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Original problem

Lets assume that we hire two persons to count passers-by for an hour. One is told to stand still at one spot on the sidewalk. The other one is told to walk 20 meters up and down the same sidewalk.

Would they get the same count? Why?

Idea reworded from this open-source(contributed by the community) -- book --

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There are many, many variables at play in this situation that pretty much makes the answer

any answer you want.

Why?

The walker counts less than the person standing still. How? Well, if someone enters the sidewalk at a point P where 0m < p < 20m along the walker's path after the walker has reached that spot, it is possible that this person will never be counted by the walker.

The walker counts the same as the person standing still. How? Assuming that people start on the sidewalk more than 20m away from the walker's starting position, the walker will always see the same commuters that the stander sees. The walker can ignore duplication, so they get the same count.

The walker counts more than the person standing still. How? Use the same scenario as the same count, but remove the assumption that the walker can differentiate between duplicates.

Of course, all of this requires that they are in the same general area. A sidewalk can be far longer than 20m, so the two can be potentially blocks away and still be on the same sidewalk. This means that they see different crowds and can have larger, smaller, or equivalent data sets.

Some variables that must be assumed to answer the question:

  1. The walker starts or ends at the same place as the stander
  2. The walker's speed.
  3. The walker's ability to match duplicates
  4. The ability of pedestrians to enter the path in the 20m range that the walker is covering.
  5. The fact that everyone is walking through the stander's field of vision, rather than behind them or exiting the path before they reach the stander.

EDIT

Taking into consideration the OPs answers to the assumptions above, as well as extrapolations taken from the original source, the parameters sit as follows:

  1. The walker starts and ends at the same place as the stander
  2. The walker's speed is unnecessary, but he makes a number of complete loops and ends at the start point when the hour is up.
  3. Yes
  4. Pedestrians can enter or exit the sidewalk at any time or location.
  5. The stander is in a doorway, so his view is completely unobstructed, and no one walks behind him.

Taking these into consideration:

We still don't have enough data to know who counted more

Why?

Let's take our three scenarios:
Walker sees more: Say fifteen people walk down the sidewalk in an hour. If three of them leave the sidewalk before they reach the stander, they will be counted by the walker and will not be counted by the stander. Thus, the walker will have a higher count.

Stander sees more: This logic holds true to my earlier statement. Again, say there are fifteen people on the sidewalk. If the walker passes the 5m mark heading toward the 10m mark and then a pedestrian enters the sidewalk at the 3m mark heading toward the 0m mark, the pedestrian will not be counted by the walker, but would be counted by the stander.

They see the same: This is the answer the book quoted in the question gives, and I would assume that it is the one OP wants. The basic premise here is that the walker will pass each pedestrian and count them once, whether it is coming or going. The stander will count each person as they pass by, thus giving them the same counts.

Here is the direct answer from the book quoted in the question:

Both of them counted the same number of passersby.
While the one who stood at the door counted all those
who passed both ways, the one who was walking counted all
the people he met going up and down the pavement.
There is another way of putting it. When the man who was
walking and counting the passers-by returned for the first
time to the man who was standing at the door, they had
counted the same number of passers-by - all those passing
by the standing man encountered the walk¬ing man either
on the way there or back. And each time the one who was
walking was returning to the one who was standing he
counted the same number of passers-by. It was the same at
the end of the hour, when they met for the last time and told
each other the final number.

Ultimately

This question goes to show how a good mathematical model needs to be very, very well defined. There are far too many variables that need to be accounted for here for us to be able to completely say who counted more, as there is too much chaos thrown in by the human element.

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  • $\begingroup$ @BinaryZebra I updated my answer to include your clarifications. $\endgroup$ – tfitzger May 17 '15 at 9:27
  • $\begingroup$ Thanks for your well thought points, quite deep. I have updated, both the goal of the question, and the rules. Hope you like it better now. $\endgroup$ – user12496 May 18 '15 at 0:54
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Assuming that the walking counter starts and finishes right next to the stationary counter: If the counters can recognize individuals and avoid double counting:

they will both get the same count. The cool thing is that this holds irrespective of how the passers-by are moving. This is because, having started in the same place, the difference in their counts at any time $t$ is less than or equal to the number of passers-by in between them. If they finish in the same place, there can be no passers-by between them and thus their counts must agree exactly.

If instead the counters may double count the same passer-by,

the walking individual may double count whereas the stationary one will not and thus the walking counter can get a higher (but not lower) count.

