UKC

I would like some new puzzles please.

As a swap, I will offer...

You have two fuses. Each has a guaranteed burn time of 1 hour. The fuses burn at non-uniform rates, in some places they burn fast, others slow. This means that you cannot just cut one in half to get a half-hour fuse.

So, with these two fuses and some matches, how can you measure 45 minutes?

In reply to lowersharpnose:

Don't have any good ones on top of my head but have the answer to yours (I think!)

There's one with switches and light bulbs in a room - know that? If not I can dig I out..

In reply to Simos:

I know that one. Three bulbs in a room connected to three switches outside the room. How can you work out which switch works which light if you are allowed only one visit to the room?

In reply to lowersharpnose:

Yes that one. I am assuming you know that old one with the tv show and the 3 doors? (opens can of worms)

In reply to Simos:

Sure do.

In reply to lowersharpnose:

Here's one I just got from google:

Three gods A, B, and C are called, in no particular order, True, False, and Random. True always speaks truly, False always speaks falsely, but whether Random speaks truly or falsely is a completely random matter. Your task is to determine the identities of A, B, and C by asking three yes-no questions; each question must be put to exactly one god. The gods understand English, but will answer all questions in their own language, in which the words for yes and no are da and ja, in some order. You do not know which word means which.

In reply to Simos:

each question must be put to exactly one god

Could I ask the first of A, the second of B etc. - or do all questions have to be asked of the same god?

In reply to lowersharpnose:
Yes I am assuming you can ask any god any question, as long as a question is addressed only to one god

Ps I think it's REALLY hard...

In reply to lowersharpnose:
You light one fuse at both ends, and at the same time you light the other fuse at only one end. When the fuse burning at both ends is finished, 30 mins has gone by. At that point you light the remaining fuse at its other end. When the second fuse is finished, 45 mins has passed.

In reply to lowersharpnose:

Question:

Would you rather have a crocodile attack you or an alligator?

<pause>

I'll give you 5 minutes to post a response.

In reply to lowersharpnose:
A king wants his daughter to marry the smartest of 3 extremely intelligent young princes, and so the king's wise men devised an intelligence test.

The princes are gathered into a room and seated, facing one another, and are shown 2 black hats and 3 white hats. They are blindfolded, and 1 hat is placed on each of their heads, with the remaining hats hidden in a different room.

The king tells them that the first prince to deduce the colour of his hat without removing it or looking at it will marry his daughter. A wrong guess will mean death. The blindfolds are then removed.

You are one of the princes. You see 2 white hats on the other prince's heads. After some time you realize that the other prince's are unable to deduce the colour of their hat, or are unwilling to guess. What colour is your hat?

In reply to sbc_10:

I'd rather have the crocodile attack the alligator please.

In reply to Beat me to it!:

Well done.

In reply to sbc_10:
a word problem? I would rather have the crocodile attack an alligator?


EDIT: mp3ferret got this first

In reply to MP3ferret and lowersharpnose.

Yes,well done.
Usually hear various theories about bite radius, jaw strength, acceleration capabilities before you have to <<emphasise>> the <<YOU>> bit.

In reply to sbc_10:

I'd rather have an alligator than have a crocodile attack me???

In reply to lowersharpnose:

OK here's another.

Baby A and Baby B were born from the same mother at midnight on 31st December on the year 2000 AD, but they are not twins, explain??

In reply to lowersharpnose:

The frame of a cube is constructed from twelve 1 ohm resistors forming the cube's edges.
What is the resistance measured across diagonally opposite corners?

In reply to lowersharpnose:

Write down a regular expression to match numbers which are multiples of three.

In reply to lowersharpnose:

You have a 3 pint jug and a 5 pint jug. How do you use only these 2 jigs to get exactly 4 pints?

In reply to lowersharpnose:

Get yourself a king snake cube puzzle

In reply to lowersharpnose:

What comes next in this sequence

1, 11, 21, 1211, 111221, 312211, 13112221, 1113213211,...

In reply to Paul Robertson:

I know the sequence one.

I have done the resistor one, can't remember it.

Not sure what the regular expression one means.

In reply to the real slim shady:

Can do that, thanks.

In reply to sbc_10:

They're two of three triplets?

In reply to lowersharpnose:


Ok, well I can recommend the king snake cube puzzle. Put it on your list for next christmas.

