• Welcome to Drummer Cafe Community Forum.

Situations: The Game

Started by Larry Lawless, January 01, 2007, 06:37 PM

Previous topic - Next topic

0 Members and 1 Guest are viewing this topic.

Dave Heim

Quote from: Bart Elliott on January 03, 2007, 02:02 PM
Although I don't think Cupid will enjoy having that red light shining up his backside. What was Santa thinking?!

Prancer has the best spot.  For everyone else, the view doesn't change much.  :)

Larry Lawless

OK, I started this puppy cause I was bored on New Year's Day. Let's see if I can end it with this:

Not a situation, but another riddle.

ALBERT EINSTEIN'S RIDDLE

ARE YOU IN THE TOP 2% OF INTELLIGENT PEOPLE IN THE WORLD? SOLVE THE RIDDLE AND FIND OUT.

There are no tricks, just pure logic, so good luck and don't give up.

1. In a street there are five houses, painted five different colours.
2. In each house lives a person of different nationality
3. These five homeowners each drink a different kind of beverage, smoke different brand of cigar and keep a different pet.

THE QUESTION: WHO OWNS THE FISH?

HINTS

1. The Brit lives in a red house.
2. The Swede keeps dogs as pets.
3. The Dane drinks tea.
4. The Green house is next to, and on the left of the White house.
5. The owner of the Green house drinks coffee.
6. The person who smokes Pall Mall rears birds.
7. The owner of the Yellow house smokes Dunhill.
8. The man living in the centre house drinks milk.
9. The Norwegian lives in the first house.
10. The man who smokes Blends lives next to the one who keeps cats.
11. The man who keeps horses lives next to the man who smokes Dunhill.
12. The man who smokes Blue Master drinks beer.
13. The German smokes Prince.
14. The Norwegian lives next to the blue house.
15. The man who smokes Blends has a neighbour who drinks water.

ALBERT EINSTEIN WROTE THIS RIDDLE EARLY DURING THE 19th CENTURY. HE SAID THAT 98% OF THE WORLD POPULATION WOULD NOT BE ABLE TO SOLVE IT.


Yes, I got this from a google riddle search, and you can google it and find the answer for yourself. But if you really want to stretch your brain today, try to solve it without help. I did, but it took me about an hour, using a grid to block things out.

Bart Elliott

Quote from: L Lawless on January 03, 2007, 02:48 PM

THE QUESTION: WHO OWNS THE FISH?


The German, in the Green house who smokes Prince and drinks coffee owns the fish.


The way I figured this out was to take small pieces of paper, writing down absolutes, such as "the bird owner also smokes Pall Mall", and began arranging them in columns and rows. I positioned the absolutes at the top of each column then began working the variables.

Took me around 30-40 minutes to complete using this method. It took me only a few seconds to realize that people of European decent tend to have a lot of bad vices!  ;) (I'm kidding of course)

Bart Elliott

QuoteALBERT EINSTEIN WROTE THIS RIDDLE EARLY DURING THE 19th CENTURY. HE SAID THAT 98% OF THE WORLD POPULATION WOULD NOT BE ABLE TO SOLVE IT.

Although it feels nice to complete the riddle, and as smart as Einstein was, I don't think he was right about 98% of the world population not being able to solve it. Perhaps I'm wrong, but I think just about anyone could solve the riddle if given enough time. Maybe he meant that only 2% of the world's population could solve this in their head (not using paper, graphs, etc.).

Okay ... bonus question ... which pet ate the missing dollar?

JeepnDrummer

Quote from: Bart Elliott on January 03, 2007, 05:05 PM
Okay ... bonus question ... which pet ate the missing dollar?
The monkey that lives in the white house?  :)

Steve Phelps (Shoeless)

My favorite of the Car Talk Puzzlers:

A warden meets with 23 new prisoners when they arrive. He tells them, "You may meet today and plan a strategy. But after today, you will be in isolated cells and will have no communication with one another.

"In the prison is a switch room, which contains two light switches labeled A and B, each of which can be in either the 'On' or the 'Off' position. I am not telling you their present positions. The switches are not connected to anything.
 
