explain xkcd - User contributions [en]

2405: Flash Gatsby

2022-03-12T13:04:25Z

162.158.62.32: /* Explanation */

{{comic
| number = 2405
| date = December 30, 2020
| title = Flash Gatsby
| image = flash_gatsby.png
| titletext = Protip: At midnight your excuse for not having read The Great Gatsby can switch from "I'm worried about violating copyright" to "I think my copy requires Flash."
}}

==Explanation==

This comic unfolds over the last few seconds of 2020 and the first few seconds of 2021. [[Cueball]] is attempting to do something requiring the overlap of two eras that only abut: creating an "unauthorized" adaptation of ''The Great Gatsby'', using the Adobe Flash plugin platform.

''{{w|The Great Gatsby}}'' is a classic novel written by {{w|F. Scott Fitzgerald}} in 1925. Copyright law in the United States of America, where ''The Great Gatsby'' was first published, was retroactively extended several times in the 1990s and early 2000s, causing the copyright on ''The Great Gatsby'' to extend [https://www.mentalfloss.com/article/595567/why-the-great-gatsby-isnt-public-domain until the end of 2020]. In 2021, it finally entered the public domain so that it became legal to make a copy without violating copyright law.

{{w|Adobe Flash}}, formerly known as Shockwave Flash, is a web plugin that was commonly used by many websites in the late 1990s and 2000s. It allowed website creators to add animations, sound, and complex logic to build games, videos, and other interactive experiences. Presumably, the Flash version of the novel is some kind of interactive reader, animated cartoon, or perhaps even a game.

Over the years, Adobe Flash was repeatedly exploited by hackers, incurring heavy costs on {{w|Adobe Inc.|Adobe}} as they tried to update Flash against these attacks after rushing features out before stabilizing them. Newer technologies are now able to provide comparable features with more compatibility, more community involvement, and less risk, so support for Flash is being phased out by most web browsers. Adobe officially [https://www.adobe.com/ca/products/flashplayer/end-of-life.html ended support for Flash] after December 31, 2020.

In line with Adobe's decision, [https://www.chromium.org/flash-roadmap#TOC-Upcoming-Changes Chrome is blocking Flash in January]. This will make [https://www.newgrounds.com/games entire internet culture histories spanning many years of making and engaging Flash experiences] unusable for most people. Therefore, Cueball's Flash version of ''The Great Gatsby'' will become legal at the very moment that everyone should stop using it.

In this comic, [[Cueball]] suggests that the withdrawal of Flash support occurs after the copyright expiration rather than simultaneously with it. This is most likely because the applicable copyright law in the United States states that the creative work becomes public domain at the end of the year 2020 and Flash gets disabled at the beginning of the year 2021. So it is conceivable (but not practical) that there is one second when the novel is public domain and Flash is still enabled.

By late 2020, Flash Player was already blocked by most browsers, but could still be whitelisted on individual sites. Using old versions of browsers, or workarounds to run blocked extensions, Cueball's Great Gatsby may still be readable after the official Flash End of Life date of January 1, 2021. Even with these workarounds, Flash Player itself will block Flash content from playing on January 12, 2021, making that the final death date for official modern versions of Flash.

After January 12, Flash content may still be accessible through older builds of Flash Player, and through various archival and emulation projects, such as the [https://archive.org/details/softwarelibrary_flash_showcase Internet Archive], [https://ruffle.rs/ Ruffle], [https://bluemaxima.org/flashpoint/ BlueMaxima's Flashpoint], and [https://www.getsupernova.com/ SuperNova].

The title wording has a number of possible meanings to it. It's the 'Gatsby' book via the medium of the electronic Flash format. Because of the briefest of availability (at best, a single moment), it appears and disappears again 'in a flash'. Being 'flash' is a very apt description of the millionaire Gatsby character himself ('Flash the cash' is being ostentatious). And, if the endeavor is not actually as legitimate as hoped, the word has also referred to felonious behaviors and forged copies.

