ASF Puzzles & Conundrums Thread

[bellenuit]

Y
If the Captain of the boat had lined the 3 Slaves up (facing away from him of course) and asked them all (both separately and/or collectively) ..... Do all the Slaves on this Island have brown eyes, they would have all said NO. Even if they heard each others answer, what does that teach them that they didn't already know? I don't see why a third party saying NO gives any added info, but I am quite prepared to admit my brain is simply not up to the "twist"[/B]

As cynic and others have pointed out, the crux of the problem (though not explicitly stated in McLovin's answer) is not what each slaves knows, but what each slave knows the other slaves know. In this case the information is not redundant. It is the reaction of the other slaves that allows the slaves to determine their own eye colour. Let's call the slaves A, B and C and look at it from the point of A trying to see what B might be thinking.

As part of his deductive reasoning, A sets up a hypothesis for testing that may or may not be correct. The hypothesis is that he, A, has brown eyes. He then tries to put himself in B's shoes. Although A knows C knows B has blue eyes, B doesn't know his own eye colour, so B doesn't know what C knows about B's eye colour.

This might make it clearer:

1 A hypothesises that he, A, has brown eyes and then looks at it from B's viewpoint.

2. So B, hypothesising that he, B, has brown eyes, must look at C and think C must have this info:
(remember this is what A is thinking B is thinking in regards to C)

3.1 C must assume that he, C, has brown eyes and thus everyone has brown eyes OR
3.2 C must assume that he, C, has blue eyes and he is the only one with blue eyes .

This nested hypothesis is in A's mind and is quite valid, though we as outside observers know isn't the case.

So 3.1 and 3.2 is assuming that 2 is true and 2 is assuming 1 is true. At this point look at 3.1. This is a valid hypothesis and it's assumption is that everyone has brown eyes. That is why the information is not redundant. That information about not everyone having brown eyes allows B to realise that with the "new"information, C must leave the island Monday (because if 1 and 2 were true, and 3.1 now ruled out with the "new" information, C would know his eye colour is blue and leave the island). When C doesn't leave the island, B (in A's mind) would know hypothesis 2 is incorrect and that he, B, must have blue eyes and then he leaves Tuesday. But when that doesn't happen, A must assume that hypothesis 1 is incorrect and thus he, A, has blue eyes, leaving Wednesday. Everything being symmetrical across a, B and C, they all leave Wednesday.

So the information isn't redundant because it allows a potential conclusion in the nested hypotheses to be ruled out which then sets off a chain reaction of conclusions. Without that information being provided, 3.1 and 3.2 would never be resolved and no conclusion could be drawn from C's lack of reaction, so 2 or 1 could also not be resolved.

 

[cynic]

Yeah its a good fun thread ..... although I'm still not convinced about the brown and blue eyed Slaves. My logic behind not being convinced is thus ....

If the Captain of the boat had lined the 3 Slaves up (facing away from him of course) and asked them all (both separately and/or collectively) ..... Do all the Slaves on this Island have brown eyes, they would have all said NO. Even if they heard each others answer, what does that teach them that they didn't already know? I don't see why a third party saying NO gives any added info, but I am quite prepared to admit my brain is simply not up to the "twist"

After reading some of the responses to this thread, I am of the opinion that you are by no means alone in that regard.

The important thing is to think about what the slaves suddenly know afterwards that they couldn't possibly know beforehand.

Whenever I try to articulate it, it looks like a typo: all slaves then know that all slaves know that all slaves know that there is at least one blue eyed slave.

Without that crucial piece of information the slaves simply cannot draw any conclusions from the reaction (or absence thereof) of their fellows on the island over the successive few days.


I'll try and formulate a series of questions, later today, that will hopefully make the distinction clearer.

Here is a simple one to start the week. I just got it off the net so hopefully a few haven't seen it. PS. If it looks too easy, try it after 3 beers!

At a recent downhill mountain bike race, four entrants entered the challenging slalom event.


Alan came first.

The entrant wearing number 2 wore red, whereas John didn't wear yellow.

The loser wore blue and Steve wore number 1.

Kev beat Steve and the person who came second wore number 3.

The entrant in yellow beat the entrant in green.

