Here is a little puzzle to think about.

The monk Gaito lives at the bottom of a hill just outside the ancient town of Huroko. One day, the monk leaves his home at 6:00am and makes his way up the hill along the narrow path that winds its way to the peak. The monk walks all day, occasionally stopping to rest and meditate, sometimes even retracing his steps for short distances, and arrives at the peak at 10:00pm. After spending the night fasting at the top, the monk starts the return journey at 6:00am the next morning and goes down the same narrow winding path, once again stopping occasionally or retracing his steps at various points along the way for contemplation. The monk returns to his home at the base of the hill at 10:00pm.

When you consider the monk’s two journeys, is itguaranteedthat there will be at least one point along the path where the monk will be located at the same time during the day for both trips?

You can put your solutions and reasons in the comments.

another stewart says

Consider two monks, at the same time one ascending the hill and one descending the hill, along the same path.They have to meet somewhere between the base and the summit. Mathematicians would probably like a bit more formalism, but this seems to demonstrate that it is guaranteed that there is a place where Gaito was present at the same time of day during both his ascent and descent.

Matt G says

If there were two monks walking both of these paths on the SAME day, they would pass each other, so yes.

Peter B says

Only at the two endpoints.

Rob Grigjanis says

Yeah, what another stewart said. To put it more mathematically, let the distance along the route be x. The position on the trip up is a monotonically non-decreasing function of t, starting at x=t=0 and ending at x=X, t=T. The position on the trip down is a monotonically non-increasing function of t, starting at x=X, t=0, and ending at x=0, t=T. The two functions must intersect.

Bruce says

The straightforward answer is yes, there is a common point, like two monks meeting, not an endpoint. But this assumes the monk’s wanderings remain on the main path. If the monk wanders on a side path to see a view, then he might not be on the main path at the same time as a return. So the reference to wandering needs clarification, if that is permitted. If we can’t know this was a rule, and don’t have other information, then there is no solution. But if we assume all detours off the main path are within experimental error of still being on the main path, then there must be a common time and place. The path may be described as narrow, which might imply this condition.

larpar says

It’s possible, but not guaranteed.

Holms says

It’s guaranteed to happen at least once. The puzzle is much clearer, obvious in fact, if we play those two opposite trips at the same time, or reformulate is as two people walking the path in opposite directions at the same time. At some point, at least once in the day but possibly more due to the backtracking and varied pace, the two people must pass each other. At which point they will be at the same point in the path as each other at the same time.

Rob Grigjanis says

larpar@6: No, it is absolutely guaranteed.Rob Grigjanis says

Re my #4: I missed the “sometimes even retracing his steps” bit. So the curves are not necessarily monotonical. But that just means they can be at the same place and time

more thanonce. But it’s still at least once.Rob Grigjanis says

Look at it this way; one function goes, in the x-t plane, from (0,0) to (X,T). The other curve goes from (X,0) to (0,T). They have to intersect at least once.

Reginald Selkirk says

To paraphrase Heraclitus, no man ever walks the same path twice. The path is different on the second day from the first. Perhaps there was an avalanche and the monk is walking 80 feet higher in altitude during the relevant portion of the return journey.

larpar says

It’s not two people, It’s one.

Say the distance is 40 miles and at a constant pace it takes 16 hours. After 8 hours you would be at the midpoint (the 20 mile mark). This would hold true on both 40 mile trips. At the midpoint it would be the same place and same time on both trips. Now say on the first 40 mile trip you took 5 minute breaks on each half of the walk. On the second 40 mile you took a 6 minute break on the first half and 4 minutes on the second half. Now, you would be a little short of the midway point after 8 hours. No longer at the same place and time.

M. Currie says

I think Bruce’s exception, while it might be more realistic, does not account for the way the problem is stated, as it’s specified that he returns on the same narrow winding path. He stops to meditate and retraces steps, but there is no suggestion that he strays from the main path while doing so.

I tried to think of any scenario, varied paces, etc., that would make it not work on a common path, but there is no way I can see where, if the two trips were begun simultaneously, the theoretical monks would not meet on the path, and if they do, then it has to be at the same time.

grahamjones says

@10: as long as the two functions are continuous. Then this theorem can be applied.

https://en.wikipedia.org/wiki/Intermediate_value_theorem

But physicists say that time and space are not continuous at tiny scales, so who knows.

