brain teaser thread

ninelven said:
Sure it does. Otherwise I could land infinitely many helicopters simply by using a shape without a center (imagine a picture frame for example) and making infinitely many infinitely smaller replicas surrounding it. Thus, if I choose a shape without a center and also get to manipulate the size of the individual heliports, I can land as many as I choose.
Click to expand...

Since I'm afraid of derailing this nice thread, I'll say only this much:
A heliport without a center is not nearest to any other because distance is from center to center.
Any finite shape (including a picture frame) has a center, it may just not be on it.

So the answer for your particular example is still the same.
 
betan said:
A heliport without a center is not nearest to any other because distance is from center to center.
Click to expand...
You didn't mention this restriction in the original problem...

betan said:
Any finite shape (including a picture frame) has a center, it may just not be on it.
Click to expand...
I know.... but the helicopter can not take off and land on "nothing." The 2d plane is restricted (let's say it is an ocean), the helicopter can not take off from or land on the "ocean" only the heliports. Thus, the center is irrelevant in this case. You must either restrict the shape (to ones with a real center) or the size.

betan said:
So the answer for your particular example is still the same.
Click to expand...
No. I can draw a picture if you would like, just name any practical number greater than
6
.

Either way, if you just say all heliports have to be the same shape and size, then you don't even need to measure from the center. This makes the problem less obvious IMHO, and is a better restriction to give the reader than "distance is from center to center" which makes the solution apparent.
 
Zaphod said:
In a similar vain:

"To be or not to be?", that is of course the question. What is the logical answer stated in decimal?
Click to expand...
:)
I'm assuming unsigned and bitwise or...

Infinity, unless of course you are implying a 32-bit register.
 
Simon F said:
:)
I'm assuming unsigned and bitwise or...

Infinity, unless of course you are implying a 32-bit register.
Click to expand...
You're on to something, but that's not the answer I'm looking for.
8 bits in a byte. We're old school here.
 
@Gerry: Yep. :yes:

PS: Please
ify your answer.

Solution:

Code: 
To be or not to be?", that is of course the question.

The question is 2B||!2B
-----------------------
HEX 2B in BIN  00101011
 (bitwise NOT) 11010100
-----------------------
  (bitwise OR) 11111111
=======================
In HEX               FF
In DEC              255
 
Yes, surely the correct answer is
-1 (2's complement), since otherwise the result would vary with computational width (8, 16, 32 etc. bits)

Cheers
 
Last edited by a moderator:
Gubbi said:
Yes, surely the correct answer is
Cheers
Click to expand...
The question really is too open in that....
  • If it is being done with "numbers base 2" then ~2B has an infinite number of 1 bits set, in which case, where is the sign bit?
  • If we are limiting ourselves to a particular (undefined) word length, then the value could also be an unsigned word, in which case the answer could also me INT_MAX.
 
Simon F said:
Then you should specify it more carefully. Some of us are no longer using 6502s and Z80s :p
Click to expand...
Well, the riddle/joke probably is that old... And, sure, it is too open in that
"To be" isn't the same as "two-B". :) I can't think of a good way to make it unambiguous without also making it obvious.

As for the Halloween and Christmas one, the reason is of course that:
Geeks think that OCT[31] == DEC[25].
 
Game theory

A TV show has a simple coin tossing game.

The host tosses a fair coin until heads come.
If the heads come at the nth throw, the player gets 2^n credits, and the game ends.**

Among a number of contestants, only highest paying one can play.

How much credits at most would you pay to enter that game?
Would your answer change if you could play the game as much as you want, given that you pay that amount per game?
What would your answers be if the reward was (2^n)/n credits instead of 2^n?

** 2^n="2 to the power of n"
 
Simon F said:
Of course. The wheels aren't driving the aircraft.
Click to expand...

If you define the speed of the wheel as angular velocity times 2pi*radius it doesn't matter how you generate thrust, you force the aircraft to remain stationary as long as the wheel doesn't slip. This because any forward movement requires the "speed" of the wheel to be higher than the conveyor which contradicts the problem description.

The moment of inertia of an aircraft wheel is small but non-zero; the acceleration of the conveyor will be obscene.

If the wheel slips it is a race against time between wheels being ripped to shreds by a super-sonic conveyor or the airplane managing to take off.
 
betan said:
A TV show has a simple coin tossing game.
Click to expand...
Without a lot of thought:
It seems to me that with a reward of 2^n and a chance of success at 1/2^n the highest rational bet would be 0.99 to as many decimals as you might prefer.
 
Zaphod said:
Without a lot of thought:
It seems to me that with a reward of 2^n and a chance of success at 1/2^n the highest rational bet would be 0.99 to as many decimals as you might prefer.
Click to expand...

The probability of success is 1, so to speak.

Player gets at least 2 credits.

Once you enter, it's a passive game for you, that is no bet on n.
Let's say, the sequence came is: t, t, t, h
Then you get 2^4=16 credits.
For t t t t h, you get 2^5, and so on.

Probabilistically the avarage or expected money player will get is:
1/2 * 2+ 1/4 * 4 + 1/8 * 8 + 1/16 * 16 ...
which is infinite.
 
betan said:
The probability of success is 1, so to speak.
Click to expand...
Right. That's what I get for reading to fast and making assumptions. (For some reason I read the rules as having the possibility of cashing out as long as the coin lands tails with heads being a game-ending failure.)
betan said:
Probabilistically the avarage or expected money player will get is:
Click to expand...

Not in your original formulation. The St. Petersburg paradox requires the player to be able to play infinite games as well (not to mention that the host must have infinite resources). To make a rational assumption as to the value of an individual game in the real world, the player must either know the number of iterations or the hosts maximum bankroll.
 
Zaphod said:
Not in your original formulation.
Click to expand...
Probabilistically, what do you think the expected outcome in my formulation, ?
The
St. Petersburg paradox
requires the player to be able to play infinite games as well (not to mention that the host must have infinite resources).
Click to expand...
Infinite resource is implied in my question. Number of games is either 1 or unbounded but not infinite of course.
To make a rational assumption as to the value of an individual game in the real world, the player must either know the number of iterations or the hosts maximum bankroll.
Click to expand...
This is a real world game, played in an open economy environment (referring to infinite resource part).
And my formulation stands still.
 
Then there is no answer. A formulation with only one game has the risk and utility increasing in step so that the player's utility expectation will be a marginal one. In an iterated version where then number of games is not infinite, a rational player must make his bid based on the expected utility of the number of games he intends to play.
 
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