explain xkcd - User contributions [en]

Talk:3015: D&D Combinatorics

2024-11-25T09:39:31Z

172.70.86.141:

The bot originally created this page as “D Combinatorics”. I renamed it to the correct title and tried to get as many of the references as possible (including a few redirects). [[User:JBYoshi|JBYoshi]] ([[User talk:JBYoshi|talk]]) 00:54, 23 November 2024 (UTC)

:The title in the Atom feed (which I'm assuming the bot consumes) is "D Combinatorics". I'm guessing something in Randall's pipeline didn't like the ampersand. --[[Special:Contributions/162.158.154.160|162.158.154.160]] 01:41, 23 November 2024 (UTC)

::Yup, if you look at [https://xkcd.com/3015/info.0.json 3015's JSON] you see that <code>title</code> and <code>safe_title</code> differ, and if you look at the HTML page source you'll see '''3''' different things: <code><title>xkcd: D Combinatorics</title></code>, <code><meta property="og:title" content="D&amp;D Combinatorics"></code>, and <code><div id="ctitle">D&D Combinatorics</div></code>! So probably what happened is Randall entered D&D but was supposed to enter D&amp;D, and the openGraph tags adder code, having to be HTML-aware, decoded & normalized D&D as HTML would, but the other parts of the pipeline just ate it for some reason.

:By raw combinatorics: 71 + 52 + 34 + 20 + 10 + 4 + 1 ways to get each of 16 - 22 respectively, for a total of 192, out of 4(6^3) = 864 total. 192/864 simplifies to exactly 2/9. I have no idea how Randall found this; if anyone has an idea, please let me know. [[User:Kaisheng21|Kaisheng21]] ([[User talk:Kaisheng21|talk]]) 01:33, 23 November 2024 (UTC)
::I used some simple python code to loop over every dice and confirm and it's 2/9 [[Special:Contributions/162.158.158.111|162.158.158.111]] 12:11, 23 November 2024 (UTC)

It seems like we edited the transcript at the same time. The odds of rolling 16 or higher in this situation seem to be 2/9? [[User:Darkmatterisntsquirrels|Darkmatterisntsquirrels]] ([[User talk:Darkmatterisntsquirrels|talk]]) 01:29, 23 November 2024 (UTC)

: There are 864 possible rolls (6 * 6 * 6 * 4). If you enumerate all of the rolls you will find that 192 are 16 or higher. 192/864 = 2/9, the value from the explanation. [[Special:Contributions/172.68.54.139|172.68.54.139]] 01:41, 23 November 2024 (UTC)

I added a table of outcomes to clarify how it works out to 2/9, anyone know how to make it pretty? -- Laurence Cheers {{unsigned ip|172.71.150.247}}

A much simpler approach: Roll two six sided dice and sum the result. You are successful if the result is 5 or 9. That happens 8 times out of 36. 8/36 = 2/9. (Or successful if the sum is 4 or 6, or 2 or 7, or 2,3,4 or 11, or several other combinations.) [[Special:Contributions/172.68.54.139|172.68.54.139]] 01:41, 23 November 2024 (UTC)
:Clever, but dice rolls in D&D involving summing all the dice, applying modifiers, if any, and then comparing to one or more threshold values. Your method makes it very difficult to apply modifiers. [[Special:Contributions/162.158.41.8|162.158.41.8]] 02:49, 23 November 2024 (UTC)
::I think you misunderstand the problem here. This is not skill, no modifiers apply, it's purely probability [[Special:Contributions/162.158.158.111|162.158.158.111]] 12:11, 23 November 2024 (UTC)

Minor quibble, arrows aren't fired (unless they're flaming or self-propelled, perhaps), they are shot. (Shotguns are fired of course.) [[Special:Contributions/162.158.41.73|162.158.41.73]] 02:52, 23 November 2024 (UTC)
:Arrows are "loosed", even more accurately. At least to avoid the confusion from how so many things may be shot, or ''a'' shot. (Many different nouns, from a physical measure of liquer/coffee/vaccine to a projectile, or an even abstract fundemental of chance; and, as verb, projectiles perhps may be shot, then so may their targets.) [[Special:Contributions/172.68.205.178|172.68.205.178]] 14:32, 23 November 2024 (UTC)

Rolling 22 or lower on percentile dice (or, equivalently, 79 or higher) is close enough, and easier to come up with. (Give or take whether 00 is treated as 100 or zero.) Or directly represent the action: roll a d10. If it's 1-5, you lose. If it's 6-10, roll again; if it's 1-5 you lose, 6-9 you win, 10 roll again. (Modify slightly if you want to distinguish the case of grabbing *two* cursed arrows.) [[User:Jordan Brown|Jordan Brown]] ([[User talk:Jordan Brown|talk]]) 03:26, 23 November 2024 (UTC)

Alternative exact solution for getting this probability using dice: Roll: 1d8, 2d6, 1d4 succeed on 19 or higher.