The title text references using excuses for not having read a book considered a classic. Before the end of 2020, a possible excuse for not trying to read it was it may not have been available in the format a person wanted it (such as via a flash program in this case) and it might have been illegal (copyright violation) to get it in that format. After 2020, the new excuse to not read it could be a technical one (flash doesn't work/nothing capable of running flash). Both excuses are quite flimsy; it's apparent the person really just doesn't care to read The Great Gatsby.

==Transcript==

:[Cueball is sitting at a desk using his laptop.]
:Off-panel voice: 3... 2... 1... ''Happy New Year!''
:Cueball: Okay, it’s up!
:Cueball: Annnnnd ... support was pulled, it’s down again.

:[Caption below the panel:]
:There's only a very short window of time in which I can post my unauthorized Flash® adaptation of ''The Great Gatsby''.

{{comic discussion}}
[[Category:Comics featuring Cueball]]
[[Category:Fiction]]
[[Category:Internet]]
[[Category:New Year]]
[[Category:Protip]] <!-- title text ->

2589: Outlet Denier

2022-03-05T00:40:27Z

162.158.62.32:

{{comic
| number = 2589
| date = March 4, 2022
| title = Outlet Denier
| image = outlet_denier.png
| titletext = There are regularly placed bumps on the underside just the right size to press the rocker switch on the power strip.
}}

==Explanation==
{{incomplete|Created by an OUTLET IN DENIAL - Please change this comment when editing this page. Do NOT delete this tag too soon.}}

This is the fifth installment in the series of [[:Category:Cursed Connectors|Cursed Connectors]] and depicts a plug that covers a whole power bar. Connectors can be difficult to plug in power strips or sockets because the shape of the connector make them unable to fit or make them cover space needed for other connectors. Some seem deliberately designed too large to share a duplex outlet so as to prevent overloading the circuit. The comic depicts an extreme case of a cumbersome connector shape designed to block an entire power strip.

Many (most?) power strips have a rocker-style power switch at one end, that can be used to turn off all the outlets. The title text says that this connector has bumps that match up with that location. It's not clear whether this will turn the power switch off or force it always on. But either way, it gets in the way of the user being able to control the power themselves.

==Transcript==
{{incomplete transcript|Do NOT delete this tag too soon.}}
Cursed Connectors #78

[Drawing of a power strip with a rocker switch and a connector that would cover the whole power strip.]

The outlet denier

{{comic discussion}}

[[Category: Cursed Connectors]]

Talk:2327: Oily House Index

2022-01-15T18:28:54Z

162.158.62.32:

Dangit Randall, this was my retirement plan & now everybody's gonna want to try it!
[[User:ProphetZarquon|ProphetZarquon]] ([[User talk:ProphetZarquon|talk]]) 00:53, 2 July 2020 (UTC)

Negative Equity (owing more than the house is worth) ''shouldn't'' be an immediate problem under most circumstances. If the householder isn't actually wanting to move and can still afford the asked-for repayments then it doesn't change the physical situation at all. The bank has no problems so long as the household has no problems, as they ride over (temporary) pricing crashes and emerge the other side. It's when banks get nervous that the home'owners' ''might'' default and thus put pressures on them (e.g. 'negotiating' for unsustainably greater repayments or 'immediate settlement' of the unforeseen temporary deficit) that they could tip their so-called customer over the edge. And an increase of defaulting further suppresses house-prices (general availability of sell-quick homes by owners/bank and/or the reduced neighbourhood value around abandoned properties not sold ''nor'' (officially) lived in) to draw more agreements into the self-creating danger-zone. Of course it aint as simple as all that. And permanently being underwater due to coastal flooding, ''probably'' won't sit well with the actuaries behind your continuing loan if your property isn't in Innsmouth... [[Special:Contributions/162.158.159.76|162.158.159.76]] 09:31, 2 July 2020 (UTC)

