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Algorithm Solves Rubik's Cubes of Any Size

timothy posted about 3 years ago | from the gordian-knot-sledgehammer dept.

Math 139

An anonymous reader writes with this excerpt from New Scientist: "Only the most hardcore puzzle-solvers ever go beyond the standard 3x3x3 Rubik's cube, attempting much larger ones. Now an algorithm has been developed that can solve a Rubik's cube of any size. It might offer clues to humans trying to deal with these tricky beasts. Erik Demaine, a computer scientist at the Massachusetts Institute of Technology has found that the maximum number of moves that will ever be required for a cube of side n is proportional to n/log n. 'It gives me a couple of ideas how to solve this thing faster,' says Stewart Clark, a Rubik's cube enthusiast who owns an 11x11x11 cube."

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139 comments

I didn't know I was hardcore (0)

atari2600a (1892574) | about 3 years ago | (#36628914)

I suppose I should inform my parents or lack of a girlfriend or something...

Pull off the stickers! (0)

Anonymous Coward | about 3 years ago | (#36628916)

That's an easy solution.

Re:Pull off the stickers! (1)

atari2600a (1892574) | about 3 years ago | (#36628970)

Fuck you. Every fucking time I have my cube on the streets HURR DURR STICKERS

Re:Pull off the stickers! (1)

Dayze!Confused (717774) | about 3 years ago | (#36629020)

I can solve a Rubik's cube in less than 30 seconds. How long does it take to pull off the stickers and put them back on?

Re:Pull off the stickers! (0)

Anonymous Coward | about 3 years ago | (#36629068)

No you can't.

Re:Pull off the stickers! (2)

Korin43 (881732) | about 3 years ago | (#36629264)

According to this [speedcubing.com] , at least 2700 people can solve a rubik's cube in 30 seconds. The fact that one of them is on Slashdot wouldn't be particularly surprising.

Re:Pull off the stickers! (2)

Galaga88 (148206) | about 3 years ago | (#36629384)

What would be particularly surprising, would be if only one of them was on Slashdot.

Re:Pull off the stickers! (1)

Joce640k (829181) | about 3 years ago | (#36631136)

He's trying to be 'funny' by saying he does it by pulling the stickers off.

(I bet it takes longer than 30 seconds though...)

Re:Pull off the stickers! (1)

jamesh (87723) | about 3 years ago | (#36630256)

I can do it in under a 90 seconds sometimes (and under 60 seconds consistently when I was a kid), and i'm nowhere near the fastest person I know at solving them. 30 seconds isn't unreasonable for someone who is actually good at it.

Re:Pull off the stickers! (1)

Joce640k (829181) | about 3 years ago | (#36631178)

I used to do it in 30 seconds back in the '80s but last time I tried on it took more like a minute.

There's quite a few people out there who can do it in about 15 seconds average using newer methods. That's probably about the limit for humans.

If you see times lower than that it's because somebody got lucky on one particular solve. Sometimes the pieces just fall into place. I once did it in 19 seconds in front of a crowd of people but it was mostly luck (not that I told anybody that at the time...)

Re:Pull off the stickers! (1)

El_Oscuro (1022477) | about 3 years ago | (#36629152)

Unfortunately, the stickers won't stick after awhile. I used to take it apart and put it back together again.

Re:Pull off the stickers! (1)

Abstrackt (609015) | about 3 years ago | (#36629250)

I used to do that too. Unfortunately the whole thing gets too loose if you pull it apart too many times.

Re:Pull off the stickers! (1)

lul_wat (1623489) | about 3 years ago | (#36629406)

I used to go out and buy a new rubix cube. Unfortunately you run out of money if you do it too many times.

Re:Pull off the stickers! (1)

Radtastic (671622) | about 3 years ago | (#36629176)

If you can remove the 54 stickers on a 3X3 cube and re-attach them in 30 seconds, I may be more impressed than I am with this guy's algorithm.

Re:Pull off the stickers! (1)

The Clockwork Troll (655321) | about 3 years ago | (#36631282)

You really only need to pull off at most 48 of them.

