Easy... yet completely and utterly wrong. With that "solution" you only save 50% on average and, worst case, can guarantee to save only one smurf.
And that was also wrong. Sorry.
brain teaser thread
- Thread starter Fred
- Start date
Of course. The wheels aren't driving the aircraft.
Very simple one:
At the end of a very long and winding corridor, there is a light bulb. You are standing in front of a switchboard with three switches (all in the off position). You can't see the light from where you are standing.
You know one of these switches turns on the light but only want to make one trip down the corridor and back to identify which is the right switch. What do you do?
At the end of a very long and winding corridor, there is a light bulb. You are standing in front of a switchboard with three switches (all in the off position). You can't see the light from where you are standing.
You know one of these switches turns on the light but only want to make one trip down the corridor and back to identify which is the right switch. What do you do?
Light bulb answer:
You turn on the first switch and wait a while. Then you turn off the first switch and turn on the second switch. Walk to the end of the corridor. If the light is on, it was the second switch. If the light is off, touch the bulb. Warm bulb = first switch, cold bulb = third switch.
There goes Kyoto out the window...
Yep.
BTW I did the 31 logicians puzzle a long time ago and have yet to go back and solve it again. I seem to recall there was an assumption I had to make... but can't remember what it was.
Bigus Dickus
Regular
does the "odd numbers" solution only work if there are 50 red and 50 blue hats to begin with?
No, it works whatever the distribution.
Think of it this way: each smurf only needs to know the even or odd "count" for each color to make his decision.
Ex: distribution unknown:
The smurf 100 calls out the odd color he sees before him - blue. All each smurf in line needs to do then is keep up with is the even or odd count for blue.
For example, say it is smurf 47's turn and blue has been called an odd (x) number of time previously. Smurf 47 then counts how many blue hats he sees before him. If y is odd then x+y=even. Since he knows there must be an odd number of hats, he knows his hat is blue. Likewise, if y=even, then x+y=odd, and he knows his hat can not be blue.
Ex: distribution unknown:
The smurf 100 calls out the odd color he sees before him - blue. All each smurf in line needs to do then is keep up with is the even or odd count for blue.
For example, say it is smurf 47's turn and blue has been called an odd (x) number of time previously. Smurf 47 then counts how many blue hats
he sees before him. If y is odd then x+y=even. Since he knows there must be an odd number of hats, he knows his hat is blue. Likewise, if y=even, then x+y=odd, and he knows his hat can not be blue.
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Making money, the Gödel way
You have two offers:
Offer A: "Make a statement. If it's correct you'll get $1000. If not you'll get some amount (still money) other than $1000."
Offer B: "Make a statement. You will get something higher than $1000 independent of the truth value of that statement"
Which would you prefer?
You have two offers:
Offer A: "Make a statement. If it's correct you'll get $1000. If not you'll get some amount (still money) other than $1000."
Offer B: "Make a statement. You will get something higher than $1000 independent of the truth value of that statement"
Which would you prefer?
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If I understand you correctly...
.. you are offering three options.
Seems that the only sensible choice is 3
- A+ Truth => $1000
- A+ False => something else. That could be $100000, a piece of string, or it could be death!!
- B => more than $1000
Seems that the only sensible choice is 3
Well I wasn't clear obviously. Whatever you pick, you always get some cash (no death, no flowers, no love).
So if you pick offer A and tell a lie (e.g. 2+2=5) you still get some money but it won't be $1000. Can be $1 or $10000.
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But what are the distributions of this versus that of B? I mean, if the A+lie choice is heavily biased towards really huge amounts but B is biased towards being just greater than $1000 then the former would be preferable. <shrug>. I can't see that there is enough information.
Since I don't understand what betan is on about I'll post another which I found quite clever.
Whilst blindfolded, you're presented with a tray of 100 coins. You're told that half the coins show heads the other half tails, and that the distribution is completely random. If you can separate the coins into two groups, each with an equal number of coins showing heads, then you get to keep the lot. How can this be done?
I'll post another which I found quite clever.Whilst blindfolded, you're presented with a tray of 100 coins. You're told that half the coins show heads the other half tails, and that the distribution is completely random. If you can separate the coins into two groups, each with an equal number of coins showing heads, then you get to keep the lot. How can this be done?
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I believe the information is enough (as teasers go) and one can argue that the solution is about the lack of information.
Here is another logical teaser that's slightly different but can be considered as a hint (no answer):
King asks the prisoner: "Tell me something, if it's true you'll be decapitated, if not you'll be hanged".
How can the prisoner survive?
How can the prisoner survive?
Since I don't understand what betan is on about
Whilst blindfolded, you're presented with a tray of 100 coins. You're told that the coins either show heads or tails, and that the distribution is completely random. If you can separate the coins into two groups, each with an equal number of coins showing heads, then you get to keep the lot. How can this be done?
Now that looks like an incomplete question.
Don't you need to tell us how many of the coins are showing heads or something?
Don't you need to tell us how many of the coins are showing heads or something?
Ahh... I think I can see that there are other options, but I have not yet worked out how to get it to my advantage.In fact, I think, so far, I've just caused deadlock :| e.g. Statement "You will NOT give me $1000" and choose option A
so close
Ah... Now that you've edited your original clues, I can solve it.A fair point.
Separate the 100 coins into two, completely arbitrary, sets of 50 each, A and B.
We know
T(A) + T(B) = 50 (eqn 1)
and T(A) + H(B) = 50 (eqn 3)
3=> T(A) = 50 - H(B)
so
1 => 50 - H(A) + T(B) = 50
=> H(A) = T(B)
i.e. number of Heads in A is the same as the number of Tails in B.
Simply turn over all the coins in set B.
We know
T(A) + T(B) = 50 (eqn 1)
and T(A) + H(B) = 50 (eqn 3)
3=> T(A) = 50 - H(B)
so
1 => 50 - H(A) + T(B) = 50
=> H(A) = T(B)
i.e. number of Heads in A is the same as the number of Tails in B.
Simply turn over all the coins in set B.