;Maths
Can someone figure out where I went wrong here?<br>
>The comic then applies dimensional analysis to this index: dividing $/sqft by $/bbl yields a result whose dimension is a linear measurement, which can be called length. 1 barrel is 5.6 cubic feet. The average price per square foot of a new single-family dwelling in the USA in 2019 was about 119 $/sqft, while the price of oil in mid 2019 was about $60/BBL or $337/cubic foot. Dividing gives 60/337 feet-1 or about 5.61 feet. (This doesn't match the value shown on the chart of around 15, so we have done something wrong here. :))<br>
Thanks. [[User:Stevage|Stevage]] ([[User talk:Stevage|talk]]) 00:54, 2 July 2020 (UTC)
:Since barrels are in the denominator, you have to divide by 5.6 to get the price per cubic foot. [[User:LegionMammal978|LegionMammal978]] ([[User talk:LegionMammal978|talk]]) 01:00, 2 July 2020 (UTC)
($/area)/($/volume)=($/sq.ft)/($/cu.ft)=1/ft? Shouldn't the result be in ft?
:($/sq.ft)/($/cu.ft)=($/sq.ft)*(cu.ft/$)=($*ft*ft*ft)/($*ft*ft)=ft I initially made the 1/ft mistake too, until I remembered to invert the fraction in the denominator[[Special:Contributions/172.69.63.155|172.69.63.155]] 14:53, 8 July 2020 (UTC)

;Units
Shouldn't area divided by volume be height, not length? It would also fit better with the graph. [[Special:Contributions/162.158.123.173|162.158.123.173]] 03:41, 2 July 2020 (UTC)
:For dimensional analysis, you don't care about the physical context of the units, just about the dimension they are associated with. Height is horizontal length, so it has the dimension of length. In the context of the comic this length can be interpreted as a height, but in another context, it could be a length in a different orientation. [[Special:Contributions/162.158.88.78|162.158.88.78]] 04:16, 2 July 2020 (UTC)

;Category
Should we start a category of dimensional analysis comics: e.g. [[687]], [[1707]], [[2312]] --[[User:WhiteDragon|WhiteDragon]] ([[User talk:WhiteDragon|talk]]) 07:41, 2 July 2020 (UTC)

;Division Error?
You can't divide by zero; which means Randall made an error. Should we update the page to reflect this? [[Special:Contributions/173.245.52.67|173.245.52.67]] 10:25, 2 July 2020 (UTC)
:"OHI briefly became infinite as oil prices reached zero in 2020" could be read as approaching both infinity and zero; that fixes the problem [[Special:Contributions/162.158.74.249|162.158.74.249]] 11:18, 2 July 2020 (UTC)
::Randall did not divide by zero. If the price went continually to zero the OHI would aproach infinity. Of course at the time the price hit zero (or negative), then the OHI breaks down, which is what infinite means. So he did not make any error. (Wrote this and had an edit conflict with the first reply.) --[[User:Kynde|Kynde]] ([[User talk:Kynde|talk]]) 11:20, 2 July 2020 (UTC)
:You can divide by zero, its just that those lazy mathematicians haven't defined it yet.
::Programmers have defined it but it isn't a number: x/0 = NaN (Not a Number) [[Special:Contributions/108.162.216.134|108.162.216.134]] 11:43, 3 July 2020 (UTC)
[[Special:Contributions/162.158.123.155|162.158.123.155]] 04:04, 3 July 2020 (UTC)
::: Actually, most programmers, including IEEE, define x/0 as Infinity if x > 0, -Infinity if x < 0, and NaN if x = 0 (for floating-point numbers; for integers, it's still undefined). [[Special:Contributions/162.158.62.32|162.158.62.32]]
:: We defined it; y'all just don't like what we came up with. https://en.wikipedia.org/wiki/Wheel_theory [[User:TobyBartels|TobyBartels]] ([[User talk:TobyBartels|talk]]) 06:45, 3 July 2020 (UTC)
:You can divide by zero, but it doesn't do the house any good: [http://i.imgur.com/SHAuVKX.jpg House divided by zero] [[User:These Are Not The Comments You Are Looking For|These Are Not The Comments You Are Looking For]] ([[User talk:These Are Not The Comments You Are Looking For|talk]]) 22:23, 5 July 2020 (UTC)
:Randall writes “became infinite”, not “became infinity”, isn’t that a difference? Dividing by zero doesn’t give something finite, so it is in-finite, but not ∞, right?

