Re: [math-fun] Miscounting errors
I recall a question from my high school AP math or math SAT exam circa 1964: Two swimmers are swimming laps of a 25' pool, but they swim at different (constant) speeds, say S1 & S2. (Assume that they bounce elastically off the pool ends!) How many times do they pass one another in 30 minutes? This problem is pretty complicated, as one swimmer could pass the other in either the same or opposite direction. The time is also long enough, that you can't simply simulate the first 1,2,3 laps, but must come up with a more general answer. I don't recall how I solved it back then, but I must have gotten the right answer, because I did very well on that test. At 05:06 AM 11/11/2017, James Propp wrote:
Anyone have a favorite puzzle in which miscounting plays a role?
My favorite is the classic bookworm puzzle (see https://math.stackexchange.com/questions/1271651/how-is-this-true-bookworm-p... ).
Another example: A man was born in 50 BC and died on the same day in 50 AD. How old was he when he died?
I'm especially interested in puzzles that lend themselves to solvers committing fencepost errors, and off-by-one errors more generally.
Jim Propp
Fun question! Assuming that each swimmer complete a whole number of laps in the allotted time (starting and ending at the same end of the pool), I think the answer is the number of laps swimmer #1 completes plus the number of laps swimmer #2 completes minus 1. Do others agree? Jim Propp On Saturday, November 11, 2017, Henry Baker <hbaker1@pipeline.com> wrote:
I recall a question from my high school AP math or math SAT exam circa 1964:
Two swimmers are swimming laps of a 25' pool, but they swim at different (constant) speeds, say S1 & S2. (Assume that they bounce elastically off the pool ends!)
How many times do they pass one another in 30 minutes?
This problem is pretty complicated, as one swimmer could pass the other in either the same or opposite direction. The time is also long enough, that you can't simply simulate the first 1,2,3 laps, but must come up with a more general answer.
I don't recall how I solved it back then, but I must have gotten the right answer, because I did very well on that test.
At 05:06 AM 11/11/2017, James Propp wrote:
Anyone have a favorite puzzle in which miscounting plays a role?
My favorite is the classic bookworm puzzle (see https://math.stackexchange.com/questions/1271651/how-is- this-true-bookworm-puzzle ).
Another example: A man was born in 50 BC and died on the same day in 50 AD. How old was he when he died?
I'm especially interested in puzzles that lend themselves to solvers committing fencepost errors, and off-by-one errors more generally.
Jim Propp
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com <javascript:;> https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
Considering a short window in which both swimmers hit the same wall at nearly the same time: If the slow swimmer reaches a pool end an instant before the fast swimmer, there are two passes: the slow swimmer hits the wall, bounces, passes the fast swimmer in the opposite direction, then very shortly after the fast swimmer bounces and overtakes the slow swimmer, passing the slow swimmer once again. If the fast swimmer reaches a pool end an instant before the slow swimmer, there is one pass: The fast swimmer hits the wall, bounces, passes the slow swimmer, and keeps going, with the slow swimmer falling further and further behind. But what if they reach a pool end at the exact same moment? Does that still count as the fast swimmer passing the slow swimmer? Certainly the fast swimmer overtakes the slow swimmer, but it's not clear that the fast swimmer "passes" the slow swimmer, since the fast swimmer remains furthest from the wall at all times except at the moment of contact, so in a sense it's not much different than if the fast swimmer turned around just before reaching the wall, in which case there would clearly be no passing. I think this case could be argued either way. Tom Henry Baker writes:
I recall a question from my high school AP math or math SAT exam circa 1964:
Two swimmers are swimming laps of a 25' pool, but they swim at different (constant) speeds, say S1 & S2. (Assume that they bounce elastically off the pool ends!)
How many times do they pass one another in 30 minutes?
This problem is pretty complicated, as one swimmer could pass the other in either the same or opposite direction. The time is also long enough, that you can't simply simulate the first 1,2,3 laps, but must come up with a more general answer.
I don't recall how I solved it back then, but I must have gotten the right answer, because I did very well on that test.
At 05:06 AM 11/11/2017, James Propp wrote:
Anyone have a favorite puzzle in which miscounting plays a role?
My favorite is the classic bookworm puzzle (see https://math.stackexchange.com/questions/1271651/how-is-this-true-bookworm-p... ).
