A few weeks after I got my first touchscreen phone, I had to google for how to answer it.
I had assumed you'd just tap the "answer" button, but that failed more often than not. It never would have occurred to me to swipe a minimum of 3cm, starting with the answer button, and I must assume this knowledge has spread to users by osmosis rather than discovery.
The accept button animates on most phones in the way you need to move it. But it's true that I've seen many first timers get confused, probably because of the inherent pressure that accompanies a phone call.
That's not so much a problem for the people who want to leave a meeting, but it is very much a problem for people who frequently toggle their mic on and off.
Shannon did not use the word intelligence to describe the mouse in this demonstration - instead, he talked about learning. That's why the second run was considered more important than whatever algorithm was used to solve the maze.
To that end, I'm curious about their cache invalidation solution. Are there timestamps, or is it a flag system?
You are being far, far, far too generous with the complexity of this design if you think there's some kind of cache invalidation. It's a purely mechanical computer, which means it is going to be very simple in abstract design, because doing anything even mildly complex would require an insane amount of space.
I can't find design documents for this, but I can make a pretty educated guess about its design.
Each square has two relays, representing the number of left turns necessary to exit the square. Each time a whisker touches a wall, a signal is sent to a mechanical adder which will add 1 to the relays in the space. When the mouse enters a square, a "register" is set with a value, based on if it entered from the left, top, right, or bottom, then the mouse is turned and the register decremented until it hit 0, then the mouse attempts to walk in the indicated direction.
Where the mice starts on x and turns the number of times in each square. You can actually put the mouse down anywhere and it will exit the maze, if the walls are left unchanged.
If my memory serves me right, you are right. I think I've read that it was implemented with two relays per cell. These encode the last cardinal direction the mouse exited the cell in.
> I'm curious about their cache invalidation solution
My guess: there would be a model somewhere (probably a binary relay map of walls) of the maze, and as soon as the mouse hits an inconsistency, this map is updated. So there isn't really a cache, it's more like a model, or perhaps you can think of collision-based cache (model) invalidation. The mouse probably then follows the solution to this modified maze, modified only insofar as it has measured modifications.
Is there a technical specification somewhere? I'd certainly be curious to read it.
This was explicitly gone over in the announcement. The short of it is that they validate further attacks by promising Trump's allegiance, assistance, and a target.
Indeed, CDC's numbers for March indicate that social isolation is reducing non-ncovid-19 deaths by twice as much as sars-cov-2 is adding them. That's short term, so not what GP was talking about, but very significant numbers nonetheless.
Weight is a weaker condition, since you can construct a polynomial weight sequence that results in linear height.
In general, height is the easiest thing to restrict, but doing so restricts dynamic performance optimizations - you can't use splay trees, for instance
> Weight is a weaker condition, since you can construct a polynomial weight sequence that results in linear height.
I'd like to hear more. The varieties of weight-balanced trees I'm aware of all have logarithmic height.
> In general, height is the easiest thing to restrict, but doing so restricts dynamic performance optimizations - you can't use splay trees, for instance
Good point.
BTW, you might be interested in Bose et al.'s "De-amortizing Binary Search Trees" <https://arxiv.org/pdf/1111.1665.pdf>, shows how to keep height logarithmic with "essentially any Binary Search Tree" (their phrase).
Limiting their size would cut down on the clickjacking annotations, but wouldn't deal with the hated "buy our merch!" annotations that were damaging to the video content and led to people turning annotations off. I think simply removing all links would be clear and effective, without damaging the error-correction use.
The third common use that I've seen is linking to other videos, and that's the one that YouTube's replacements systems tackle.
Another somewhat noteworthy thing is that the annotation system is tied to YouTube's channel logo overlay. Once the annotations go, it's likely that the logo will not be hideable anymore.
My understanding of the argument is that you have to assume you keep playing after you lose, so your stake is dictated by the risk. For the example in the article, the potential upside is 110% profit and the potential downside is 100% loss, so the optimal stake is relatively conservative in order to prevent any loss from impacting your future ability to invest.
Imagine you get the expected result of one win and one loss and had staked 5%: you'll end up with 1.055×0.95=1.002 wealth. If you had picked a 50% stake, you'd be sitting at 1.55×0.5=0.775 wealth instead, since the loss more than erases your winnings.
The unstated contrast is to models where e.g. there are only a limited number of investment opportunities, losses aren't total, or there are capital infusions.
Apologies for the confusion but I was talking about Samuelson's argument against the Kelly criterion.
If I was to make one gamble per day and wanted to maximise my profit after 50 years, I don't know if I'd use the Kelly criterion. I need to think about it.
If you just want to maximise the average amount of money that you end up with then you should bet your entire fortune every day. Of course you will be bankrupt if you ever lose your bet, which is very likely over the course of 50 years. But in the ridiculously unlikely event that you won every bet your fortune would be even more ridiculously large. So the average amount of money would be very large, even though you would very probably be bankrupt.
I see. What if I wanted to maximise the expected profit, given that the chance of bankruptcy should not exceed 1%? Does that sort of problem become difficult to solve?
Sorry for the late reply (don't know if you'll see this!) but I happened to be thinking about this again and the question popped into my head: what if I wanted to maximise a specific percentile (e.g. the 5th percentile) of my total winnings?
Fixing the percentile seems akin to the gambler's fallacy: since losses below a certain point don't matter, you just have to bet everything when you fall off, like you would if you thought after a big loss streak you were "owed" a win.
