Four Snails Traveling on a Plane -- Do they cross paths?

[pennystuck]

Homework Statement

Four snails travel in uniform, rectilinear motion on a very large plane surface. The directions of their paths are random, (but not parallel, i.e. any two snails could meet), but no more than two snail paths can cross at any one point. Five of the (4 × 3)/2 = 6 possible encounters have already occurred. Can we state with certainty that the sixth encounter will also occur?

Relevant Equations

N/A

Since the snails are all in the same plane and their paths are not parallel, shouldn’t the solution simply be yes since the non-parallel lines in the same plane will intersect at some point? This answer seems too simple so I’m unsure if I‘m missing out on a detail.

 

[PeroK]

Homework Statement:: Four snails travel in uniform, rectilinear motion on a very large plane surface. The directions of their paths are random, (but not parallel, i.e. any two snails could meet), but no more than two snail paths can cross at anyone point. Five of the (4 × 3)/2 = 6 possible encounters have already occurred. Can we state with certainty that the sixth encounter will also occur?
Relevant Equations:: N/A

Since the snails are all in the same plane and their paths are not parallel, shouldn’t the solution simply be yes since the non-parallel lines in the same plane will intersect at some point? This answer seems too simple so I’m unsure if I‘m missing out on a detail.

Have you tried to find a pattern where the last encounter does not occur?

 

[Delta2]

Hmmm, sorry for this question but this problem is from a physics book or a combinatorial analysis book?

 

[PeroK]

I assumed it was from a malacology textbook. That being the study of molluscs.

 

[Delta2]

I assumed it was from a malacology textbook. That being the study of molluscs.

Yes ok you don't have to be ironic e hehe, it isn't exactly combinatorial analysis but some sort of combinatorics/probability theory.

 

[berkeman]

Yes ok you don't have to be ironic e hehe

You should have seen the thread title before I fixed it up a bit... Reminded me of this:

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

 

[jbriggs444]

What is meant by "encounter"? Are we talking about an intersection of two straight line snail paths on the plane or are we talking about a mid-plane collision where the two snails arrive at the same point at the same time?

 

[kuruman]

You should have seen the thread title before I fixed it up a bit...

What did you fix? Shouldn't the title be "Do their paths cross? instead of "Do their cross paths?"

What is meant by "encounter"? Are we talking about an intersection of two straight line snail paths on the plane or are we talking about a mid-plane collision where the two snails arrive at the same point at the same time?

Also, what does "cross" mean? Are we to assume that the paths are semi-infinite and begin at the starting point of each snail or are they infinite, extending in both directions?

 

[jbriggs444]

What did you fix? Shouldn't the title be "Do their paths cross? instead of "Do their cross paths?"

Also, what does "cross" mean? Are we to assume that the paths are semi-infinite and begin at the starting point of each snail or are they infinite, extending in both directions?

If my crystal ball is working right, exactly five intersection points are of one-way infinite trajectories into the past and the question focuses on the potential existence of a sixth intersection point somewhere in the infinite future. It's not just paths crossing. It's trajectories intersecting at events.

I think I can see my way clear toward a useful intuition from which the answer can be read off. The key is that it is not a problem in two dimensional geometry.

 

[berkeman]

What did you fix? Shouldn't the title be "Do their paths cross? instead of "Do their cross paths?"

Oopsies. Fixed now, thanks.

 

[Delta2]

If my crystal ball is working right,

Would like to have one too sir, do they sell any in Amazon or anywhere else in web?

 

[kuruman]

think I can see my way clear toward a useful intuition from which the answer can be read off. The key is that it is not a problem in two dimensional geometry.

In my naive way of thinking this as a Euclidean plane geometry problem (the statement of the problem does mention a plane surface), I drew two snapshot pictures (see below). In each, the snails start at the circles and proceed with equal speeds. The arrowheads are their locations a very long time later when the snapshots were taken and the lengths of th arrows are equal.

The picture on the left shows that there is no crossing in the future. The picture on the right shows that merely by moving the starting position of one snail, there could be a sixth crossing. I can't believe it is that simple and I suspect that I missed something. That is why I asked if the paths are semi infinite.

 

[Delta2]

Hmm if the problem was asking for the probability that the 6th crossing will happen given that 5 crossings have happen then is the answer easy? Is it something like 0.5?
@kuruman something doesn't look quite right with your diagrams, the snails can't move all with the same velocities or at least don't all start moving at the same time. Because for example if we mark X the point where A and C meet, then it is $AX>CX$ so if they started at the same time $$\frac{AX}{t}>\frac{CX}{t}\Rightarrow v_A>v_C$$. But the problem isn't clear if all the snails have the same speed and if they start at the same time.