In case you haven't already seen it, here's one which I posted here about 18 months ago:
For a positive whole number N, define rotate(N) as 'transfer the leftmost digit of N to the rightmost position'

So:
rotate(23) = 32
rotate(677190) = 771906
etc.

Now find a number N, greater than zero such that:
rotate(N) = 1.5 * N

In reply to sbc_10:



One was an anal baby

In reply to Byronius Maximus:

Yes. Baby C is the third of triplets. Good work.


!!....<not an anticipated response>

In reply to Beat me to it!:
Thanks. I know the hats and those sort of 'gap/wait logic problems.

My favourite is...

On a (small) island, if you know you have green eyes, you have to shave your hair off. No mirrors and no telling. Some white guy explorer turns up and says, whilst addressing a crowd, "hello green eyes". Twenty seven days later, twenty-seven folk kill themselves.

Why?

In reply to Beat me to it!:

White.

If a prince can see two black hats, then since there are only two black hats they would immediately know they are wearing a white hat.

If you are wearing a black hat the other two princes would both see one black and one white hat. They would conclude that if they were wearing a black hat the prince they could see with the white hat would see two black hats - one on their head and one on yours and would have immediately solved the problem. Therefore they cannot be wearing a black hat.

Therefore, since they have not solved the problem the other two princes must also be seeing two white hats so the hat on your head must be white.

But the other princes have not solved the problem. So the conclusion is that everyone is seeing two white hats so all three hats are white.

In reply to the real slim shady:
to get exactly 4 pints?

Fill up the 5 pint jug.

Fill up the three pint jug from the five pint jug. 2 pints left in 5 pint jug.

Empty 3 pint jug. Pour contents of 5 pint jug into three pint jug. 2 pints in three pint jug.

Fill up 5 pint jug. Top off the last pint in 3 pint jug from 5 pint jug. 4 pints remain in 5 pint jug.

In reply to Paul Robertson:


Had a think about this and I'm intrigued that such a thing might be possible. Please put me out of my misery.

In reply to lowersharpnose:

You have a hole 2 x 12 m and some MDF that is 3 x 8m . Ignoring depth , how can you cut the MDF into 2 pieces so it fits the hole perfectly?

In reply to tom_in_edinburgh:


Is it more likely that due to centuries inbreeding , and not having to beat competition to retain their privileged position, that it unlikely that you would find 3 intelligent princes as described.
( just saying

In reply to lowersharpnose:

One a friend gave me recently, which he got from reddit:

A king has 1,000 barrels of wine stored in his cellar. Unfortunately the guards found an assassin that had snuck in last night and poisoned exactly one of the barrels. Even 1 drop of poisoned wine would be enough to kill a man, but the poison takes 30 days before becoming fatal, plus or minus a few hours. The poison is colorless, odorless, and victims show no signs of being poisoned in the 30 days prior to dying. There is absolutely no way to detect the poison.
In 31 days time, there will be a grand feast in which the king will need as much of the wine as possible. He has 10 prisoners on which to test the wine. His first idea is to give each prisoner a drop from 100 barrels of wine, and then when 1 prisoner dies, throw out all 100 barrels of which the dead prisoner drank from. However his jester speaks up and explains a solution that will pinpoint exactly which barrel is poisoned using only the 10 prisoners.
What was the jester's solution?

In reply to lowersharpnose:

And another shorter one I remember from a maths teacher:

4 people arrive at a bridge that can only bear the weight of two of them at a time and they cannot see the other side. As a result of this two must crossover and then one must return so the others know it's safe to cross. One of the people takes 1 minute to cross, one of them takes 2 minutes, the third 5 minutes and the 4th person 10 minuets. When crossing together they must stick together. They need to get across in 17 minuets, how do they do it?

In reply to RBonney:

Since 2 ^ 10 = 1024 I assume it involves dividing the 1000 barrels into two and testing one half with one prisoner. If he dies you know the poison is in one barrel in that half, so you divide that half by 2 into 250 vs 250 etc. and test one half again with another prisoner. You have less than 24 hours to do this testing though , so you'd have to be quick.

In reply to Beat me to it!:

I thought you had it with 2^10=1024 but then you lost me, but the answer doesn't involve halfs.

