## Search found 119 matches

- Mon Oct 09, 2017 2:40 pm UTC
- Forum: Logic Puzzles
- Topic: Could there still be new hat puzzles?
- Replies:
**11** - Views:
**2426**

### Re: Could there still be new hat puzzles?

Proof of a theoretical maximum, assuming a deterministic strategy: EDIT: Actually, I don't even think we need to assume a deterministic strategy. I'll edit my argument in. There are 128 configurations of hats. Each player, in any given deterministic strategy, will be right in 64 cases and wrong in ...

- Wed Sep 27, 2017 1:03 pm UTC
- Forum: Logic Puzzles
- Topic: Could there still be new hat puzzles?
- Replies:
**11** - Views:
**2426**

### Re: Could there still be new hat puzzles?

An absurd strategy where the prime-numbered players (2,3,5,7) pick the minority color and the other players pick the majority color As I said in the OP, "They can devise a strategy, but it can't be nominative. The player's can't take into account their own identity, the whole strategy has to b...

- Thu Sep 21, 2017 4:05 pm UTC
- Forum: Logic Puzzles
- Topic: Could there still be new hat puzzles?
- Replies:
**11** - Views:
**2426**

### Re: Could there still be new hat puzzles?

It is a good start, but you are right, it is possible to get less than half this error rate.HonoreDB wrote:This fails for one of the two most common splits, and doesn't use relative positions, so it's probably not optimal.

- Tue Sep 19, 2017 3:17 pm UTC
- Forum: Logic Puzzles
- Topic: Could there still be new hat puzzles?
- Replies:
**11** - Views:
**2426**

### Could there still be new hat puzzles?

Here's a hat puzzle inspired by this one . Seven perfectly rational players stand in a circle. Everyone gets assigned, randomly and independently, a black hat or a white hat. Nobody can see the color of his own hat but the colors of all other players. All the players must try to guess the color of t...

- Tue Sep 19, 2017 11:34 am UTC
- Forum: Logic Puzzles
- Topic: Another hat puzzle
- Replies:
**9** - Views:
**1037**

### Re: Another hat puzzle

I don't have a formal proof that this will work for any n>2, but it worked smoothly for the values I tried, and it seems that increasing n just gives more margin of error when you build the strategy. Here it is: First, I have to determine in advance the lower bound of the number of correct guesses. ...

- Wed Sep 06, 2017 8:20 am UTC
- Forum: Logic Puzzles
- Topic: 538 hats riddle
- Replies:
**20** - Views:
**1390**

### Re: 538 hats riddle

It may be possible to do better with two rules than SirGabriel has with one, but your posted win rate is less than his. To keep the thread alive, I think it is relevant to confirm that it is indeed possible to do better than SirGabriel, under the assumptions pointed out by jaap. For any number of p...

- Wed Apr 12, 2017 1:12 pm UTC
- Forum: Logic Puzzles
- Topic: Two secrets
- Replies:
**20** - Views:
**4030**

### Re: Two secrets

I think I have an O(log(N)) solution. Let A initially be {1, 2, 3, ..., N-1}. At all times, we know that at least one of the numbers is in A. Divide A into three roughly equal sets A 1 , A 2 , and A 3 , and ask about A 1 ∪ A 2 , A 1 ∪ A 3 , and A 2 ∪ A 3 . If we get a "yes" response to on...

- Wed Apr 12, 2017 9:40 am UTC
- Forum: Logic Puzzles
- Topic: Two secrets
- Replies:
**20** - Views:
**4030**

### Re: Two secrets

Ok, I took a chance to work on my solution again, even if now it is just a correction of Nitrodon's solution, but here is where I landed: Let the initial state be state I, where we have one set containing both numbers. We cut the set in n parts and test every pair of parts. If there is only one yes,...

- Wed Apr 12, 2017 8:31 am UTC
- Forum: Logic Puzzles
- Topic: Two secrets
- Replies:
**20** - Views:
**4030**

### Re: Two secrets

I think I have an O(log(N)) solution. Let A initially be {1, 2, 3, ..., N-1}. At all times, we know that at least one of the numbers is in A. Divide A into three roughly equal sets A 1 , A 2 , and A 3 , and ask about A 1 ∪ A 2 , A 1 ∪ A 3 , and A 2 ∪ A 3 . If we get a "yes" response to on...

- Mon Apr 03, 2017 9:53 am UTC
- Forum: Logic Puzzles
- Topic: Two secrets
- Replies:
**20** - Views:
**4030**

### Re: Two secrets

So we just have to show that for any distinct a, b, c and d we can find g, h, i, j and T with a and b but neither c nor d in S(g,h,i,j,T). Let g be the position of any digit where a and c differ, h the position of a digit where a and d differ, i the position of a digit where b and c differ ...

