## Search found 136 matches

- Mon Oct 29, 2012 10:07 am UTC
- Forum: Logic Puzzles
- Topic: (Hopefully) new balance puzzle
- Replies:
**7** - Views:
**2776**

### 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:
**2776**

### 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:
**2776**

### (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:
**4408**

### 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:
**4408**

### 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:
**4408**

### 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:
**5273**

### 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:
**19720**

### 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:
**8031**

### 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:
**5414**

### 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:
**5672**

### 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:
**11986**

### 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:
**11986**

### 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:
**11986**

### 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:
**4031**

### 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:
**4031**

### 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:
**3516**

### 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.

- Wed Mar 16, 2011 9:10 pm UTC
- Forum: Logic Puzzles
- Topic: golf ball problem
- Replies:
**5** - Views:
**3516**

### Re: golf ball problem

Yeah. Too bad the original thread uses solutions involving different cases instead of just giving one clean set of weight comparisons.

- Wed Mar 16, 2011 5:38 pm UTC
- Forum: Logic Puzzles
- Topic: golf ball problem
- Replies:
**5** - Views:
**3516**

### Re: golf ball problem

Name the 12 balls A, B, C, D, E, F, G, H, I, J, K and L Weight A, H, I and K versus E, F, G and J Weight B, G, I and J versus D, F, H and L Weight C, G, H and L versus D, E, I and K Then just look at the results, finding the answer will be pretty straightforward. I am not going to describe the 24 p...

- Wed Mar 09, 2011 10:08 am UTC
- Forum: Logic Puzzles
- Topic: Hooray for Hats!
- Replies:
**9** - Views:
**4596**

### Re: Hooray for Hats!

I get this result too, but I think my solution is much easier than the one proposed in the thread. It also can be generalised. Each player has an id between 1 and 7, agreed upon beforehand. When a player sees the hat on the other player's heads, he justs xor's the ids of people with blue hat...

- Wed Feb 16, 2011 4:15 pm UTC
- Forum: Logic Puzzles
- Topic: Cars in Garages
- Replies:
**17** - Views:
**2605**

### Re: Cars in Garages

Just keep in mind that letting the presenter open 6 doors and swapping to the other one at the end is equivalent to opening the 7 other doors yourself : If the car is behind one of these 7 doors, you'll get it at the end.

- Wed Feb 16, 2011 3:36 pm UTC
- Forum: Logic Puzzles
- Topic: Cars in Garages
- Replies:
**17** - Views:
**2605**

### Re: Cars in Garages

Well, I thought I did : When you first chose your door, you have a probability of 1/8 of having the right door. This probability does not change until the end, because the doors that the presenter opens don't affect the probability of your door. At the end, you have two choices : keep your door or s...

- Wed Feb 16, 2011 3:32 pm UTC
- Forum: Logic Puzzles
- Topic: Cars in Garages
- Replies:
**17** - Views:
**2605**

### Re: Cars in Garages

No, no. 7/8. Try it with cards.

- Wed Feb 16, 2011 3:28 pm UTC
- Forum: Logic Puzzles
- Topic: Cars in Garages
- Replies:
**17** - Views:
**2605**

### Re: Cars in Garages

On your first decision, you have a probability of 1/8 of chosing the door with the ferrari. Don't swap until the very end. When you are stuck with your door, 6 opened doors and one other closed door, then you swap. Bingo, you have a probability of 7/8 to win the ferrari. If you have several choices ...

- Sun Jan 23, 2011 11:12 am UTC
- Forum: Logic Puzzles
- Topic: Dividing an Orchard
- Replies:
**4** - Views:
**2856**

### Re: Dividing an Orchard

**Spoiler:**

- Thu Jun 10, 2010 11:25 am UTC
- Forum: Logic Puzzles
- Topic: I can't find a specific puzzle on here. Help please?
- Replies:
**9** - Views:
**2468**

### Re: I can't find a specific puzzle on here. Help please?

If you followed the rules and the result is 24, then you have the answer. Seems like a good way of checking your work.

By the way, are you sure that you can only use the operators once ? The only solution I see uses / twice

By the way, are you sure that you can only use the operators once ? The only solution I see uses / twice

- Wed Mar 17, 2010 12:03 pm UTC
- Forum: Logic Puzzles
- Topic: Bear Puzzle
- Replies:
**23** - Views:
**17936**

- Fri Feb 26, 2010 4:51 pm UTC
- Forum: Logic Puzzles
- Topic: A fairly simple puzzle.
- Replies:
**11** - Views:
**2177**

### Re: A fairly simple puzzle.

I don't think it is because it is too difficult. Maybe after having manually converted binary to ascii twice, people lost motivation. You should use a different type of puzzle at each step.

- Fri Feb 26, 2010 3:12 pm UTC
- Forum: Logic Puzzles
- Topic: What Do These Numbers Have in Common?
- Replies:
**23** - Views:
**16980**

### Re: What Do These Numbers Have in Common?

Every integer above 2 has this propriety. It makes much more sense as a sequence.Carlington (The Aussie) wrote:Yes. The result I was looking for wasSpoiler:

though.

