Brain teaser thread

[Guest]

ok no "real" posted for a long time
I like this one:
"In a rectangular array of people, which will be taller, the tallest of the shortest people in each column, or the shortest of the tallest people in each row?"
(with an explanation plz as there are only 2 choices)

Im going to sleep within the hour, so if you feel relatively confident in your answer, or by popular demand, post your own...

[One_Brow]

Originally posted by LulThyme
ok no "real" posted for a long time
I like this one:
"In a rectangular array of people, which will be taller, the tallest of the shortest people in each column, or the shortest of the tallest people in each row?"
(with an explanation plz as there are only 2 choices)

Im going to sleep within the hour, so if you feel relatively confident in your answer, or by popular demand, post your own...

The shortest person amon the tallest peope from each row will be at least as tall as the tallest person among the shortest person in each column.

Let's say we have m rows and n columns. We will use "i" to denote rows and "j" to denote columns. Each position is identified by (i,j).

Definitions:
t(i,j) is the height of the person in (i,j).
maxr(q) is the largest value of any of t(q,j) for a specific q.
The min-max is smallest value of maxr(q).
minc(q) is the smallest value of any of t(i,q) for a specific q.
The max-min is the largest value of minc(q).
Note the answer above is saying that the max-min <= min-max.


Proof:
We must have some individual (m,n) where the min-max = maxr(m) = t(m,n). This means that for all j, t(m,j) <= t(m,n)

Assume max-min > t(m,n). This means there exists some p such that minc(p) > t(m,n). Thus, there exists some p such that for any i, t(i,p) > t(m,n). In particular, t(m,p) > t(m,n). This is a contradiction.

Thus, the max-min <= min-max.

[One_Brow]

I don't reaaly have time to look up one right now, so here's a quickie:

You start of game of Civ3 with 7 opponents, all civs chosen randomly (including yours), no cxultural groupings (so every Civ has an equal chance of appearing independently of the others).

Without regard to which civ is yours, how many different combinations of civs are possible in the original (16 civs) and in PTW (24 civs)?

[Winston]

Those last two sound more like high school assignments than brain teasers.

What's the matter, dwarfs with multi-coloured hats not good enough for ya' anymore?

[Zero-Tau]

Originally posted by One_Brow
You start of game of Civ3 with 7 opponents, all civs chosen randomly (including yours), no cxultural groupings (so every Civ has an equal chance of appearing independently of the others).

Without regard to which civ is yours, how many different combinations of civs are possible in the original (16 civs)?

16*15*14*13*12*11*10*9/(8*7*6*5*4*3*2*1) = 12870

Simple mental arithmetic...

and in PTW (24 civs)?

24*23*22*21*20*19*18*17/(8*7*6*5*4*3*2*1) = 735471

Needed a calculator for this one...

[Guest]

OK the another fun one
family
In Apolytonia, boys and girls are as likely results of a pregnancy. All males are called Ming.
There is also a custom that states, that every family must have exactly one Ming, so each family has kids until they have exactly one boy which they call Ming.
IN the long run, what will be the ratio of Ming to Non-Mings in Apolytonia.

[Dauphin]

50:50

I did the maths too, but I'll let someone else show their summation of probability dens... yada yada.

summation [x * 0.5^x+1] for x =0 to infinity

[Zero-Tau]

Originally posted by Sagacious Dolphin
50:50

I did the maths too, but I'll let someone else show their summation of probability dens... yada yada.

summation [x * 0.5^x+1] for x =0 to infinity

Actually, there's a much simpler argument: Each birth has a 50% chance of being a boy, and a 50% chance of being a girl. It stays this way, regardless of whether some families stop having children. Therefore the answer is 1 to 1.

[Guest]

Yes to both of you...
Of course Zero-Tau's argumetn is much nicer...
No kids get discarded, all newborns are kept, and there are 50% of each... why would the ratio be any different in the population?
This one was easy but it was fun to check if someone would post the easy way or hard way first...
I guess you want another one?

[Dauphin]

It wasn't that hard a way. I did it whilst drunk.

Plus it is very similar to puzzles involving coin tosses for which the answers are pretty much imbedded in my brain.

[Guest]

I did say it was sorta of easy and I can sum geometrical series in my head too...
But being a math major myself, I like trying to ask puzzles that anyone can do, and sometimes layman even do better on...
Ok then how about this one...

Sagacious Dolphin has just made a daring escape from Mingapulco. If he can just reach the secret "Rebel Against Ming the Tyran" (or wasnt I suppose to mention it?) base at the North Pole, he surely will be safe. Mingapulco is of course on the equator at an undisclosed location. He decides he has better chance of evading would be pursuers by heading straight NorthWest for the duration of the trip.
What is the length of his path?
Assume a spherical Earth and that the distance from the equator to the North Pole is about 20000km.

[Dauphin]

[20,000² + 20,000²]^0.5 = ~28,300km

Better start now, may take a while.

[Zero-Tau]

Originally posted by Sagacious Dolphin
[20,000² + 20,000²]^0.5 = ~28,300km

Better start now, may take a while.

Not that simple, I'm afraid. That formula only works for a plane, not for a sphere. The actual distance would be shorter. Not that I know the correct formula, though...

Oh and LulThyme, what planet are you living on? On mine, the distance from equator to the North Pole is about 10000 km, 1/4 of the circumference of the entire planet, 40000 km...

[Guest]

Of course the actual number used is not very important...
But you are correct...
Oh maybe Mingapulco is not where we think it is

Now about Dolphin's answer.
At any point on Earth, if your speed NW is Speed, then your Speed North IS Speed / square root of 2
If you think about it, the fact that globally you are travelling on a sphere makes no difference.
So dolphin's answer is correct.
Whats interesting, is that you will circle around the pole an infinite amount of times, but the length of the path and thus the time take is still finite (a sort of round version of Zeno's paradox I would say).

Go ahead Master Escapee, your turn

[Dauphin]

is that you will circle around the pole an infinite amount of times

My alternative answer to your question would have involved Archimedian spirals, but the maths was less easy.

Let me find one and tailor it.