1337
Last week’s story arc was a lot of fun for me. I hope you enjoyed it! A few notes for the more pedantic readers:
- In Part 2, Elaine and Dr. Knuth are not competing to calculate the time-complexity of a known algorithm. They’re each trying to develop an algorithm to solve a particular problem. His “lower bound” comment is referring to his lower bound on how fast the problem can be solved (akin to saying “the lower bound on comparison sorts is n log n”), and she proves him wrong with her counterexample.
- In the interest of time-continuity, panel #2 in the same strip is a montage showing the two children at different times. Bobby is quite a bit younger than Elaine (although he doesn’t strictly have to be — “drop table” has been in the SQL standard since at least the start of the 90’s).
- Nick Oliver writes in to suggest that Elaine should have played in the band Perl Jam.
- By 2002, Richard Stallman had grown frustrated with the direction of the blogosphere’s construction. He detached his airship and took it to parts unknown.
- Upside-Down-Ternet is here. It seems to me that you could do an awful lot of pranks with it — turning everything upside down is just the beginning.
If you want to order a shirt for Christmas, you may want to do it soon. We can guarantee Christmas delivery to the US if you order by December 7th (and probably for the week after that) but we may go out of stock in some shirts/sizes before that.
November 19th, 2007 at 3:20 am
Like a-ringin’ a bell.
November 19th, 2007 at 5:30 am
I have a theorem that states that the entire point of that mini-series was to make the “Dread Pirate Roberts” joke at the very end. Please comment. The publishers are expecting a full proof of this theorem tomorrow morning.
November 19th, 2007 at 6:00 am
That was the best story arc….ever. I was heart broken when it ended! I’m seriously contemplating getting them printed in A3 and framed!
November 19th, 2007 at 6:14 am
Apologies if this is too pedantic (just kidding). But the story isn’t finished…
A: is on neighbour’s WiFi
B: “you’ve been pwned pretty hard, man”
B: tells story of Elaine. tells epilogue.
What happens next? Just how hard has A been “pwned”? I guess the arc resembles Little Harmonic Labyrinth in this way, except, you know, with only one inner story.
November 19th, 2007 at 7:50 am
An friend and I used the Upside-Down-Ternet idea some time ago in our laboratory proxy to change the Google logo for our own Google doodle, with one of our teachers photoshopped into a famous singer. Some of our colleages and teacher freaked out so bad…
BTW, awesome comic strips last week.
November 19th, 2007 at 9:51 am
After reading the comic, I coded my own squid for the purpose of annoying my brother who’s grounded from the internet for whatever reason. I also replaced his wifi antenna with a coax lead to the basement, so it only picks up our router and pretty much nothing else [he likes to hop networks when I block him].
First day I reversed all text between tags, except script and style of course.
Second day I replaced all images with kittens, in honor of the kittenwar idea on Upside-Down-Turnet. It didn’t look as pretty as I thought it might though, darn site logos.
Third day I replaced all non-html & non-image files with pi to 10 million digits. May have wanted to look up the size of the downloads and calculated it to that length, but I wasn’t that bored at the time.
November 19th, 2007 at 11:46 am
awesome story, and I hope to get in time to order a few t-shirts!
November 19th, 2007 at 12:26 pm
I hope you realize that this is the second time you’ve been mentioned on LanguageLog, an impressive feat.
November 19th, 2007 at 1:08 pm
Im a huge fan of your comics … you seem wake up in the morning with a stroke of genius behind your eyes, and I digg that, a lot.
A brief question if you have the time, like myself you most likely spend lots of time online, have you any clues or ideas as to what this thing is? What are we surrounded by? Who are these wrinkled slow people that built the computer but that do not know how to use it, at all, and where the hell are all of the adults and authority figures that do not require protection with a pistol?
Nicholas
November 19th, 2007 at 5:25 pm
I think the Upside-down-ternet is more fun from the target’s perspective (assuming they have some reference material). That is, imagine the fun in trying to get around the MAC filtering (try the MAC addresses of common ethernet cards, or at least use the different brands to narrow down your brute force attack). Or running your own server to flip the images back. Or, maliciously request many enormous files to create your DOS attack. Think, man, think!
November 19th, 2007 at 6:20 pm
I think he’s been on Language Log more than twice . . .
P.S. I miss the sum of two numbers as my proof of humanity. It kept away the fools so well.
November 19th, 2007 at 10:38 pm
Way back in the day - early 90s, 92ish or so - someone created the USEnet group alt.fan.perl-jam. They *meant* to create alt.fan.pearl-jam, of course. Almost immediately some wit newgrouped alt.fan.sed-jam and alt.fan.awk-jam.
It still makes me smile to remember that.
November 20th, 2007 at 3:38 am
The first day I brought home my laptop from Office Depot, I tried to steal wireless access from my neighbors, too. After 4 hours of Facebooking and streaming porn, everything redirected to Goatse.