- Fri Feb 26, 2010 10:54 am UTC
- Forum: Logic Puzzles
- Topic: A fairly simple puzzle.
- Replies:
**11** - Views:
**2177**

### Re: A fairly simple puzzle.

Well, after some pretty boring work I got stuck with some hexadecimal number which did not mean much to me, but when I came back to the thread to say where I was, I saw it was simply Is that it ?

**Spoiler:**

- Mon Feb 15, 2010 10:23 am UTC
- Forum: Logic Puzzles
- Topic: Traveler's Dilemma: What would YOU do?
- Replies:
**166** - Views:
**19577**

### Re: Traveler's Dilemma: What would YOU do?

If I were in a situation like this, I would choose $98 and looking at the poll it seems like the best decision. No. Looking at the poll, the best decision is 99. If you exclude all possible opponents below 98 (which will give the same return for both 98 and 99), then 99 gives an expected value of 9...

- Thu Feb 04, 2010 5:15 pm UTC
- Forum: Logic Puzzles
- Topic: birdwatchers
- Replies:
**14** - Views:
**3542**

### Re: birdwatchers

I think that if the thing must be balanced, then as everyone must call two people, everyone can also receive a call from only two people. Therefore, we have N pairs of people which have to contain at least one reliable person. This seems to be independant from the method used, provided it is makes s...

- Fri Oct 23, 2009 3:42 pm UTC
- Forum: Mathematics
- Topic: Conic intersections
- Replies:
**9** - Views:
**1505**

### Re: Conic intersections

Well, I was either too lazy or too stupid to check the english version of the wikipedia article. The paragraph you link gives away the solution : The first step is to find a linear combination of the equations that can be put in the form AX²+BXY+CY², using any linear transformation of x and y to get...

- Fri Oct 23, 2009 1:11 pm UTC
- Forum: Mathematics
- Topic: Conic intersections
- Replies:
**9** - Views:
**1505**

### Re: Conic intersections

Okay sir, This leads to something like S(x+T)² + U(y+V)² = W, which is still a third conic. I know that if the first two intersect, then this third one will intersect both in their intersection points. I have been turning around things like that for a few days now, and I still have no clue about how...

- Fri Oct 23, 2009 12:00 pm UTC
- Forum: Mathematics
- Topic: Conic intersections
- Replies:
**9** - Views:
**1505**

### Re: Conic intersections

I assume this is homework. So here's a hint: First, eliminate the xy term. Nope, wrong assumption. Collision detection. I will update my profile to avoid such misunderstanding. But if this can be given as homework, then this must actually be simple. Good news. That means I am dumb. Bad news. When y...

- Fri Oct 23, 2009 11:00 am UTC
- Forum: Mathematics
- Topic: Conic intersections
- Replies:
**9** - Views:
**1505**

### Conic intersections

Hello I have a problem which seems very simple, and yet I am totally stuck : I have two 2D conic curves, wich equations are Ax²+Bxy+Cy²+Dx+Ey+F=0 and Gx²+Hxy+Iy²+Jx+Ky+L=0. I need to find out wether these two lines intersect or not, ie wether there is at least a {x;y} which satisfies both equations....

- Fri Oct 16, 2009 11:45 am UTC
- Forum: Logic Puzzles
- Topic: Tic Tac Toe With a Twist
- Replies:
**39** - Views:
**6156**

### Re: Tic Tac Toe With a Twist

@skep: I'm sorry, I must be missing something. How does X force a win after: . . . X . O . . . or: . . O . . . X . . Yes, I think you are missing something in the first case, and in the second one you must just make a mistake somewhere. What you seem to be missing is that in the first case the boar...

- Tue Sep 01, 2009 1:54 pm UTC
- Forum: Logic Puzzles
- Topic: Traveler's Dilemma: What would YOU do?
- Replies:
**166** - Views:
**19577**

### Re: Traveler's Dilemma: What would YOU do?

I agree with Pianissimo, I'll write $100 which will at least net me $98. @Goldstein, I think the manager will pay both of them, it's only fair and logical to do so. But that's just me. :wink: Pianissimo did not get the problem statement right. Goldstein is right. If you write 100 and I write 37, I ...

- Thu Jul 30, 2009 12:08 pm UTC
- Forum: Logic Puzzles
- Topic: Spherical Poker
- Replies:
**6** - Views:
**2646**

### Re: Spherical Poker

I am not sure I get the rules right. If I follow your explanations, there is no way to raise. The second player puts 1, the first either puts 1 or folds, and if he put 1, then the second player also has the choice to put 1 or fold. After the second one puts 1, the betting part ends. This seems unfai...

- Fri Jul 10, 2009 12:05 pm UTC
- Forum: Logic Puzzles
- Topic: An easy puzzle: Russian Roulette
- Replies:
**18** - Views:
**4270**

### Re: An easy puzzle: Russian Roulette

Yeah pretty much. Want to do the math and check that my situation does turn out counter-intuitively? Okay, doing the math is fun, and counter-intuitive situations are even more. Let's do it. First of all, I disagree with your conclusions : on the third shot, we know that the last two slots were emp...