big-Theta,

Wrong again.


You can state that quicksort has run time O(n^2). The function you're describing with O(n^2) is f : S -> R, where S is the set of input values and R is the set of real numbers, and f(x) is the running time of running quicksort on input value x, and n : S -> N is the size of input value x. The notation O(g(n)) describes the function f in terms of the function n.

That f is in O(g(n)) means there exist constants C and N_0 such that for all x in S, provided that n(x) >= N_0, it is true that |f(x)| <= |C g(n(x))|.

And that f is in Omega(g(n)) means there exist C > 0 and N_0 such that for all x in S, provided that n(x) >= N_0, it is true that |f(x)| >= |C g(n(x))|.


Perhaps it is you that need to be educated.

They are independent only if you explicitly state that you are computing them for best/average/worst case independently. Is it commonly done? Sure. But there is nothing in the definition that forces us to do that.

If the things were as you state would you have any use for Omega and Omicron then? Wouldn't just Theta suffice?


> But there is nothing in the definition that forces us to do that.

That's true. You're right.

> If the things were as you state would you have any use for Omega and Omicron then? Wouldn't just Theta suffice?

Couldn't there be cases where we don't have a known tight asymptotic bound but do have an upper and/or lower bound? And although it's an abuse of notation, you do often see big-O used in place of Theta. From CLRS:

"In the literature, we sometimes find O-notation informally describing asymptotically tight bounds, that is, what we have defined using Theta-notation."


My interpretation is that "O(...)" informally means "Theta(...) in the average/amortized case". In the average case, Quicksort is Theta(n log n).


I would argue it's not the difference in wealth, but the cost of living, that entices these people to move to poorer areas.

The greater difference in wealth may in fact hold them back, for fear of criminals with even less to lose than the criminals in their home country.


How would you implement analog parity? Parity doesn't translate well as a concept into the analog space.

You can take an analog signal and "quantize" it into sixteen possible values so that you can apply a parity algorithm that returns sensible results and doesn't fail with expected noise, but you're digitizing the signal.


I don't understand what analog has to do with it? The video is digitized first, then error-correction (parity?) information added before transmission, so all parity would be related to the digital bitstream -- unless I missed something?


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