## Search found 56 matches

- Mon Jan 17, 2011 12:55 am UTC
- Forum: Computer Science
- Topic: What would you do if you could solve the Halting Problem?
- Replies:
**33** - Views:
**4286**

### Re: What would you do if you could solve the Halting Problem

Well, presumably the box needs to specify some sort of formal language it will accept, as well as an execution environment, to avoid paradox. If the box only accepts valid descriptions of Turing machines, and solves the halting problem only for such descriptions, then there's no paradox, and you ca...

- Sun Jan 16, 2011 5:58 pm UTC
- Forum: Computer Science
- Topic: What would you do if you could solve the Halting Problem?
- Replies:
**33** - Views:
**4286**

### What would you do if you could solve the Halting Problem?

Imagine you have a black box that you know can solve the Halting problem. The input is a text reader and it only has a binary output (accept/reject). You also know for sure that if you try in any way to open the box, it will stop working and it will self-destruct. There's no way to find out how it w...

- Mon Feb 15, 2010 11:46 am UTC
- Forum: Individual XKCD Comic Threads
- Topic: 0702: "Snow Tracking"
- Replies:
**135** - Views:
**35356**

### Re: “Snow Tracking” discussion

I have to agree that the longcat panel is a bit dodgy. Even if the cat is long it has normal length legs, so its stride should be like that of a regular cat, maybe with only the space between two consecutive tracks being actually shorter than usual.

- Mon Jan 25, 2010 3:59 pm UTC
- Forum: Mathematics
- Topic: Taylor expansion
- Replies:
**3** - Views:
**570**

### Re: Taylor expansion

You're right, it does diverge at 0. I've never checked that before. I guess that means that what I was computing there wasn't the actual Taylor series. But the two expressions (f(n) and the series of series) are equivalent, right?

- Mon Jan 25, 2010 2:19 pm UTC
- Forum: Mathematics
- Topic: Taylor expansion
- Replies:
**3** - Views:
**570**

### Taylor expansion

Right, so I'm trying to find the nth coefficient of the Taylor expansion around 0 for f(x) = (1+1/x)^x by using a computer. The way I've expanded the function is by computing the expansion for log(1+x) and e^x: e^{x} = \sum \frac {x^k} {k!} log(1+x) = \sum \frac{(-1)^{k+1}} {k} * x^{...

- Sun Jan 03, 2010 3:24 pm UTC
- Forum: Mathematics
- Topic: Math Books
- Replies:
**378** - Views:
**243044**

### Re: Math Books

Hi! I'm looking for a book about the relation between maths and music. I've been searching my college's library, but the only books I've managed to find already assume I have a rather high level of music knowledge. Is there a book that starts from a lower level, perhaps even explaining the basics of...

- Sun Dec 20, 2009 12:01 pm UTC
- Forum: Logic Puzzles
- Topic: Three secretive neighbors
- Replies:
**12** - Views:
**3307**

### Re: Three secretive neighbors

Bob asked Carol whether the number of her house is a perfect square. Carol answers, Bob thinks she's telling the truth. Alice overhears and correctly thinks she is lying. This is fairly fun puzzle, but feels ruined by inexact wording. Does Alice "overhearing" Carol's answer mean that Bob ...

- Fri Oct 16, 2009 7:49 pm UTC
- Forum: Mathematics
- Topic: Why isn't this problem easy? (ants in a box)
- Replies:
**12** - Views:
**1193**

### Re: Why isn't this problem easy?

You don't have to actually take the box apart. Imagine unfolding the box and bringing all faces in the same plane. Try solving the problem there. After you find the path you're looking for, you can fold the box back the way it was.

- Fri Oct 16, 2009 7:48 pm UTC
- Forum: Mathematics
- Topic: Parity of n choose k
- Replies:
**2** - Views:
**954**

### Parity of n choose k

Apparently, Lucas proved that n choose k is odd <=> (whenever a bit is on in the binary form of k, then that same bit has to be on in the binary form of n) This is a pretty neat theorem and it's supposed to have a nice proof. I haven't got around trying to solve it yet, but I thought I might as well...

- Fri Oct 02, 2009 8:39 pm UTC
- Forum: Logic Puzzles
- Topic: Jumping beetle on projectEureka
- Replies:
**8** - Views:
**1862**

### Re: Jumping beetle on projectEureka

Hmm, I think I may have misunderstood what diagonally adjacent means. Sorry for that. Edit: It can be proven that, for a 2k+1 x 2k+1 board, the beetles in even positions can swap places without any problems, and it can also be proven that in the same type of boards, there will always be at least one...