I couldn’t remember the formula for binomial coefficients (“n choose k”), but there’s an easy way to calculate that the probability of drawing no cursed arrows is 2/9 without that formula. You just need to multiply the probabilities that each of the arrows drawn is not cursed. Since only two arrows are drawn, you only have to multiply two numbers.

The probability that the first arrow is not cursed is 5/10 – there are 5 non-cursed arrows and 5 cursed arrows out of 10 total. After taking out one non-cursed arrow, there are 4 non-cursed arrows and 5 cursed arrows out of 9 total, so the probability that the second arrow is not cursed is 4/9. Multiplying the two probabilities, the probability of drawing two non-cursed arrows is (4*5)/(10*9) = 20/90 = 2/9.

I was considering writing this observation in the Explanation section of the page, but I’m not if it belongs there. This solution avoids using formulas from combinatorics, so it might not be connected enough to the comic.—[[User:Roryokane|Roryokane]] ([[User talk:Roryokane|talk]]) 06:02, 23 November 2024 (UTC)

My simple-minded approach:
* Roll d10 once for your first arrow: if 1 to 5, the arrow is cursed, otherwise not;
* Roll d10 again for your second arrow: same rules, but repeat until you have a different number from the first one (so d10 is in fact only a d9 this time)
* I won't calculate probabilities – these are your arrows, live with it ;-) [[Special:Contributions/172.69.109.51|172.69.109.51]] 07:33, 23 November 2024 (UTC)
:That has the benefit (over 3d6+1d4) of telling you which arrow(s) (if either) was cursed. [[User:RegularSizedGuy|RegularSizedGuy]] ([[User talk:RegularSizedGuy|talk]]) 07:52, 23 November 2024 (UTC)
:If you don't like re-rolls, you can make d9 out of 2d3. Nine possibilities, so just assign one of them (perhaps by rolling them one at a time) to be the more significant digit. Don't have a d3 handy? Use d6 and modulo off the extra! (1=1, 2=2, 3=3, 4=1, 5=2, 6=3) [[Special:Contributions/172.68.150.91|172.68.150.91]] 05:59, 24 November 2024 (UTC)

There seems to be doubt that a "N locks and M keys to unlock them" system could be easily accomplished. I think it could be trivial, with strategically interlocking locked-restraints. A chain formed of bike-locks can give a larger locked loop that can be unlocked by just unlocking any ''single'' one of the constituent locks, leaving the other locked loops to not matter (or you could also try the {{w|Borromean rings}} system, whereby it is again secure against itself, until just one ring is opened up to reveal that the rest now ''aren't even locked at all''...). With almost arbitrary ability to cross-link (or, if you will, repeated/alternating-reflected Borromean triplet connections), you can extend the requirements to more than one unlocking being required (by looping chain elements to mre than just the 'adjacent' loops, sideways onto a parallel meta-loop or up/down the chain, all you might do is allow some slack (could be sufficient to get a thing held directly closed by the taut loop-of-loops, but not enough if the passage of the loop through a hasp/sneck actually prevents the otherwise free movement of the final slide-to-unlock action to occur), but a second (or third, or fourth) unlocking can be required to open-end the whole metaloop of locks. At the top end, M=N solutions are also trivial (e.g. two keys, two locks popularly of safety deposit boxes or [[2677: Two Key System|other things]]). Which is not to say that a specific M-of-N puzzle (where 1<M<N) might not need a ''little'' bit of thought to actually design and implement, but there's no obvious reason why all such combinations shouldn't be nicely doable. [[Special:Contributions/172.69.79.165|172.69.79.165]] 14:56, 23 November 2024 (UTC)
:Can we first confirm that the M-of-N Encryption was what Randall was referencing in the first place? [[Special:Contributions/172.71.154.140|172.71.154.140]] 03:17, 24 November 2024 (UTC)
::No, first confirm that this is what the explanation treats as what Randall was referencing. As it was, "complicated lock mechanics" and/or "magic" were suggested as the only ways of doing this, when this (or what we thought this was) just needs a little thought and N bike-locks suitably entangled. [[Special:Contributions/172.70.58.45|172.70.58.45]] 13:17, 24 November 2024 (UTC)
:I'm glad someone else chimed in on this, because it is definitely ''not'' difficult to require unlocking of multiple discrete locks! I can't even figure out why one might think it would be? [[User:ProphetZarquon|ProphetZarquon]] ([[User talk:ProphetZarquon|talk]]) 15:55, 24 November 2024 (UTC)