In reply to RBonney:



The 1 and 10 minutes guys cross together. The 1 minute guys returns. - 11 mins elapsed.
They now know it is safe to cross. The 2 and 5 min guys. cross. 16 mins elapsed.
The 1 minute guy waits 5 mins then crosses. 17 mins elapsed.

Of course the 1 minute guy could cross, and then 2 and 5 min guys wait a min and then cross together. So at least 2 solutions.

In reply to RBonney:


Wait 31 days
Divide the 1000 barrels in two batches of 500 barrels + 500 barrels
Put a drop from each barrel in batch of 500 into glass, one prisoner drinks from glass. If he dies the poisoned barrel is in that batch, otherwise it's the other batch.
Divide the poisoned batch into 2 batches of 250 barrels. Perform same test with one of the prisoners and a batch of 250
Repeat for batches of 125 barrels
Then 62 or 63 etc.
After 10 prisoners you have tested for 2 batches of exactly 1 barrel each and fond the poisoned barrel.

In reply to lowersharpnose:
I suppose the alternate way is to label the barrels in binary, adding leading zeros. You then mix into 10 glasses. For instance if the 10 digit of a barrels number is a 1 then you put a drop of it in glass 10. If the 9th digit is also a 1 then you put a drop in glass 9 etc. the prisoners then drink their mixes, I.e. Prisoner 10 and only prisoner 10 drinks from glass 10. Based on which prisoners die you have the 1's (and their positions) of the number of the barrel that is poisoned.

In reply to Beat me to it!:

Make a stepped cut so you end up with two L-shaped pieces

In reply to RBonney:
The solution will involve prisoners being given drops from multiple barrels, where each barrel is tasted by a unique combination of prisoners. I'll have a think on the exact solution.

EDIT Beat Me To It's binary numbering method works.

Barrel 769: 1100000001

Prisoners 1, 9 & 10 each get a drop from this barrel etc.

In reply to Beat me to it!:


I think that's the only way to do it. You need to run all 10 tests simultaneously rather than waiting for the result of the first test before starting the second because the poison takes 30 days to have an effect.

In reply to Beat me to it!:

The bridge crossing one that was explained to me involved a torch. Which means that someone has to come back with it. That would mean in your version, the 2 min guy would have to come back with the torch before the 1 minute guy sets off.

There is a still a 17 minutes solution, though.

In reply to RBonney:

I think the plus/minus a few hours is key. Give prisoner 1 a drop from barrels 1 to 100, prisoner 2 a drop from 101 to 200 etc. Wait 8 hours, then give p1 a drop from 1 to 10, 101 to 110, 201 to 210 etc. P2 gets 11 to 120 and so on. Wait another 8 hours, then give p1 1, 11, 21, and so on.
First one to die reveals the 100 digit, second the tens and third the units.

In reply to lowersharpnose:


Here is one my son gave me. I haven't cracked it yet:

You have twelve coins which look identical but one is counterfeit. The coins all weigh the same except for the counterfeit. The counterfeit is either heavier or lighter than the other coins but you don't know which.

Using balance scales you have to determine which coin is the fake, and whether it is lighter or heavier than the others.

You are allowed three weighings where you can compare the weight of the coins in one pan with the weight of the coins in the other pan.

In reply to Paul Robertson:

I can do that one. It is one of the best puzzles ever, I'll leave it with you.

There is only just enough information possible to do this with the three weighings.

In reply to Paul Robertson:

I am looking at your rotate puzzle.

I think the answer is above 10,000,0000 and am using brute force.

In reply to Beat me to it!: If I were being ultra-pedantic, I could say that it just mentions the burn time, no mention of only lighting one end.

In reply to Paul Robertson:
12 coins puzzle.

Divide in 3 piles of 4 coins.
Put 4 coins on one side, and 4 coins on the other side of the scale.

If the the scale remains level then the 8 coins on the scale are genuine.
Counterfeit coin is in the pile of 4 not weighed.
Take 3 of coins not weighed and weigh them against 3 coins known to be genuine.
If lighter then counterfeit is in 3 on scale, one final weighing of one each of those coins will determine the counterfeit. If final weighing shows balanced then counterfeit is the 3rd coin.
If the same then counterfeit was 4th coin not weighed.

If the scale is not balanced then take 3 coins off lighter side and move to heavier side. Take 3 coins from the pile not weighed and put on lighter side.