- Fri Mar 31, 2017 1:02 pm UTC
- Forum: Logic Puzzles
- Topic: Two secrets
- Replies:
**20** - Views:
**4030**

### Re: Two secrets

So we just have to show that for any distinct a, b, c and d we can find g, h, i, j and T with a and b but neither c nor d in S(g,h,i,j,T). Let g be the position of any digit where a and c differ, h the position of a digit where a and d differ, i the position of a digit where b and c differ ...

- Wed Jan 18, 2017 10:33 am UTC
- Forum: Logic Puzzles
- Topic: New puzzle : Equalize
- Replies:
**13** - Views:
**2848**

### Re: New puzzle : Equalize

a solution is impossible for the following example "random" starting position: r r bb bb r r rrrr rbbr rbbr rbbr gives 1 1 1 1 1 -2 -2 1 -1 2 2 -1 -1 1 1 -1 with the original system, and 2 3 3 2 3 3 3 3 2 5 5 2 1 3 3 1 with yours. In the first case, both sums are 2, in the second case bot...

- Wed Aug 05, 2015 2:28 pm UTC
- Forum: Computer Science
- Topic: the halting problem
- Replies:
**16** - Views:
**6115**

### Re: the halting problem

Change your definition of x to this. x takes just one input, and: x(p) internally uses h and works this way: -if program p halts with input p, loop forever -if program p does not halt with input p, return 1 Now x(x) halts if and only if x(x) does not halt. There is an important point here: this is ...

- Wed Aug 05, 2015 1:55 pm UTC
- Forum: Computer Science
- Topic: the halting problem
- Replies:
**16** - Views:
**6115**

### Re: the halting problem

It sounds like you're not familiar with quines , which are programs that can print their own source code. Well, I actually am, wrote one years ago in C, discovered recently that it was possible with compressed files too. The very last part is the one that really blew my mind. I mean, the fact that ...

- Wed Aug 05, 2015 1:43 pm UTC
- Forum: Computer Science
- Topic: the halting problem
- Replies:
**16** - Views:
**6115**

### Re: the halting problem

I'm not sure where you got this proof from, but the explanation on Wikipedia seems pretty clear to me (and different from yours). Well, from various places, including the wikipedia article. Can you tell me how it is different from mine ? Wikipedia's h is the same as my h, my x(i,i) would be wikiped...

- Wed Aug 05, 2015 1:24 pm UTC
- Forum: Computer Science
- Topic: the halting problem
- Replies:
**16** - Views:
**6115**

### Re: the halting problem

For example, you could be passing x as a string (the source code), and x-as-a-program runs some checks on the string to decide whether it halts or not. Ok, but with what inputs? By definition, x halts or not depending on its input. Here, x is supposed to decide wether x halts or not. But with which...

- Wed Aug 05, 2015 1:06 pm UTC
- Forum: Computer Science
- Topic: the halting problem
- Replies:
**16** - Views:
**6115**

### Re: the halting problem

That is a function that does not use its input. What does it have to do with my question?

- Wed Aug 05, 2015 12:16 pm UTC
- Forum: Computer Science
- Topic: the halting problem
- Replies:
**16** - Views:
**6115**

### the halting problem

Hello I apologize in advance for how trivial this may seem for anyone that truly understands the halting problem and/or if this question has already been answered here, but my searches were not conclusive. There is something I don't get in the halting problem, and none of the explanations I've read ...

- Fri Apr 24, 2015 7:22 am UTC
- Forum: Logic Puzzles
- Topic: Simple math word puzzle
- Replies:
**9** - Views:
**2131**

### Re: Simple math word puzzle

I would prefer from any student to see the full solution. 1. y=x+20 2. xy=wz+d 3. z=w+26 4. x+y=w+z Putting 1 into 4 4a. 2x+20 = w+z Putting 3 into 4a 4b. 2x+20=2w+26 ---------- -26 4c. 2x-6=2w --------------- /2 5. x-3=w Putting 1 into 2 2a. x(x+20)=wz+d Putting 3 into 2a 2b. x(x+20...

- Thu Apr 23, 2015 11:49 am UTC
- Forum: Logic Puzzles
- Topic: Simple math word puzzle
- Replies:
**9** - Views:
**2131**

### Re: Simple math word puzzle

Here's my method:

**Spoiler:**

- Tue Feb 03, 2015 12:30 pm UTC
- Forum: Logic Puzzles
- Topic: Six statements. How many are true?
- Replies:
**20** - Views:
**5643**

### Re: Six statements. How many are true?

Of course I solved tis with the classic method, but considering how some of the statements were designed, I wanted to try something else. I consider the nature of a statement depends only on how this statement matches reality, and not on what other statements say about it. For example "This sta...