Re:Pull off the stickers! (1)

G3ckoG33k (647276) | about 3 years ago | (#36631354)

But then it would be no challenge.

This can lead to impossible cubes (1)

istartedi (132515) | about 3 years ago | (#36629214)

Back in the 80s when the cube was new and popular, I was really into it. IIRC, I got my times down under 3 minutes. Yeah, I know. Pretty pathetic by today's standards.

Anyway, somebody gave me a cube to solve once. After about 15 or 20 minutes it dawned on me that I had an impossible cube. Somebody who thinks they're moving closer to the solution by swapping stickers can do something like put the white sticker on a corner cubie with the blue sticker, not realizing that white and blue are always on opposite sides.

Re:This can lead to impossible cubes (1)

starofale (1976650) | about 3 years ago | (#36629486)

Anyway, somebody gave me a cube to solve once. After about 15 or 20 minutes it dawned on me that I had an impossible cube. Somebody who thinks they're moving closer to the solution by swapping stickers can do something like put the white sticker on a corner cubie with the blue sticker, not realizing that white and blue are always on opposite sides.

White is opposite yellow. Blue is opposite green. Are you sure you hadn't just forgotten how to solve a Rubik's Cubes?

Re:This can lead to impossible cubes (1)

istartedi (132515) | about 3 years ago | (#36629534)

White is opposite yellow. Blue is opposite green. Are you sure you hadn't just forgotten how to solve a Rubik's Cubes?

In jr. high, I knew what I was doing. Now? I forgot which color was opposite which and didn't bother to google-check it.

Re:This can lead to impossible cubes (2)

Trunklebob (457750) | about 3 years ago | (#36629546)

White is opposite yellow. Blue is opposite green. Are you sure you hadn't just forgotten how to solve a Rubik's Cubes?

White/yellow is the Western color scheme; White/Blue was more common in Japan, but I've had one or two of them here in the States, too.

http://www.speedsolving.com/wiki/index.php/Japanese_Color_Scheme [speedsolving.com]

Re:This can lead to impossible cubes (1)

istartedi (132515) | about 3 years ago | (#36629700)

Ah, so it's possible that my memory is not so flawed. I could have sworn blue was opposite white. Come to think of it, I do seem to recall that the colors were not rigidly standardized, and having seen what I thought were knock-offs. Maybe some of the Japanese cubes got sent to the US to meet demand, or I had an early cube made before the "Western" color scheme was adopted.

Re:This can lead to impossible cubes (1)

Joce640k (829181) | about 3 years ago | (#36631192)

I've seen both schemes...

Re:This can lead to impossible cubes (1)

jamesh (87723) | about 3 years ago | (#36630292)

You can make a cube unsolvable just by rotating a single piece (eg not by completely disassembling it, just twisting one of the corners around). No amount of moving the cube around by the correct means can rotate only a single corner piece or flip a single edge piece - they have to be done in pairs or multiples of two.

Re:This can lead to impossible cubes (1)

Joce640k (829181) | about 3 years ago | (#36631202)

Built-in parity checking!

From TFA... (3, Informative)

Lightborn (7556) | about 3 years ago | (#36628938)

It's n^2/log n.

Re:From TFA... (5, Informative)

RussellSHarris (1385323) | about 3 years ago | (#36628984)

Actually, it's n²/log n, but since Slashdot doesn't like Unicode it just eats that ² character and you end up with n/log n. Which, of course, is wrong.

Re:From TFA... (4, Funny)

arth1 (260657) | about 3 years ago | (#36629664)

Yeah, I kind of wondered about that. It would imply that a standard 3x3 cube would always be solvable in 7 moves or less, which is clearly wrong, unless these are gentoo moves.

Re:From TFA... (1)

Anonymous Coward | about 3 years ago | (#36629766)

That wouldn't imply that anyways since we're talking asymptotics here. It doesn't say anything about how fast you can solve a specifically sized cube, only the relationship between the difficulties in solving different sized ones.