The comic should not use the word "mortgage," because the calculations are based on sale price. The size of the mortgage depends on the down payment. [[User:Cellocgw|Cellocgw]] ([[User talk:Cellocgw|talk]]) 11:40, 2 July 2020 (UTC)
:My understanding is that if you fully mortgage your house (so that you now have in your hands enough money to buy the house again) and convert the total amount of money that the house is worth into oil, you can then fill your house with X feet of oil. [[Special:Contributions/141.101.98.56|141.101.98.56]] 15:11, 2 July 2020 (UTC)
::Find me a mortgage that will give me 100% of the value of my house. Please! [[Special:Contributions/162.158.123.155|162.158.123.155]] 04:02, 3 July 2020 (UTC)
:::Hint: Start by not writing in the "Reason for mortgage" box anything like "I want to entirely fill my home with viscous and/or flammable liquid". Notwithstanding what I believe the US calls 'zoning laws'... [[Special:Contributions/162.158.159.14|162.158.159.14]] 20:38, 3 July 2020 (UTC)

It's a pity the graph doesn't go back as far as the [https://en.wikipedia.org/wiki/1973_oil_crisis 1973 oil crisis]. [[Special:Contributions/141.101.98.26|141.101.98.26]] 12:02, 3 July 2020 (UTC)

I belatedly wonder if the title is meant to be similar to some other famous Index that I happen to have not heard of. Searching around, Freedom House is an American... thing, which apparently publishes a "Freedom House Index" which is a sort of measurement of how "democratic" a country is from year to year. So something like that... Robert Carnegie rja.carnegie@excite.com [[Special:Contributions/162.158.159.14|162.158.159.14]] 21:25, 3 July 2020 (UTC)

Shouldn't the OHI be higher because the footprint of a multiple story house is less than its square footage? - 162.158.74.173

I sent this to Matt Levine, and he included it in his Money Stuff newsletter. [https://www.bloomberg.com/opinion/articles/2020-07-02/free-wirecard-money-costs-softbank] [[User:Benjaminikuta|Benjaminikuta]] ([[User talk:Benjaminikuta|talk]]) 07:01, 8 July 2020 (UTC)

Blue Eyes

2022-01-07T12:22:13Z

162.158.62.32: No mention of hobby, removing Category:My Hobby

{{comic
| date = October 11, 2006
| title = Blue Eyes
| image = Blue Eyes.jpg
| before = The Hardest Logic Puzzle in the World
| lappend = blue_eyes.html
| extra = yes
}}
{{TOC}}
==Explanation==
xkcd's [http://xkcd.com/blue_eyes.html Blue Eyes] puzzle is a logic puzzle posted around the same time as comic [[169: Words that End in GRY]]. [[Randall]] calls it "The Hardest Logic Puzzle in the World" on its page; whether or not it really is the hardest is up to speculation.

The page contains two comics. On the top is [[82: Frame]], and at the bottom is [[37: Hyphen]]. These particular comics may have been chosen intentionally, as 82 involves a mind screw (and formal logic can be pretty mind-screwy to the uninitiated) and 37 involves linguistic ambiguity, which Randall has explicitly gone out of his way to avoid (interestingly, [[169]] involves the infuriating ambiguity caused by misquoting riddles). That said, Randall could have simply picked those comics out of a hat to plug for his comic (which he also does explicitly), and the date of release could also be completely random.

Randall cites "some dude on the streets in Boston named Joel" as his source for the comic idea (although he's rewritten it).

==The Puzzle==

A group of people with assorted eye colors live on an island. They are all perfect logicians -- if a conclusion can be logically deduced, they will do it instantly. No one knows the color of their eyes. Every night at midnight, a ferry stops at the island. Any islanders who have figured out the color of their own eyes then leave the island, and the rest stay. Everyone can see everyone else at all times and keeps a count of the number of people they see with each eye color (excluding themselves), but they cannot otherwise communicate. Everyone on the island knows all the rules in this paragraph.

On this island there are 100 blue-eyed people, 100 brown-eyed people, and the Guru (she happens to have green eyes). So any given blue-eyed person can see 100 people with brown eyes and 99 people with blue eyes (and one with green), but that does not tell him his own eye color; as far as he knows the totals could be 101 brown and 99 blue. Or 100 brown, 99 blue, and he could have red eyes.