Another example: A man was born in 50 BC and died on the same day in 50 AD. How old was he when he died?
I'm especially interested in puzzles that lend themselves to solvers committing fencepost errors, and off-by-one errors more generally.
Jim Propp
Consider a plot of distance v. time. This is a long strip 25' wide and 30 minutes long. But you can *unfold* it -- e.g., make it 50' wide and 30 minutes long, which means that the wall doesn't exist and the swimmer keeps going. The elastic bounce has become a straight line. At 01:41 PM 11/11/2017, Tom Karzes wrote:
Considering a short window in which both swimmers hit the same wall at nearly the same time:
If the slow swimmer reaches a pool end an instant before the fast swimmer, there are two passes: the slow swimmer hits the wall, bounces, passes the fast swimmer in the opposite direction, then very shortly after the fast swimmer bounces and overtakes the slow swimmer, passing the slow swimmer once again.
If the fast swimmer reaches a pool end an instant before the slow swimmer, there is one pass: The fast swimmer hits the wall, bounces, passes the slow swimmer, and keeps going, with the slow swimmer falling further and further behind.
But what if they reach a pool end at the exact same moment? Does that still count as the fast swimmer passing the slow swimmer? Certainly the fast swimmer overtakes the slow swimmer, but it's not clear that the fast swimmer "passes" the slow swimmer, since the fast swimmer remains furthest from the wall at all times except at the moment of contact, so in a sense it's not much different than if the fast swimmer turned around just before reaching the wall, in which case there would clearly be no passing. I think this case could be argued either way.
Tom
Henry Baker writes:
I recall a question from my high school AP math or math SAT exam circa 1964:
Two swimmers are swimming laps of a 25' pool, but they swim at different (constant) speeds, say S1 & S2. (Assume that they bounce elastically off the pool ends!)
How many times do they pass one another in 30 minutes?
This problem is pretty complicated, as one swimmer could pass the other in either the same or opposite direction. The time is also long enough, that you can't simply simulate the first 1,2,3 laps, but must come up with a more general answer.
I don't recall how I solved it back then, but I must have gotten the right answer, because I did very well on that test.
At 05:06 AM 11/11/2017, James Propp wrote:
Anyone have a favorite puzzle in which miscounting plays a role?
My favorite is the classic bookworm puzzle (see https://math.stackexchange.com/questions/1271651/how-is-this-true-bookworm-p... ).
Another example: A man was born in 50 BC and died on the same day in 50 AD. How old was he when he died?
I'm especially interested in puzzles that lend themselves to solvers committing fencepost errors, and off-by-one errors more generally.
Jim Propp
Understood, but it seems to me that that's a different problem. Under that model, there is no passing at all, since the fast swimmer immediately overtakes the slow one and remains in the lead until the end. The question is asking about one swimmer physically moving past the other swimmer, which can occur in two circumstances: If they are moving in the same direction, or if they are moving in the opposite direction. I don't see bouncing at the same time as either of those cases. Tom Henry Baker writes:
Consider a plot of distance v. time.
This is a long strip 25' wide and 30 minutes long.
But you can *unfold* it -- e.g., make it 50' wide and 30 minutes long, which means that the wall doesn't exist and the swimmer keeps going.
The elastic bounce has become a straight line.
At 01:41 PM 11/11/2017, Tom Karzes wrote:
Considering a short window in which both swimmers hit the same wall at nearly the same time:
If the slow swimmer reaches a pool end an instant before the fast swimmer, there are two passes: the slow swimmer hits the wall, bounces, passes the fast swimmer in the opposite direction, then very shortly after the fast swimmer bounces and overtakes the slow swimmer, passing the slow swimmer once again.
If the fast swimmer reaches a pool end an instant before the slow swimmer, there is one pass: The fast swimmer hits the wall, bounces, passes the slow swimmer, and keeps going, with the slow swimmer falling further and further behind.
But what if they reach a pool end at the exact same moment? Does that still count as the fast swimmer passing the slow swimmer? Certainly the fast swimmer overtakes the slow swimmer, but it's not clear that the fast swimmer "passes" the slow swimmer, since the fast swimmer remains furthest from the wall at all times except at the moment of contact, so in a sense it's not much different than if the fast swimmer turned around just before reaching the wall, in which case there would clearly be no passing. I think this case could be argued either way.