Rob Grigjanis says

larpar@12: Draw axes on a piece of paper. Horizontal axis is t. Vertical axis is x. Now draw an arbitrary continuous curve from (0,0) to (T,X), where T and X are arbitrary positive values. Draw another curve from (0,X) to (T,0). The only restrictions on the curves are that they be continuous, totally within 0≤t≤T and 0≤x≤X, and that each value of t corresponds to a single value of x (i.e. you can’t be in two places at the same time). The two curveshaveto intersect at least once.Rob Grigjanis says

grahamjones@14:Physicists say “we don’t know if it’s continuous at tiny scales”.

Holms says

#5 Bruce, the monk is stated to walk forwards and backwards along this path, but is not stated to make side trips. The text makes no mention of ‘wander-‘.

#12 Larpar, yes it’s one person, but that person is making two journeys. The question being asked is whether those two journeys ever reach the same point in the path at the same time of day, and the best way to interrogate that question is to play both journeys at the same time, as if walked by two people coming in opposite directions.

larpar says

In my first example the monk is at the same place at the same time of day at the halfway point.

In my second example, tell me where the monk is at the same place at the same time of day.

Possible, not guaranteed.

Matt G says

grahamjones@14- Imagine a spherical monk with radius Planck length in a vacuum….

LykeX says

No longer at the same place and time,

at the midpoint, but then what?At eight hours, the first trip would be at the theoretical midpoint, but the second trip would be a bit short. So, they haven’t passed each other yet. Monk A is at the theoretical midpoint and Monk B is a bit short. Some distance remains between them and now they continue walking. What happens? At some point, they meet. Whatever time that is.

The question is not whether they intersect at any particular point, but whether they intercept at ANY point. And they must, at some point and time.

Reginald Selkirk says

@18 We don’t have to tell you where that point is, just that it exists.

I am probably wrong @11, but if I do win, I don’t have to split the jackpot with umpteen people.

Since it is a day later, the Earth has moved to a different spot in its annual orbit around the Sun, and the Sun will have moved in its orbit about the galactic center, so to call it the same “point” is a matter of semantics.

ardipithecus says

At the start of the monk’s return journey, he passes each point at an earlier time than he did the day before. As he nears the end of his journey, he passes each point at a later time than he did the day before. There must be a point in the journey where earlier than changes to later than. That is the point where he is at the same place at the same time.

There may be more than one such place/time because he sometimes retraces his steps.

larpar says

LykeX @20

There is no Monk B.

Yes, two people walking the same route but from opposite directions will pass each other, buts that’s not the scenario here.

Reginald Selkirk says

As a physicist, Mano understands that time is only another component of space-time, and that the “same path” on the second day is a different region of the chrono-synclastic infundibulum, making the premise of the question absurd.

OverlappingMagisteria says

larpar @23:

You are right that there is only one monk, but it works out the same way. Consider this modified scenario:

Imagine 2 Monks. Monk A spends the day walking up the hill. At the same time, Monk B walks down the hill. It should be obvious that at some point in time their paths cross. The next day, Monk A walks down the hill, but exactly replicates Monk B’s descent from the previous day. Do you see that, at some point in the journey, Monk A will be at the same place that he was 24 hours ago? At, the point where Monk A and Monk B met the day before?

Now here’s the key: remove Monk B from the story above. It makes no difference. Monk A will still be at the same point as he was 24 hours ago at some point in the downward journey.

Holms says

#23 larpar

It doesn’t matter whether one monk or two are doing this. That is the distraction the question sets up by talking about this as one monk on two days. Rather than looking at it as two monks A and B, look at it as walks/trips/journeys A and B. They are walked on different days, but the question only asks about the time of day and so they can take place on the same day or on different days with no impact on the question.

Inevitably, the two walks

mustpass each other; at that moment, the people are at the same place at the same time.Deepak Shetty says

The answer to the logic question is yes. The answer to the Geometric question is no -- You dont have to be at the same “point” at any time since the monk will have a stride

tuatara says

I don’t think that it is possible to

The puzzle states that on day one, the monk leaves his home at 06:00 and arrives at the peak at 22:00, then on day 2 he leaves the peak at 06:00 and arrives at his home at 22:00. It also states that he randomly stops for breaks and at times retraces some of his steps.

My reasoning is -but I am probably wrong):

For the monk to be at the same point along the path at the same lapsed time of his journey, unless the monk covers exactly half the distance in exactly half the time on both journeys, the distance or time for each journey will need to be inverted.