Only one of the entrants wore the same number as their final position.


Can you determine who finished where, the number and colour they wore?

Yes I can, but that's only because I've run out of beer!!

Edit: Just noticed that Bellenuit has beaten me to the punch on clarification of the slaves puzzle. Thanks Bellenuit.

 

[barney]

LOL ..... Thanks for such a detailed response "Bell" ........

Leave it with me for a while ..... My brain only runs in Stereo, not Quadraphonic!

 

[barney]

After reading some of the responses to this thread, I am of the opinion that you are by no means alone in that regard.

Even after Bells detailed response I still have questions to be answered ..... Its almost as hard to write a question to articulate the problem as it is to try and solve the problem (in my brain )

Yes I can, but that's only because I've run out of beer!!

I should have posted it up after 5pm

 

[Tisme]

This is a subset of a puzzle that has proved controversial over the years.

A king has dominion over several islands each filled with slaves. The slaves have either blue or brown eyes. Some islands have just blue eyed slaves, some have just brown eyed slaves and some have a mixture of both.

The slaves on these islands are very smart and logical. Each has a burning ambition to be made free and travel to the mainland where the king and all the free people live.

The only way the slaves can become free is by deducing the colour of their own eyes and having done so they must depart the island that very day for the mainland on one of the many boats that go back and forth each day. This is not so easy as although the slaves can see the colour of everybody else's eyes, there are no reflective surfaces that allow them to see the colour of their own eyes. Additionally, slaves are never allowed discuss other slaves' eye colours so cannot tell or indicate in any way to another slave what eye colour that other slave has.

On one island there are just 3 slaves, all with blue eyes. One Monday morning, a king's official is visiting the island to check on things. Later that morning, as he is about to get on one of the many ferries back to the mainland, he is asked by the captain: "Does everyone on this island have brown eyes?". He answers: "No". Both the question and the response are overheard by the 3 slaves.

The question is, what (apart from the normal mundane daily drudgery of slaves) happens that week on the island once the slaves hear this seemingly redundant information?

I didn't want to think about this because I would presume that if someone asks if everyone has brown eyes, it would be based on observing someone with brown eyes, but disregarding that logic:

Each one would look at the other two and see blue eyes. Each would suppose by intuition the other two would take off the next day, but because neither do, surety rules apply hence and logically mean (if people = x and if days = y) y = x-1 days would pass to make sure all three could be sure they all blue eyes. Therefore it would be Wednesday before they pack up the esky and pink zinc and head for the ship.

 

[trainspotter]

Yeah its a good fun thread ..... although I'm still not convinced about the brown and blue eyed Slaves. My logic behind not being convinced is thus ....

If the Captain of the boat had lined the 3 Slaves up (facing away from him of course) and asked them all (both separately and/or collectively) ..... Do all the Slaves on this Island have brown eyes, they would have all said NO. Even if they heard each others answer, what does that teach them that they didn't already know? I don't see why a third party saying NO gives any added info, but I am quite prepared to admit my brain is simply not up to the "twist"

Here is a simple one to start the week. I just got it off the net so hopefully a few haven't seen it. PS. If it looks too easy, try it after 3 beers!

At a recent downhill mountain bike race, four entrants entered the challenging slalom event.


Alan came first.

The entrant wearing number 2 wore red, whereas John didn't wear yellow.

The loser wore blue and Steve wore number 1.

Kev beat Steve and the person who came second wore number 3.

The entrant in yellow beat the entrant in green.

Only one of the entrants wore the same number as their final position.


Can you determine who finished where, the number and colour they wore?

It should be simple enough to create a Matrix with names, colours, numbers and finishing positions?

1) Alan number 2, red
2) Kev number 3, yellow
3) Steve number 1, green
4) John number 4, blue

 

[cynic]

It should be simple enough to create a Matrix with names, colours, numbers and finishing positions?

1) Alan number 2, red
2) Kev number 3, yellow
3) Steve number 1, green
4) John number 4, blue

View attachment 62608

Yes! That's what I got. But I cheated by not drinking beer beforehand.

 

[bellenuit]

It should be simple enough to create a Matrix with names, colours, numbers and finishing positions?

