Klye @khird@qoto.org
Mod
Joined Sep 2018
@realcaseyrollins Oh, it sounds like a Seagrams/Schweppe's/Canada Dry type of drink, something like that?
My go-to is Reeds, kind of unusual in that the sweeteners are honey and pineapple juice rather than sugar/HFCS. So definitely has a unique taste. I could see a comparison to Sprite, but probably more juicy rather than less.
Blenheim #3 gets my vote for "most flavourful" but also "why on earth would you want that much ginger!?" Tastes absolutely nothing like a Sprite.
@realcaseyrollins Mind if I ask what it is? I used to drink quite a lot of ginger beer & you have me curious
@louisrcouture yes, but you can't stop them from selling or giving away the source once they have it. So if your business model is to charge for a copy of the software, it won't last long in an environment where people upload your source to github or whatever.
@freemo s/complex field/Gaussian rationals/
@stardust Because of federation, I believe. Once other servers have local copies of the toot, those local copies could fall out of sync (ActivityPub doesn't guarantee consistency). So with Delete+Redraft, everyone who knows of the existence of object X agrees on what it is. They might be unaware it's been deleted, or not see replies from certain instances, or whatever; but there's no disagreement about what it is. But if you allow editing, that's no longer the case - for example, maybe a poll was removed, and remote instances will try to send votes to what's now a plain-text post, confusing the server.
What might work would be to make an "edit" a self-reply with an annotation to that effect in the metadata. Then clients could grab the parent (older-version) post, and display replies in that thread with a warning that they are responses to an older version, like what Wikipedia does if you look at earlier revisions of a page. Clients unaware of the new feature would fail reasonably gracefully and show it as a normal self-reply, because under the hood it's a new status with a new object ID, not really an edit. It just looks that way by adding an annotation with those semantics.
@freemo Oh man, hope you feel better soon!
I just spent a bit looking at the sequential case and I think I've proved it's impossible. Obviously the penultimate state can't have four heads or four tails, because then you're just hoping that by luck you happen to get the odd one out. So it's got to be a three-two split, of which there are two types.
Your first choice is completely blind, so it's possible you choose one of the majority (which I've labelled black in the diagram, but could be heads or tails). Let's stipulate that's the case, so there's no new information gained after the first move and we can predetermine the second move. If you flip a second black coin, you have to know exactly where the third black coin is to guarantee a win. If you flip a white coin, you have to know exactly where the second white coin is to guarantee a win. But in either case, if you draw one of the pairs shown in the diagrams, there are two possible orientations of the board - you know you're on one of the lines of a certain colour, but not which, and consequently not where the final piece you need is.
@freemo Could you please type up your strategy? I haven't really looked at the sequential choice version, but it might offer clues to solving the general case.
> People are going to get excited and shout from the roof tops when they are winning.
People are going to shout from the rooftops if they stand to benefit from increased demand. If I think the price of X will continue to go up, I'll quietly buy up X and not brag about how much it's already gone up. If I'm not so sure, then maybe I start trying to stoke public demand to get it on an upward swing again.
@freemo You're half right. Your test is necessary but not sufficient, so it'll sometimes answer "possible" when there isn't really a solution.
For example, five boxes three choices. Your selection modes are:
00111
01011
Their bitwise union is:
01111
which has a single zero, so by your algorithm it should be solvable. But in fact it isn't. If it were, you'd have a nearly-solved state, where, no matter how the table was spun, you'd be able to guarantee you either get *all* the remaining heads or *all* the remaining tails.
There's another failure mode too in that "standard form" isn't necessarily the worst-case form. Consider six boxes three choices:
000111
001011
001101
010101
Their bitwise union is 011111. But if we rewrite the selection modes, rotating the second and third ones, as
000111
010110
010011
010101
then we get a bitwise union of 010111 which is clearly unsolvable.
@mushambo because the behaviour of cryptocurrency enthusiasts is so much like that of scammers that it sets off our BS detectors every time we try to read up on it. Every time bitcoin goes up, there are thousands of people bragging about how well their portfolios are doing, and when it goes back down they all disappear. It's *exactly* the behaviour of someone who is trying to convince me to buy an asset when it's already overpriced.
Crypto might not actually be such a terrible investment but anyone who spends significant time on the internet develops such a strong nope reflex to "I made X thousand dollars this week doing Y" posts that by default our assumption is:
- I wouldn't make close to X doing Y
- you somehow stand to profit off me doing Y
@freemo some further observations:
For any number of boxes:
If you're allowed to choose all the boxes, a win is trivially possible: flip all to heads.
If you're allowed to choose all but one of the boxes, a win is trivially possible: flip all to heads; if you then see a tails turn it to heads, if not turn all coins to tails.
For five boxes:
There's no solution with three choices, because there's no penultimate state from which you can guarantee that you'll choose all the remaining heads or all the remaining tails. So five boxes has only the trivial solutions.
For six boxes:
I have a solution for four choices in five moves, which may not be optimal (I haven't searched exhaustively for other strategies).
The number of patterns for given numbers of boxes and choices is from the OEIS: https://oeis.org/A047996, under "example". I don't understand the notation they use for the generating formula, but they call it a "circular binomial coefficient" which makes sense.
@freemo I don't think three flips solves six boxes. Your patterns are 1-2-3, 1-2-5, 1-3-5, 2-3-5, unless I'm forgetting one. If box four doesn't match box six you're out of luck.
@freemo jumping right to the general case!
I think it's going to be exponential difficulty in at least flips-per-round, and probably also in boxes.
My algorithm is this:
Go through all known cases: if I were told the state of the boxes, for each pattern I could choose, what possible combinations would I find?
For each state-choice-combination triplet, and each possible combination I could set the coins to, what state (or superposition of states) could I leave it in?
Go through the unknown states the same way, by superposing the known ones, building all the way up to completely unknown (all states possible).
Make the decision tree by minmaxing, like in two-player game theory. If the starting state of "completely unknown" has a path to either all-heads or all-tails, count the steps in the path. If there isn't a path linking the starting state to the winning states, it's unsolvable.
@freemo only opening two boxes per turn, you can't do five or six. I think there might be a solution if you allow the player to open three per turn, at least for five.
For five boxes, choosing two: whenever you choose adjacent, suppose you choose boxes one and two; whenever you choose skip-one, suppose you choose one and three. Box four contains a heads, and box five contains a tails. You'll never be able to get all five the same.
For six, choosing two: suppose your adjacent choices are one and two, your skip-one are one and three, and your opposite are one and four. You'll never be able to get five to match six.
@freemo exactly!
For similar reasons, even under the original instant-win rules, a version with five boxes is unsolvable. No matter how many times you open any arrangement of two boxes, fate can always allow there to be two boxes that don't match and you never see.
I think five boxes, opening three per turn, should work, but I haven't figured out what the target number of moves should be.
@freemo proof of impossibility could be a solution to the puzzle 😎
@freemo unless you can logically deduce from the the coins you've seen that it must be solved or must not be solved, you cannot make the assumption in either direction.
@freemo I don't believe you can guarantee a win in five without the instant-win condition in the original problem.
@freemo No. If you know they're all the same, you can just leave them for the remaining turns. Or if you got them all the same without realising it, you can flip them out, flip them back, whatever you like until your N turns are up. I'm just removing the assumption that the fact the game hasn't ended implies all four are not the same.
@freemo Let's extend the riddle by saying you have to set the coins the same after exactly N moves. You're allowed to have all four coins the same before then, but you won't win unless they're still the same (or again the same) at exactly N moves.
What's the smallest N for which there's a guaranteed winning strategy?
🇳🇱