"other polyhedral dice, with the number of faces denoted by dX (e.g., d10 is a 10-sided die, with numbers from 1 to 10 on it)." - the d10 may be a poor choice as exemplar here; Back in the last century, when I was playing D&D, d10 were typically (and uniquely) numbered 0-9, not 1-10. This may no longer be the case, and I may be showing my age, but if it is still the norm, the d8 or d20 might be a better choice of example. [[Special:Contributions/172.68.210.6|172.68.210.6]] 02:40, 24 November 2024 (UTC)
:Typically, I've only seen 0-9 d10s, as part of a "d100" dice pair, with one reading 0-9 & the other reading 0⁰-9⁰... Single d10, mostly seem to come in 1-10? Maybe it depends which reseller one shops at... [[User:ProphetZarquon|ProphetZarquon]] ([[User talk:ProphetZarquon|talk]]) 15:49, 24 November 2024 (UTC)
::They are usually numbered 0-9, but the 0 represents 10, since writing 10 would require that face to have a different font size. It is still a d10, since the die has ten sides, and still cannot roll at 0. The d100 variant does the same thing with 100, but for the added reason that the 00 face actually does mean 0 when the other die rolls a 1-9. This is the convention, so a die that actually writes 10 on it instead of 0 will be rare. [[User:Stardragon|Stardragon]] ([[User talk:Stardragon|talk]]) 23:14, 24 November 2024 (UTC)

You've all been nerd-sniped. [[User:CalibansCreations|'''<span style="color:#ff0000;">Caliban</span>''']] ([[User talk:CalibansCreations|talk]]) 10:53, 24 November 2024 (UTC)

Combinatorics degree? Does such a degree really exist? --[[Special:Contributions/162.158.130.37|162.158.130.37]] 17:19, 24 November 2024 (UTC)
:There are degrees for all kinds of things. A quick search reveals a number of "Combinatorics" or "Combinatorics and <Foo>" (e.g. "Optimisation") degrees. Some of them are marked as Masters degrees, and I haven't dug into the others to see if there are any 'pure' undergraduate ones (apart from anything else, I know there are crucial differences between the structures and scopes of UK and US 'degree courses' to consider, in particular), but there seems to be representation on both sides of the Atlantic (and elsewhere, e.g. Oceana).
:At the very least, it could be a selected specialised segment of an even wider mathematical degree course, or a cross-disciplinary one (like my own, which was part under Physics and part under Computing, but could have included a Stats-based element). [[Special:Contributions/162.158.74.49|162.158.74.49]] 19:07, 24 November 2024 (UTC)

3012: The Future of Orion

2024-11-17T12:12:57Z

172.70.86.141: /* Explanation */

{{comic
| number = 3012
| date = November 15, 2024
| title = The Future of Orion
| image = the_future_of_orion_2x.png
| imagesize = 740x300px
| noexpand = true
| titletext = Dinosaur Cosmics
}}

==Explanation==
{{incomplete|Created by a TYRANNOSTARUS REX - Please change this comment when editing this page. Do NOT delete this tag too soon.}}

Stars in the night sky change over time. Some, like {{w|Betelgeuse}}, a star in the constellation {{w|Orion (constellation)|Orion}}, is expected to go {{w|supernova}} between [https://astrobites.org/2023/07/01/betelgeuse-betelgeuse-betelgeuse-is-it-supernovatime/ tens of] and [https://earthsky.org/brightest-stars/betelgeuse-will-explode-someday/a thousand] years, and then disappear from the night sky. Further, all stars are moving relative to us and each other, which results in apparent movement in our sky, called {{w|proper motion}}, a function of a star's relative movement in three dimensions and its distance from us.