For example, if at 10:00 after 4 hours of travel (or 25% of the total time taken for the journey) the monk while ascending had covered 25% of the distance between home and peak, the descending monk would need to have covered 75% of the distance after 25% of the time to meet the guaranty. But the puzzle does not give us any indication that this state of inverted “velocity” (as a measure of distance travelled over time) is guaranteed.

While he will most certainly be at the same point along the path (as distance between home and peak, not as a point in space-time) at some time during each of his journeys, I don’t see how it can be

guaranteedthat this will occur after the same amount of elapsed time from the beginning of each journey.They can be described as two simultaneous journeys in opposite directions, can be plotted as such and the plots indeed intersect, but I cannot help but think of the adage, the origin of which I cannot remember, that states that the description of an elephant is not an elephant.

But as I said, I am probably wrong.

Intransitive says

I saw a similar puzzle that is about space, not time:

Take two pieces of square paper with numbers 1 to 100 in equal size squares. Place one on a surface, crumple the other into a ball, loose or tight doesn’t matter.

Is it possible to place the ball on the flat paper so that no number is directly above its counterpart? It turns out, you can’t. There will always be at least one number above itself. Sorry, I can’t find a link to the puzzle.

lochaber says

This comment thread is more amusing than I expected.

another stewart says

I didn’t expect anybody to have trouble seeing the equivalence between two monks on one day and one monk on two days, but ardipithecus @ 22 has given an explanation without the distraction of the two monks,

Dan Schwartz says

The reasoning expressed above, for example in messages 15 and 22, seems at first to be ironclad. There must be a point on the path that is passed at the same time. Not to spoil the fun, I will just say that there are a couple of loopholes to be considered. The loopholes do not involve stepping off the path, consideration of the earth’s motion, or weird quantum phenomena.

Reginald Selkirk says

Wait a minute…

1) What if the two consecutive days involved are the last day of Standard Time and the first day of Daylight Savings Time? This gives us a one hour offset in the two days. DST always takes effect at 2:00 AM, so there is no discontinuity during either day’s journey. So it still seems reasonable that there is an isochronous point, but does it affect the

guarantee?—

2) How about this: Gaito’s home at the bottom of the hill and the resting point at the top of the hill are in different time zones. In fact, the path may wander back and forth between the two time zones. This gives us at least one discontinuity of an hour during each journey to work with. However, time zones are decided by geography, so for any point at which the upward and downward journeys met, the same time zone would apply.

—

3) Combine (1) and (2). I don’t know if it makes any real difference, but it sure confuses matters.

Reginald Selkirk says

Does the specification that the return journey follows “the same… path” rule out a change in the path due to an earthquake?

Rob Grigjanis says

Dan Schwartz@32:If, instead of a monk walking a narrow path, we consider a bead sliding along a string, do the same loopholes apply?

Reginald Selkirk says

Loopholes:

We will use the “two monks in one day” framing, which seems to be valid, for clarity.

We are told the monk “leaves his home” at a certain time, and that he “arrives” at a certain time. On the way down, he “starts his journey” at the specified time, and “returns to his home” at a specified time. But sometimes he stops and sometimes he might reverse. So is it possible that he actually leaves his starting point more than once during the day, or arrives at the destination more than once during the day? It does not seem to be ruled out by a strict reading.

So then, Monk A could start out, return to his home for a time, during which time Monk B starts and finishes his opposing journey. Then Monk A starts out again. After that Monk B reverses for a while.

There is still a point at which they meet, but that point is one of the endpoints, either at the monk’s home or on the top of the peak. The answer would then hinge on whether the two endpoints are considered to be “along the path” as specified by a strict reading of the riddle.

Holms says

But that would have them at the same position on the track at the same time: the start point for A / the end point for B i.e. home.

Rob Curtis says

you can generalize this problem to higher dimension. Walking the path is essentially the 1-D example. The intermediate value theorem proves it’s true.

another example, you can show that there exists, somewhere on earth, two antipodal points that have the same temperature.

google this and go down a rabbit hole:

Borsuk-Ulam Theorem

mnb0 says

“is it guaranteed that ….

No. A point doesn’t have dimensions. A path has two; the other one is width.