1) Alan number 2, red
2) Kev number 3, yellow
3) Steve number 1, green
4) John number 4, blue

Same with me

 

[barney]

Yes! That's what I got. But I cheated by not drinking beer beforehand.

LOL You are a funny man Mr Cynic.

Indeed all are correct ...... I knew that would be easy for you guys

 

[bellenuit]

A not too difficult one.

You are being interviewed for a job in a factory that makes fuses for explosives (the type that you light and will burn from end to end for a specific length of time). As an intelligence test, the interviewer gives you this problem.

He shows you two of the fuses they make that are identical. Both are yellow at one end and green at the other. They are flexible and can be twisted any way you want. The fuses can be lit from either end and burn for exactly two minutes. They do not burn at a constant rate and the burn rate (cms/sec) will vary along their lengths, sometimes going faster and sometimes going slower. However, because they are manufactured identically, the two fuses, if lit from the same coloured end, will burn exactly at the same rate as each other, both increasing or decreasing in speed at the same time. As they burn they completely vaporise and leave no residue trail of any sort. Fuses can ignite each other if the burning point of one comes into contact with the other.

The interviewer hands you the two fuses and shows you the heat resistant table to use for your test. This table also won't stain or show a trail of a burnt fuse. He gives you a box of matches with just two matches, each which only burns for a couple of seconds. He says: "Choose your begin time and then tell me when 90 seconds has elapsed after that". You are not allowed use any device other than the fuses and matches.

You can assume that the fuse will ignite the instant it is touched by the flame.

Extra points for being able to gauge with the most accuracy when 90 seconds is up.

 

[cynic]

I would lay the two fuses on the table so that they intersect each other 3/4 of the way along with the green end of one fuse and yellow end of other a 3/4 length away from the intersection.

I'd then light each fuse simultaneously from the end that is farthest from the intersection (i.e. the green end on one fuse and yellow end on the other) and declare the 90 seconds to have begun.

When the flame of one fuse first meets the intersection, I'd observe the now burning residual portions of the 1/4 lengths. As soon as they appeared to be approximately equal length I'd declare the 90 seconds over.

Edit: It just occurred to me that the 90 seconds could be declared over when the residual portion of the slower 3/4 length burnt out.

 

[bellenuit]

I would lay the two fuses on the table so that they intersect each other 3/4 of the way along with the green end of one fuse and yellow end of other a 3/4 length away from the intersection.

I'd then light each fuse simultaneously from the end that is farthest from the intersection (i.e. the green end on one fuse and yellow end on the other) and declare the 90 seconds to have begun.

When the flame of one fuse first meets the intersection, I'd observe the now burning residual portions of the 1/4 lengths. As soon as they appeared to be approximately equal length I'd declare the 90 seconds over.

Edit: It just occurred to me that the 90 seconds could be declared over when the residual portion of the slower 3/4 length burnt out.

When the flame of one fuse first meets the intersection, I'd observe the now burning residual portions of the 1/4 lengths. As soon as they appeared to be approximately equal length I'd declare the 90 seconds over.

That wouldn't work, if I am understanding you correctly. Assume as an example, that A burns really fast for the first 3/4 of its length and really slow for the rest. The other, B, since it is in reverse, will be slow for the 1st quarter and then fast for the remaining 3/4. Then A will obviously burn to the intersection first and ignite B. But since A now drops to its slow speed and B is at its top speed, B will obviously burn its remaining 1/4 at a faster rate than A, so the residual portions will never be equal length.

Edit: It just occurred to me that the 90 seconds could be declared over when the residual portion of the slower 3/4 length burnt out.

Again I think this is wrong. If we assume A burns 120 times faster in the 1st 3/4 than the last 1/4, then it will reach the intersection in under a second, where it ignites B. But B is now in the fast portion of its burn too, so will complete the full burn in less than a second as well. All up under 2 seconds.

 

[Tisme]

A not too difficult one.

You are being interviewed for a job in a factory that makes fuses for explosives (the type that you light and will burn from end to end for a specific length of time). As an intelligence test, the interviewer gives you this problem.