This comic shows changes in Orion from Betelgeuse disappearing and three of its fastest moving stars, and recommends revising the {{w|constellation}} (or creating a new {{w|Asterism (astronomy)|asterism}}) from one which depicts a hunter to another matching the {{w|Tyrannosaurus}} from Ryan North's [https://www.qwantz.com Dinosaur Comics]. The proper motion of {{w|Chi1 Orionis|χ¹ Orionis}} shown near the top at the end of Orion's arm (and the back of the dinosaur's head) is 0.2 arcseconds per year, so it will traverse the depicted angular distance of 0.84 arc degrees in about 15,000 years. {{w|Pi1 Orionis|π¹ Orionis}} at the top of Orion's bow (and the end of the dinosaur's tail) has a proper motion of 0.14 arcseconds per year, so it will traverse its distance of 0.87° in about 23,000 years. However, with a proper motion of 0.46 as/yr, {{w|Pi3 Orionis|π³ Orionis}}, in the middle of the bow, will take only about 9,600 years to traverse its longer depicted distance of 1.24°. (The angular distance traversed by the stars was calculated relative to the distance between Orion's two outermost belt stars, {{w|Alnitak}} and {{w|Mintaka}}, which are shown becoming the dinosaur's hips.) Thus, the new constellation won't form until its current name has lasted more than three times as long as it already has.

There are no official constellations currently depicting dinosaurs. The process of recognizing constellations started around 3000 BC for the northern hemisphere, continued with the investigations like those of {{w|Ptolemy}} (in the 2nd century AD) who used Greek mythology for visible 'southern' constellations and was more or less set in stone after voyages to the southern hemisphere by European navigators, like {{w|Johann Bayer}}, in the early 17th century. The first fossil to be later identified as a dinosaur was found in 1676, and the term "dinosaur" was not introduced until 1842 to describe them. As the {{w|International Astronomical Union}} did not establish the current official list of constellations until 1922, though, they could have recognized a dinosaur constellation had one been proposed and widely accepted. There is, however, a constellation of another large, fearsome reptile, albeit mythological (a {{w|Draco (constellation)|dragon}}, one of Ptolemy's), and "the lizard" ({{w|Lacerta}}) was defined in 1687.

The title text is another joke regarding Dinosaur Comics, replacing "comics" with "cosmics" because we're talking about a dinosaur in the sky.

Orion is also mentioned in [[1020: Orion Nebula]]. T-Rex is also featured in [[1452: Jurassic World]]. In 2006, Randall emulated the style of Dinosaur Comics with [[145: Parody Week: Dinosaur Comics]].

==Transcript==
{{incomplete transcript|Do NOT delete this tag too soon.}}
:Orion Today:
:[Star map of Orion constellation 2024]

:Predicted Changes:
:[Scribbled on]: Star movement
:[Scribbled on]: Star Death (Betelgeuse)
:[Star map's predicted changes over next couple centuries]

:Orion in the future:
:[Scribbled on]: Suggested lines
:[New lines are drawn overlaying the future changes]

:[[https://www.qwantz.com/ Dinosaur Comics] dinosaur overlayed]

{{comic discussion}}

[[Category:Comics with color]]
[[Category:Dinosaurs]]
[[Category:Astronomy]]
[[Category:Space]]
[[Category:Comics with red annotations]]

Talk:3009: Number Shortage

2024-11-11T17:21:09Z

172.70.86.141:

I bet there's plenty of 9s left. They obviously didn't get a proper range of digits at Benford's Discount Number Store. [[Special:Contributions/172.69.195.113|172.69.195.113]] 05:53, 9 November 2024 (UTC)
:There are so many 9s because they get used the least on microwaves. [[User:N-eh|N-eh]] ([[User talk:N-eh|talk]]) 22:27, 10 November 2024 (UTC)
:I expect pricing psychology uses up most of the 9's that Benford stocks, using the first-in-last-out method.