It’s exactly the two monks example that makes this clear. Two monks can’t occupy the same point at the same time. So even when they pass each other they are not on the same point. In the case of one monk going up and down it is possible, but if the monk systematically walks at the right (or left) side of the path they never will pass the same point.

Mathematically speaking: as the path has two dimensions (length and width) we can use a coordinate system; one of the axes may not be a straight line. The coordinate that expresses the length of the path is guaranteed to be the same at the same time at least once, but the other not. And in a twodimensional coordinate system this means not the same point.

Rob Grigjanis says

mnb0@39: You missed some crucial wording (my bolding).In other words, it’s referring to one of the two dimensions of the path, the length.

The nit-picking about the problem is just silly, IMO.

file thirteen says

The answer is yes. Not even daylight saving’s time (Reginald #33) screws things up.

The monk walks up the path on day 1. On the same path, the monk walks down on day 2. Even if daylight saving starts and monk-day-2 starts an hour earlier than monk-day-1, monk-day-2 does not finish his journey until after monk-day-1 has started; that is to say, their paths still overlap during the day during a 15 hour window (6am to 9pm -- if daylight saving ended it would be 7am to 10pm) because each (in a temporal sense) monk’s start and end times are fixed (meaning one can’t finish before the other begins).

So mapping the monk’s journey going up in day 1 onto the journey going down on day 2, the point at which monk-day-1 passes monk-day-2, which must exist for each to reach the other end, is by definition at the same time and point. Therefore, guaranteed. 🙂

Reginald Selkirk says

@37 Yes, there would still be a meeting at one of the endpoints, and the success of the loophole would rely on whether one of the endpoints meets a strict reading of the relevant verbiage “along the path.”

I think this loophole fails, but I thought I would give it a try.

Thanks for reading it so closely.

file thirteen says

Reginald #42

I think the loophole fails anyway, but a stronger reason is that the monk doesn’t remain motionless at either endpoint. The puzzle says the monk

leaves his homeat 6am andarrives at the peakat 10pm. If the monk were to wait for an hour at the start point before leaving, that would be leaving at 7am, not 6am. I think it’s defined that as soon as the monk leaves his home he’s considered to be on the path. Logic is similar for the other end.Reginald Selkirk says

@43: He leaves the starting point (at either end) at 6 AM, but he sometimes reverses and sometimes rests. So my reasoning was that he could leave at 6 AM, reverse to the starting point, and rest for a while. He could in fact do this several times, we have no idea how long the journey would be w/o reversing and resting, perhaps it is only a 5 minute trip. That takes care of the objection about 7 AM.

I do agree that the endpoints are probably considered part of the path.

Khakaure Senusret says

Going downhill should be quicker than going uphill, so Gaito is taking more time to rest and meditate on his downhill journey.

So there is no guarantee that Gaito would be located at the same time and space on both journeys.

file thirteen says

Reginald #44

Maybe, but if the monk returns home to get his cellphone after starting out, can you really say he must have started his journey from the first time he left, not restarted it? Similarly, when he arrives at the end, and goes back to get said cellphone where he left it on the path while photographing a wild mathematician, is that still part of his total journey? Semantics I suppose.

——————————————————————-

As usual, I think of more things once I’ve made a comment. Another way to view this problem is with the monk’s journeys plotted on a graph, ranging from (bottom of hill) to (top of hill) on the (say) horizontal axis and (time of journey start in a day) to (time of journey end in the same day) on the vertical axis. The first day, the monk starts at (bottom of the hill, 6am) and proceeds to (top of the hill, 10pm). The second day, the monk starts at (top of the hill, 6am) and proceeds to (bottom of the hill, 10pm). You can draw lines, as convoluted as you like, connecting the start and end points for each journey, but it should soon become evident that the lines must intersect.

Rob Grigjanis says

Khakaure Senusret@45: Imagine there is a spot from which one can see the entire path. From there, film the monk’s journey from home to the hilltop. The next morning, film the monk’s journey down. Now superimpose the two films. You will see a monk journeying from home to hilltop, and another journeying from hilltop to home. Their speeds don’t matter at all. The two images of the monk (one going up, one going down)mustmeet at some point.bargearse says

Aaand, now I’m in a massive argument with my boss over this puzzle. I say it is guaranteed for the reasons stated by others here, he’s adamant it’s impossible. Lunch today should be fun (for us at least, the others who eat lunch with us will not enjoy it)

Reginald Selkirk says

@48 Sometimes it doesn’t pay to be smarter than the boss.

