ASF Puzzles & Conundrums Thread

[cynic]

It looks to me like you might've gotten that job, SKC.

 

[bellenuit]

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.

also,

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).

Both of these solutions are right and will give an approx 90 second duration. As Mc Lovin noticed with his solution, there may be an issue with lighting the 2nd end of the fuse that is lit at both ends. Will the match stay lit. Assuming it does, then there will be some small inaccuracy with the time lag in moving the match from one end to the other (the lengths weren't specified so we are not sure how big this lag is). SKCs solution doesn't seem to have this problem, but could be logistical difficult trying to place the now burning 2nd fuse over the same exact path as that section of fuse 1 without some sort of fumbling (for instance, the short portion of fuse 2 is on the table and the long portion will overhang: so how do you stop it falling off?). It also assumes that when lit, the fuses stay perfectly still and don't move or curl in anyway, because if they do, then the end point where fuse 1 burns out may not be where that exact point on fuse 1 originally started, meaning fuse 2 placed on that burn out point may not take exactly 60 seconds to reach the table edge.

I would think SKCs solution would give the most accurate result if those factors don't come into play, though McLovin's has the element of simplicity.

Both realised a key element in the solution in that if one of the fuses is lit at both ends, irrespective of the burn rate at different parts of the fuse, the burns will always meet at the 60 second point (which may be quite different to the half length point).

The solution that I like has elements of both of the above and also an idea that cynic suggested but didn't pursue.

Make a circle with one fuse so that both ends of it meet at one point. Place the 2nd fuse in a line with one end (it doesn't matter which end) starting from this same point but moving outwards from the circle. Light the three ends that meet with one match and start your count. When the circle fuse burns out (which will be at 60 seconds) light the other end of the 2nd fuse. At this time, the 2nd fuse will already have been burning for 60 seconds from the circle end, so this new burn will meet up at the old burn in 30 seconds, which gives your count.

It is very similar to McLovin's and has the same simplicity, but avoids the lag between having to light both ends of one fuse when the ends are not together. It also doesn't require any fuse placement action be taken once the fuses are initially laid out, which make it simpler that SKCs.

 

[Tisme]

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?

I thought that would be self evident:
bring both ends together and light first fuse
lay lit fuse next to (parallel) unlit fuse
place burnt match on burnout point
have second match lit in readiness

 

[skc]

It is very similar to McLovin's and has the same simplicity, but avoids the lag between having to light both ends of one fuse when the ends are not together. It also doesn't require any fuse placement action be taken once the fuses are initially laid out, which make it simpler that SKCs.

McLovin's solution was much more elegant. All he needed was place the 3 ends of the 2 fuses together at the start without the 2 fuses touching each other (except at the 3 ends). Light the 3 ends with match 1, then light the last remaining end with match 2 when fuse 1 burns out.

 

[McLovin]

McLovin's solution was much more elegant. All he needed was place the 3 ends of the 2 fuses together at the start without the 2 fuses touching each other (except at the 3 ends). Light the 3 ends with match 1, then light the last remaining end with match 2 when fuse 1 burns out.

Yeah good idea. That gets around the lit match problem.

 

[Tisme]

McLovin's solution was much more elegant. All he needed was place the 3 ends of the 2 fuses together at the start without the 2 fuses touching each other (except at the 3 ends). Light the 3 ends with match 1, then light the last remaining end with match 2 when fuse 1 burns out.

Not sure that will work due to the non linear fashion of the burn, but worth doing the maths I suppose....later on.

 

[cynic]

Not sure that will work due to the non linear fashion of the burn, but worth doing the maths I suppose....later on.

Good point.

If the slowest burning minute length of the fuse is sufficiently long, it would be possible to light the remaining unlit end by curling it to touch its burning end. This would leave one spare match and show the interviewer that one has an appreciation for economy as well as accuracy.

 

[luutzu]

Good point.

If the slowest burning minute length of the fuse is sufficiently long, it would be possible to light the remaining unlit end by curling it to touch its burning end. This would leave one spare match and show the interviewer that one has an appreciation for economy as well as accuracy.

Place two fuse next to each other, same coloured ends together.

Use match 1 to light both fuse at, say Green ends and at the same time use match 2 to lit the other end - but only lit 1 fuse.