If the same side is heavier then either the remaining coin on the heavier side is heavier or the remaining coin on the lighter side is lighter. Weigh on of them again a genuine coin to determine which is counterfeit.

If they are now the same. Then one of the 3 coins removed from the heavy side is counterfeit. Weigh two if them again each other. If one us heavier it's the counterfeit. If they are the same, then the unweighted one is counterfeit.

If the lighter side is now heavier. One of the three coins that went from the lighter side to the heavier side is lighter. Weigh two of these coins against each other. If they weigh the same then counterfeit is the one remaining. If they do not weigh the same then the lighter one is counterfeit.

Think that's it.

In reply to Simos:



There needs to be some kind of trick because there are only 8 possible outcomes from 3 yes or no questions but there are 6 possible permutations of gods and 2 possible permutations of yes/no words so 12 possible permutations total. Therefore my guess is you need a question that some gods can't answer to create a 3rd possible alternative i.e. yes/no/no answer.

Ask the first (left hand) god "Will the god on your right tell the truth?" If the god on the right is random then the True and False gods wont be able to answer. The random god will say 'da' or 'ja' at random.

If the first god does not reply then the possible orders are T R F and F R T. Ask the same god 'Do you always lie'. Both T and F gods will reply with the word for no - so you now know whether da=yes, ja=no or vice versa. Ask the god on the right 'Does the god in the middle ever lie' you know the god in the middle is the random god so the true answer is yes. If the god answers with the word for 'true' then the order is F R T otherwise it is T R F.

If the first god does reply the possible orders are T F R, F T R, R T F, R F T. Ask the god in the middle 'Will the god on the right tell the truth'. If it does not answer then the possible options are T F R and F T R. If it does answer it will be using the word for 'no' so we can now translate Da/Ja into yes/no.

If the middle god answered then ask it 'is the god on the left random' we know the answer is yes so we can tell if it is lying and therefore distinguish between R T F and R F T.

If the middle god did not answer the options are F T R and T F R. We now know that the reply from the first question to the left hand god was the word for 'no' since this would be the response in both possible cases. Therefore we just have to ask the middle god 'Is the god on your right random' to determine if it is lying and distinguish between F T R and T F R.

In reply to tom_in_edinburgh:

White - yes.

In reply to Paul Robertson:

Got to a billion with no answer.
Closest I have had is 352941176, which rotates to 529411763

The factor is ~ 1.499999997

I don't know if brute force is going to work. I might need to work out some rules for the search above a billion.

In reply to lowersharpnose:




Interesting one.

I'll posit the following.

There are 27 people on the island. Not a single one of those people knows if they have green eyes. Up to the point the explorer turns up no one has been told they have green eyes and none has shaved their head. So after the explorer visits each person wonders if they have green eyes, but there can be no telling and no one has shaved their head. Since no one has ever shaved their head, and since one has ever said they have green eyes, none of the 27 know what green eyes look like.

So a day passes and no one has shaved their head. They know at least one of them has green eyes, but not who or how many. They don't know if they themselves have green eyes. Another day passes and still no one has shaved their head. A third day passes and still no one has shaved their head. They get to the 27th day and no one has shaved their head. At this point each person on the island reasons that they can't see any shaved heads amongst the other 26 because no one else has green eyes and they are the one with green eyes. The shame of having green eyes but not having shaved their heads makes them commit suicide.

Flaky I know .

In reply to Beat me to it!:

It was late last night when I posted. Forget suicide, it should all be about shaving heads.

There are more people on the island than 27, everyone knows everyone else.

They all know what green eyes look like.

Your reasoning was correct.

In reply to lowersharpnose:

There are ten unique ways of picking one prisoner from the ten.
There are 45 unique ways of picking two prisoners from the ten.
And so on...
There is one way of picking ten prisoners from ten.
Add all those together and you have (I think) 1023 unique ways of picking groups of prisoners, which can use to test each barrel with a unique combination of prisoners.
The binary explanation is more elegant though.

In reply to lowersharpnose:

another maths one.

What is negative frequecy?

In reply to lowersharpnose:

You'll need a fair bit of computing power to brute force it. But it can be solved with pencil paper and a bit of GCSE-level algebra.