Tom
Henry Baker writes:
I recall a question from my high school AP math or math SAT exam circa 1964:
Two swimmers are swimming laps of a 25' pool, but they swim at different (constant) speeds, say S1 & S2. (Assume that they bounce elastically off the pool ends!)
How many times do they pass one another in 30 minutes?
This problem is pretty complicated, as one swimmer could pass the other in either the same or opposite direction. The time is also long enough, that you can't simply simulate the first 1,2,3 laps, but must come up with a more general answer.
I don't recall how I solved it back then, but I must have gotten the right answer, because I did very well on that test.
At 05:06 AM 11/11/2017, James Propp wrote:
Anyone have a favorite puzzle in which miscounting plays a role?
My favorite is the classic bookworm puzzle (see https://math.stackexchange.com/questions/1271651/how-is-this-true-bookworm-p... ).
Another example: A man was born in 50 BC and died on the same day in 50 AD. How old was he when he died?
I'm especially interested in puzzles that lend themselves to solvers committing fencepost errors, and off-by-one errors more generally.
Jim Propp
The notion of "physical passing" can be formalized as follows: Assume the pool lane runs from left to right. Denote a swimmer's position at any moment in time as the swimmer's distance from the left end of the lane. So the positions increase from 0 to L, then decrease from L down to 0, and the pattern repeats until the race ends (assuming the swimmers start at the left end). Let d1(t) and d2(t) denote the positions of swimmers 1 and 2 at time t. Let d(t) = d1(t) - d2(2). Then d(t) can range from -L to L. If during some interval of time the sign of d(t) changes, either one pass occurred, or some other odd number of passes occurred. If it does not, then either no pass occurred, or else an even number of passes occurred. If you narrow your intervals to those in which there was a single sign change throughout the entire interval, then those intervals correspond to the passes. Under this definition, my example holds: A simultaneous bounce is *not* a pass. On the other hand, suppose at the instant the swimmers touch the wall, they swap swimming speeds. In this case, the fast swimmer catches up to the slow swimmer, they bounce, then the slow swimmer speeds up and the fast swimmer slows down, so the formerly slow swimmer starts moving back more quickly than the formerly fast swimmer. They have changed positions, and have physically passed each other, even though they have not "logically" passed each other. It's the exact reverse of the normal case, in which they have not physically passed each other, but have logically passed each other (ignoring the difference in lap counts). Tom Tom Karzes writes:
Understood, but it seems to me that that's a different problem. Under that model, there is no passing at all, since the fast swimmer immediately overtakes the slow one and remains in the lead until the end. The question is asking about one swimmer physically moving past the other swimmer, which can occur in two circumstances: If they are moving in the same direction, or if they are moving in the opposite direction. I don't see bouncing at the same time as either of those cases.
Tom
Henry Baker writes:
Consider a plot of distance v. time.
This is a long strip 25' wide and 30 minutes long.
But you can *unfold* it -- e.g., make it 50' wide and 30 minutes long, which means that the wall doesn't exist and the swimmer keeps going.
The elastic bounce has become a straight line.
At 01:41 PM 11/11/2017, Tom Karzes wrote:
Considering a short window in which both swimmers hit the same wall at nearly the same time:
If the slow swimmer reaches a pool end an instant before the fast swimmer, there are two passes: the slow swimmer hits the wall, bounces, passes the fast swimmer in the opposite direction, then very shortly after the fast swimmer bounces and overtakes the slow swimmer, passing the slow swimmer once again.
If the fast swimmer reaches a pool end an instant before the slow swimmer, there is one pass: The fast swimmer hits the wall, bounces, passes the slow swimmer, and keeps going, with the slow swimmer falling further and further behind.
But what if they reach a pool end at the exact same moment? Does that still count as the fast swimmer passing the slow swimmer? Certainly the fast swimmer overtakes the slow swimmer, but it's not clear that the fast swimmer "passes" the slow swimmer, since the fast swimmer remains furthest from the wall at all times except at the moment of contact, so in a sense it's not much different than if the fast swimmer turned around just before reaching the wall, in which case there would clearly be no passing. I think this case could be argued either way.
Tom
Henry Baker writes:
I recall a question from my high school AP math or math SAT exam circa 1964:
Two swimmers are swimming laps of a 25' pool, but they swim at different (constant) speeds, say S1 & S2. (Assume that they bounce elastically off the pool ends!)
How many times do they pass one another in 30 minutes?