Re:From TFA... (0)

Anonymous Coward | about 3 years ago | (#36630002)

But still his math is wrong. He's trying to state it in Big O notation, which would mean that a 3x3x3 cube should be soved in 3^2/log 3 moves, or ~6.2 moves. Heck, even without the "Advancement" by the team, the paper states that any cube can already be solved in n^2, or in this case 9 moves. But the "God Number" is 20?

Am I missing something?

Re:From TFA... (1)

artor3 (1344997) | about 3 years ago | (#36630608)

That's not how big O notation works. If something is O(n) it doesn't mean that it can be solved in n calculations. It means it can be solved in C*n+D calculations, where C and D are constants.

Re:From TFA... (1)

FrootLoops (1817694) | about 3 years ago | (#36631168)

Sort of. The D is unnecessary, and to be clear C*n is an upper bound on the number of calculations needed. That bound also only has to hold for n larger than some threshold value n0, though this is an unimportant constraint in CS since you can just increase C to cover all positive integer n. The constant C can sometimes be found explicitly in cases like this, though it's often a horrific mess. From the God's Number calculation, one can conclude a lower bound on C, though, to prevent the false contradiction that started this thread :) (if you have it work for all positive integers).

Re:From TFA... (0)

Anonymous Coward | about 3 years ago | (#36629896)

Not 7. "proportional to n^2/log n"

Re:From TFA... (1)

SuperMonkeyCube (982998) | about 3 years ago | (#36630134)

4/log(2) is 13.28, more than 11. 9/log(3) is 18.86, less than 20. This curve doesn't fit so great. Anybody do the math for the 4x4x4 already?

Is that base ten or base e? (1)

colinrichardday (768814) | about 3 years ago | (#36630350)

In the latter case, log 3=1.09, and 11.9/1.09=10.9. . . Although the article says the upper bound is proportional to n^2/log n.

3x3x3 cube... 11x11x11 cube... (0)

Anonymous Coward | about 3 years ago | (#36628950)

Isn't giving all 3 dimensions a little redundant if you're specifying that it's a cube?

Re:3x3x3 cube... 11x11x11 cube... (1)

QuasiSteve (2042606) | about 3 years ago | (#36628994)

Only if you assume each cell is of the same dimension. You could very well have a cube that is 3x4x5 cells. Although it wouldn't work as a Rubik's Cube.

Back on topic.. I'm pretty sure I saw a 50x50x50 or something solved by software on YouTube.. is this algorithm bringing something new to the table, or...?

Re:3x3x3 cube... 11x11x11 cube... (0)

Anonymous Coward | about 3 years ago | (#36629218)

But then it wouldn't be a cube.

Re:3x3x3 cube... 11x11x11 cube... (1)

praxis (19962) | about 3 years ago | (#36629334)

Why can a cube not be composed of 3x4x5 cells so long a the cells are sized such that the cube is well, a cube?

Re:3x3x3 cube... 11x11x11 cube... (1)

Obfuscant (592200) | about 3 years ago | (#36629502)

Why can a cube not be composed of 3x4x5 cells so long a the cells are sized such that the cube is well, a cube?

Because when you try to rotate the five-cell face, e.g., you'll be trying to move some of the cubelets from the four-cell face to the 3-cell face and vice versa. You'll have four cells trying to match up to 3 cells and vice versa. That may work, but if you then try to rotate the four-cell face, you'll be trying to move fractional cubelets.

Personally, I'll be impressed when the algorithm handles 50x50x50x50 Rubick's Hypercubes (patent pending).

Re:3x3x3 cube... 11x11x11 cube... (1)

kmoser (1469707) | about 3 years ago | (#36631314)

Yes, but theirs goes up to 11x11x11.

Yes, but... (3, Interesting)

Anonymous Coward | about 3 years ago | (#36628972)

Does it scale to >3 dimensions?

Solving != best solution (0)

MrEricSir (398214) | about 3 years ago | (#36628996)

The headline claims they can "solve" any Rubik's cube, but who cares? You can solve it just by performing random moves.

The import part is NOT solving it, it's that they can do it in the minimum number of moves.

Re:Solving != best solution (0)

Anonymous Coward | about 3 years ago | (#36629204)

I have to disagree with you there -- the MOST important thing for any problem is to first get it solved. Performance can then be optimized afterwards.