Is this not an error? "15 2s and 12 3s" uses up one 3. So shouldn't it next be 11 3s left, not 10? -- [[Special:Contributions/172.69.144.152|172.69.144.152]] 10:41, 9 November 2024 (UTC)
: No - they started with fifteen 2s and twelve 3s. They used two 2s saying '15 '''2'''s and 1'''2''' 3s' which leaves thirteen 2s. They used two 3s saying ...12 '''3'''s / No wait, 1'''3''' 2s...' which leaves ten 3s. However, by the next line they still have nine 3s (having used up one saying '... and 10 '''3'''s') - not 8, as the explanation previously said.[[Special:Contributions/172.70.160.134|172.70.160.134]] 17:06, 11 November 2024 (UTC)

the consequences of our actions /ref [[User:CalibansCreations|'''<span style="color:#ff0000;">Caliban</span>''']] ([[User talk:CalibansCreations|talk]]) 10:45, 9 November 2024 (UTC)

The above mentioned “error” is not an error. When she says there are 13 2s left, that uses up one 3. [[User:PedanticMan|PedanticMan]] ([[User talk:PedanticMan|talk]]) 11:13, 9 November 2024 (UTC)

: There's no pause, no "No wait" after "13 2s". Is she reevaluating numbers instantly realtime midsentence? Did she start the sentence planning to say one thing and instantly altered it partway through? I guess that's what Randall is going for. -- [[Special:Contributions/172.69.144.162|172.69.144.162]] 12:00, 9 November 2024 (UTC)
::That's what I assumed, and I already included it in the explanation. But I'm not sure if the title text is consistent with that interpretation. [[User:Barmar|Barmar]] ([[User talk:Barmar|talk]]) 17:35, 9 November 2024 (UTC)
:::Title text is perhaps semi-consistent. Regardless of what was said/used beteen "ten minutes ago" and now, the statement (ten minutes ago) of "2 0s" used one zero. The statement (now) of "10 minutes ago" used another and ''technically'' used a third (but it could be considered recycled from the original statement being quoted).
:::Whichever way the counting works (and presuming that any quotable re-use principle doesn't allow just "Ns" to be requoted as "X Ns" in a way that preserves the N stock even whilst depleting Xs), we're certainly down to the stage where we can no longer say there are "0" of something, nor that the something involved is the 0s.
:::...to put it another way, a different TT might be "10 minutes ago we were down to only 2 ... oh darnit!". But that wouldn't have made itself quite so obvious (the Ns could have been 0, 1 or 2). I suppose having 3 0s ten minutes ago might only have led to the necessity of that logic being explained (then: "3 0s {0s=>2}"; now: "10 minutes {0s=>1} .. 3 0s {0s=>0}" "now?" " {0s=>0 ∴ unable to even begin to answer} "), however...
:::But, much like the TT question posed, wise use of "them”/etc might be useful in ''some'' (not-title text) circumstances. "I earlier said there were 9 9s, but now there are 7 of them. ...still 7. Yep, definitely 7 of them! (7s, on the other hand are now...)", vs. "there were 9 9s, but now there are 7 9s. 6 9s! 5 9s! <...> 1 9! <curses>" [[Special:Contributions/172.69.195.114|172.69.195.114]] 12:19, 10 November 2024 (UTC)
::I believe the title text is clear. There where still 2 like there was when she checked ten minutes ago. She can only say the sentence because she still had those two left. But after it is not possible to day there are 0 0 left. And thus the Idk. --[[User:Kynde|Kynde]] ([[User talk:Kynde|talk]]) 08:43, 11 November 2024 (UTC)
::: Presumably the scenario that resulted in this was: There were three 0s remaining; Person A said "How many 0s do we have left?" (reducing them to two); Person B then said "2". If, instead, someone had said "We have only 2 0s left", that would have reduced them to one, leaving insufficient to complete the sentence in the title text.[[Special:Contributions/172.70.86.141|172.70.86.141]] 17:21, 11 November 2024 (UTC)

If anyone wants to see self-referentiality of numbers taken even further, here is a series of posts that I wrote on "self-describing numbers": https://atmos.warplight.dev/profile/1p8WCZnqqG6N3ZOsJxBgUTo/p1cNCw1OTsioTQBRk [[User:Fabian42|Fabian42]] ([[User talk:Fabian42|talk]]) 12:09, 9 November 2024 (UTC)