This problem is pretty complicated, as one swimmer could pass the other in either the same or opposite direction. The time is also long enough, that you can't simply simulate the first 1,2,3 laps, but must come up with a more general answer.
I don't recall how I solved it back then, but I must have gotten the right answer, because I did very well on that test.
At 05:06 AM 11/11/2017, James Propp wrote:
Anyone have a favorite puzzle in which miscounting plays a role?
My favorite is the classic bookworm puzzle (see https://math.stackexchange.com/questions/1271651/how-is-this-true-bookworm-p... ).
Another example: A man was born in 50 BC and died on the same day in 50 AD. How old was he when he died?
I'm especially interested in puzzles that lend themselves to solvers committing fencepost errors, and off-by-one errors more generally.
Jim Propp
I think both swimmers simultaneous bouncing off the wall should count as a "double pass", since that's what it turns into if you perturb the history of the system ever-so-slightly in either direction (having the slow swimmer reach the wall first or having the fast swimmer reach the wall first). Jim On Saturday, November 11, 2017, Tom Karzes <karzes@sonic.net> wrote:
The notion of "physical passing" can be formalized as follows: Assume the pool lane runs from left to right. Denote a swimmer's position at any moment in time as the swimmer's distance from the left end of the lane. So the positions increase from 0 to L, then decrease from L down to 0, and the pattern repeats until the race ends (assuming the swimmers start at the left end).
Let d1(t) and d2(t) denote the positions of swimmers 1 and 2 at time t.
Let d(t) = d1(t) - d2(2). Then d(t) can range from -L to L.
If during some interval of time the sign of d(t) changes, either one pass occurred, or some other odd number of passes occurred. If it does not, then either no pass occurred, or else an even number of passes occurred.
If you narrow your intervals to those in which there was a single sign change throughout the entire interval, then those intervals correspond to the passes.
Under this definition, my example holds: A simultaneous bounce is *not* a pass.
On the other hand, suppose at the instant the swimmers touch the wall, they swap swimming speeds. In this case, the fast swimmer catches up to the slow swimmer, they bounce, then the slow swimmer speeds up and the fast swimmer slows down, so the formerly slow swimmer starts moving back more quickly than the formerly fast swimmer. They have changed positions, and have physically passed each other, even though they have not "logically" passed each other. It's the exact reverse of the normal case, in which they have not physically passed each other, but have logically passed each other (ignoring the difference in lap counts).
Tom
Tom Karzes writes:
Understood, but it seems to me that that's a different problem. Under that model, there is no passing at all, since the fast swimmer immediately overtakes the slow one and remains in the lead until the end. The question is asking about one swimmer physically moving past the other swimmer, which can occur in two circumstances: If they are moving in the same direction, or if they are moving in the opposite direction. I don't see bouncing at the same time as either of those cases.
Tom
Henry Baker writes:
Consider a plot of distance v. time.
This is a long strip 25' wide and 30 minutes long.
But you can *unfold* it -- e.g., make it 50' wide and 30 minutes long, which means that the wall doesn't exist and the swimmer keeps going.
The elastic bounce has become a straight line.
At 01:41 PM 11/11/2017, Tom Karzes wrote:
Considering a short window in which both swimmers hit the same wall at nearly the same time:
If the slow swimmer reaches a pool end an instant before the fast swimmer, there are two passes: the slow swimmer hits the wall, bounces, passes the fast swimmer in the opposite direction, then very shortly after the fast swimmer bounces and overtakes the slow swimmer, passing the slow swimmer once again.
If the fast swimmer reaches a pool end an instant before the slow swimmer, there is one pass: The fast swimmer hits the wall, bounces, passes the slow swimmer, and keeps going, with the slow swimmer falling further and further behind.
But what if they reach a pool end at the exact same moment? Does that still count as the fast swimmer passing the slow swimmer? Certainly the fast swimmer overtakes the slow swimmer, but it's not clear that the fast swimmer "passes" the slow swimmer, since the fast swimmer remains furthest from the wall at all times except at the moment of contact, so in a sense it's not much different than if the fast swimmer turned around just before reaching the wall, in which case there would clearly be no passing. I think this case could be argued either way.
Tom
Henry Baker writes:
I recall a question from my high school AP math or math SAT exam circa 1964:
Two swimmers are swimming laps of a 25' pool, but they swim at different (constant) speeds, say S1 & S2. (Assume that they bounce elastically off the pool ends!)