The Guru is allowed to speak once (let's say at noon), on one day in all their endless years on the island. Standing before the islanders, she says the following:

"I can see someone who has blue eyes."

Who leaves the island, and on what night?

==Solution==

Randall's solution is at [http://xkcd.com/solution.html xkcd.com/solution.html].

Here are some observations that help simplify the problem.

No one without blue eyes will ever leave the island, because they are given no information that can allow them to determine which non-blue eye color they have. The presence of the non-blue-eyed people is not relevant at all. We can ignore them. All that matters is when the blue eyed people learn that they actually are blue-eyed.

There are two ways in which blue-eyed people might leave the island. A lone blue-eyed person might leave on the first night because she can see that no one else has blue eyes, so the guru must have been talking about her. Or an accompanied blue-eyed person can leave on a later night, after noticing that other blue-eyed people have behaved in a way that indicates that they have noticed that her eyes are blue too.

The problem is symmetrical for all blue-eyed people, so this means they will either all leave at once or all stay forever.

'''Theorem:''' If there are N blue-eyed people, they will all leave on the Nth night.

'''Dual Logic.'''

Blue eyed people leave on the 100th night.

If you (the person) have blue eyes then you can see 99 blue eyed and 100 brown eyed people (and one green eyed, the Guru).
If 99 blue eyed people don't leave on the 99th night then you know you have blue eyes and you will leave on the 100th night knowing so.

'''Intuitive Proof.'''

Imagine a simpler version of the puzzle in which, on day #1 the guru announces that she can see at least 1 blue-eyed person, on day #2 she announces that she can see at least 2 blue eyed people, and so on until the blue-eyed people leave.

So long as the guru's count of blue-eyed people doesn't exceed your own, then her announcement won't prompt you to leave. But as soon as the guru announces having seen more blue-eyed people than you've seen yourself, then you'll know your eyes must be blue too, so you'll leave that night, as will all the other blue-eyed people. Hence our theorem obviously holds in this simpler puzzle.

But this "simpler" puzzle is actually perfectly equivalent to the original puzzle. If there were just one blue-eyed person, she would leave on the first night, so if nobody leaves on the first night, then everybody will know there are at least two blue-eyed people, so there's no need for the guru to announce this on the second day. Similarly, if there were just two blue-eyed people, they'd then recognize this and leave on the second night, so if nobody leaves on the second night, then there must be a third blue-eyed person inspiring them to stay, so there's no need for the guru to announce this on the third day. And so on... The guru's announcements on the later days just tell people things they already could have figured out on their own.

It's obvious that our theorem holds for the "simpler" puzzle, and this "simpler" puzzle is perfectly equivalent to the original puzzle, so our theorem must hold for the original puzzle too.

Another way of looking at it is to use selective attention. Although each blue-eyed person can see each other blue-eyed person on the island, she doesn't need to. The only thing she needs to know in order to determine whether to leave on night N is whether or not she can see an Nth person with blue-eyes. On night 1, she only needs to see 1 other blue-eyed person to not leave; on night 2, she can see 2 other blue-eyed people, so she doesn't leave; and so on and so on until night 100 when she can't see a 100th blue-eyed person, and then leaves.

'''Formal Proof.'''

To prove this more formally, we can use mathematical induction. To do that, we'll need to show that our theorem holds for the base case of N=1, and we'll need to show that, for any given X, *if* we assume that the theorem holds for any value of N less than X, then it will also hold for N=X. If we can show both these things, then we'll know the theorem is true for N=1 (the base case), for N=2 (using the inductive step once), for N=3 (using the inductive step a second time) and so on, for whatever value of N you want.

Base case: N=1. If there is just one blue-eyed person, she will see that no one else has blue eyes, know that the guru was talking about her, and leave on the first night.