Re:Solving != best solution (0)

Anonymous Coward | about 3 years ago | (#36629222)

I, too, can Not solve a Rubik's cube of any size, in the minimum number of moves. Would you like to see the 100-page paper I wrote about this?

Re:Solving != best solution (2)

barrtender (1930830) | about 3 years ago | (#36629322)

The headline claims they can "solve" any Rubik's cube, but who cares? You can solve it just by performing random moves.

The import part is NOT solving it, it's that they can do it in the minimum number of moves.

Can you prove that?

Hint: Random moves has a number of infinite move cases that never get solved.

Re:Solving != best solution (0)

Anonymous Coward | about 3 years ago | (#36629538)

If your infinite sequence of moves doesn't visit each position equally often in the limit, then those moves are not random.

Re:Solving != best solution (0)

Anonymous Coward | about 3 years ago | (#36629594)

> Random moves has a number of infinite move cases that never get solved.

However, probability of actually selecting such a sequence is zero.

Re:Solving != best solution (1)

mark-t (151149) | about 3 years ago | (#36629908)

Only if the sequence is infinitely long.

Re:Solving != best solution (0)

Anonymous Coward | about 3 years ago | (#36629786)

Exactly. Just because you can make random moves doesn't mean any of those sequences (for an n-sized cube) will necessarily result in a solution. Plenty of similar puzzles have arrangements that are in fact insoluble. Now, it makes sense that any Rubik's cube can be solved, but as you say, proving that mathematically is a whole different story.

Re:Solving != best solution (1)

mark-t (151149) | about 3 years ago | (#36629936)

Now, it makes sense that any Rubik's cube can be solved

You should probably qualify that a little... for example, any normal rubik's cube, or any rubik's cube that hasn't been tampered with, or something similar... it is actually very easy to make any rubik's cube completely and permanently unsolvable.

Re:Solving != best solution (1)

RussellSHarris (1385323) | about 3 years ago | (#36629956)

it is actually very easy to make any rubik's cube completely and permanently unsolvable

Only if "make" involves operations that "solve" does not. For instance, a freight train can very easily "make" a Rubik's cube completely and permanently unsolveable.

Re:Solving != best solution (1)

mark-t (151149) | about 3 years ago | (#36629966)

So can gluing it with crazy glue in an unsolved position... as can taking it apart and switching the orientation of exactly piece. There are many ways to make a rubik's cube that looks perfectly fine impossible to solve.

Re:Solving != best solution (1)

RussellSHarris (1385323) | about 3 years ago | (#36630036)

My point was more that any cube in any position that can be reached by standard moves is solveable by those same standard moves. Obviously, rearranging stickers, disassembling the cube, or introducing outside forces (crazy glue, freight trains) can make the cube "unsolveable" by any possible combination of standard moves.

But that all begs the question: is a Rubik's cube still a Rubik's cube if it's been disassembled, modified, and/or re-assembled in a condition that would be impossible for a Rubik's cube to ever reach? Is a Rubik's cube just a toy with a set of specially interlocking cubes, or is it also a mathematical and geometrical set of rules? I'd say the latter. If you violate those rules, you don't have a Rubik's cube; you have something that superficially resembles a Rubik's cube, but isn't one. Therefore there's no reason to make special exclusion for tampered cubes when you make general statements such as:

Now, it makes sense that any Rubik's cube can be solved

-because it goes without saying.

Re:Solving != best solution (0)

Anonymous Coward | about 3 years ago | (#36629942)

Exactly. Just because you can make random moves doesn't mean any of those sequences (for an n-sized cube) will necessarily result in a solution. Plenty of similar puzzles have arrangements that are in fact insoluble. Now, it makes sense that any Rubik's cube can be solved, but as you say, proving that mathematically is a whole different story.

A Rubik's cube of finite size has a finite number of states. QED.

Re:Solving != best solution (1)

MrEricSir (398214) | about 3 years ago | (#36630272)

A Rubik's cube has a finite number of states and a finite number of moves. So I don't understand how performing infinite moves wouldn't eventually result in the solution.