How many times do they pass one another in 30 minutes?
This problem is pretty complicated, as one swimmer could pass the other in either the same or opposite direction. The time is also long enough, that you can't simply simulate the first 1,2,3 laps, but must come up with a more general answer.
I don't recall how I solved it back then, but I must have gotten the right answer, because I did very well on that test.
At 05:06 AM 11/11/2017, James Propp wrote:
Anyone have a favorite puzzle in which miscounting plays a role?
My favorite is the classic bookworm puzzle (see https://math.stackexchange.com/questions/1271651/how-is- this-true-bookworm-puzzle ).
Another example: A man was born in 50 BC and died on the same day in 50 AD. How old was he when he died?
I'm especially interested in puzzles that lend themselves to solvers committing fencepost errors, and off-by-one errors more generally.
Jim Propp
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Well, if you perturb it by moving the fast swimmer ahead slightly, then you get (a) the fast swimmer overtaking the slow swimmer on the way up, followed by (b) the swimmers swimming past each other. If you perturb it by moving the slow swimmer ahead slightly, then you get (a) the swimmers swimming past each other, followed by (b) the fast swimmer overtaking the slow swimmer on the way back. In other words, the two types of passes occur in the opposite order. I argue that these disappear entirely if they bounce at the exact same time. Tom James Propp writes:
I think both swimmers simultaneous bouncing off the wall should count as a "double pass", since that's what it turns into if you perturb the history of the system ever-so-slightly in either direction (having the slow swimmer reach the wall first or having the fast swimmer reach the wall first).
Jim
On Saturday, November 11, 2017, Tom Karzes <karzes@sonic.net> wrote:
The notion of "physical passing" can be formalized as follows: Assume the pool lane runs from left to right. Denote a swimmer's position at any moment in time as the swimmer's distance from the left end of the lane. So the positions increase from 0 to L, then decrease from L down to 0, and the pattern repeats until the race ends (assuming the swimmers start at the left end).
Let d1(t) and d2(t) denote the positions of swimmers 1 and 2 at time t.
Let d(t) = d1(t) - d2(2). Then d(t) can range from -L to L.
If during some interval of time the sign of d(t) changes, either one pass occurred, or some other odd number of passes occurred. If it does not, then either no pass occurred, or else an even number of passes occurred.
If you narrow your intervals to those in which there was a single sign change throughout the entire interval, then those intervals correspond to the passes.
Under this definition, my example holds: A simultaneous bounce is *not* a pass.
On the other hand, suppose at the instant the swimmers touch the wall, they swap swimming speeds. In this case, the fast swimmer catches up to the slow swimmer, they bounce, then the slow swimmer speeds up and the fast swimmer slows down, so the formerly slow swimmer starts moving back more quickly than the formerly fast swimmer. They have changed positions, and have physically passed each other, even though they have not "logically" passed each other. It's the exact reverse of the normal case, in which they have not physically passed each other, but have logically passed each other (ignoring the difference in lap counts).
Tom
Tom Karzes writes:
Understood, but it seems to me that that's a different problem. Under that model, there is no passing at all, since the fast swimmer immediately overtakes the slow one and remains in the lead until the end. The question is asking about one swimmer physically moving past the other swimmer, which can occur in two circumstances: If they are moving in the same direction, or if they are moving in the opposite direction. I don't see bouncing at the same time as either of those cases.
Tom
Henry Baker writes:
Consider a plot of distance v. time.
This is a long strip 25' wide and 30 minutes long.
But you can *unfold* it -- e.g., make it 50' wide and 30 minutes long, which means that the wall doesn't exist and the swimmer keeps going.
The elastic bounce has become a straight line.
At 01:41 PM 11/11/2017, Tom Karzes wrote:
Considering a short window in which both swimmers hit the same wall at nearly the same time:
If the slow swimmer reaches a pool end an instant before the fast swimmer, there are two passes: the slow swimmer hits the wall, bounces, passes the fast swimmer in the opposite direction, then very shortly after the fast swimmer bounces and overtakes the slow swimmer, passing the slow swimmer once again.
If the fast swimmer reaches a pool end an instant before the slow swimmer, there is one pass: The fast swimmer hits the wall, bounces, passes the slow swimmer, and keeps going, with the slow swimmer falling further and further behind.