In reply to Jonny2vests:


http://quaxio.com/triple/

In reply to Paul Robertson:
You have to find a prime number p with a repeating fraction for 1/p containing p-1 repeating digits. With this property, every digit in the repeating fraction appears in each place exactly once (i.e., every repeated digit appears as the first digit after the decimal exactly once for n/p where p > n > 0).

You are looking to find any occurrences where the nth digit is followed by the 1.5nth digit. 'n' will obviously be even.

Prime 7 is the first but doesn't gave the 1.5 n property. Second the next prime is 17 with the p- 1 repeating digits property. It also has the 1.5n property for 2/17 , 4/17, 6/17 and 8/17.

Remove the decimal and I end up with 1,176,470,588,235,294 (derived from 2/17 sequence) as the first number with rotate property you want. As the rotation derives the 3/17 sequence of 1,764,705,882,352,941. 3/17 divided by 2/17 = 1.5 but you need to get rid of decimal or you are always rotating 0 in the sequence.

In reply to RBonney:


1 and 2 cross - 2 min.
1 returns - 3 min
5 and 10 cross - 13 min
2 returns - 15 min
1 and 2 cross - 17 min.

In reply to Beat me to it!:

I have read your explanation a couple of times and I can't make sense of it yet. Why do I need a prime number p...

In reply to lowersharpnose:

lowersharpnose got my barrel one correct and tom_in_edinburgh gt the bridge one correct, I'm sure others did too but these are the correct ones I've seen just scrolling though.

In reply to lowersharpnose:
Fractions of the form n/p, where p is a prime number and n is a positive integer less than p, and p is not a factor of 10 (i.e., the primes 2 and 5), result in repeating decimals with very definite patterns.

For any prime number p above, the set of all repeating decimals derived from the fractions 1/p, 2/p, 3/p, ... (p-1)/p is made up of equal-length "chains" of digits that repeat themselves at fixed intervals.

You don't get the same patterns (as in number if relating digits being one less than p) if not using prime numbers. At least I don't think you do. Ling time since I did the group / number theory at uni.

In reply to Simos:



Do all the gods reply in the same language?

In reply to lowersharpnose:
Either I've mis-understood the question, or you're trying too hard. The first one I've found is four digits long.

In reply to Paul Robertson:

31131211131221

The kind of puzzle that's easy to overthink...

In reply to Paul Robertson:



Split the coins into three groups of 4 - A, B C. Call them A1,A2,A3,A4, B1,B2,B3,B4 and C1,C2,C3,C4.

Weigh A1,A2,A3,A4 against B1,B2,B3,B4. If = then one of C1,C2,C3,C4 is counterfeit. If A > B then C1,C2,C3,C4 are genuine and either one of group A is counterfeit and too heavy or one of group B is counterfeit and too light. The situation where A < B is symmetrical with A > B (just rename the groups A -> B and B -> A).

In the situation A = B to find which of C is counterfeit. Weigh C1 C2 against C3 A1. If = then C4 is counterfeit, weigh C4 against A1 to see if it is too heavy or too light. If < then (situation with > is symmetrical and can be solved the same way)) either C1 or C2 is too heavy or C3 is too light. Weigh C1 against C2. If = then C3 is too light. If > then C1 is too heavy. If < C2 is too heavy.

In the situation A > B we know that if a coin in A is counterfeit it is too heavy and if a coin in B is counterfeit it is too light. Weigh A1 A2 B1 B2 against A3 B3 C1 C2.
If = then either A4 or B4 is counterfeit. Weigh A4 against B4 if > A4 is counterfeit and to heavy. If < B4 is counterfeit and too light.

If > then either A1 or A2 is too heavy or B3 is too light. Weigh A1 against A2. If = then B3 is counterfeit and too light. If > then A1 is counterfeit and too heavy. If < A2 is counterfeit and too heavy.

If < then either B1 or B2 is too light or A3 is too heavy. Weigh B1 against B2. If = then A3 too heavy. If < then B1 too light. If > B2 too light.

In reply to Paul Robertson:


Should be 31131211131221

In reply to Paul Robertson:




This is an excellent and classic problem. Most people get there eventually with a bit of haphazard experimentation (See Tom's solutiuon above!) A more sophisticated challenge is to prove that it is possible to find the odd one out of (3^n-3)/2 coins with n weighings.