November 20th, 2007 at 8:19 am
My neighbours have an unsecured wireless access point with default passwords and everything. I know this because although my home network is fully wired (although I cheated and re-purposed the blue and brown pairs to get 2 sockets on one cable, so it will never do 1GB) a friend came round with his spiffy new wireless laptop, and he found their network straight away.
It did make me think of getting another router, same make and model as theirs, wiring it up to my own proxy (set up long ago to filter out advertisments), disabling their router altogether and setting them up with their own upside-down-ternet. If I could tap into their telephone wiring and replace their exchange line with an ATA running through my Asterisk server (creating the fabled “sky-blue pink box with yellow spots on”), I’d even be able to listen in on the tech support calls ….. it would be fun to, say, change all the graphics to mono and claim they hadn’t got a valid colour TV licence.
November 20th, 2007 at 3:04 pm
Lower bounds are denoted not with an omicron (big-oh or little-oh) but with an omega. Your first note doesn’t address this objection.
Signed,
One Pedantic Reader
November 20th, 2007 at 3:11 pm
But, um, great story arc.
Signed,
788004892cbbee3a84007b618a5eddc1
aka c79676bfccc8cf6f412088fcd24cc537f0585376
aka T25lIFBlZGFudGljIFJlYWRlcg==
aka:
___________
| o o.ooo|
| oo o.oo |
| oo .o o|
| o . |
| o o . |
| oo .o o|
| oo .o |
| oo . o|
| oo o.oo |
| ooo .o |
| oo o. o|
| oo . oo|
| o . |
| o o . o |
| oo .o o|
| oo . o|
| oo .o |
| oo .o o|
| ooo . o |
___________
November 20th, 2007 at 4:17 pm
And yet more pedantically: if they’re discussing the true lower bound as indicated above in the note, then an omega (big or little depending on strictness) is the right letter. If they’re speaking loosely, as in “the best algorithm I found has this worst case bound” and not about the problem’s algorithm-independent bound, then they’re using “my lower bound” to mean “the worst case performance of my best guess”. Which would seem like OK usage. But the note seems to clarify against this interpretation.
Signed,
One Pedantic
aka:
IC0tLQogLS4KIC4KIAogLi0tLgogLgog
LS4uCiAuLQogLS4KIC0KIC4uCiAtLi0u
CiAKIC4tLgogLgogLi0KIC0uLgogLgog
Li0uCiAuLi4tLi0K
November 20th, 2007 at 4:46 pm
A Dread Pirate Roberts joke would, indeed, have been enjoyed.
Perhaps Bobby Roberts was the original Dread Pirate.
November 21st, 2007 at 2:01 am
I’m sorry but XKCD really seems to have lost it’s way. I’m sorry to say that I’m removing it from my rss feeds. I really hope you’re able to recapture what you once had.
November 21st, 2007 at 2:58 pm
Jeff: Great; more bandwidth for the rest of us. nom nom nom.
November 22nd, 2007 at 12:02 am
Can I buy a t-shirt at the store? The one that has Elaine in it?
November 23rd, 2007 at 10:10 pm
I want a t-shirt that says “Cease this affront to freedom or stand and defend yourselves!”. Don’t make me make one myself, last time I tried it went horribly, horribly wrong.
November 23rd, 2007 at 10:49 pm
So when are you going to either a) start accepting Google checkout or b) accept credit cards through PayPal WITHOUT requiring an actual PayPal account?
Many sites use PayPal, but don’t force the buyer to create a PayPal account. They’re just “powered by” PayPal if you will.
There are 3 shirts, at least, that I’m looking to buy.
November 24th, 2007 at 10:15 pm
I wish this wasn’t my first post, but I kinda agree with Jeff, minus the rss-suicide-note hysterics. I like jokes about self-referentiality a lot more than actual self-referentiality… I winced when I saw the cory-doctorow-balloon thing. The latest comics have been good, but they aren’t stay-up-reading-the-archives-til-5-am material. I don’t know you nearly well enough to diagnose, but it feels like a fanbase feedback loop is forming, and is sucking xkcd into meme-land. Not that there’s anything wrong with being a meme, some of my best friends are memes, but still…
November 24th, 2007 at 11:04 pm
Mm, actually, re-read the last 40 posts or so, and I think I need to take that back. Sadly, wp does not include a delete button…
November 27th, 2007 at 7:21 pm
Who’s Dread Pirate Roberts?
November 28th, 2007 at 4:15 pm
Oh Christ, Jesus.
Go watch The Princess Bride.
Now.
At any rate, Randall — don’t listen to those “this shit ain’t good” folks. They don’t know what stories are.
’twas brilligly beautiful.
GMP
November 29th, 2007 at 12:09 am
I laughed, I cried… I don’t know where to begin.
When you publish xkcd, the baak, this would be a good ender - although I know you will have many more brilliant things to come that would work too, of course. I look forward to them all.
When Stallman arrived, my life was made complete. You have my undying admiration. (Turkey vultures, what?)
November 29th, 2007 at 12:44 pm
I considered removing XKCD from my webcomic list. It was in a word, terrible.
November 29th, 2007 at 6:50 pm
Oh give me a break.