But what if they reach a pool end at the exact same moment? Does that still count as the fast swimmer passing the slow swimmer? Certainly the fast swimmer overtakes the slow swimmer, but it's not clear that the fast swimmer "passes" the slow swimmer, since the fast swimmer remains furthest from the wall at all times except at the moment of contact, so in a sense it's not much different than if the fast swimmer turned around just before reaching the wall, in which case there would clearly be no passing. I think this case could be argued either way.
Tom
Henry Baker writes:
I recall a question from my high school AP math or math SAT exam circa 1964:
Two swimmers are swimming laps of a 25' pool, but they swim at different (constant) speeds, say S1 & S2. (Assume that they bounce elastically off the pool ends!)
How many times do they pass one another in 30 minutes?
This problem is pretty complicated, as one swimmer could pass the other in either the same or opposite direction. The time is also long enough, that you can't simply simulate the first 1,2,3 laps, but must come up with a more general answer.
I don't recall how I solved it back then, but I must have gotten the right answer, because I did very well on that test.
At 05:06 AM 11/11/2017, James Propp wrote: >Anyone have a favorite puzzle in which miscounting plays a role? > >My favorite is the classic bookworm puzzle (see >https://math.stackexchange.com/questions/1271651/how-is- this-true-bookworm-puzzle >). > >Another example: A man was born in 50 BC and died on the same day in 50 AD. >How old was he when he died? > >I'm especially interested in puzzles that lend themselves to solvers >committing fencepost errors, and off-by-one errors more generally. > >Jim Propp
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I agree with Jim's interpretation. I don't remember the exact speeds from the original problem, but I would imagine that they were chosen in such a way that the *only* time both swimmers hit the wall simultaneously was at the very start. This would eliminate the necessity to explain in great detail exactly what a "pass" entailed. At 08:03 PM 11/11/2017, Tom Karzes wrote:
Well, if you perturb it by moving the fast swimmer ahead slightly, then you get (a) the fast swimmer overtaking the slow swimmer on the way up, followed by (b) the swimmers swimming past each other. If you perturb it by moving the slow swimmer ahead slightly, then you get (a) the swimmers swimming past each other, followed by (b) the fast swimmer overtaking the slow swimmer on the way back. In other words, the two types of passes occur in the opposite order. I argue that these disappear entirely if they bounce at the exact same time.
Tom
James Propp writes:
I think both swimmers simultaneous bouncing off the wall should count as a "double pass", since that's what it turns into if you perturb the history of the system ever-so-slightly in either direction (having the slow swimmer reach the wall first or having the fast swimmer reach the wall first).
It would certainly have been smart of the problem designers to avoid that case. I would approach the limit from a different perspective, and argue that there's a discontinuity when they bounce simultanrously: Suppose we shorten the fast swimmer's lane by moving the wall in question closer to the opposite wall. Now if they both bounce simultaneously, there is clearly no passing around the time of the bounces, since at no time are they at the same distance at the same time. Now slowly increase the length of the fast swimmer's lane by moving the wall back out, almost to the same distance as the slow swimmer's wall. There is still clearly no passing: The fast swimmer is behind the slow swimmer as they approach their walls. They both bounce simultaneously and start to head back. At no point is the fast swimmer as far to the right as slow swimmer (assuming the bounce walls are on the right). If you continue to move the fast swimmer's wall to the right, you reach the point in question, where the walls are both the same distance from the opposite wall. And if you continue past that, so that the fast swimmer's wall is to the right of the slow swimmer's, then you have a double-pass case: the fast swimmer passes the slow swimmer just before they both bounce, and again immediately after they both bounce (and both passes are "overtaking" passes, as opposed to "swimming past each other" passes. So there is a discontinuity at the point where both walls are at the same distance, with the passing count jumping from 0 to 2. Tom Henry Baker writes:
I agree with Jim's interpretation.
I don't remember the exact speeds from the original problem, but I would imagine that they were chosen in such a way that the *only* time both swimmers hit the wall simultaneously was at the very start. This would eliminate the necessity to explain in great detail exactly what a "pass" entailed.