So you have 1 fuse (F1) burning at both ends (started at the same time with 2 matches); and the other fuse also burning at but only one end (was lit when you burn both fuses).

When the two ends of F1 meet, that's 60sec; at this point it's also been 60sec on F2 - with remaining 60sec to burn through for F2.

At point where F1 is about to burnt out, you hold F2 and place unburnt end there to be ignited by F1's 2 ends.

with F1 now gone and F2 burning at both ends... knowing there's only 60sec left when F2 was lit by F1.. .when F2 burnt out that's 30 sec. Hence 90.


Or you lit both and count 1 mississippi, 2 mississippi to 90.

 

[skc]

Or you lit both and count 1 mississippi, 2 mississippi to 90.

Nailed it.

 

[bellenuit]

Make a circle with one fuse so that both ends of it meet at one point. Place the 2nd fuse in a line with one end (it doesn't matter which end) starting from this same point but moving outwards from the circle. Light the three ends that meet with one match and start your count. When the circle fuse burns out (which will be at 60 seconds) light the other end of the 2nd fuse. At this time, the 2nd fuse will already have been burning for 60 seconds from the circle end, so this new burn will meet up at the old burn in 30 seconds, which gives your count.

I still think the solution I offered last night is still the simplest and most accurate of everything I have read so far.

SKCs amendment to McLovin's solution (McLovin's solution was much more elegant. All he needed was place the 3 ends of the 2 fuses together at the start without the 2 fuses touching each other (except at the 3 ends). Light the 3 ends with match 1, then light the last remaining end with match 2 when fuse 1 burns out) is just this restated.

Tisme's potential issue (Not sure that will work due to the non linear fashion of the burn, but worth doing the maths I suppose....later on) is not a problem. The math is fine. A fuse lit at both ends will always burn out in half the time that the same fuse lit at just one end will. This applies to the first fuse burnt in full (yielding 60 seconds) and the second fuse that is already underway for 60 seconds and then has the other end lit. That remaining segment of fuse will burn out 30 seconds after the other end is lit.

 

[bellenuit]

Two easy ones.

#1

Swiss Rail run a non stop goods train on a single line track between two towns separated by a large mountain range (part of the Alps). The train leaves every morning at 9:00 am and reaches its destination at 5:00 pm. Because of the danger of fallen rocks and snow drifts, the train only travels during the day, so it makes the return journey the following morning at 9:00 am reaching its destination at 5:00 pm. Because of the nature of the path, it doesn't travel at a constant speed, but will be slow going up hill and a lot faster going down hill and of course these same hills will assist or resist in the opposite way on the return journey.

Both towns are in flat valleys that give the driver sufficient time to moderate the speed upwards or downwards so that the train always arrives exactly on time, should the train be behind or ahead of schedule. Being exact to the scheduled arrival time is more important to Swiss Rail than being early

The question. One Monday morning the train makes its outbound journey and the following day it returns. Without the driver intentionally trying to do so, will the train (using its exact middle point as reference) ever be at the same point on the tracks at the same clock time on that Tuesday's return journey as it was on Monday's outward journey?

and

#2.

A blind person is given a standard (52) pack of playing cards that have 10 cards facing up and the remaining facing down. The cards have been shuffled, so the upward facing cards are distributed randomly throughout the deck. The pack is brand new and there is no way to tell if a card is facing up or down by feel.

The blind person is asked if he could deal out two separate stacks of cards (not necessarily with the same number of cards in each), so that both stacks have exactly the same number of upward facing cards.

Without the assistance of anyone else or a device of any sort, can he do it?

 

[barney]

Nailed it.

LOL ..... great thread!! Humour and brain power

 

[cynic]

Two easy ones.

#1

Swiss Rail run a non stop goods train on a single line track between two towns separated by a large mountain range (part of the Alps). The train leaves every morning at 9:00 am and reaches its destination at 5:00 pm. Because of the danger of fallen rocks and snow drifts, the train only travels during the day, so it makes the return journey the following morning at 9:00 am reaching its destination at 5:00 pm. Because of the nature of the path, it doesn't travel at a constant speed, but will be slow going up hill and a lot faster going down hill and of course these same hills will assist or resist in the opposite way on the return journey.

