## Search found 131 matches

- Fri Nov 30, 2018 3:54 pm UTC
- Forum: Logic Puzzles
- Topic: Rome
- Replies:
**1** - Views:
**883**

### Re: Rome

As I understand it, it is the classic switch puzzle, (http://forums.xkcd.com/viewtopic.php?f=3&t=73 http://forums.xkcd.com/viewtopic.php?f=3&t=13669 http://forums.xkcd.com/viewtopic.php?f=3&t=69986), the unlocked door playing the role of the switch. If we can't count on the room to stay ...

- Mon Nov 26, 2018 1:15 pm UTC
- Forum: Logic Puzzles
- Topic: Poker
- Replies:
**1** - Views:
**957**

### Re: Poker

Brutally arithmetical solution: Call A, B and C the number of coins each initially takes, translate what we know into three equations (for example, 2B/3+42 = (A+B+C)/3, because after Paul removed 1/3 of his initial stock and getting 42 coins back, he has 1/3 of the total), then solve...

- Tue Nov 20, 2018 9:37 am UTC
- Forum: Logic Puzzles
- Topic: Protection fee
- Replies:
**2** - Views:
**1270**

### Re: Protection fee

An easy answer:

**Spoiler:**

- Tue Nov 20, 2018 9:31 am UTC
- Forum: Logic Puzzles
- Topic: Penalty fee
- Replies:
**1** - Views:
**977**

### Re: Penalty fee

Calling x the smallest number, the others are x+1 and x+2, so the sum of all three is 3x+3 and the product is x*(x+1)*(x+2) = x^3+3x²+2x. So we need to solve x^3+3x²+2x-(3x+3) = 0, which is x^3+3x²-x-3 = 0. 1 and -1 are trivial solutions, we easlily get to this: The three va...

- Sun Nov 11, 2018 2:12 pm UTC
- Forum: Logic Puzzles
- Topic: Last Call
- Replies:
**1** - Views:
**1022**

### Re: Last Call

So, if I get the explanations correctly, there are 6 oarsmen including me, their crossing times are 5, 10, 25, 30, 30 and 30 minutes. That is, assuming I can still able to row the boat, which doesn't seem to be very clear. Again, it seems implied that the boat can be handled by a single oarsman when...

- Tue Nov 06, 2018 9:48 am UTC
- Forum: Logic Puzzles
- Topic: Sacrifice
- Replies:
**8** - Views:
**5182**

### Re: Sacrifice

Go back to your smartass solution for a second - there's more to be done there to get a solution with at most 4 chicken risked, and if you're lucky you lose none But then, we can do better, with the "break the game" solution: Take one chicken, make it sip from the first canister. If it di...

- Thu Apr 19, 2018 9:57 am UTC
- Forum: Individual XKCD Comic Threads
- Topic: 1982: "Evangelism"
- Replies:
**109** - Views:
**22988**

### Re: 1982: "Evangelism"

"Oh, Hey, I didn't see you guys all the way over there" should have been the punchline.fibonacci wrote:If you point a telescope out to the right off the edge of the chart, on a clear night you can spot "Vegans".

https://xkcd.com/435/

- Fri Feb 02, 2018 8:59 am UTC
- Forum: Logic Puzzles
- Topic: The Goddess, the Villagers, and the Traveler
- Replies:
**16** - Views:
**7687**

### Re: The Goddess, the Villagers, and the Traveler

Well, the villagers are supposed to have the power to selectively erase information from their own memory on command. Forget about limitations not stated in the problem.Cradarc wrote:The only exception is if you put in additional limitations to what the villagers can do which are not stated in the problem.

- Wed Jan 31, 2018 10:04 am UTC
- Forum: Logic Puzzles
- Topic: The Goddess, the Villagers, and the Traveler
- Replies:
**16** - Views:
**7687**

### Re: The Goddess, the Villagers, and the Traveler

I have something that is not very satisfying because it implies that the villagers are very gullible, but it doesn't require magically or hypnotically removing information they already have. Just updating it in a way you control, so that every time they change their mind, they actually have more inf...

- Tue Jan 30, 2018 8:54 am UTC
- Forum: Logic Puzzles
- Topic: The Goddess, the Villagers, and the Traveler
- Replies:
**16** - Views:
**7687**

### Re: The Goddess, the Villagers, and the Traveler

Actually that is a completely valid command to give. You can ask the villagers to forget and remember any bits of information on a whim so long as it's in the form of command and follows in accordance with the three rules set for them I have a real problem with this. Forgetting something on command...

- Fri Jan 26, 2018 11:50 am UTC
- Forum: Logic Puzzles
- Topic: The Goddess, the Villagers, and the Traveler
- Replies:
**16** - Views:
**7687**

### Re: The Goddess, the Villagers, and the Traveler

There's something that I don't get here. Please consider this first theoretical situation: The goddess actually is just a phantom flying around the village. She is not specifically linked to any of the villagers. The villagers think that she is walking among them, but they're wrong. apart from that,...

- Mon Oct 16, 2017 12:17 pm UTC
- Forum: Logic Puzzles
- Topic: Could there still be new hat puzzles?
- Replies:
**12** - Views:
**8908**

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

Well, I was right - there is a much more elegant maximal strategy. You don't even need to divide it up into cases to use the strategy (though checking it will divide into cases)! Check the players that are first, second, and fourth from your left. Guess whichever color hat is worn by an even number...

- Mon Oct 09, 2017 2:40 pm UTC
- Forum: Logic Puzzles
- Topic: Could there still be new hat puzzles?
- Replies:
**12** - Views:
**8908**

### 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:
**12** - Views:
**8908**

### 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:
**12** - Views:
**8908**

### 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:
**12** - Views:
**8908**

### 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:
**5478**

### 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:
**6846**

### 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:
**9390**

### 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:
**9390**

### 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:
**9390**

### 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:
**9390**

### 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:
**9390**

### 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:
**5865**

### 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:
**7460**

### 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:
**7460**

### 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:
**7460**

### 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:
**7460**

### 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:
**7460**

### 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:
**7460**

### 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:
**2825**

### 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:
**2825**

### 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:
**7055**

### 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:
**5058**

### 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:
**2614**

### 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:
**2614**

### 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:
**2614**

### 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:
**2614**

### (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:
**4225**

### 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:
**4225**

### 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...