At 08:03 PM 11/11/2017, Tom Karzes wrote:
Well, if you perturb it by moving the fast swimmer ahead slightly, then you get (a) the fast swimmer overtaking the slow swimmer on the way up, followed by (b) the swimmers swimming past each other. If you perturb it by moving the slow swimmer ahead slightly, then you get (a) the swimmers swimming past each other, followed by (b) the fast swimmer overtaking the slow swimmer on the way back. In other words, the two types of passes occur in the opposite order. I argue that these disappear entirely if they bounce at the exact same time.
Tom
James Propp writes:
I think both swimmers simultaneous bouncing off the wall should count as a "double pass", since that's what it turns into if you perturb the history of the system ever-so-slightly in either direction (having the slow swimmer reach the wall first or having the fast swimmer reach the wall first).
Henry and I say that the number of passes exhibits a removable discontinuity if you change the swimmers' paths. To see why, create a diagram in which the x-axis says how far swimmer #1 has gone and the y-axis says how far swimmer #2 has gone, measured in pool-lengths. The graph of the relation "both swimmers are at the same location" is a union of lines of the form x-y=2n and x+y=2n+1 that form a square grid (obtained from the usual grid by dilating by sqrt(2) and rotating by 45 degrees). Assuming that swimmer #1 always swims faster than swimmer #2, asking "How many times does swimmer #1 pass swimmer #2?" is the same as asking "At how many times t is the point (x(t),y(t)) on that grid?", and as long as the curve {(x(t),y(t)} doesn't pass through a place where grid-lines intersect, the answer is independent of the cuve: it's the number of grid-lines that path must cross. So in fact one doesn't need to know that the swimmers swim at constant speeds; if one knows how many laps each swimmer completes, and one knows that the faster swimmer is faster throughout, and one knows that they never reach a wall simultaneously (or, alternatively, if one agrees that such occurrences count as double-passes), then the number of times the fast swimmer passes the slow swimmer is a topological invariant. I don't know if Tom is right in saying that the number of passes exhibits a jump discontinuity if one changes the length of the pool, but I'm pretty sure this kind of diagram will help us understand Tom's claim as well. Jim Propp On Sunday, November 12, 2017, Tom Karzes <karzes@sonic.net> wrote:k
It would certainly have been smart of the problem designers to avoid that case.
I would approach the limit from a different perspective, and argue that there's a discontinuity when they bounce simultanrously:
Suppose we shorten the fast swimmer's lane by moving the wall in question closer to the opposite wall. Now if they both bounce simultaneously, there is clearly no passing around the time of the bounces, since at no time are they at the same distance at the same time.
Now slowly increase the length of the fast swimmer's lane by moving the wall back out, almost to the same distance as the slow swimmer's wall. There is still clearly no passing: The fast swimmer is behind the slow swimmer as they approach their walls. They both bounce simultaneously and start to head back. At no point is the fast swimmer as far to the right as slow swimmer (assuming the bounce walls are on the right).
If you continue to move the fast swimmer's wall to the right, you reach the point in question, where the walls are both the same distance from the opposite wall. And if you continue past that, so that the fast swimmer's wall is to the right of the slow swimmer's, then you have a double-pass case: the fast swimmer passes the slow swimmer just before they both bounce, and again immediately after they both bounce (and both passes are "overtaking" passes, as opposed to "swimming past each other" passes.
So there is a discontinuity at the point where both walls are at the same distance, with the passing count jumping from 0 to 2.
Tom
Henry Baker writes:
I agree with Jim's interpretation.
I don't remember the exact speeds from the original problem, but I would imagine that they were chosen in such a way that the *only* time both swimmers hit the wall simultaneously was at the very start. This would eliminate the necessity to explain in great detail exactly what a "pass" entailed.
At 08:03 PM 11/11/2017, Tom Karzes wrote:
Well, if you perturb it by moving the fast swimmer ahead slightly, then you get (a) the fast swimmer overtaking the slow swimmer on the way up, followed by (b) the swimmers swimming past each other. If you perturb it by moving the slow swimmer ahead slightly, then you get (a) the swimmers swimming past each other, followed by (b) the fast swimmer overtaking the slow swimmer on the way back. In other words, the two types of passes occur in the opposite order. I argue that these disappear entirely if they bounce at the exact same time.
Tom
James Propp writes:
I think both swimmers simultaneous bouncing off the wall should count as a "double pass", since that's what it turns into if you perturb the history of the system ever-so-slightly in either direction (having the slow swimmer reach the wall first or having the fast swimmer reach the wall first).
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