- Mon Dec 17, 2012 6:00 pm UTC
- Forum: Logic Puzzles
- Topic: Create a perfect lossless compression algorithm
- Replies:
**21** - Views:
**4539**

### Re: Create a perfect lossless compression algorithm

Does Alice have a way of telling Bob the transfer is over, or do we need to establish a protocl which allows Bob to know when he has read the last bit, and stop expecting new information from Alice ? If Bob can know when the transfer is over wothout having to figuring it out with the data, then we c...

- Mon Oct 29, 2012 5:26 pm UTC
- Forum: Logic Puzzles
- Topic: (Hopefully) new balance puzzle
- Replies:
**7** - Views:
**2314**

### Re: (Hopefully) new balance puzzle

Thank you for proving me that this can be solved with a non computer-enhanced brain!

- Mon Oct 29, 2012 10:07 am UTC
- Forum: Logic Puzzles
- Topic: (Hopefully) new balance puzzle
- Replies:
**7** - Views:
**2314**

### Re: (Hopefully) new balance puzzle

Snark: I think we're supposed to specify all three trials in advance, before we have the results of any of them. Here's my solution, retaining your labeling of the coins as ABCDEFG: Test ABC vs. DEF, AD vs. BE, and AF vs. CD. ABC < DEF, AD < BE, AF < CD: AG are fake ABC < DEF, AD < BE, AF = CD: AC ...

- Sun Oct 28, 2012 10:58 pm UTC
- Forum: Logic Puzzles
- Topic: (Hopefully) new balance puzzle
- Replies:
**7** - Views:
**2314**

### Re: (Hopefully) new balance puzzle

you need to tell the weights beforehand. Just checking for clarification of this point. Does it mean that you know what weights both types of coins are before you start weighing them (eg: you know the real coins are (say) 30g and the fake ones are (say) 29 grams)? Or does it mean something else? I'...

- Sun Oct 28, 2012 9:53 pm UTC
- Forum: Logic Puzzles
- Topic: (Hopefully) new balance puzzle
- Replies:
**7** - Views:
**2314**

### (Hopefully) new balance puzzle

I have never seen this variation of the classic balance puzzles before, so I tried to solve it and found the solution. I am not going to tell a story around this because it is a classic type of problem, the deal is that you have 7 coins, but 2 of these coins are slighly lighter than the other 5. Of ...

- Wed Oct 24, 2012 4:51 pm UTC
- Forum: Logic Puzzles
- Topic: Another gambling game
- Replies:
**17** - Views:
**3781**

### Re: Another gambling game

Even given superrationality, I disagree with the superrational solutions posted so far... Say every one of the million people chooses to play with probability p. Then we would, between them all, expect a total of 10 8 p dollars be paid in entry fees, and a total of 10 6 (1 - (1-p) 10000...

- Tue Oct 23, 2012 11:00 am UTC
- Forum: Logic Puzzles
- Topic: Another gambling game
- Replies:
**17** - Views:
**3781**

### Re: Another gambling game

I totally forgot about mixed strategies: if the perfectly logical strategy is to play with a probability p, then there are 1,000,000p players to split the $1,000,000, so the gain is 1/p-100. As this happens with probability p, that makes p(1/p-100), that is 1-100p. That means that if p is be...

- Tue Oct 23, 2012 9:40 am UTC
- Forum: Logic Puzzles
- Topic: Another gambling game
- Replies:
**17** - Views:
**3781**

### Re: Another gambling game

If all other players are perfect logicians, then they will all come to the same conclusion : "Either I play and everybody plays, and we all lose $99, or I don't play, and nobody plays, and we are all even. I don't play." Now, I am not a perfect logician, I am just human, so my reas...

- Mon Aug 06, 2012 9:08 am UTC
- Forum: Logic Puzzles
- Topic: Rubik's Cube false solution
- Replies:
**11** - Views:
**4618**

### Re: Rubik's Cube false solution

What ?Tirian wrote:Spoiler:

Do you have an example on how to do that from a situation which is not one of the 4 obvious ones pointing directly to the center ?

- Tue Mar 20, 2012 10:10 am UTC
- Forum: Logic Puzzles
- Topic: Imposter Puzzle
- Replies:
**69** - Views:
**18005**

### Re: Imposter Puzzle

Two solutions which came to my mind: 1) Use an asymmetric encryption method. Each member i has a public key A_i which is known by all, and a private key B_i only known to them. In a meeting of 2 persons, both exchange their number i. Both generate a random string, encypt it with the public key ...