He shows you two of the fuses they make that are identical. Both are yellow at one end and green at the other. They are flexible and can be twisted any way you want. The fuses can be lit from either end and burn for exactly two minutes. They do not burn at a constant rate and the burn rate (cms/sec) will vary along their lengths, sometimes going faster and sometimes going slower. However, because they are manufactured identically, the two fuses, if lit from the same coloured end, will burn exactly at the same rate as each other, both increasing or decreasing in speed at the same time. As they burn they completely vaporise and leave no residue trail of any sort. Fuses can ignite each other if the burning point of one comes into contact with the other.

The interviewer hands you the two fuses and shows you the heat resistant table to use for your test. This table also won't stain or show a trail of a burnt fuse. He gives you a box of matches with just two matches, each which only burns for a couple of seconds. He says: "Choose your begin time and then tell me when 90 seconds has elapsed after that". You are not allowed use any device other than the fuses and matches.

You can assume that the fuse will ignite the instant it is touched by the flame.

Extra points for being able to gauge with the most accuracy when 90 seconds is up.

Light one fuse at both ends and the 60 second mark will be measurable when it burns out. Immediately light the second fuse and allow to burn to the same length that the first fuse burnt itself out.

 

[bellenuit]

Light one fuse at both ends and the 60 second mark will be measurable when it burns out. Immediately light the second fuse and allow to burn to the same length that the first fuse burnt itself out.

Your getting hot. One logistical problem though. The first leaves no residual marks, so how will you determine the same length as the first with what you have available. Also, since you have just matches, how do you light the first at both ends? Will there be a lag in moving the lit match from one end to the other or do you use both matches to do this?

 

[cynic]

When the flame of one fuse first meets the intersection, I'd observe the now burning residual portions of the 1/4 lengths. As soon as they appeared to be approximately equal length I'd declare the 90 seconds over.

The wouldn't work, if I am understanding you correctly. Assume as an example, that A burns really fast for the first 3/4 of its length and really slow for the rest. The other, B, since it is in reverse, will be slow for the 1st quarter and then fast for the remaining 3/4. Then A will obviously burn to the intersection first and ignite B. But since A now drops to its slow speed and B is at its top speed, B will obviously burn its remaining 1/4 at a faster rate than A, so the residual portions will never be equal length.

Yes I do see your point. I didn't think that one through with sufficient care.

Edit: It just occurred to me that the 90 seconds could be declared over when the residual portion of the slower 3/4 length burnt out.

Again I think this is wrong. If we assume A burns 120 times faster in the 1st 3/4 than the last 1/4, then it will reach the intersection in under a second, where it ignites B. But B is now in the fast portion of its burn too, so will complete the full burn in less than a second as well. All up under 2 seconds.

Again I see your point.

How about intersecting them as previously described but with the intersection exactly halfway this time and declaring the 90 seconds up when the north and west pointing 1/2 portions (presuming that the fuses were lit at the ends pointing east and south) appear to have combined residual lengths equal to 1/2.

 

[cynic]

Following on from Tisme's logic I think I might know an answer (thanks Tisme).

Curl the first fuse into a loop so that both ends can be lit by the first match at once.

Curl the second fuse into a loop and then twist it into a double loop (again so that both ends can be lit by a single match).

Light the first looped fuse and a minute later when it has burnt out light the double looped fuse. A further 30 seconds will have elapsed when the second fuse has burnt out.

Adjust for the seconds it took to light each of the fuses in the 90 seconds estimate.

 

[bellenuit]

Yes I do see your point. I didn't think that one through with sufficient care.

Again I see your point.

How about intersecting them as previously described but with the intersection exactly halfway this time and declaring the 90 seconds up when the north and west pointing 1/2 portions (presuming that the fuses were lit at the ends pointing east and south) appear to have combined residual lengths equal to 1/2.

I think the math of the combined residual lengths equal 1/2 is probably correct, but the logistics of doing it are horrendous. Firstly, since there is no residue left from what has already being burnt, gauging not just what 1/2 of the full length is would be very difficult, but trying to visually combine two differently orientated non-touching and decreasing lengths to assess whether they are the same length as 1/2 something that has already vanished would be a bit much. You could possibly place the matches at the intersection and one end of the fuse to gauge what 1/2 the full length is, but I still think visually comparing the sum of the lengths of the two decreasing residual fuses to that length would not be easy and inaccurate in comparison to other solutions.