- Fri Oct 02, 2009 6:38 pm UTC
- Forum: Logic Puzzles
- Topic: Jumping beetle on projectEureka
- Replies:
**8** - Views:
**1862**

### Jumping beetle on projectEureka

Here 's the link to the problem. Now, the answer is 3 but I think it can be done in just 1 . Please tell me if I'm wrong. First of all, in a 2x2 board, you can have 0 squares occupied by more than one beetle; just have each beetle jump to the square diagonally placed from them. They all swap places...

- Mon Sep 21, 2009 9:54 pm UTC
- Forum: Logic Puzzles
- Topic: Infinite Balls and Jugs [solution]
- Replies:
**611** - Views:
**97716**

### Re: Infinite Balls and Jugs [solution]

My thinking on this problem as a whole is that there are compelling arguments for both sides. "Every ball that is added is later removed, so there cannot be anything in the urn" - check. "The number of balls in the urn strictly increases in each step" - also check. The fact that...

- Mon Sep 21, 2009 4:16 pm UTC
- Forum: Logic Puzzles
- Topic: Infinite Balls and Jugs [solution]
- Replies:
**611** - Views:
**97716**

### Re: Infinite Balls and Jugs

The only version of Thompson's Lamp I know is the one from Wikipedia . It says there that a being toggles the switch. The being's actions are determined up until two minutes have passed, so it might just as well be a device. Or, imagine that the being toggles the device by placing and extracting bal...

- Mon Sep 21, 2009 3:35 pm UTC
- Forum: Logic Puzzles
- Topic: Infinite Balls and Jugs [solution]
- Replies:
**611** - Views:
**97716**

### Re: Infinite Balls and Jugs

Also, considering the 1-1+1-1... series, what happens if at each odd iteration you add a ball (numbered, consecutive different balls used for each odd step) and at each even iteration you take it out, similar to Thompson's lamp ? Same argument can be made here: at the 2nth step, ball n will be remo...

- Mon Sep 21, 2009 10:49 am UTC
- Forum: Logic Puzzles
- Topic: Competition, Number game
- Replies:
**13** - Views:
**2573**

### Re: Competition, Number game

This is similar to Traveler's Dilemma, except here you have 100 players. It's still the same game about predicting how your adversary predicts you predicting him predicting you etc. Thread's locked now, but there's a serious discussion about what being rational means.

- Sun Sep 20, 2009 9:42 pm UTC
- Forum: Mathematics
- Topic: Mathematical limits for computer programs
- Replies:
**59** - Views:
**4523**

### Re: Mathematical limits for computer programs

This website says that The proof of Gödel's Incompleteness Theorem is so simple, and so sneaky, that it is almost embarassing to relate. His basic procedure is as follows: 1. Someone introduces Gödel to a UTM, a machine that is supposed to be a Universal Truth Machine, capable of correctly answering...

- Sun Sep 20, 2009 9:16 pm UTC
- Forum: Logic Puzzles
- Topic: Infinite Balls and Jugs [solution]
- Replies:
**611** - Views:
**97716**

### Re: Infinite Balls and Jugs

A slight variation to case A: Each step, you add nine balls (not ten), and instead of removing the lowest numbered ball, you get out a pen and add a '0' to the end of its number. So you add 1-9, and the 1 becomes 10; second step add 11-19, and the 2 becomes 20. So, at every step, the state of the j...

- Fri Sep 18, 2009 3:25 pm UTC
- Forum: Logic Puzzles
- Topic: Infinite Balls and Jugs [solution]
- Replies:
**611** - Views:
**97716**

### Re: Infinite Balls and Jugs

For the random one: I think you need to determine the probability for each ball to be extracted. I'm guessing P = 1/10 + 1/19 + 1/ 28 +... which means P = \sum \frac {1} {9k + 1} That can't be right... \sum \frac {1} {9k +1} > \sum \frac {1} {9k+9} = \frac {1} {9} \sum \frac {1} {k+1} = \infty I th...

- Thu Sep 17, 2009 9:37 pm UTC
- Forum: Logic Puzzles
- Topic: Infinite Balls and Jugs [solution]
- Replies:
**611** - Views:
**97716**

### Infinite Balls and Jugs [solution]

Problem thread. For the random one: I think you need to determine the probability for each ball to be extracted. I'm guessing P = 1/10 + 1/19 + 1/ 28 +... which means P = \sum \frac {1} {9k + 1} That can't be right... \sum \frac {1} {9k +1} > \sum \frac {1} {9k+9} = \frac {1} {9} \sum \frac {1} {k+...