Re:Solving != best solution (1)

Nerdos (1960936) | about 3 years ago | (#36630956)

Easier example. Say you're 10 meters away from your house. If there's exactly a 50% chance to easier take a step forward or backward, the probability that you will never reach your house is greater then 0. If the chance to take a step forward is 50% + epsilon, then that probability goes down to zero.

Re:Solving != best solution (1)

wisty (1335733) | about 3 years ago | (#36631220)

No. That's simply not true. If there is a 99% chance of you stepping *back* and a 1% chance of you stepping *forward*, the probability of you *eventually* reaching your house approaches 100%, eventually. The average time for you to reach your house, however, is really quite long.

Re:Solving != best solution (0)

Anonymous Coward | about 3 years ago | (#36631264)

Only because you'd be walking backwards all the way around the planet.

finite number of states != finite number of moves (0)

Anonymous Coward | about 3 years ago | (#36631522)

take a move, revert it with the next move, repeat ad libitum. You get an infinite sequence of moves that does not ensure the reachability of the solution.

IS my Brother hardcore? (0)

Anonymous Coward | about 3 years ago | (#36629002)

MY youngest brother is 11 and he can do rubik's cubes up to a 7x7.

Not hardcore (0, Troll)

Anonymous Coward | about 3 years ago | (#36629012)

Crazy. And unlikely of ever getting laid.

Re:Not hardcore (0)

Anonymous Coward | about 3 years ago | (#36629418)

Why are you on /.?

Well that's not good (2)

Psychotria (953670) | about 3 years ago | (#36629246)

I've been working on my 60x60x60 cube for 17 years now and have 5 cublets in place on one side. This news has ruined my day :-( At my current rate I am never going to finish according to TFA, especially since I am essentially just doing random moves. Damn.

Re:Well that's not good (0)

Anonymous Coward | about 3 years ago | (#36629970)

At my current rate I am never going to finish according to TFA

RTFA ruined your life?

Fool, you could've solved it in six moves! Paint, paint, paint...

Re:Well that's not good (0)

Anonymous Coward | about 3 years ago | (#36630048)

What's funny about this is that when I was in college, my roommate was obsessed with Rubik's cubes. He wrote a program to simulate an arbitrarily sized cube, and would then sit there and solve it. The largest I remember him doing was 42x42x42.

Re:Well that's not good (1)

Psychotria (953670) | about 3 years ago | (#36630152)

I've been working on my 60x60x60 cube for 17 years now and have 5 cublets in place on one side. This news has ruined my day :-( At my current rate I am never going to finish according to TFA, especially since I am essentially just doing random moves. Damn.

What's funny about this is that when I was in college, my roommate was obsessed with Rubik's cubes. He wrote a program to simulate an arbitrarily sized cube, and would then sit there and solve it. The largest I remember him doing was 42x42x42.

Your roommate is amazing. Anyway, after some deep contemplation I've gotten over the devastation I experienced when reading TFA. I shall now focus my efforts on my continuing efforts (12 years so far) to solve my 100 disk Towers of Hanoi puzzle that I've set up next to my bed in the basement. At least that should be solvable within a convenient amount of time, unlike the silly 60x60x60 Rubik's cube (what could I have been thinking?!) The cube is a bit cumbersome to use anyway as I constructed it out of wood and it weighs too much (8kg). Moving the disks for TOH is much easier and less likely to cause me to have a hernia or develop RSI.

Re:Well that's not good (2)

FrootLoops (1817694) | about 3 years ago | (#36631194)

If you started the Towers of Hanoi puzzle in the usual state, the minimum solution uses 2^100 - 1 moves. One move per second would take about 4*10^22 years, or about 3*10^12 times the age of the universe.

Re:Well that's not good (1)

Joce640k (829181) | about 3 years ago | (#36631224)

He's fooling you. The number of twists needed to solve that would take years - longer if you're pointing and clicking with a mouse.