 

[kuruman]

Yes, you are correct. I didn't think the drawing through properly.

 

[jbriggs444]

Hmm if the problem was asking for the probability that the 6th crossing will happen given that 5 crossings have happen then is the answer easy? Is it something like 0.5?
@kuruman something doesn't look quite right with your diagrams, the snails can't move all with the same velocities or at least don't all start moving at the same time. Because for example if we mark X the point where A and C meet, then it is $AX>CX$ so if they started at the same time $$\frac{AX}{t}>\frac{CX}{t}\Rightarrow v_A>v_C$$. But the problem isn't clear if all the snails have the same speed and if they start at the same time.

My reading of the problem differs from that of @kuruman. My interpretation is that it is not enough that the paths cross. The snails must actually meet simultaneously at the events where the paths intersect.

We have the set of four snails each proceeding in a straight-line path and each at a constant speed but with no requirement that the four speeds are the same.

The fact that we have a two dimensional spatial plane plus one dimension of time suggests that we could model the trajectories as lines in a three dimensional space. But I do not want to spoil the answer to this interpretation of the problem by arguing too much further.

 

[Delta2]

Yes, you are correct. I didn't think the drawing through properly.

I think the drawing cases are still correct if we assume that the speeds are not all equal or that the snails don't start moving at the same time.

 

[willem2]

The fact that we have a two dimensional spatial plane plus one dimension of time suggests that we could model the trajectories as lines in a three dimensional space. But I do not want to spoil the answer to this interpretation of the problem by arguing too much further.

That works indeed. Of course you can still prove that the snails all move in the same plane in 3 dimensional space.

 

[pbuk]

Homework Statement:: Four snails travel in uniform, rectilinear motion

The motion of each snail can either be uniform or rectilinear, but not both.

 

[jbriggs444]

The motion of each snail can either be uniform or rectilinear, but not both.

"Definition of rectilinear motion
: a linear motion in which the direction of the velocity remains constant and the path is a straight line."

What stops straight line motion at a constant speed from being rectilinear?

 

[pbuk]

Ah, another example of being divided by a common language. In the UK https://dictionary.cambridge.org/dictionary/english/rectilinear
"moving in or formed from straight lines" i.e. a series of line segments.

Also by Occam's razor, movement in a straight line is "linear motion": rectilinear motion must be something else.

 

[Steve4Physics]

Encounter = two snails at the same place at the same time.
Call the snails A, B, C and D.
The 6 possible encounters are: AB, AC, AD, BC, BD and CD.

If there have been 5 encounters, one of the snails - say A - must have encountered the 3 others.

Now imagine being in A’s frame of reference. B, C and D (moving in straight lines) encounter A, though at different times. In A's frame, the trails of B, C, and D could look like this for example:

Clearly, B, C and D can’t encounter one another before or after they pass through A because their trails intersect only at A – their trails diverge as we move away from A.

The only way I can see to achieve more than 3 encounters is if the trails are colinear. But the question says the trails are 'random' and 'not parallel' – and surely colinear trails can't be considered random or parallel.

It’s really irritating me. Is there a mistake in the question or in my logic?

 

[kuruman]

It’s really irritating me. Is there a mistake in the question or in my logic?

I will send you a PM with the way I see it.

 

[pbuk]

Encounter = two snails at the same place at the same time.

This would depend on the size of the snails, which is not given. If the snails are of zero size then they can be at the same place at the same time with zero probability.

 

[pbuk]

It seems that a lot of people have put in a lot of work thinking about this poorly defined question, however the OP is not one of them.

 

[jbriggs444]

This would depend on the size of the snails, which is not given. If the snails are of zero size then they can be at the same place at the same time with zero probability.

The conditional probability of this, given the provisions of the problem statement is unity.

 

[jbriggs444]

Encounter = two snails at the same place at the same time.
Call the snails A, B, C and D.
The 6 possible encounters are: AB, AC, AD, BC, BD and CD.

If there have been 5 encounters, one of the snails - say A - must have encountered the 3 others.

Now imagine being in A’s frame of reference. B, C and D (moving in straight lines) encounter A, though at different times. In A's frame, the trails of B, C, and D could look like this for example:
View attachment 302886

Clearly, B, C and D can’t encounter one another before or after they pass through A because their trails intersect only at A – their trails diverge as we move away from A.