Doable, but not the best solution.

 

[bellenuit]

Following on from Tisme's logic I think I might know an answer (thanks Tisme).

Curl the first fuse into a loop so that both ends can be lit by the first match at once.

Curl the second fuse into a loop and then twist it into a double loop (again so that both ends can be lit by a single match).

Light the first looped fuse and a minute later when it has burnt out light the double looped fuse. A further 30 seconds will have elapsed when the second fuse has burnt out.

Adjust for the seconds it took to light each of the fuses in the 90 seconds estimate.

I have issues with this part.

Light the first looped fuse and a minute later when it has burnt out light the double looped fuse. A further 30 seconds will have elapsed when the second fuse has burnt out.

When you say a double loop, do you mean a figure 8 or two zeros on top of each other. I don't believe either would work.

Again thinking about a situation where the burn rate for the first 3/4's is 120 times (or some big number) faster than the last quarter. For the stacked zero configuration, it would zip around in under a second (as one of the loops is entirely at the top speed and that will ignite the other). For the figure 8 config, the intersection is where the 1/4 and 3/4 points cross. The fast end will reach that intersection and continue around the outside loop back to that intersection in under a second. The other quarter will have been lit at both ends, firstly at the start (when the joined slow and fast ends were lit) and secondly when the fast end first crossed the intersection. But that 1/4 segment is really really slow, even though lit from both ends. It takes about 119 seconds (approx) to burn from 1 end, so will take almost a minute to burn if lighting at both ends.

 

[McLovin]

Lay the two fuses next to each other with the same coloured ends. Light one of the fuses at both ends and one at one end, with the first match. When the first fuse extinguishes, light the remaining unlit end with the second match.

So the first fuse will go for 1 minute and then once the unlit end of the second fuse is lit it will run for a further 30 seconds. Giving 90 seconds all up.

This won't work if the matches go out very quickly (ie you're not supposed to be able to light both ends with the same match).

Thanks for posting these bellenuit, it really gets the grey matter firing!

 

[skc]

A not too difficult one.

You are being interviewed for a job in a factory that makes fuses for explosives (the type that you light and will burn from end to end for a specific length of time). As an intelligence test, the interviewer gives you this problem.

He shows you two of the fuses they make that are identical. Both are yellow at one end and green at the other. They are flexible and can be twisted any way you want. The fuses can be lit from either end and burn for exactly two minutes. They do not burn at a constant rate and the burn rate (cms/sec) will vary along their lengths, sometimes going faster and sometimes going slower. However, because they are manufactured identically, the two fuses, if lit from the same coloured end, will burn exactly at the same rate as each other, both increasing or decreasing in speed at the same time. As they burn they completely vaporise and leave no residue trail of any sort. Fuses can ignite each other if the burning point of one comes into contact with the other.

The interviewer hands you the two fuses and shows you the heat resistant table to use for your test. This table also won't stain or show a trail of a burnt fuse. He gives you a box of matches with just two matches, each which only burns for a couple of seconds. He says: "Choose your begin time and then tell me when 90 seconds has elapsed after that". You are not allowed use any device other than the fuses and matches.

You can assume that the fuse will ignite the instant it is touched by the flame.

Extra points for being able to gauge with the most accuracy when 90 seconds is up.

1. Place fuse 1 with both ends at one edge of the table. The two halves of the fuse shall be parallel and as close to each other without touching.

2. Ignite both ends of the fuse at the same time.

3. Hold the second fuse at the ready. Note where fuse 1 is about to burn out and ignite the correct end of fuse 2 using the very last spark (or as much as possible) from fuse 1. The correct end is the end closest to the burnt out location for fuse 1. At the same time, place remainder of fuse 2 towards the edge of the table where the ends of fuse 1 were placed.

4. Declare 90 second count begins.

5. When fuse 2 burns to the edge of the table, 60 seconds would have passed. At this time, ignite the remaining end of fuse 2. Another 30 seconds shall pass when fuse 2 is burnt out. Making a total of 90 seconds.