- Mon Sep 14, 2009 3:21 pm UTC
- Forum: Mathematics
- Topic: Chances question
- Replies:
**4** - Views:
**717**

### Chances question

I've just seen on tv an ad saying that for a missing child, the chances of him or her being found drop by 80% after the first 24h. As sad as this statistic might be, I wonder how did they come up with this number and if this is actually relevant. I think it's sensible to assume that the numbers they...

- Sat Sep 12, 2009 6:51 pm UTC
- Forum: Mathematics
- Topic: Where did I go wrong on this question?
- Replies:
**26** - Views:
**2262**

### Re: Where did I go wrong on this question?

Hi guys, I did a test yesterday, and we graded it. I got a 92 (plus some extra credit points) on it, but the one question that I got wrong completely was this: Find the inverse function of y=13x 2 and determine if it is a function I know your question has already been answered (by the way, you coul...

- Wed Sep 09, 2009 11:00 pm UTC
- Forum: Mathematics
- Topic: Practical Explanation for calculating the area under curves
- Replies:
**17** - Views:
**2064**

### Re: Practical Explanation for calculating the area under curves

Oh, right, it was just a misunderstanding. I thought the rate of change for the area under the curve meant the way the function changed (like a derivative), not how much you add to the already computed area.

Thanks for clearing that up for me!

Thanks for clearing that up for me!

- Wed Sep 09, 2009 10:41 pm UTC
- Forum: Mathematics
- Topic: Practical Explanation for calculating the area under curves
- Replies:
**17** - Views:
**2064**

### Re: Practical Explanation for calculating the area under curves

The value of a function at a point describes the height of a small rectangle that describes the rate of change of the area under the curve. Sorry for bumping this, but I'm afraid I'm misinterpreting your statement. Even with the risk of sounding silly, I must ask how does the rectangle describe the...

- Tue Sep 08, 2009 12:56 pm UTC
- Forum: Mathematics
- Topic: Inf and Sup
- Replies:
**8** - Views:
**1096**

### Re: Inf and Sup

Sorry, wrote the question wrong. Edited now. \sup_y \inf_x g(x,y) = \inf_x g(x,y_0) , and \inf_x \sup_y g(x,y) = \sup_y g(x_0,y) . This, I don't get. Shouldn't it be \sup_y \inf_x g(x,y) = \sup_y g(x_0,y) ? I mean, it's the highest value, after you've ...

- Mon Sep 07, 2009 8:18 pm UTC
- Forum: Mathematics
- Topic: Inf and Sup
- Replies:
**8** - Views:
**1096**

### Inf and Sup

I have no idea how to approach this. It's sorta homework (autumn has barely begun, so no school).

Consider g : X x Y -> R, bounded.

Prove that

[math]\sup_y \inf_x h(x,y) <= \inf_x \sup_y h(x,y)[/math]

Consider g : X x Y -> R, bounded.

Prove that

[math]\sup_y \inf_x h(x,y) <= \inf_x \sup_y h(x,y)[/math]

- Fri Aug 21, 2009 4:48 am UTC
- Forum: Logic Puzzles
- Topic: brain in a vat
- Replies:
**214** - Views:
**24664**

### Re: brain in a vat

If the computer has a logic system by which it works (and not just a lame near-exhaustive list of questions and answers), couldn't you use Godel's theorem against it?

- Thu Jun 25, 2009 4:51 pm UTC
- Forum: Logic Puzzles
- Topic: Yet another hat problem
- Replies:
**10** - Views:
**2303**

### Re: Yet another hat problem

Some questions: Does each clone know its position in the line? I mean, if at first they are all numbered from beginning to last do they know their number and the total number of clones? (So does the last clone know it's last?) "Announcing" means that all clones know what was guessed and wh...

- Mon Jun 22, 2009 3:01 pm UTC
- Forum: Logic Puzzles
- Topic: Spider
- Replies:
**23** - Views:
**6706**

### Re: Spider

They're not? I thought they were...

Wiki says:

The Giant Panda (Ailuropoda melanoleuca, literally meaning "cat-foot black-and-white") is a bear native to central-western and southwestern China

It even has a citation.

Wiki says:

The Giant Panda (Ailuropoda melanoleuca, literally meaning "cat-foot black-and-white") is a bear native to central-western and southwestern China

It even has a citation.

- Sun Jun 21, 2009 6:23 pm UTC
- Forum: Forum Games
- Topic: Maximize the post:views ratio! (if you look, you must post)
- Replies:
**36617** - Views:
**1911670**

### Re: Maximize the post:views ratio! (if you look, you must post)

The ratio is WAAY off.