Inductive step: Here we assume that the theorem holds for any value of N less than some arbitrary X (integer greater than 1), and we need to show that it would then hold for N=X too. If there are X blue-eyed people, then each will reason as follows: "I can see that X-1 other people have blue eyes, so either just those X-1 people have blue eyes, or X people do (them plus me). If there are just X-1 people with blue eyes, then by our assumption, they'll all leave on night number X-1. If they don't all leave on night number X-1, then that means that there is an Xth blue-eyed person in addition to the X-1 that I can see, namely me. So if they all stay past night number X-1, then I'll know I have blue eyes, so I'll leave on night number X. Of course, they'll also be in exactly the same circumstance as me, so they'll leave on night number X too."

This suffices to prove our theorem. The base case tells us the theorem holds for N=1. That together with the inductive step tells us that it therefore holds for N=2, and that together with the inductive step again tells us that it holds for N=3, and so on... In particular, it holds for the case the original puzzle asked about, N=100, so we get the conclusion that the 100 blue-eyed people will leave on the 100th night.

==Randall's thought-provoking questions==

After giving his solution, Randall posed three questions for further thought about the puzzle. (I'll answer them in a different order than he asked.)

'' '''Question 2.''' Each person knows, from the beginning, that there are no less than 99 blue-eyed people on the island. How, then, is considering the 1 and 2-person cases relevant, if they can all rule them out immediately as possibilities?''

Blue-eyed people can't see their own faces, so blue-eyed people can see one less blue-eyed face than non-blue-eyed people can. Even though I can see that there are at least 99 blue-eyed people, I don't know that they can see that, so I need to imagine people who see only 98, who would base their actions in part by imagining people who can see only 97 who would base their actions in part by imagining people who can see only 96, and so on... All the levels are relevant. (It's like [https://www.youtube.com/watch?v=U_eZmEiyTo0 the Princess Bride scene] where Vizzini is trying to think about what Wesley would choose in part based upon Wesley thinking about what Vizzini would choose in part based upon... "So clearly I cannot choose the one in front of me!") Each layer of thinking about what someone else might be thinking about can decrement by 1 the number of blue-eyed people visible to the lattermost imagined person, so it turns out that even the base case with N=1 blue-eyed person is relevant. As the days go by, some of the more far-fetched "he might be thinking that I might be thinking that he might be thinking that I might be thinking that..." hypotheses get ruled out. But it's only after night N-1 that the blue-eyed people rule out all the possibilities in which they have brown eyes, whereas the brown-eyed people only learn on night number N that they don't have blue eyes.

It might help to think of all the different situations people might be in. (Remember brown-eyed people always are situated where they can see one more blue-eyed face than blue-eyed people can.)

'''Situation 0.''' If I see 0 blue-eyed people, I can leave right after the announcement on night 1.
'''Situation 1.''' If I see 1 blue-eyed person, then she might be in situation 0 and about to leave on night 1; or else she might be in situation 1 just like me, in which case we'll both leave together on night 2.
'''Situation 2.''' If I see 2 blue-eyed people, they might each be in situation 1 watching to see whether anyone in situation 0 leaves the first night (I know nobody will leave that night, but they wouldn't know this), in which case they would leave together on night 2; or else they might be in situation 2 just like me, in which case we'll all leave together on night 3.

:
:
:
'''Situation N.''' If I see N blue-eyed people, they might be in situation N-1 watching to see whether any people in situation N-2 leave on night N-1 (I know nobody will leave that night, but they wouldn't know this), in which case they would leave together on night N; or else they might be in situation N just like me, in which case we'll all leave together on night N+1.
:
:

Even though I start out in situation 99, I need to worry that the blue-eyed people might be in situation 98, so I need to wait long enough for people in situation 98 to figure out what's going on, and then see whether they act like they are indeed in situation 98. But if they're in situation 98, then they're worrying about whether all the blue-eyed people might be in situation 97, so they're going to need to wait long enough for people in situation 97 to figure out what's going on. Of course, that requires waiting long enough for people in situation 96 to figure out what's going on, and so on, down all the way to situation 0. All the levels are relevant, and it takes a separate day to eliminate each level, which is why the whole process takes N days.