The only way I can see to achieve more than 3 encounters is if the trails are colinear. But the question says the trails are 'random' and 'not parallel' – and surely colinear trails can't be considered random or parallel.

It’s really irritating me. Is there a mistake in the question or in my logic?

My take:

Changing the frame of reference effectively projects the paths of B, C and D onto a two dimensional coordinate system stationary relative to A. It is a rotation in three dimensions followed by a projection into two dimensions. This can make non-parallel paths parallel. In the same way the rotation/projection makes A's path into a point.

It is then possible for A, B and C to be colinear and for the encounter events to be overtake events.

Rather than rotate and project in this manner and thereby losing information, the approach that I recommend is the one that @willem2 alludes to. If snails A, B and C all encounter one another at distinct events, their trails define a two dimensional plane in three dimensional space-time. Since D encounters at least two of these, it must be coplanar. Use this plane.

 

[Steve4Physics]

My take:

Changing the frame of reference effectively projects the paths of B, C and D onto a two dimensional coordinate system stationary relative to A. It is a rotation in three dimensions followed by a projection into two dimensions. This can make non-parallel paths parallel. In the same way the rotation/projection makes A's path into a point.

Yes, that explains it! Many thanks.

 

[pbuk]

I don't think this is correct. I don't think we can state with certainty that the sixth encounter will occur because a counter-example can easily be found - @kuruman almost had it in the right hand diagram in #12 but the origin of trail A needs to be closer to trail B so that the extension of trail D does not intersect trail A.

 

[kuruman]

I think I have a handle on this problem. Here is the streamlined version omitting all the hair-tearing and going back and forth. First I will restate the question as I understand it.

Four snails, A, B, C and D are moving with constant velocities on an infinite plane. No paths are parallel to each other. There are 6 possible intersections, AB, BC, CD, DA, AC and AD and 6 possible encounters, i.e. 6 different times at which two snails are simultaneously at the intersection of their paths. According to the statement of the question "Five of the (4 × 3)/2 = 6 possible encounters have already occurred. Can we state with certainty that the sixth encounter will also occur?" The short answer to the question the way it's pharesed is "yes and no". The uncertainty that goes with "no" is that because I think that the question should have been phrased "Five and only five of the (4 × 3)/2 = 6 possible encounters have already occurred. Can we state with certainty that the sixth encounter will also occur?" Then the answer is "yes". I explain why and then provide a method for writing the snail path equations that provide the desired results.

Assumptions
1. We are given a drawing of five points on a plane representing the five encounter points that have already occurred.
2. The sixth intersection point to be found must occur at a later time than any of the five earlier points.
3. We are free to choose the speed and direction of motion of each snail on the path it follows.
The goal is to find the sixth encounter point and demonstrate that the time associated with it is greater than the previous five.

Reasoning
If one pair of the six is removed, the five pairs have two snails (letters) with three encounters and two snails with two encounters. For example, if CD is removed, then A and B appear three times and C and D appear twice.
Conclusion: In the given pattern of five encounters, one can always draw two intersecting lines with three points each. In the example above, one line could be labeled "A" and the other "B" with points (AB, AC, AD) and {BA, BC, BD) intersecting at AB=BA.

With this in mind, there are three possible patterns that can be produced. They resemble the letters X, T and V. These are shown below. In black are the two three-encounter path segments for easy recognition of V, X or T. Red lines indicate how to add the remaining two paths. Numbers indicate the temporal order of encounters and arrows indicate the direction of motion. It should be clear by inspection that no solution can be found for a "T" pattern without situating the sixth point between either arm and the foot of the "T". Thus, five given points in a "T" pattern imply that the sixth intersection point has already occurred. This violates Assumption 2.

Thus, the answer to the question posed by the problem is that
1. If there is no discernible V, X, or T pattern in the given five points, there can be no five encounters.
2. If there is a T pattern, the sixth crossing is one of the given five points.
3. If there is a V or X pattern one can construct a solution and find the sixth point as outlined below.