If you’re not liking XKCD at the moment, WHY express that when there are OBVIOUSLY people who are liking it?
It’s not our problem. Just kindly disappear, for the rest of us, lurker, or usual speaker to enjoy ourselves.
In other words.
I’ve got my eye on a few XKCD shirts, I might order them soon?
November 30th, 2007 at 11:04 pm
Randall fucks goats!
December 6th, 2007 at 9:27 pm
COME ON!!
Jesus fucking christ.
XKCD is like my favorite webcomic!
If all you bitches don’t like it, then fuck off, seriously!
And hey Randall, if you would like to, i would like to have you come to my high school to have a wiked time and give a short presentation or something
December 12th, 2007 at 3:04 am
You guys rock… That is so awesome.
Um… I know this might not be the appropriate place to post a question about the most recent comic but Thevenin’s theorum states something like you can summarise any unknown circuit as a sum of it’s voltage sources and resistances. I have only Wikipedia to go on as I had no clue what it was before.
So the solution is 3 ohms, isn’t it? Electricity always follows the shortest path and thus wouldn’t traverse any of the superflouous infinite 1 ohm resistors other than the shortest path. (In theory)…
Am I missing something? I really don’t know much about physics or electronics yet. I’m sorry for being such a n00b.
- trying to learn
December 12th, 2007 at 6:16 pm
xkcd has been and still is my favorite comic. I found this arc to be particularly tasty. That said, the few people that left a comment to the effect that xkcd no longer tickled them are doing a great service, particularly since they all voiced their opinions in a respectful manner.
While an author can’t and shouldn’t try to please everyone it is valuable information to learn that once dedicated fans no longer take interest.
Constructive negative feedback is much more useful than destructive positive feedback.
December 18th, 2007 at 9:48 am
Confused: I’m afraid it’s not even remotely that simple. When you have resistors in parallel, the total resistance is equal to 1 over the sum of 1 over each individual resistance.
For instance, a 20? resistor in parallel with a 30? resistor is equal to 1/(1/20? + 1/30?) = 12?, far less than 20?. The current splits across *both* of them simultaneously.
If you simplify the stated problem to a rectangular grid containing 6 nodes and 7 resistors (2 across the top, 2 across the bottom, 3 vertically), then the resistance is, I’m fairly sure, 1/(1/3? + 1/3? + 1/3?) = 1?, since there are three different 1? paths to choose from (and paths that form loops by touching the same node twice are invalid, so those are the only permitted paths).
However, when you make the grid big enough to allow 5? paths, you now have 1/(1/3? + 1/3? + 1/3? + 1/5? + 1/5? + …), which is obviously a bit smaller than 1? since it’s the reciprocal of a number slightly larger than 1.
So, to compute the overall answer for an infinite grid of resistors, you have to take the limit as the size of the grid approaches infinity. And there’s a combinatorial explosion in the number of possible paths, so it’s something you have to put some significant thought into.
December 18th, 2007 at 9:50 am
Confused: I’m afraid it’s not even remotely that simple. When you have resistors in parallel, the total resistance is equal to 1 over the sum of 1 over each individual resistance.
For instance, a 20 Ohm resistor in parallel with a 30 Ohm resistor is equal to 1/(1/20 + 1/30) = 12 Ohms, far less than 20 Ohms. The current splits across *both* of them simultaneously.
If you simplify the stated problem to a rectangular grid containing 6 nodes and 7 resistors (2 across the top, 2 across the bottom, 3 vertically), then the resistance is, I’m fairly sure, 1/(1/3 + 1/3 + 1/3) = 1 Ohm, since there are three different 1 Ohm paths to choose from (and paths that form loops by touching the same node twice are invalid, so those are the only permitted paths).
However, when you make the grid big enough to allow 5 Ohm paths, you now have 1/(1/3 + 1/3 + 1/3 + 1/5 + 1/5 + …), which is obviously a bit smaller than 1 Ohm since it’s the reciprocal of a number slightly larger than 1.
So, to compute the overall answer for an infinite grid of resistors, you have to take the limit as the size of the grid approaches infinity. And there’s a combinatorial explosion in the number of possible paths, so it’s something you have to put some significant thought into.
December 18th, 2007 at 9:52 am
D’oh, sorry about the double-post. That’ll teach me for assuming that WP throwing up an error about using UTF-8 meant it had rejected my comment.
January 7th, 2008 at 2:53 am
I wish I had a Linksys WRT54GL router just so that I can implement really crazy fun on all unauthorized wireless devices (RST flood, anyone?).
February 7th, 2008 at 4:14 pm
get uncut Paris Hilton sex porn tape here…
Where can i download paris hilton sex tape? …
February 24th, 2008 at 4:50 pm
The 1337 and Blagofaire comics are definitely inspiring me to create an xkcd-based RPG.
As for the resistance problem, it appears to me that the number of parallel n-length paths increases faster than n, causing the resistance to approach 0. I need someone to verify me on that, though.
March 11th, 2008 at 5:16 am
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