Re:Well that's not good (1)

ccabanne (1063778) | about 3 years ago | (#36631536)

I've been working on my 60x60x60 cube for 17 years now and have 5 cublets in place on one side. This news has ruined my day :-( At my current rate I am never going to finish according to TFA, especially since I am essentially just doing random moves. Damn.

hahahahahahahaha aaaaahhhhhhh

General algorithm already known (3, Informative)

Michael Woodhams (112247) | about 3 years ago | (#36629454)

Once you can solve a size 4 and size 5 cube, all larger sizes are obvious generalizations of the same algorithm. (At least, for the algorithm I use it is so.) I've seen an edited video of someone solving a (computer simulated) size 100 cube. So the fact of a "general algorithm" is not news.

That it is an efficient algorithm and sets a new* upper bound on how many moves you need is interesting. (This upper bound is proportional to (n^2/log n), not (n/log n) as stated in the summary.)

* I don't follow cubology that closely, so I'm taking their word for it that this is a new upper bound.

Re:General algorithm already known (2)

Mogusha (1091607) | about 3 years ago | (#36629564)

It's not even a generalization. It's the exact same algorithms. Once you can solve a 4x4x4 there is no extra algorithms needed at all to solve any cubes of any higher degree.

The typical idea is to solve the cube to the point of being a 3x3x3 with all the centers and edges solved. When solving the edges if you're trying to get the edges in place any portion of the edges can be grouped to look like a cube with a smaller degree.

Someone could argue that the 5x5x5 and 4x4x4 algorithms are needed, because of the extra center edge pieces, but with proper edge pairing that's usually a moot issue. But, I could always be wrong, the 4x4x4 and 5x5x5 algorithms might be needed for any cube larger than a 3x3x3. The real point is that anything above a certain point is the exact same algorithms.

Re:General algorithm already known (2)

kallisti (20737) | about 3 years ago | (#36629920)

Reducing to a size 3 cube is only one way of solving the cube. It has the drawback of a parity error where your reduced 3 cube can't be solved by using standard methods. For instance, on a size 4 cube you can get to a position where two adjacent edge pieces are swapped and inverted, as if the size 3 cube had one edge flipped, which is impossible. You can also have two corners swapped. There are some known moves to fix these errors, and most (all?) speed cubers use reduce-to-3 so it isn't a major problem. But there are at least two new algorithm required to solve bigger cubes than 3 using this method.

Now, if you don't use reduce-to-3, then you can solve any size cube with just 3 algorithms: one to cycle 3 edges with flipping, one to cycle 3 corners and one to swap center pieces, which are actually very easy because there are always 4 of them that look identical. The really trippy part is that this method was expanded to work on 4 and 5 dimensional cubes, too.

Re:General algorithm already known (1)

ewanm89 (1052822) | about 3 years ago | (#36631424)

There are parity solve algorithms that can be reduced to.

Re:General algorithm already known (0)

Anonymous Coward | about 3 years ago | (#36630662)

Once you can solve a size 4 and size 5 cube, all larger sizes are obvious generalizations of the same algorithm. (At least, for the algorithm I use it is so.) I've seen an edited video of someone solving a (computer simulated) size 100 cube. So the fact of a "general algorithm" is not news.

That it is an efficient algorithm and sets a new* upper bound on how many moves you need is interesting. (This upper bound is proportional to (n^2/log n), not (n/log n) as stated in the summary.)

* I don't follow cubology that closely, so I'm taking their word for it that this is a new upper bound.

Thanks, and mod parent up please. I solved a 5x5x5 cube once, and recall that this puzzled reduced pretty much to two instances of what was essentially the 3x3x3 case (with some minor caveats). Most any size of cube, I have assumed, should be solvable recursively with the same approach.

Patent Pending (2, Funny)

cultiv8 (1660093) | about 3 years ago | (#36629478)

This system and method for solving a Rubik's cube is a social media networking plug-in widget that synergizes, optimizes, and rightsizes the Rubik cube solving experience, utilizing HTML5/CSS3/JS code in a novel and innovative way to provide end-users with a single, solidified and parsimonious User Interface (UI), optimized for the User Experience (UX), utilizing Information Architecture (IA), over 256bit SSL security. It is built on a 100% cloud-based, distributed OS, independent architectural framework and uses the actual "internet" to facilitate communication between said end-user's "keyboard", to our proprietary "software", and back to end-user's "monitor".

