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Riddle me this.. Riddle me that.

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Re: Riddle me this.. Riddle me that.

Postby Torquemada » Sat Oct 01, 2011 6:32 pm

The rocks are 1, 3, 9, and 27 lbs. For each weight below, the stones will be listed vs the weight to be measured, X.

1 = 1 vs X.
2 = 3 vs X + 1.
3 = 3 vs X.
4 = 3 + 1 vs X.
5 = 9 vs X + 3 + 1.
6 = 9 vs X + 3.
7 = 9 + 1 vs X + 3.
8 = 9 vs X + 1.
9 = 9 vs X.
10 = 9 + 1 vs X.
11 = 9 + 3 vs X + 1.
12 = 9 + 3 vs X.
13 = 9 + 3 + 1 vs X.
14 = 27 vs X + 9 + 3 + 1.
15 = 27 vs X + 9 + 3.
16 = 27 + 1 vs X + 9 + 3.
17 = 27 vs X + 9 + 1.
18 = 27 vs X + 9.
19 = 27 + 1 vs X + 9.
20 = 27 + 3 vs X + 9 + 1.
etc. etc.

I'm not sure if it works out because they're all powers of 3 (3^0=1, 3^1=3, etc.), but it does.
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Re: Riddle me this.. Riddle me that.

Postby bldavis » Sat Oct 01, 2011 7:19 pm

pulls one out, hands the hat back to the king and asks the king to show the remaining peice of paper to the court, showing death
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Re: Riddle me this.. Riddle me that.

Postby Alacia » Sun Oct 02, 2011 3:16 am

Torquemada wrote:I'm not sure if it works out because they're all powers of 3 (3^0=1, 3^1=3, etc.), but it does.

Indeed, that would be the reason. Each rock has 3 different possible "states": being on the left side of the scale, not being on it at all or being on the right side of it, corresponding to a multiplier of 1, 0 or -1 (the total weight could be written on the form X = a + b*3 + c*9 + d*27, where a,b,c,d could each take on the values -1,0,1)
Basically, what we have is a Ternary number system, where instead of using 2 as a number, we use -1 (2 ≡ -1 mod 3) giving us a similar set of possible values as a regular 4-digit ternary number, only offset by -40 (-1111 in ternary).

A similar problem, but probably slightly harder (Edit: doublechecked answer, slightly rephrased the question for a "prettier" answer):
Assume instead that you have a scale where the weights are set so that they can only be placed on one side or the other (i.e they cannot be taken off or put somewhere where they don't affect the result). What would be the minimum number of weights needed to be able to weigh each integer from 0 up to 32, and what would those weights be?

On the notes in the hat: basically what bldavis said, but in order to actually force someone to show the other note, he could destroy (rip apart) the one he takes out too.
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Re: Riddle me this.. Riddle me that.

Postby Xenix » Sun Oct 02, 2011 6:54 am

Alacia wrote:On the notes in the hat: basically what bldavis said, but in order to actually force someone to show the other note, he could destroy (rip apart) the one he takes out too.


The answer given on the show was that he slaps a big grin on his face when he reads the one he pulls out of the hat, and then eats it before proceeding to celebrate. :lol: That way, there's no way to prove what was on the one he picked aside from looking at the one in the hat.



Let's see, as for the 0->32 without taking any off the balance scale, that'd be like binary numbers except all the values have to be (-1) or (1), and to be able to get zero you can't use only powers of some number since (-1 -1 -1 ... 1) or vice versa will add up to 1, never 0.

So, to all have them add up to zero, yet still be able to get a 1, you'll need a 0.5 to start out and another 0.5 somewhere to cancel that out. Since we've got two values, it'll have to be some take on binary, and you can assume the numbers will have to be of a similar form as binary for the minimum values.

If you assume that you have to add a final number that cancels out the negative of all the previous ones to get zero, you can write out a series of equations as follows:

[ -1 -1 -1 -1 -1 -1 1 ] = 0
[ -1 -1 -1 -1 -1 1 1 ] = 1
[ -1 -1 -1 -1 1 -1 1 ] = 2
[ -1 -1 -1 -1 1 1 1 ] = 3

etc., and do a quick matrix inverse to find the values of x0 x1 x2 x3 x4 x5 x6 that will make it correct. Note that if you had said from 0 to 31 it would have been 6 bits, but including 32 sets it at 7.

Anyways, doing this shows that the values are

x0 - x1 - x2 - x3 - x4 - x5 - x6
16 - 8 - 4 - 2 - 1 - .5 - 31.5

As I expected, the smallest number is 0.5 and the biggest has another 0.5 to cancel out with it. Each number is that same number in 6 bits of binary (using -1's instead of 0's) with another 1 tacked on at the end.

So, as for the answer: 7 rocks. If you didn't want to include the number 32, it would be 6 rocks of size 15.5, 6, 4, 2, 1, .5 instead.


All right, one final puzzler from the same source - no cheating and looking it up, or answering if you've heard it before:

-------- New Riddle ------------

A warden has a group of 15 pirates who are to be executed the next day, but he wants to have some fun first by giving them a "fair" chance. He marches the group outside the day before the execution and makes them stand in a line where they can only see the people in front of them, and tells them what will happen the next day.

"Tomorrow, I will march you out here in a line, just like now, and place either a white or a black hat on each of your heads. Then, starting with the person at the back of the line, I will ask you what color the hat on your head is. If you are correct, you go free. If you're wrong, you go off for execution. I then will move on to the next person, etc. If anyone turns around, says something either than "black" or "white" or does something that looks like cheating, you'll all be executed.

Feel free to spend tonight talking amongst yourselves to try and come up with a plan that's better than guessing - I'll hold the executions first thing in the morning."