Rob Grigjanis says

bargearse@48: I had protracted and sometimes heated arguments with colleagues about the Monty Hall problem. Luckily, not with my boss.bargearse says

49 & 50

For the record while it may get heated the argument will be mostly good natured, we’ve been friends for over 30 years before I started working for him.

larpar says

In my earlier stab at this problem, I made an error, but I still think it’s possible to not be in the same place at the same time of day.

Here’s how I visualized it.

a——————--b—c—————-d

A is the start (the monk’s home) and d is end point (the top of the hill). In my earlier example I said that if the monk takes the same amount of meditation/backtracking time during each half of the one-way trip then the center point (b) would be where the monk is at the same place at the same time of day during the round trip. Before, I used 5 minute breaks (meditation/backtracking time) on each half of the one way trip, but it could be any amount of time (within reason). The key is that it needs to be the same amount of time during each half of the walk in order for center point be the place where you are at the same place at the same time of day on the round trip.

In my second example, I made the break times uneven. Same amount of break time but distributed unevenly between halves. A slightly longer break time on the first half of the return trip and you will end up at point c rather than at point b.

This is where I made my mistake. I asked where in my second example was the mesh point, thinking there wasn’t an answer. I was wrong. The answer was some where between points b and c.

Here’s where I throw the wrench back into the gears. Between b and c holds true if the monk continues walking right away, but let’s say the monk uses some of his remaining break time before he precedes from point c. More time than it takes to walk from point c to point b. That mesh point no long exists and I don’t see another one.

I probably messed up something again. I still object to using the two monk model because the situations are different. I don’t know if the difference has a formal name and I could be wrong about this too, but it’s the difference between simultaneous events and consecutive events.

Side note: This is the longest comment I’ve ever made anywhere. Sorry.

tuatara says

I just want to offer my thanks to you all here.

I considered all the comments above, and after reformulating the puzzle in a way that made it possible for me to play the two journeys through in my head more clearly, I now see where my reasons for believing it is not possible to guaranty that the two journeys will be at the same point between a and b at the same time of day were leading me off the path.

Hey, I learned something about the elephant!

This is one of the reasons why I like participating in FTB!

Rob Grigjanis says

larpar@52: You’rewayoverthinking this. Imagine the scenario in my #47, where you film both the journey up and the journey down, and then superimpose the two films. The two ‘versions’ of the monk have to meet at least once, regardless of any breaks/retracing. One starts at the bottom and ends up on the top, the other starts at the top and ends up at the bottom. There’s no way around that. Under what circumstances do you imagine they wouldn’t meet?bargearse says

Further to me @48 & 51

The stairs leading up to my office are now numbered with post it notes and the boss has cornered someone else so we can film them going up and down the stairs. I’m not sure how this is supposed to prove his point but it is a terrific distraction from actual work.

larpar says

Rob @54

I think that everyone else is over thinking and my model is pretty simple. : )

I can’t think of circumstance where they don’t meet using your method, however I don’t think that method is valid. I can’t articulate why. But something in the back of my head, something I’ve heard before, tells me there is something wrong with comparing simultaneous and consecutive events. Sorry I can’t be clearer.

Dan Schwartz says

So here’s what I had in mind when I mentioned loopholes. Reginald brought up daylight saving time. In the U.S., the clocks change at 2 am. But maybe the monk lives somewhere in the world where daylight saving time officially starts at, say, noon. Now imagine what that looks like on the graph described by Rob in message 15. The clocks get set one hour ahead, so that means there’s a discontinuity in that day’s curve — it jumps horizontally (along the time axis) by 1 hour. There is a 1 hour gap in the curve. It would not be correct to represent that part of the curve as a horizontal line through the gap — there’s simply a gap in which there’s no curve. This is true because, if daylight saving time starts at noon, the times from noon to 1 pm do not exist at all on that day. And now the other curve can pass through that gap, and there’s no intersection.

If there’s no place in the world where daylight saving time starts during the 6 am -- 10 pm window, there’s still one other possibility — a leap second. It would have to be a negative leap second (in which clocks are advanced by 1 second), which so far has not occurred. But if we have one, and it occurs during the monk’s walk, it will create a 1 second gap through which the other day’s curve can pass.