"After today, from time to time whenever I feel so inclined, I will select one prisoner at random and escort him to the switch room. This prisoner will select one of the two switches and reverse its position. He must move one, but only one of the switches. He can't move both but he can't move none, either. Then he'll be led back to his cell.

"No one else will enter the switch room until I lead the next prisoner there, and he'll be instructed to do the same thing. I'm going to choose prisoners at random. I may choose the same guy three times in a row, or I may jump around and come back.

"But, given enough time, everyone will eventually visit the switch room as many times as everyone else. At any time anyone of you may declare to me, 'We have all visited the switch room.'

"If it is true, then you will all be set free. If it is false, and somebody has not yet visited the switch room, you will be fed to the alligators."

Here's the question:

What is the strategy the prisoners devise?



I have hints...


Larry Lawless

Quote from: Bart Elliott on January 03, 2007, 05:05 PM
Although it feels nice to complete the riddle, and as smart as Einstein was, I don't think he was right about 98% of the world population not being able to solve it. Perhaps I'm wrong, but I think just about anyone could solve the riddle if given enough time. Maybe he meant that only 2% of the world's population could solve this in their head (not using paper, graphs, etc.).

Okay ... bonus question ... which pet ate the missing dollar?

well done, Bart, and I agree. I don't think it's as much a matter of intelligence as perseverence. Most people would just give up too easily.
http://www.manbottle.com/popular/einstein_s_riddle/einstein_s_riddle_answer

Dave Heim

I need to wrap my brain around this more.  I'll do that in the morning.  But in the mean time I guess the obvious deductions (at least for me) are these:

The prisoners, once they have their one and only strategy meeting, can really only communicate via the settings of the switches.  So they'd need to agree on some sort of switch configurations that tell them things.  Exactly what, I haven't thought out yet.

Having "anyone of you" just declare that they've all visited the switch room is really risky, so it would make sense to also agree, at that first meeting, which of them will do the declaring.  It'd have to be someone they all trust - and that's tricky.  After all, they're all prisoners - in the slammer for something. . .  but its late and I digress.

That's my baseline thought.  I'm working from there.


Steve Phelps (Shoeless)


Louis Russell

This is easy.  They elect a counter or spokesman.  When a prisoner comes into the switchroom for the first time they make a small scratch beside the switches. When the counter/spokesman counts 22 then evenyone has been in the room.

Robyn

Quote from: JeepnDrummer on January 03, 2007, 06:48 PM
The monkey that lives in the white house?  :)

Ooooo!!  :oGood answer!!  ;D

robyn

Dave Heim

Quote from: Louis on January 03, 2007, 10:35 PM
This is easy.  They elect a counter or spokesman.  When a prisoner comes into the switchroom for the first time they make a small scratch beside the switches. When the counter/spokesman counts 22 then evenyone has been in the room.

The only problem I see with this, Louis, is that the spokesman may only get into the switch room once.  There's no guaruntee he'll be brought back in there at the end to count up the scratches.

Louis Russell

Quote from: Dave From Chicago on January 04, 2007, 06:57 AM
The only problem I see with this, Louis, is that the spokesman may only get into the switch room once.  There's no guaruntee he'll be brought back in there at the end to count up the scratches.

No, the riddle stated that everyone would get to the switch room the same number of times, given enough time.  There is also another way but would take a lot more visits.  They still elect a spokesman/counter.  Select one switch as the count switch and each person can only move it twice and only when it is off.  When the counter finds it on he turns it off and increments his count one.  The problem is that no one knows the initial position of the switch and the counter may not be the first one in.  This way when the count reaches 42 everyone has been there at least once.

Dave Heim

Quote from: Louis on January 04, 2007, 08:25 AM
No, the riddle stated that everyone would get to the switch room the same number of times, given enough time.  There is also another way but would take a lot more visits.  They still elect a spokesman/counter.  Select one switch as the count switch and each person can only move it twice and only when it is off.  When the counter finds it on he turns it off and increments his count one.  The problem is that no one knows the initial position of the switch and the counter may not be the first one in.  This way when the count reaches 42 everyone has been there at least once.

