Riddle

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Posted: Fri Jul 20, 2007 2:37 am

Maybe another puzzle?

The top ot the picture is the floor plan of a bar 9x12 = 108 sq.yards minus the bar 1x8= 8 sq.yards total of 100 sq.yards floor space.
Bottom is a piece of high quality carpet 10x10 yards= 100sq. yards.
How can I make the carpet fit by cutting it in only two pieces?

Posted: Fri Jul 20, 2007 2:51 am

I don't know but I went to kitchen and found the biggest pair of scissors that I could find and for the life of me I don't think it will work. My PC is not a laptop or a flat screen. It is a fat screen. Confused

I'll get me jammies! Laughing

Posted: Fri Jul 20, 2007 8:20 am

Baiting Guru Joined: 04 Jan 2006 Posts: 3193

Quote:

You have two doors
One is labeled "The door to freedom is engraved with lies, the other with the truth."
The other is labeled "The door to freedom is engraved with the truth, the other with lies."

This sounds like the guards problem, but it is very different.

Neither door actually tells you which way freedom may be found. They merely make a statement about themselves and each other.

Our assumption is that someone is trying to help us to make a choice, and therefore that the labels on the doors are relevant. For them to be relevant they must impart information.

Consider if freedom lies behind the "lies" door. Then both doors can be paraphrased "I am lying, he is telling the truth." This is the "liars paradox". If I say I am lying, then either I am lying, in which case I am telling the truth, or I am telling the truth, in which case I must be lying, and so forth ad infinitum. Both doors are espousing the same paradox and impart no information.

Consider if freedom lies behind the "truth" door. Both doors can be paraphrased as "I am telling the truth, he is lying".

We first assume that the "truth" door is telling the truth. If so then it's first clause is true. It also says that the "lies" door is lying. If so then the statement on the lies door must be false, and that can be paraphrased as "the lies door is telling lies, the truth door is telling the truth". That is consistent with what we have been assuming.

We now assume that the "truth" door is telling lies. If so then it's statement is false, which can be paraphrased as "the truth door is telling lies, the lies door is telling the truth". That is the same as the statement on the lies door, so the lies door is telling the truth. That is consistent with what we have been assuming.

So whichever door is telling the truth, freedom is to be found behind the "truth" door.

Hello I'm New here! Joined: 09 Apr 2005 Posts: 6

angusmactavish wrote:

YouTube is your friend, 7x13=28!

YouTube is your buddy, 5x14=25

( Ma & Pa Kettle Math )

http://www.youtube.com/watch?v=Bfq5kju627c

Posted: Fri Jul 20, 2007 4:30 pm

@bill Am I allowed to fold the carpet before I make my cut?

Posted: Fri Jul 20, 2007 4:48 pm

Anything you wish for. Rules are, knife goes in and cuts to another side of the carpet, then both pieces should fit seamless as 100 sq yard=100sq yard Wink

You're close, good job, if you think this is too easy I got a real good one for you Laughing

Posted: Fri Jul 20, 2007 7:27 pm

I think this was basicly answered, but on the guards and the doors:

The problem is you don't know which guard lies and which tells the truth, so you have to ask a question that provides information on the doors, and the answer must be the same no matter which guard you asked.

Ask either of the guards which door the other guard would say leads to death, and take that door.

The truthtelling guard would say "he'll say that (life) door (cause he's a liar)"

The liar guard would say "he'll say that (life) door" because the truthtelling guard would have actually said the death door.

Alternately, you could ask either which door the other guard would say leads to life, and take the opposite door.

Master Baiter Joined: 04 May 2007 Posts: 139

And now, the solution. Drumroll please...
(apologies to the mods for the long post, but I want people to have a fair chance of not seeing it by accident)

Posted: Fri Jul 20, 2007 7:51 pm

Baiting Guru Joined: 04 Jan 2006 Posts: 3193

Posted: Fri Jul 20, 2007 9:43 pm

SumYunGai Very Good, but you blew it for the rest Smile

Never mind
wanna try a difficult one or do you have one for me?