Rob Grigjanis says

larpar@56: In the superimposed film, you’ll agree that you will see the ‘two monks’ apparently meet at some point on the path? We’ll call that point C.Both films start at 6 am and end at 10 pm, albeit on different days. So the films are synchronized. Let’s say they meet when the clocks say 1 pm.

That means that, on the first day, the monk reached C at 1 pm, on the way up. On the second day, the monk reached point C at 1 pm, on the way down. Same point C, same time. There’s no ‘method’ in that conclusion.

What do you think is wrong with that?

John Morales says

Dan:

Or imagine a part of the world where there is no daylight saving time.

(I live in such a place)

Sheesh, what a weak objection. I’d’ve hoped you’d at least gone with sidereal time considerations or suchlike.

—

I was gonna comment, but I read #1 and figured it rendered any further comments moot.

—

Anyway, if one wants to get truly picky, one could refer to the Ship of Theseus paradox.

(Yeah, I know. A bit philosophical)

file thirteen says

Dan #57

It would be a very strange place that implemented the daylight saving clock adjustment during the day! That said, even if that were the case, if the monk were viewed from a location which didn’t have a discontinuous time function then a precise meeting time would still be guaranteed… there. But I guess it’s true that if you measure time with a discontinuous function, that the paths wouldn’t have to meet at the same time. In fact all the monk has to do to easily avoid that is to adjust his timekeeping device forward or backward by half a day once he reaches the peak.

KG says

It’s worth considering that if you can’t be clearer,

it may be because you’re wrong.mnb0 says

@40 RobG: mathematicians think it quite important to clarify how many dimensions the space has we are talkling about. If you call that nitpicking it says more about you than about me, especially as I already made clear that the answer is yes in a onedimensional space.

Not that I’m surprised, given the content of many of your comments.

Reginald Selkirk says

Does such a place exist?

No. Boom!

And leap seconds always appear near midnight.

Reginald Selkirk says

@57 @63 -- My mistake. The leap seconds take place at one instant around the world, so they occur at different times of day in different time zones.

Wikipedia

Dan Schwartz says

Hey, we’re just trying to have some fun here, right? To me the most interesting aspect of the DST/leap second loophole is not whether it meets some legal definition of solving the problem. It’s more interesting to work through the head-screw of determining whether it actually works. I think the graphical approach (the gap in the curve) is helpful to visualize the result.

Exercise for the reader (if anyone cares) — Consider the superimpose-the-two-films approach mentioned earlier. Can you use it to show that this loophole works? I claim the answer is yes, if you do it right.

Rob Grigjanis says

mnb0@62: For someone so pedantically concerned with dimensionality, you seem to have forgotten that the monk himself is a 3D extended body, not a point particle. If one defines his instantaneous position as, say, his centre of mass (including robe, sandals and tote bag), then one can say with near certainty that his positions going up and down will never be the same.That’show you do proper nit-picking.Reginald Selkirk says

I think we have probably exhausted the issue, so it is about time for the “official” solution.

larpar says

KG @61

I have considered being wrong. I even admitted being wrong about one aspect of my reasoning, but I think I have resolved that error. Maybe not. I could still be wrong.

One other reason that I’m not yet ready to admit defeat is that this is a “logic problem”. Usually in these types of questions there is some kind of twist that makes the obvious answer wrong. I’m holding out for the twist. It may never come. : )

larpar says

Another thought on this. The two monk or one monk superimposed solutions only guarantee than the monk will be at the same point during the two legs of his journey. It doesn’t necessarily need to be at the same time of day. It can be at the same time of day, but it’s not guaranteed. You can vary your pace and avoid being at the same place at the same time of day.

That’s it. My brain hurts. I’m done. (until an “official” solution is posted : )

[You many want to look at my new post where I discuss why some may find the arguments that the paths are guaranteed to be at the same point at the same time unpersuasive. I’d be interested in your response. -- Mano]

Reginald Selkirk says

Wrong. They are walking the same path in opposite directions, so there are many points (an infinite number, or if you want finite subdivisions: all of them) where they overlap positions. The place where the two curves overlap is where they are at the same position at the same time of day.

Rob Grigjanis says

larpar@69:Yes it does! In the superimposed films, the time is synchronized; they both start at 6 am, and finish at 10 pm. For each film, imagine the time displayed on the bottom, like subtitles; for the first film, the time is displayed at the bottom left, and for the second, it’s at the bottom right. The two times will be the same throughout the screening. So if you see the ‘two monks’ meet, it is

at the same timeas well as the same place.file thirteen says

The meta-puzzle is how to explain things in such a way that even the most doubtful agrees it’s guaranteed!