- Tue Feb 14, 2012 2:27 pm UTC
- Forum: Logic Puzzles
- Topic: Cutting an unfolded cube from a square of paper.
- Replies:
**20** - Views:
**7135**

### Re: Cutting an unfolded cube from a square of paper.

tomtom2357 wrote:Okay. what is the highest percentage of the rectangle you can use, I have 2/3 so far.

I get 3/4 : draw crosses (connecting opposite corners) on two opposite faces, cut along the lines, then cut one of the edges linking those two faces. Other cuttings of the opposite faces get the same result.

- Mon Jan 16, 2012 1:22 pm UTC
- Forum: Logic Puzzles
- Topic: Numbering the insides of a Rubiks Cube uniquely
- Replies:
**23** - Views:
**4825**

### Re: Numbering the insides of a Rubiks Cube uniquely

WarDaft wrote:Consider:Code: Select all

`X1 1X3 3X`

2 2 2

2 2 2

X1 1X3 3X

4 4 4

4 4 4

X1 1X3 3X

Does this not satisfy your placing criteria?

Using the same pieces, you could also get this :

Code: Select all

` X1 1X3 3X`

2 2 4

2 2 4

X1 1X3 3X

4 4 2

4 4 2

X1 1X3 3X

- Thu Jul 28, 2011 5:02 pm UTC
- Forum: Logic Puzzles
- Topic: Volumetrically Maximized A4 Cylinder
- Replies:
**8** - Views:
**4665**

### Re: Volumetrically Maximized A4 Cylinder

My attempt was based on what felt like the most obvious solution, assuming the discs were not required to be each on one side of the rectangle. It consists of one vertical rectangle 29.7cm long, this detemines the radius of the discs, both discs are put on the same side of the rectangle, thus determ...

- Tue Jul 19, 2011 10:19 am UTC
- Forum: Logic Puzzles
- Topic: A hundred tourists
- Replies:
**28** - Views:
**10948**

### Re: A hundred tourists

Imagine you are one of the prisonners, and you see 27 red hats, 31 white hats and 41 blue hats. The value you see is 31+2*41 = 113 (congruent to 2 modulo 3). Now you are lucky and you are not the first one to be picked. The guy has a white hat, so the value he sees is 112 + the value of your own hat...

- Wed Jul 13, 2011 8:47 am UTC
- Forum: Logic Puzzles
- Topic: A hundred tourists
- Replies:
**28** - Views:
**10948**

### Re: A hundred tourists

I think the day to strategise means that they're okay actually. Yeah, I guess they could come up with a strategy to avoid the zero survivors scenario, but without having additionnal information from what the others say or what the cannibals do to them, I don't really see how they could individually...

- Tue Jul 12, 2011 2:42 pm UTC
- Forum: Logic Puzzles
- Topic: A hundred tourists
- Replies:
**28** - Views:
**10948**

### Re: A hundred tourists

If all the tourists can hear what the others answer : let red=0, white=1 and blue=2. The first one sums all the hats he sees, and answers the color which matches this number modulo 3. He has 2/3 chances of dying, but now everybody else, knowing what is the color of the first tourist's hat, can d...

- Wed Jun 15, 2011 8:46 am UTC
- Forum: Mathematics
- Topic: 100 boxes, one prisoner
- Replies:
**6** - Views:
**3694**

### Re: 100 boxes, one prisoner

I like this puzzle. Well, thank you. And thank you all for these solutions. I had been struggling with this problem for a couple of days, and the only leads I had were brute force (which took forever for N=10), and a twisted 2D recursive method based on sets of boxes pointing to other sets of boxes...

- Tue Jun 14, 2011 7:32 pm UTC
- Forum: Mathematics
- Topic: 100 boxes, one prisoner
- Replies:
**6** - Views:
**3694**

### 100 boxes, one prisoner

Hello I am not really sure it is in the right section, because it is a problem I don't have an elegant solution for, but i find it very close to the puzzle with the prisoners and the boxes (http://forums.xkcd.com/viewtopic.php?f=3&t=7164), and I am pretty sure this one will be a joke for some pe...

- Thu Mar 17, 2011 9:28 am UTC
- Forum: Logic Puzzles
- Topic: golf ball problem
- Replies:
**5** - Views:
**3171**

### Re: golf ball problem

Your solution has some algebraic structure.

Actually, that's exactly how I designed it. Except your columns were my rows... That's the first time I see someone else using this type of solution.