- Sun Jun 21, 2009 6:09 pm UTC
- Forum: Forum Games
- Topic: Simple Question = Silly Answer
- Replies:
**4335** - Views:
**250803**

### Re: Simple Question = Silly Answer

Because the Achievement you're aiming for is not receiving an Achievement.

What am I asking?

What am I asking?

- Sun Jun 21, 2009 6:03 pm UTC
- Forum: Logic Puzzles
- Topic: Spider
- Replies:
**23** - Views:
**6706**

### Re: Spider

**Spoiler:**

- Sun Jun 21, 2009 3:42 pm UTC
- Forum: Logic Puzzles
- Topic: Spider
- Replies:
**23** - Views:
**6706**

### Re: Spider

Concerning the bear: This is a rather stab-worthy question because The time is measure from when the bear enters the hole. so, It's not the gravitational acceleration that is different from what you'd expect. The bear is on Planet Earth. Now, if the acceleration is 1g, what else ca must be d...

- Fri May 15, 2009 8:35 pm UTC
- Forum: Logic Puzzles
- Topic: Number of balls
- Replies:
**10** - Views:
**1459**

### Re: Number of balls

The problem you've suggested is basically the inverse of the one I presented. You start with n and need to find the k. It's a lot easier that way. I'm having problems visualizing the relation going the other way. k dictating what n could be. While the problem I suggested is much easier to solve tha...

- Thu May 14, 2009 4:50 pm UTC
- Forum: Logic Puzzles
- Topic: Number of balls
- Replies:
**10** - Views:
**1459**

### Re: Number of balls

How would you go about calculating the probabilities for n? For simplicity's sake, I'll assume n is between 3 and 5, inclusive, and k is 2. Then, if n=3, there is P(3) that the ball will repeat in 2 iterations (you can find out it's value if you really want to). If n = 4...

- Thu May 14, 2009 3:00 pm UTC
- Forum: Logic Puzzles
- Topic: Number of balls
- Replies:
**10** - Views:
**1459**

### Number of balls

Consider this: You have an urn which, according to your knowledge, can hold an arbitrarily large number of balls. You also know that these balls are numbered from 1 to n. n is finite, obviously, but it can be as large as you wish. (I believe there is a difference between an infinite number of balls ...

- Sat May 02, 2009 3:57 pm UTC
- Forum: Logic Puzzles
- Topic: About this function...
- Replies:
**18** - Views:
**1601**

### Re: About this function...

Jesus. Worst written proof ever. Sorry about that.

Anyway, I thought limits can always be swapped as long as there are different variables for each limit. I can't really think of a double limit where the result can't be switched.

Anyway, I thought limits can always be swapped as long as there are different variables for each limit. I can't really think of a double limit where the result can't be switched.

- Sat May 02, 2009 10:13 am UTC
- Forum: Logic Puzzles
- Topic: About this function...
- Replies:
**18** - Views:
**1601**

### Re: About this function...

jestingrabbit wrote:What you'rs summing isn't varying with i, to begin with.

Yeah, sorry, the argument was supposed to be i/k. Fixed it now.

@skeptical scientist:

Well, I'm not that familiar with Lebesgue spaces and what-not but thanks for pointing that out.

- Fri May 01, 2009 3:38 pm UTC
- Forum: Logic Puzzles
- Topic: About this function...
- Replies:
**18** - Views:
**1601**

### Re: About this function...

I think you can also prove it by using Riemann sums. \int_0^1 f^n(x) dx = \lim_{k\rightarrow \infty} \frac{1}{k} \sum_{i = 1}^{k} f^n(\frac{i}{k}) so \lim_{n\rightarrow\infty}[\int_0^1 f^n(x) dx] = \lim_{n\rightarrow\infty}[\lim_{k\rightarrow \infty} \frac{1}{k} \sum_{i = 1}^...

- Fri May 01, 2009 3:11 pm UTC
- Forum: Logic Puzzles
- Topic: Biggest value
- Replies:
**27** - Views:
**3159**

### Re: Biggest value

Notch wrote:It might be interesting to try to solve this for some clearly defined space.

For example, with just basic math and one letter, the best I can do is "9". I've got a feeling this is optimal.

What qualifies under basic math? I can say "z" in base 50.

- Fri May 01, 2009 11:28 am UTC
- Forum: Logic Puzzles
- Topic: About this function...
- Replies:
**18** - Views:
**1601**

### Re: About this function...

I think the best it can do is hold for functions continuous over intervals.