Yes.  I misspoke.  Thanks for the correction.  Seems to me, though, that its not a sure thing that the counter person will be brought in at "the end". 

Louis Russell

Quote from: Dave From Chicago on January 04, 2007, 09:06 AM
Seems to me, though, that its not a sure thing that the counter person will be brought in at "the end". 
Since the counter is the one who would end the thing he would always be the one brought in at the end.  All the counter has to do is wait until everyone else has made their scratch then wait until he comes in again.  Actually this way they do not need a counter, everyone counts scratches and when a prisoner makes a mark that totals 23 he could speak up.

Steve Phelps (Shoeless)

The prisoners all meet, and the leader of the prisoners says, "Okay, guys, here's our strategy. First, there's only one guy who can count past two, so we're naming him 'The Counter.' He's going to be responsible for telling the warden we've all been in the switch room when the time comes."

He then proceeds to give instructions to the other inmates. He says, "We're going to designate Switch A -- the switch on the left - as the "real switch." That's the only switch that matters to The Counter. The other switch, Switch B, is a dummy. It won't tell us anything, and you just use it when you have to move a switch, but don't want to move Switch A. You got it? So Switch A is the meaningful switch and Switch B is a placeholder."
   

So, each of the 22 prisoners is told, "When you go into the switch room, we want you to move Switch A to the "On" position. If Switch A is already in the "On" position, then leave it there, flick switch B and walk out." All the prisoners nod.
   

Now I want each of you to flick Switch A to the "On" position twice, and only twice. So if you go in there and Switch A is already on, that doesn't count. I want each of you to actually flick it "On" two times. You got that? "
   

All the prisoners nod. One of them raises his hand, tentatively.

"Yes, Berman?"

"Who's going to be flicking Switch A off, he asks.

"Good question," says the leader. "The Counter is the only one with the authority to turn off Switch A."
   

So, each time The Counter is taken into the switch room, finds the switch in the "On" position, he knows that at least one prisoner has been in there.
   

It could be one prisoner who came in and turned it on, or it could be six prisoners -- the first one turning it on and the next five leaving it on. But when The Counter walks in and finds Switch A in the "On" position, he knows at least one prisoner has been in the room since the last time The Counter turned the switch off.
   

And when you work it all out, The Counter has to turn off Switch A 44 times in order to know that all 23 prisoners have been in the switch room. And the reason he has to count that high is that he doesn't know what the original position of the switch is, and therefore he has to wait for everyone to go in twice.

In other words, if the warden started with Switch A in the "On" position, and The Counter was brought in first, he could be fooled into thinking that another prisoner had been in there. And that's why it's 44 instead of 22. .

If you have trouble understanding the strategy, try it for two prisoners -- Prisoner 1 and Prisoner 2.
   

Let's say Prisoner 2 is The Counter. Prisoner 1 is brought in first. He sees Switch A is in the "Off" position, so he flicks it on. Let's say Prisoner 1 is brought in again. Remember -- the warden picks prisoners randomly. So Prisoner 1 leaves Switch A on and flicks Switch B, which doesn't matter. At some point, The Counter comes in, finds Switch A on and flips it off. That's once, he says. Eventually, Prisoner 1 is brought back in, and he turns Switch A on again. When Prisoner 2 comes back, he sees Switch A on, and now he knows that the other prisoner has been there. Last time he was there, he personally turned off Switch A, and now it's back on, so he knows the other prisoner did it, and he says, "Warden, our work here is done."



And interestingly, the number of times The Counter has to turn Switch A off is exactly twice the number of the other prisoners -- not counting him -- in the group. So, for two prisoners, it's two times. For 23 prisoners, it's 44 times.

Bart Elliott

Quote from: JeepnDrummer on January 03, 2007, 06:48 PM
The monkey that lives in the white house?

Oh, you mean the monkey with the cigar?
No, he doesn't live in the white house anymore.  ;D

Dave Heim

Works for me.

But if these guys were that smart, they wouldn't have been caught in the first place!  :)