Construction
We will now see how to construct a path given a possible diagram. Construction means finding four speeds and six crossing times and verify the results. The given quantities are the five points in the plane and, clearly, there is no uniques solution. It is advisable to develop a method that will make the task easy to accomplish. I chose the first diagram on the left. Then
1. I drew a horizontal line below the diagram, called it the x-axis, and extended the four segments until they intersected it.
2. I labeled the intersection, left to right, A, B, C and D. Encounter 1 is AB, encounter 2 is AC and so on.
3. I measured the angles that the segments formed relative to the positive x-axis, $0 \leq \theta_{\!}i \leq 2\pi$ and the ditances $x_{0,i}$ from the leftmost intersection.
4. The guarantee for encounters is that all four $y$-components be equal. I set the speed $v_{\!A}=1$ so that $v_{y,i}=v_A\sin\theta_{\!A}$. I calculated the remaining three speeds using $v_i=\dfrac{v_{\!A}\sin\theta_{\!A}}{\sin\theta_{i}}$ and $x$-components $v_{x,i}=v_{\!A}\sin\theta_{\!A}\cot\theta_{i}$.
5. The positions as functions of time are $$\mathbf{r_i}=(x_{0,i}+v_{x,i}~t)\mathbf{\hat x}+v_{y,i}~t~\mathbf{\hat y}.$$6. To verify that the temporal sequence is correct, the encounter times are found from $~~x_{0,i}+v_{x,i}~t_{ij}=x_{0,j}+v_{x,j}~t_{ij}.$

The figure below shows a 3d plot of the result. The third axis is time. I used the typical snail speed of 1 cm/min for $v_{\!A}$.

 

[mital]

Homework Statement: Four snails travel in uniform, rectilinear motion on a very large plane surface. The directions of their paths are random, (but not parallel, i.e. any two snails could meet), but no more than two snail paths can cross at any one point. Five of the (4 × 3)/2 = 6 possible encounters have already occurred. Can we state with certainty that the sixth encounter will also occur?
Relevant Equations: N/A

Since the snails are all in the same plane and their paths are not parallel, shouldn’t the solution simply be yes since the non-parallel lines in the same plane will intersect at some point? This answer seems too simple so I’m unsure if I‘m missing out on a detail.

Why to consider same speed of all snails?

 

[kuruman]

Why to consider same speed of all snails?

It does not have to be the same. See post #29.

 

[mital]

The occurrence of sixth collision will depend on the velocities of the snails since it may happen that although their paths intersect, they do not meet as they can be at different positions at that time. Also, another case may arise when t=0 is chosen such that the two of the paths have crossed at t<0. Then only 5 collisions will occur(t>0).

 

[haruspex]

The occurrence of sixth collision will depend on the velocities of the snails since it may happen that although their paths intersect, they do not meet as they can be at different positions at that time. Also, another case may arise when t=0 is chosen such that the two of the paths have crossed at t<0. Then only 5 collisions will occur(t>0).

It depends on each snail having a constant velocity. As proved in post #29, it does not depend on all having the same speed.

 

[TSny]

Unless I'm overlooking something, I think @Steve4Physics has a nice approach in post #21.

Given that the 5 encounters AB, AC, AD, BC, and BD occur, we want to prove that the sixth encounter CD must occur. Steve4Physics has essentially shown that in the frame of reference of snail A, snails B, C, and D must move along the same straight line. To recap the reasoning, suppose that in the A-frame, B and C do not move along the same straight line:

The black dot is snail A at rest in this frame, the brown circle is B, and the blue circle is C. B encounters A at the instant shown. C cannot also encounter A at this instant since that would imply that A, B, and C occupy the same spatial point in the original reference frame at this instant. But we are given that no more than two snail paths can cross at any one point of the original frame. It should then be clear that for any speeds of B and C in the picture above, B and C can never encounter each other if they are moving along different lines in the A-frame. We conclude that the only way that B and C can encounter each other is for B and C to move along the same straight line in the A-frame. The same argument holds for B and D. Thus, all three snails B, C, and D move along the same line in the A-frame. (They don’t necessarily move in the same direction along the line.)

@jbriggs444 already pointed out in post #26 that nonparallel paths in the original frame can become parallel in the A-frame. We see that the nonparallel paths of B, C, and D in the original frame become the same line in the A-frame.

We can also conclude that in the A-frame the velocity vectors of B, C, and D must all be different. If two of the snails had the same velocity in the A-frame, then the two snails would have the same velocity in the original frame. This would imply parallel paths for these two snails in the original frame. However, the problem statement assumes that none of the paths are parallel in the original frame.

So, in the A-frame, snails B, C, and D move along the same straight line with different velocities. In particular, C and D move along the same line with different velocities. Hence C and D must have an encounter, which is what we wanted to show. We conclude that 5 encounters imply 6 encounters.