How do the pirates increase their odds of survival the most? (e.g. if you come up with a way that half are guaranteed to survive, that's only correct if there is no way to guarantee more than half will survive - feel free to post any ideas though.)
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Re: Riddle me this.. Riddle me that.

Postby Alacia » Sun Oct 02, 2011 8:12 am

Nice, very different from (and more thorough than) my solution of the math problem, and I did kinda miss that you'd only need 6 rocks for 0-31 (math at 4AM is bad :P). My original answer was 16, 8, 4, 2, 1, 0.5, 0.5, but I see the benefit of subtracting 0.5 from a power of 2 instead

First take on the pirate question: The first person says "white" if the two people standing in front of him have the same color hats, and "black" otherwise, meaning the next two will know what to say, repeating the procedure for every third person -> 2/3 are guaranteed to live.
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Re: Riddle me this.. Riddle me that.

Postby Xenix » Sun Oct 02, 2011 8:24 am

Alacia wrote:First take on the pirate question: The first person says "white" if the two people standing in front of him have the same color hats, and "black" otherwise, meaning the next two will know what to say, repeating the procedure for every third person -> 2/3 are guaranteed to live.


That's a good one - you're on the right track, but if there's more than three pirates you can improve the odds further than that.
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Re: Riddle me this.. Riddle me that.

Postby Alacia » Sun Oct 02, 2011 8:46 am

Right, of course :P Expanding on my previous method, the first person would say "white" if there's an even number of white hats in front of him, and "black" otherwise. The next person would then compare that to what they can see, giving them the correct answer. The rest keep track of how many times someone has said "white", and would then know whether there's an even or odd number of white hats left. Hence, everyone except the first one would be guaranteed to live.
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Re: Riddle me this.. Riddle me that.

Postby Passionario » Mon Oct 03, 2011 12:00 am

Xenix wrote:The answer given on the show was that he slaps a big grin on his face when he reads the one he pulls out of the hat, and then eats it before proceeding to celebrate. :lol: That way, there's no way to prove what was on the one he picked aside from looking at the one in the hat.


Sure there is.

"Guards! Slice his stomach open and retrieve the paper!"

And once it is determined that the swallowed note did, in fact, say 'death', you can just leave him to bleed to death.
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Re: Riddle me this.. Riddle me that.

Postby fuzzygeek » Mon Oct 03, 2011 10:22 am

Xenix wrote:That's a good one - you're on the right track, but if there's more than three pirates you can improve the odds further than that.


Is there a constraint on the numbers of white or black hats? I didn't see any in the original text,
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Re: Riddle me this.. Riddle me that.

Postby Xenix » Mon Oct 03, 2011 12:02 pm

fuzzygeek wrote:Is there a constraint on the numbers of white or black hats? I didn't see any in the original text,


Nope - Alacia was correct on his second post. That method will give the first person a 50/50 chance, and everyone else will go free, regardless of how many hats of each color were used.


Hrm, since no one else has posted another riddle, let's go with another hat one!

You are one of a king's three wise men. He decides that he only wants to keep one wise man and calls the three of you in for a test. He seats you all in a circle so everyone can see the other two and proceeds to put a white or black hat on each person's head. You see the other two people are wearing black hats.

He then says "Raise your hand if you can see a black hat." Everyone raises their hand. He then says "Stand up if you know the color of your own hat." The other two wise men sit there scratching their heads, while you think for a minute, then get up and announce what color your hat is.

The question is: Are you wearing a white hat, or a black hat, and how do you know?
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Re: Riddle me this.. Riddle me that.

Postby Alacia » Mon Oct 03, 2011 12:18 pm

Assuming the other two really are somewhat wise, you're wearing a black one. If you'd been wearing a white one, the other two only would've been able to see a black one on top of eachother's heads, and since everyone raised their hands they'd know there's at least one other black hat - on themselves.

Don't really know any other riddles off the top of my head I'm afraid, might edit something in if I come up with anything.
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Re: Riddle me this.. Riddle me that.

Postby Xenix » Mon Oct 03, 2011 1:16 pm

k, seems the hat ones are getting too easy with so many in a row. Time for a change of pace:

You are blindfolded and sat down in front of a pile of 32 pennies. You are told that 11 of them have heads up, and your job is to divide them into two groups with the same amount of coins that have heads up in each group. So that you can't cheat and feel which side is heads, you have to wear gloves (and not put them in your mouth or anything like that), but you can pick up the coins, shake them in your hands, move them around, flip them over, stack them up, whatever.

How do you divide them into two groups with the same number of heads-up coins in each one?
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Re: Riddle me this.. Riddle me that.

Postby Passionario » Tue Oct 04, 2011 1:16 am

Xenix wrote:How do you divide them into two groups with the same number of heads-up coins in each one?


Pick any 11 coins, flip them, then move them to Pile 2.
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Re: Riddle me this.. Riddle me that.

Postby Xenix » Tue Oct 04, 2011 9:00 am

Correct.

All right - one last riddle, and then I'm done giving them out - I want to answer some, damn it. :D

Two roommates leave a bar at night, with one of them too drunk to drive. Both of them drove to the bar, though, and are parked in front of it. The first guy says "Well, let's just have you drive your car a block, then you walk back to my car and you drive it a block, you get out and go to your car and we repeat all the way home." The second guy says "Nah - I have a better idea. Let's just drive your car home, then I'll walk back to my car and drive it home. That way I only have to walk a third as far as if I did it your way."

The second guy was correct - how is that possible?
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Re: Riddle me this.. Riddle me that.

Postby lythac » Tue Oct 04, 2011 9:15 am

Xenix wrote:That way I only have to walk a third as far as if I did it your way."

The second guy was correct - how is that possible?

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