Posted: Fri Jul 20, 2007 9:49 pm

^^^^^^

Some others who blew it for the rest:

Euclid
Galileo
Newton
Kepler
Einstein
Hubble
Heisenberg
Hawking
Monica Lewinsky

Why didn't they just shut up and let us keep guessing? Wink

Posted: Fri Jul 20, 2007 10:03 pm

LOL and your Escher gives me a headache, but the lego thing almost seems possible, photoshoped?

Master Baiter Joined: 04 May 2007 Posts: 139

Looks like a very good photoshop.

I'll take that harder one you had mentioned.

Posted: Fri Jul 20, 2007 10:12 pm

Posted: Fri Jul 20, 2007 10:14 pm

Sure?
You have twelve bags with gold dust, 11 have the same weight and 1 is heavier or lighter, we don�t know yet. You have a basic scale (seesaw=beam and two baskets) to weigh them, but you are only allowed to weight three times. Can you determine the faulty bag? And is it lighter or heavier? And don�t be too fast, I have to refigure this one out for myself again. Laughing

Master Baiter Joined: 04 May 2007 Posts: 139

Hmmm...

This would be easy if we knew that the faulty bag was heavier or lighter than the rest, or if there were only nine bags. Without knowing the relative weight of the faulty bag to the others you have to "waste" a weighing to determine this, by determining that two groups of bags balance and two more do not.

6v6 is right out as a first weighing; because you do not know whether the bag is heavier or lighter, weighing them all at once tells you nothing. Same for 5v5; if they balance you know it's one of the ones you didn't weigh and it's easy in two more weighings to identify your bag, but there's only a 1 in 6 chance the faulty bag isn't being weighed, and in the other cases you have "wasted" a weighing to only discount two bags.

OK, start by dividing the bags into three groups of four. Weigh group 1 and group 2. Then weigh group 2 and group 3. You now know 2 groups that balance, and one group that is heavier or lighter. You now know that the faulty bag is heavier or lighter and is in the group that didn't balance. For example:

1=2
2>3

You now know that the faulty bag is in group 3 and is lighter than all the others. However, you now have only one weighing left and four possibilities. The best worst-case is that we narrow it down to 2 bags; we have a 50-50 shot of knowing exactly which bag is off-weight by weighing any two of them, and if they don't balance you have found your faulty bag. But if they balance, the best you can do is know it's one of the two you didn't weigh and whether the faulty bag is heavier or lighter.

So you are guaranteed to know the relative weight of the faulty bag, but only a 50% chance of identifying it; the rest of the time you know it's one of two.

Let's try four groups of 3 along similar logic, just for grins: weigh 1 and 2, then 2 and 3. If either of those misbalance, you're home free; you've identified a group of three, not four, that has the faulty bag, and you know if it's heavier or lighter. You then use your last weighing to weigh two of those three bags, and if they balance, it's the one you didn't weigh, otherwise it's the heavier or lighter of the ones you did. You have a 75% chance that the faulty bag is in the nine you weigh on the first two weighings.

However, if all three groups you weigh balance, on the plus side you've narrowed your search to the three bags in the unweighed group, but you now have only one weighing and do not know if the faulty bag is heavier or lighter. If two of them balance, the faulty bag is the unweighed one and you do not know if it is heavier or lighter. If the two bags you weigh misbalance, it's one of those two but you can tell neither which it is nor the relative weight.

So, this method gives you a 75% chance of knowing the entire answer, but failing that you only have a one-in-three shot at identifying the faulty bag and no chance of knowing its relative weight.

Hmm. you're right; this IS a toughy. While I'm working on it, here's one back... (next post)

Master Baiter Joined: 04 May 2007 Posts: 139

This is an easier one than the one you gave me:

After an art theft, police are interviewing 6 suspects. The police are sure it was one of these six people. Below are summaries of the statements recorded by each suspect:

Alan said that it wasn't Brian, Dave or Eddie.

Brian said that it wasn't Alan, Charlie or Eddie.

Charlie said that it wasn't Brian, Freddie or Eddie.

Dave said that it wasn't Alan, Freddie or Charlie.

Eddie said that it wasn't Charlie, Dave or Freddie.