HINT: on each weighing,one of three things happens: either the Left side of the scales goes down or the Right side goes down or the scales Balance Balance, so a sequuence of weighings generates a sequence such LRBBRLLB..... There are 3^n possible such sequences for n weighings............................

In reply to Beat me to it!:





Ingenious, but I think the solution I gave the last time Paul put this problem on here was somewhat simpler!

In reply to lowersharpnose:

The king decides to cull his wise men (there are 10 of them). He tells them that he is going to line them up all facing the same way along the line and then put either a red or blue hat on each of them so that each can see the colour of the hats of all those in front of them but not their own or those behind them. Each will then be asked the colour of their hat and (starting with the one at the end of the line who can see all the others' hats and working along the line). Those getting it wrong will not be told at the time but will be shot afterwards.

The wise men are given ample time to come up with a plan.
How many can they be sure to save?

The problem generalises to any number of wise men and (much more interestingly ang challengingly) any number of possible colours for the hats.

In reply to Paul Robertson:


Any chance you could tell us what this means?!

In reply to lowersharpnose:


I give up. I am either stupid or it is a trick question. I am assuming nothing of the room can be seen from the location of the switches. Please put me out of my misery!

In reply to Robert Durran:
Wise men - I think they can save 9.

The first wise man can deduce nothing about the colour of his hat. So whether he says red or blue makes no difference to his odds of being shot. So he may as well save the others. So he counts the number of red hats he can see. If there are an odd number of red hats then he says red else he says blue.

The second man now knows there is either an odd number of reds or blues (amongst the remaining 9) including his own hat. He counts the red hats he can see and deduces the colour of his own. He states the colour he has deduced.

The third man can now deduce the colour of his hat as can the fourth then fifth etc.

You save 9 who definitely answer correctly and the 1st has a 50% chance if being saved as well.

Alternately the first one ignores what they agreed. He counts number of red hats he can see and if odd says blue. The other 9 will then all get the colour wrong. The king seeing that at least 9 out if 10 wise men got it wrong decides not to shoot any of them as he can't afford to lose 9 of them.

In reply to Beat me to it!:


Cunning!

Have you generalised?

In reply to Robert Durran:

It's a trick question. The answer is further up the thread. The trick is to turn a bulb on and off again before you go in so there is one on bulb, one off and cold bulb and one off but warm bulb. There's an unstated assumption you can get close enough to the bulbs to sense their temperature.

In reply to tom_in_edinburgh:


I hate that..... but at least I'm not stupid (I'd convinced myself it must be an annoying trick and lost interest).

In reply to Robert Durran:

Turn two switches on.

Wait a while, then turn one of those two off and go into the room.

One of the lights that is off will be warm from being on for a while.

In reply to tom_in_edinburgh:

Maybe this helps:

If I asked you <question>, would you answer 'ja'?

In reply to Beat me to it!:

Yes

In reply to Simos:



I'm not convinced I need help (yet) since as far as I know the solution I presented is valid. My guess is there are many possible solutions.

In reply to Robert Durran:

It only looks like haphazard experimentation to the untrained eye It is actually recursive decomposition followed by exhaustive search when the problem size is small. Classic computer science.

In reply to tom_in_edinburgh:


So will it prove the generalised result I mentioned.

Or does that need proper mathematics

In reply to Hairy Pete:

What is your four digit solution?

In reply to lowersharpnose:
4356
Oh, and you can keep putting 9s in the middle like 43956

In reply to tom_in_edinburgh:

Maybe not sure - I would guess though that where your solution is not valid is the fact that you are introducing a 3rd state in the answers where a god simply does not answer if Random is sitting next to them. Apart from stating clearly that gods answer only yes or no n their language, I would have also thought that the answer would be simply NO if you asked say Truth whether Random will tell the truth. Random will just give random answers, not the true answers, so why would the other gods not answer?

In reply to Hairy Pete:



You have misread the question.

Rotate(N) turns 4356 into 3564 not 6534.

It's a badly named function although the original poster did explain it. It's not so much rotate as translate MSD to LSD.

In reply to lowersharpnose:

You are sat in a small boat on a pond. In the boat is a large lead block. You throw this into the pond. What happens to the water level in the pond?

Cheers

Toby

In reply to Toby_W:

It goes down. When the lead is in the boat it displaces its weight in water. When it's in the pond it displaces its volume in water.