The solution? (4, Funny)

tool462 (677306) | about 3 years ago | (#36629524)

-After the researchers solve the 3x3x3-

Buttercup: We'll never succeed. We may as well die here.

Westley: No, no. We have already succeeded. I mean, what are the three terrors of the General Cube Solution? One, the pieces coming off - no problem. There's a popping sound preceding each; we can avoid that. Two, the stickers peeling off, which you were clever enough to discover what that looks like, so in the future we can avoid that too.

Buttercup: Westley, what about the R.O.U.S.'s?

Westley: Rubik's Of Unusual Size? I don't think they exist.

-- Immediately, Westley is attacked by a 4x4x4 cube --

solve (1)

Anonymous Coward | about 3 years ago | (#36629588)

I once solved a 1x1x1 rubik's cube!
(I cheated and took the stickers off ;-;)

The cube problem solved? (1)

byteherder (722785) | about 3 years ago | (#36629630)

Has someone notified the Borg?

It's time for Slashdot to support Unicode properly (4, Insightful)

eobanb (823187) | about 3 years ago | (#36629638)

Lack of support for 20 year-old standard is usually just annoying as hell, but in this case it's actually caused the summary to be wrong. For a site that frequently discusses such topics as technology, math and language (for all of which Unicode is an important part—at least insofar as even being able to TALK about these subjects) there is absolutely no excuse for not doing Unicode.

As far as I'm concerned Slashdot ought to be able to render MathML too.

Re:It's time for Slashdot to support Unicode prope (0)

Anonymous Coward | about 3 years ago | (#36630116)

You know what else should work? When they change the UI, it should ruin such things like being able to actually click links in comments.

Re:It's time for Slashdot to support Unicode prope (0)

Anonymous Coward | about 3 years ago | (#36630756)

Unfortunately fixing text encoding issues would do nothing about the leading cause of incorrect summaries around here, better known as "slashdot editors".

Re:It's time for Slashdot to support Unicode prope (1)

tlhIngan (30335) | about 3 years ago | (#36630950)

Lack of support for 20 year-old standard is usually just annoying as hell, but in this case it's actually caused the summary to be wrong.

It did, at one point. The problem was the trolls went and ruined it for all of us by embedding all sorts of control characters in post that ended up destroying the flow of a web page and making the text unreadable.

Hell, I think most blogs probably aren't immune to the problem. There used to be one common indication of this because people would force 4 or 5 unicode control characters together followed with a normal character (some strange millions-comma thing).

And it's a pain in the butt, too - you can't search for it, certain operations destroy the characters, so you get very strange behavior.

Re:It's time for Slashdot to support Unicode prope (0)

Anonymous Coward | about 3 years ago | (#36631276)

But shouldn't there be some decent libraries that filter control characters? You might just allow a few more planes than Basic Latin...
UTF-8 is ASCII-compatible, why shouldn't it be possible to search for it?

Re:It's time for Slashdot to support Unicode prope (1)

Archibald Buttle (536586) | about 3 years ago | (#36631528)

Agreed entirely.

Slashdot's codebase often feels like a joke to me. My own pet peeve is the discussion threshold slider - it's numbers are *never* correct. Right now on a different browser tab a discussion is saying there's "10 Full, 3 Abbreviated" - whereas it's actually displaying 7 full, and 0 abbreviated. Another is telling me "16 Full, 0 Abbreviated" - that one has the abbreviated count right, but there's only 6 comments visible. The numbers are often so wildly out that they're completely meaningless.

If they can't even manage to do basic counting, I fear there's bugger all chance they'll manage to do unicode any time soon.

side n (1)

Memroid (898199) | about 3 years ago | (#36629798)

required for a cube of side n

I don't even want to think about solving a cube with n sides. Not to mention, how a cube ever got n sides to begin with...

Mr Rubik (4, Funny)

pinballer (655113) | about 3 years ago | (#36629874)

I could only ever manage to get 5 out of the 6 sides :(

Re:Mr Rubik (0)

SuperMonkeyCube (982998) | about 3 years ago | (#36630084)

Since my sarcasm meter is clearly broken, or I fail to see the humor in this at all, please describe in detail any state of a solvable Rubik's cube where only 5 of the six sides are solved.