Freddie said that it wasn't Charlie, Dave or Alan.

The officers questioning the suspects know that four of the suspects told exactly one lie each and that all other statements are truthful. Who is the art thief?

Baiting Guru Joined: 04 Jan 2006 Posts: 3193

Where I've got to so far.

To determine which of 4 bags is faulty in two weighings given a known good bag.

Weigh 1+2 vs 3+good
If the same then weigh 4 vs good
If different weigh 1 vs 2, you know from the previous step if the fault is light or heavy. If they are the same then 3 is the faulty bag.

So we only need to weigh 8 out of the 12 in the first step. Similarly you can remove two bags before the second weighing and another one before the third, (replacing it with a one you removed previously, which is now known good.)

However that still leaves five bags to distribute over three weighings, such that each bag has been on the same side and on the opposite side from every other bag in at least one weighing. I haven't found a suitable arrangement yet and I'm not sure it's possible.

Posted: Sat Jul 21, 2007 12:51 am

It is possible to solve all possible solutions by logic, there is an answer, but first I'll try the one SumYunGai posted.
Thanks guys, I like impossible problems Laughing

Edit:
SumYunGai I'm sorry, I'll download the picture and study it, if you manage to do it in Lego you're a wizz

Baiting Guru Joined: 04 Jan 2006 Posts: 3193

I realised my mistake after I'd gone to bed. You can remove three bags before the second weighing.

Weigh 1 2 3 4 vs 5 6 7 8
If unequal, label all bags on the light side L, and on the heavy side H
If equal proceed as in my previous post with bags 9-12, using bag 1 as a known good weight.

Otherwise bags 9-12 are now known good and can be used to make up weight.

Weigh 2 4 5 vs 1 3 9
If unequal label the light and heavy bags as before
If equal weigh 6 vs 7, if unequal then the faulty bag is the one whos label agrees with this weighing. If equal then 8 is faulty and its label says how.

Weigh 1 2 vs 3 4
If unequal then label all bags as light or heavy.
If equal then bag 5 is faulty and its label says how

Otherwise only one bag will be labeled HHH or LLL; that is the faulty bag.

Posted: Sat Jul 21, 2007 4:04 pm

Here's one for all the mathematicians. I don't remember the answer for this and I'm hoping someone can remind me. It's a great bar trick and you can use it to win free drinks all night long. The problem: Lay out three rows of coins thusly

00000

0000000

00000

5 in the top, 7 in the middle, 5 in the bottom.

The challenge is to tell your vic, uh, opponent that you will either pick up the last coin(s) or force him/her to pick up the last coin. (it works either way). The rules are that each player must pick up either one, two or three coins in each turn, but they must be touching. (coins touch horizontally only, not vertically). Either player can go first. The secret is there is a simple mathematical formula that tells you how many coins to pick up. You plug in data from the opponents last move and get an infallible answer. A co-worker and I spent the afternoon once deriving the formula. We NEVER lost this challenge. Try as I may, I can't seem to get it again. This was 25 years ago! Anybody want to try to derive the formula?

OK, so this is less a puzzle and more a scam. Embarassed

But I'd still like to remember how it's done.

Posted: Sat Jul 21, 2007 4:39 pm

@thud419
Very good and close, you have a PM in a sec.

@angusmactavish

00000 0000000 00000
Each oponent picks up (connected)coins max three at once and he who takes the last one is the looser. ?
Should be fun Laughing

Thanks

Edit:
Let the other party start and make sure that per turn 4 coins are gone after four turns he is left with 1 coin [4 x 4 = 16 total 17 coins]

Baiting Guru Joined: 04 Jan 2006 Posts: 3193

Nah, I am there. My method works just as well as yours. I will of course eat humble pie if you can find a scenario my method gets wrong.

One for the mathematicians. There is no simple answer and I don't know the answer, but if you want to try it's a good work-out and you can be justly proud if you find the answer.

You have a circular lawn. You wish to tether the goat to one edge of the lawn so that he can eat exactly half of the grass. How long should the chain be in terms of the radius of the lawn? You can make the standard assumptions, such as that the goat is infinitely small and so forth.