the first stage of grief is denial [[user talk:lettherebedarklight|youtu.be/miLcaqq2Zpk]] 12:19, 9 November 2024 (UTC)
: So you saw that the Harris banner is still up too, eh? There may be no shortage of absolute numbers, but numbers of <em>things</em>, yeah, there are shortages. Like, chances to act to avert disaster, like weren't taken in 2016, and we got lucky ... [[Special:Contributions/172.68.23.92|172.68.23.92]] 17:37, 9 November 2024 (UTC)
::Just heard on Not The Bee that Trump is planning to help the Democrats out by paying Kamala's campaign debt at a rate of $10/mention in speeches over the past 180 days. Should be plenty to pay off the $20 million.[[Special:Contributions/172.71.142.75|172.71.142.75]] 21:02, 10 November 2024 (UTC)
::: [https://www.dude-n-dude.com/2024/11/05/amoebas-lorica-experimentum-finita/ When logic and proportion][https://www.dude-n-dude.com/2024/11/03/ai-elect/ have fallen sloppy dead], {{w|It_Can%27t_Happen_Here|remember}} {{w|White_Rabbit_(song)|what the dormouse said}}.[[Special:Contributions/172.71.142.30|172.71.142.30]] 05:48, 11 November 2024 (UTC)

I'm pretty sure "15" uses up one 3 (3*5) and "12" uses up two 2's and one 3 (2*2*3) [[Special:Contributions/172.71.222.213|172.71.222.213]] 15:09, 9 November 2024 (UTC)
:No, I think the "3s" in the first statement uses one, and the ones place of "13" referring to the number of 2s left in the second statement uses another. [[User:Laneymarie96|Laneymarie96]] ([[User talk:Laneymarie96|talk]]) 03:13, 10 November 2024 (UTC)

"Well? How many numbers do we have left?"
"Oh great! There's one more!"
(Yes, I know this goes against the logic of the original comic)
[[User:Turquoise Hat|Turquoise Hat]] ([[User talk:Turquoise Hat|talk]]) 17:36, 9 November 2024 (UTC)

Why don't we try using Roman numerals while we wait for the shortage to get fixed? [[Special:Contributions/172.69.135.53|172.69.135.53]] 04:05, 10 November 2024 (UTC)
:We could always go back gto tally marks...wait also who says this is Miss Lenhart? {{unsigned ip|172.68.70.56|12:41, 11 November 2024}}

Seems very related to the {{w|Look-and-say sequence}}! [[Special:Contributions/172.71.170.93|172.71.170.93]] 21:48, 10 November 2024 (UTC)Bumpf

Instead of having to say "I don't know" in the title text, one could just say "none". [[Special:Contributions/172.70.110.170|172.70.110.170]] 19:46, 9 November 2024 (UTC)
:It's funnier to imagine that they forgot how to articulate "zero" as a concept. [[User:Psychoticpotato|P?sych??otic?pot??at???o ]] ([[User talk:Psychoticpotato|talk]]) 20:56, 9 November 2024 (UTC)
:By that logic, they could (partially) escape the "10 minutes ago 0s" issue by saying "ten minutes ago" and not depleting certain digits. Or, as it says "numbers", depleting 10s (but neither 1s nor 0s). But you have to consider if "one" is a "1" (also "a(n)", maybe), and if you can get away with "a couple of" or dozens, scores, grosses, ton(ne)s of, a "pony" or a "grand" or even a googol/googolplex. Then one (1s--) may won(1s--?)der what really ''are'' the uses to be strictly atone(1s--?)d for? Still, I've got 33×3 problems, but the number of 3²s aint one! (1s--). [[Special:Contributions/172.70.160.218|172.70.160.218]] 12:47, 10 November 2024 (UTC)

So, is this comic related to the Google incident? Google seems to be suffering from money '''shortage''' after being fined in large '''number'''s. [[User:CategoryGeneral|CategoryGeneral]] ([[User talk:CategoryGeneral|talk]]) 01:51, 10 November 2024 (UTC)
: ...no? Big leap in logic here. [[Special:Contributions/172.69.22.243|172.69.22.243]] 04:06, 10 November 2024 (UTC)

This one needs a nerd snipe warning! Anyone else get sniped with this one? [[User:Fephisto|Fephisto]] ([[User talk:Fephisto|talk]]) 14:02, 11 November 2024 (UTC)
:I'll let you know. I have only fifteen vowels left. No, nine. Or is it six? No, two. Grrrrr! [[Special:Contributions/172.70.160.135|172.70.160.135]] 15:39, 11 Nvmbr 2024 (TC)