Re:Mr Rubik (1)

rubycodez (864176) | about 3 years ago | (#36630206)

I'll bet you're a hit at parties, like an opened canister of Zyklon B lobbed into the room

What about a 3x3x3x3x3 "cube"? (2)

Rich0 (548339) | about 3 years ago | (#36629972)

It seems boring to me to just extend the size in 3 dimensions. What about extending it beyond three dimensions?

Re:What about a 3x3x3x3x3 "cube"? (1)

jamesh (87723) | about 3 years ago | (#36630306)

I think you could do this by having an array of cubes somehow interlinked... or http://www.superliminal.com/cube/cube.htm [superliminal.com]

Re:What about a 3x3x3x3x3 "cube"? (1)

Garble Snarky (715674) | about 3 years ago | (#36630734)

Why stop there? Why not extend it to other platonic solid puzzles? Archimedean solid puzzles? Don't forget the square one [wikimedia.org] .

So what is the algorithm? (0)

Anonymous Coward | about 3 years ago | (#36630408)

Is it a secret? I couldn't find it linked off of or within the article.

The algorithm (0)

Anonymous Coward | about 3 years ago | (#36630654)

Step 1: Remove all stickers
Step 2: Reapply stickers of one color to one face of cube
Step 3: Repeat step 2 with remaining colors/faces

What's so hard about that?

Re:The algorithm (1)

Garble Snarky (715674) | about 3 years ago | (#36630740)

Does anyone actually think this is funny?

Re:The algorithm (1)

Joce640k (829181) | about 3 years ago | (#36631238)

Not to anybody with an IQ larger than their shoe size.

Demaine's first task is completed? (1)

KenT9 (2328180) | about 3 years ago | (#36630886)

The article says "the maximum number of moves that will ever be required for a cube of side n is proportional to n^2/logn". It also says "standard Rubik's cube can be solved in 20 moves or less" and then it says Demaine's "first task is to work out how to turn that into an exact number for given sizes of cube." but if the number of moves for n=3 is 20 and the number of moves is proportional to n^2/logn then the number of moves for cubes of any length =kn^2/logn where k=20/(3^2/LOG(3)) = 1.1 . So maximum number of moves = 1.1 n^2/logn (approximately) and Demaine's first task is completed unless he needs exact numbers for each n.

The CERN life challenge (0)

G3ckoG33k (647276) | about 3 years ago | (#36631160)

The Demaine father (Martin) and son (Erik) are a very impressive couple.

I congratulate both of them.

Let me suggest that they tackle "The current status of work on the origin of life" http://indico.cern.ch/conferenceDisplay.py?confId=137302 [indico.cern.ch]

"Work on the Origin of Life is poised to converge onto a fourth phase and, many of us hope, success. The first phase concerned prebiotic synthesis of the small molecules, amino acids, nucleotides, lipids and others, essential for life and spanned some forty years. The second overlapping phase was inspired by the symmetric of the DNA or RNA double helix, presumed that life must necessarily be based on some form of template replication of one strand by ligation of free nucleotides to create the second strand, melting of the two strands and cycling again. Spearheaded by L. Orgel, but with many others, this effort has, to date, failed. The third phase begins with the discovery that RNA molecules can act as enzymes, and posited the RNA world, in which RNA molecules dominated. This has led to slightly successful efforts to evolve an RNA sequence able to template replicate itself. Current success is an evolved ribozyme able to do so for 14 nucleotides."

How can the Demaines not do better than those CERN guys, where "The author has gathered some 17 scientists from around the world to collaborate and compete with one another, CERN/LHC experiments style, in a generative scientific environment."? It was after all a folding competition, wasn't it...

Already done! (1)

ewanm89 (1052822) | about 3 years ago | (#36631414)

If one can solve the 3x3, 4x4 and 5,5 it's relatively simple to solve any sized cube by splitting it into overlapping subcubes and expanding the algorithms. The only real thing, is time, it takes a lot of it.
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