Both towns are in flat valleys that give the driver sufficient time to moderate the speed upwards or downwards so that the train always arrives exactly on time, should the train be behind or ahead of schedule. Being exact to the scheduled arrival time is more important to Swiss Rail than being early

The question. One Monday morning the train makes its outbound journey and the following day it returns. Without the driver intentionally trying to do so, will the train (using its exact middle point as reference) ever be at the same point on the tracks at the same clock time on that Tuesday's return journey as it was on Monday's outward journey?

Are both towns at the same level?

and

#2.

A blind person is given a standard (52) pack of playing cards that have 10 cards facing up and the remaining facing down. The cards have been shuffled, so the upward facing cards are distributed randomly throughout the deck. The pack is brand new and there is no way to tell if a card is facing up or down by feel.

The blind person is asked if he could deal out two separate stacks of cards (not necessarily with the same number of cards in each), so that both stacks have exactly the same number of upward facing cards.

Without the assistance of anyone else or a device of any sort, can he do it?

Do both stacks have to add up to the entire deck?

 

[Tisme]

Nailed it.

yeah both of you

 

[keithj]

A blind person is given a standard (52) pack of playing cards that have 10 cards facing up and the remaining facing down. The cards have been shuffled, so the upward facing cards are distributed randomly throughout the deck. The pack is brand new and there is no way to tell if a card is facing up or down by feel.

The blind person is asked if he could deal out two separate stacks of cards (not necessarily with the same number of cards in each), so that both stacks have exactly the same number of upward facing cards.

Without the assistance of anyone else or a device of any sort, can he do it?

Deal 6 equal stacks each containing 4 cards (& chuck the rest in the bin)

There is guaranteed to be at least two stacks containing the same number of cards.

eg worst possible scenario is 1st, 2nd, 3rd & 4th contains 1,2,3 and 4 upward facing cards, which means the 5th & 6th contain none.

Does this solution satisfy the the terms of the problem statement ?


Or the trivial solution is to deal 52 stacks each containing 1 card..... ???

 

[bellenuit]

Are both towns at the same level?

Not necessarily.

Do both stacks have to add up to the entire deck?

Yes

 

[bellenuit]

Deal 6 equal stacks each containing 7 cards.

There is guaranteed to be at least two stacks containing the same number of cards.

eg worst possible scenario is 1st, 2nd, 3rd & 4th contains 1,2,3 and 4 upward facing cards, which means the 5th & 6th contain none.

Does this solution satisfy the the terms of the problem statement ?

Its meant to be just two stacks in total. Though I must concede that you have had him deal two stacks each with the same number of upward facing cards (0). Technically that is a solution, but I think pushing the boundaries of what the wording said:

The blind person is asked if he could deal out two separate stacks of cards (not necessarily with the same number of cards in each), so that both stacks have exactly the same number of upward facing cards.

A couple of marks for lateral thinking though.

 

[keithj]

Its meant to be just two stacks in total. Though I must concede that you have had him deal two stacks each with the same number of upward facing cards (0). Technically that is a solution, but I think pushing the boundaries of what the wording said:

The blind person is asked if he could deal out two separate stacks of cards (not necessarily with the same number of cards in each), so that both stacks have exactly the same number of upward facing cards.

A couple of marks for lateral thinking though.

hmmm... thought so .

I think the bolded bit above is the key. You'd expect exactly 2 decks with an equal number of upturned cards to be exactly 5 in each, however....








... by dealing 26 cards into one stack & 26 into the other stack, but flipping each 2nd card, you'd end up with around half the face up cards now facing down, and around half the face down ones now facing up...

... and closer to an equal number of face up cards in each stack.


getting closer ?

 

[bellenuit]

getting closer ?

Depends where you are going.

Just to clarify for all doing this. He deals out all the 52 cards into two separate heaps so that each have the same number of upward facing cards. There need not be the same number of cards in each heap.

 

[skc]

The question. One Monday morning the train makes its outbound journey and the following day it returns. Without the driver intentionally trying to do so, will the train (using its exact middle point as reference) ever be at the same point on the tracks at the same clock time on that Tuesday's return journey as it was on Monday's outward journey?[/B]

Yes... Visualise that you have 2 trains both departing at the same time, one from either town, running towards each other. They will definitely run into each other at one moment in time through the journey.

This would have been harder to work out but the fuse question before was too similar.