In reply to Hooo: top marks, too easy I think.

Cheers

Toby

In reply to lowersharpnose:

Ok, here's my last offering: It's called the 'mass-weight experiment'. You have a weight suspended from the ceiling by a thread. A second, identical thread is attached to the bottom of the weight and hangs down conveniently so that you can pull down on it.

If you pull down on the hanging thread, which thread will break? The lower one or the upper one?

In reply to Simos:


Can you prove the question is answerable without a third state?

There are only 2^3 = 8 possible outcomes from asking 3 true/false questions.

There are 6 possible arrangements of gods and 2 possible arrangements of yes/no words so 12 possible permutations which must be distinguished. Unless there is some trick in the wording of the question which eliminates some of these 12 options I claim it is impossible to distinguish between 12 possible options based on 8 possible outcomes of the questions so a third type of response such as not answering is required.

In reply to Philip:

DOH! Didn't spot that, though I did read the question many many times.

In reply to Paul Robertson:

The upper one as it supports the weight and the puller, the lower one only support the puller. No doubt it is more complicated and I have taken a pratfall.

In reply to Paul Robertson:

The upper one as it is supporting the weight and me

In reply to Paul Robertson:

It depends.

If the threads are inelastic and/or the extra pulling force is applied gradually, the upper string will break first. The tension in the upper string is the pulling force plus the weight (m*g) of the mass, the tension in the lower string is the pulling force only.

With a slightly stretchy string and a very quick change from zero force to full force pulling, it might be possible to break the lower string. This is a much more complicated problem to work out because it is dynamic not static. The key point is the mass must move down some distance before the upper string can break, but it will take time to do so because it has inertia. If you yank the lower string, you might be able to snap it before this happens.

In reply to Jack B:

Good answer.
The experiment was done to demonstrate the difference between mass (i.e. inertia or a tendency to stay put) and weight (a force acting on a mass).
Most of the audience (maths undergraduates) voted for the upper thread. Predictably, the lecturer gave a sharp tug and broke the lower thread.

With a 500g weight and cotton thread it's quite easy to break whichever thread you choose.

In reply to tom_in_edinburgh:

Have a look at my hint above - using that 'lemma' basically allows you get around the extra states introduced by the other language,

Eg "if I asked you whether the god on your right is Random, would you say 'ja'?"

If you ask True this question and Random is next to him he will say 'ja'. (irrespective of whether ja means 'yes' or 'no'). Basically this lemma relies on the fact that a double negative makes a positive

In reply to Simos:

It doesn't matter what the questions are. If the only possible answers are Ja and Da and I'm only allowed 3 questions the responses are one of:

Ja Ja Ja
Ja Ja Da
Ja Da Ja
Ja Da Da
Da Ja Ja
Da Ja Da
Da Da Ja
Da Da Da

There are only 8 possible outcomes from the questions. There are 12 possible cases to distinguish between.

True False Random Ja = True
True Random False Ja = True
False True Random Ja = True
False Random True Ja = True
Random False True Ja = True
Random True False Ja = True
True False Random Ja = False
True Random False Ja = False
False True Random Ja = False
False Random True Ja = False
Random False True Ja = False
Random True False Ja = False

No matter how clever the questions it can't be done.

In reply to Robert Durran:


A regular expression is a kind of shorthand used in programming to search text.

In reply to Jonny2vests:


Ok. I'll bin that question then!

In reply to Simos:

OK I see your point. The idea is you don't have to decode the Ja/Da thing because of the way the questions are structured so there are only 6 possible options.

Still a bit dodgy though since you need to make assumptions about how the False god will evaluate the two-part sentence. The results will be different if it does each clause one after the other (in the way you expect True to work) or figures out what True would say to the whole statement and says the opposite.

In reply to tom_in_edinburgh:

Some interesting history here since you are into puzzles:

http://en.wikipedia.org/wiki/The_Hardest_Logic_Puzzle_Ever

In reply to Simos:


Thanks for that - it's really interesting. I'm relieved to see some of the real mathematicians went down the same path as my solution of finding a question that True and False couldn't answer, as well as the double or triple question idea

In reply to tom_in_edinburgh:

Yeah actually you were pretty close!!! Maybe you should have studied philosophy!