'' '''Question 3.''' Why do they have to wait 99 nights if, on the first 98 or so of these nights, they're simply verifying something that they already know?
''

Consider an analogy. I've heard that miners used to take canaries down into mines because canaries pass out more quickly in poor air than miners do. Suppose you know the canary will do fine for 98 or so seconds, and then pass out if the air is bad. As you watch the canary for those 98 seconds, there's a sense in which you're just verifying something you already know (it'll do fine), but it seems more accurate to say that your best detector for the quality of the air takes 98 seconds to give you a reading, and you're waiting 98 seconds to see what that reading is.

When the blue-eyed people wait 98 or so days to leave, that's because their best available detector of their own eye-color takes 98 or so days to give a reading. (This detector involves watching what the other blue-eyed people do, and of course they themselves are waiting on a detector that takes 97 or so days to yield its result...) There's a sense in which they're "simply verifying something that they already know", but it seems more accurate to say that they're waiting for their best available detector of their own eye-color to deliver its reading.

'' '''Question 1.''' What is the quantified piece of information that the Guru provides that each person did not already have?''

Before the Guru speaks, the hypothetical chain of A imagining B imaging C imagining D...imagining Z seeing N blue eyed people cannot terminate uniquely. Z seeing no blue eyed people can consider whether or not they are blue eyed. This means it is not {{w|Common knowledge (logic)|common knowledge}} that there are blue eyes. Once the guru makes their pronouncement it is common knowledge and every chain of reasoning must terminate at 1 blue eyed person and Z above would have to conclude that they had blue eyes. From then on every midnight the common knowledge that there are N blue eyed people increments by 1 as everyone sees nobody leaving on the ferry.

Stated another way, there's only one stable set of beliefs for the blue eyed people that would allow them to have so many exist on the island indefinitely. That is if each blue eyed person believed not only that they have brown eyes, but also that every other blue-eyed person believed, incorrectly, that they had brown eyes. Logic reduces this to "all blue-eyes believe that all blues-eyes have brown eyes". The Guru eliminates that particular possibility.

Another simple way to understand why the Guru's information is important is thus. Each blue-eyed person knows two sets of information: what the actual situation is on the island (both now and in the past), and what would happen in a hypothetical situation. Each blue-eyed person then needs only to compare the actual situation to a known hypothetical one, and if it matches up, then they take the corresponding action. Consider this: If there were only one blue-eyed person, and the guru never made the announcement, she would not leave on day 1 because she would not know that N is greater than or equal to 1. Now let's add a 2nd blue-eyed person. Blue-eyes 2 would not be able to inductively determine whether or not to leave on night 2, because blue-eyes 2's knowledge of whether or not to leave on night 2 is dependent on what blue-eyes 1 does on night 1 if and only if blue-eyes 1 knows what to do on night 1. If blue-eyes 1 doesn't know that N is greater than or equal to 1, then blue-eyes 1 doesn't know what to do on night 1, so her lack of leaving gives blue-eyes 2 no new information, since it was an uninformed action and blue-eyes 2's inductive reasoning was dependent on blue-eyes 1 knowing what to do, and so the inductive process never takes off for the hypothetical situation. This means a hypothetical situation for N people cannot be induced. As such, blue-eyes 100 does not have certain knowledge of the hypothetical situation that would occur on nights 99 and 100, and so even though she knows N = either 99 or 100, she can't take action on either of those nights, because she has no certain hypothetical situation to compare reality to, and as such cannot have certainty about the actions she should take.

==Trivia==
*The web page which contains the puzzle has no {{w|CSS|style sheet}}. The font size of the heading and subheading is increased with deprecated HTML tags, rather than the heading tags. The way the page is displayed therefore depends on the browser's settings. Despite this fact, due to a similarity of default settings between computers, most computers will by default display the page similarly to the way it is displayed in this page's screenshot, with a white background, black text and the {{w|Times New Roman}} font or a similar one. However, it has two line breaks after every paragraph instead of HTML paragraph breaks, meaning that paragraph spacing will not vary between browsers, relative to the font size.

{{comic discussion}}
[[Category:Multiple Cueballs]]
[[Category:No title text]]
[[Category:Comics with lowercase text]]