My best attempt is as follows.

Imagine two monks, one walking up the hill, one down. Each monk can’t get to the other end without meeting each other on the way. There’s no avoiding them passing each other if the monks want to get to the other end. Thus a meeting point is guaranteed to exist. And in that

momentof the monks passing each other, well that has to happensomewhere, and amomentis a time of day andsomewhereis a point on the path!As for one monk on two days, I think Rob Grigjanis nailed it in #47 when he wrote about superimposing films of both walks of the monk. There’s really no difference between filming two monks on different days and filming the same monk twice. Therefore, see above.

larpar says

I said I was done but I was wrong about that. My brain won’t stop working.

I think I’ve found the flaw with the 2 monks or the superimposed monk methods. It’s not really a flaw, it’s more of a not showing the entire picture kind of thing. The results are just as described, but the problem is the results are just one data point. You need more data points. You need to repeat the experiment.

Now if you do the experiment again without changing anything you will get the same results. So in order to get different results something has to change. The only thing we can change is the pace that the monk covers various sections of the path.

This is where the mentation/backtracking (breaks) inclusion enters into the picture. If there were no breaks (or if the breaks are evenly distributed) the results would stay the same in all the trials. However if you change distribution of breaks (change the pace) you can get different results. I can explain this further if I need to, but it’s already a long comment.

Let’s start the trials.

Trial 1. Just like Rob described above. You end up with a point some where along the path at the same time of day.

Trial 2. The first leg is just like it was in trial 1. You could even use the same film from trial 1. The change of pace is on the downhill leg. The monk decides to spend some extra time meditating somewhere on the upper portion of the trail.( He compensates for this by skipping some meditation time somewhere on the lower portion of the trail.) This would result in the crossing point being a little further up the hill than it was in trial 1. It would probably be at a different time than trial 1, but could still be at the same time of day as the uphill leg. The conclusion I reach from trial 2 is that by varying the pace in

various sections of the trail you can move the intersection point up or down the hill.

Trial 3. This time we are going to change the pace on both legs and pick the crossing point.

Pick a point some where along the path, say a third of the way up. You can design a walking/resting schedule to get to that point at just about any time with some practical restrictions (you need to leave time to get to the other end of the trail). Say you pick a time some where around mid mourning. Now plan out the return (downhill) leg. You can make a schedual that will get you to that at mid afternoon.

You could even devise schedules that would get you to the trial 3 crossing point at the same time of day as the crossing point in trial 1.

There now I’m done……….maybe. : )

another stewart says

@73: without retracing there is exactly one point on the path that is passed at the same time of day on both journeys. For any point on the path, you can devise a schedule for which that point is not the one point. What you can’t do is devise a schedule for which that is the case for every point.

Let the point which is passed at the same time of day be x. If you change the descent to arrive at x earlier then there will be a new point downslope of x which is passed at the same time of day. If you change the descent to arrive at x later then there will be a new point upslope of x which is passed at the same time of day. If you change the ascent to arrive at x earlier then there will be a new point upslope of x which is passed at the same time of day. If you change the ascent to arrive at x later then there will be a new point downslope of x which is passed at the same time of day.

To create a counterexample you can’t consider points in isolation; you have to consider the whole ensemble. That is you have to create two graphs of position against time of day where the two graphs don’t cross. Try that on a piece of graph paper and you should realise that provided that graphs are continuous you can’t do that.

file thirteen says

I guess it wasn’t good enough.

larpar #73

You seem to be attempting to solve a different problem than the one described. It’s trivial to demonstrate that if the monk was shown to “cross” at a particular time and place, that if he were to repeat the walks on subsequent days that it wouldn’t have to happen at the same time or place. But that’s not the point. The point is that there cannot be a scenario in which

nocrossing occurs, andanycrossing is a solution to the problem.Holms says

#73 larpar

You can randomise the walks a thousand times, a million, infinity, and you can choose from them any upward journey against any downward journey. There will always be a point at which the playback of those two randomly selected random walks cross, that is there will always be a point where the upward journey and the downward occupy the same point of the path at the same time of day. Always.

Mano Singham says

This thread is interesting for many reasons. I have created a new post where I try to explain why some may find the argument that there is a guaranteed crossing point unpersuasive.