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    $\begingroup$ Surely they can be different? Imagine that there's an extremely slow passerby at the end of the walker's route, who the walker would pass twice with every trip. $\endgroup$ – xnor May 16 '15 at 7:08
  • $\begingroup$ @xnor see my updated answer for assumptions under which they can be different. $\endgroup$ – Corvus May 16 '15 at 7:37
  • $\begingroup$ That's exactly my concern. Should a walking counter (the person) count people he pass? Then the walking counter should count twice a person standing at the sidewalk. Is this description valid? $\endgroup$ – user12496 May 16 '15 at 7:38
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    $\begingroup$ Surely if the difference at any time $t$ is at most the number of passers-by between them, then the walking counter has to un-count a person every other time he passes them? $\endgroup$ – Fax May 16 '15 at 19:27
  • $\begingroup$ @Fax you're right. Good point! $\endgroup$ – Corvus May 16 '15 at 20:19
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Counts would be same. They are humans, they will obviously count right number of people, given that they are inclined towards the work.
Bonus:

If a machine (which just counts the number of bodies passing by) is made to move like that, the count may be different, because a person may walk too slow or too fast.

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  • $\begingroup$ So, maybe the definition of "passer-by" needs to be better stated, so that a machine could also perform the same (equal?) count. Maybe we need to substract persons that the walker pass (if he is walking faster than the passer-by). But, then, how to count a person standing on the sidewalk? By-definition, such standing foe is not a passer-by. But if we add a rule to "not count" some individuals, would the final resulting count be correct? $\endgroup$ – user12496 May 16 '15 at 8:05
  • $\begingroup$ Let me clarify, I am talking about a people counter. So in that case, it would be incorrect. $\endgroup$ – Aditya Agarwal May 16 '15 at 8:08
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Assumptions:

  1. Since it's not specified otherwise in the question, assume that every passer-by who walks along the sidewalk will walk the complete length of the sidewalk, in a single direction, without stopping, changing speed, or turning around.
  2. Since it's not specified otherwise in the question, assume the counters count passer-bys as they move past. (That is, the counting occurs as a passer-by changes state from "behind the counter" to "in front of the counter", or vice-versa)
  3. Since it's not specified otherwise in the question, assume the walking counter walks slightly slower than the slowest passer-by (and therefore will not pass a single person twice, while that person is only traversing the sidewalk once).
  4. Since it's not specified otherwise in the question, assume that when the surveying period starts, there are no pedestrians already on the sidewalk.
  5. Similarly, assume that when the surveying period ends, there are no pedestrians left on the sidewalk.
  6. Assume that the OP felt that the question was solvable as initially posted, and assume that they still feel that way, as no new information or clarifications have been added to it even as every answer thus far has included at least one assumption. Therefore, assume that the (unspecified) mechanics of how the counting system works and how the pedestrians behave are not relevant to the puzzle. This means that this is not actually a mathematics question, despite being tagged that way – it is instead a puzzle about finding the cheeky trick which renders moot all those missing mechanical details.

So after some thought, here's the trick which renders everything else moot:

The walking counter is a "passer-by" from the point of view of the stationary counter, since the walking counter is a person walking past them on the sidewalk.

So regardless of any of the assumptions listed above, over any sufficiently long survey duration the stationary counter will show a higher count than the walking one; they will both have counted approximately the same number of pedestrians, but the stationary counter will also have counted the walking counter moving back and forth.

QED.

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  • $\begingroup$ Excellent write-up. Good review of the needed factors to get an answer. I agree to some and disagree to some others. I do not think that the question is about finding any cheeky trick. The question was presented and solved in the linked book. In fact, I do not agree with the solution presented by the book author and I am in search of a better/more reliable answer. I am concern of starting to add some (personal) limits, the question will be faithful to the original one. However, it seems that there is no other way to find a (reasonable) answer. Probably I should start adding rules. $\endgroup$ – user12496 May 17 '15 at 3:31
  • $\begingroup$ @BinaryZebra Ah. So it turns out that there was one more assumption that I didn't even notice I was making: "Assume that the OP knows the correct answer to the puzzle they've posed, and has provided all the information which is required to arrive at that correct answer." $\endgroup$ – Trevor Powell May 17 '15 at 7:34
  • $\begingroup$ Yes :) Exactly. I do not know of a good answer, Yet!. $\endgroup$ – user12496 May 17 '15 at 23:52
  • $\begingroup$ I sure hope that I could choose your answer as well. But I am forced to only select one. I hope you still will help us. Thanks. $\endgroup$ – user12496 May 18 '15 at 1:19
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Both persons count the same number of passers-by. While walking along the pavement, each passer-by must pass both the stationary person, and also the walking person, so both will count the same number of passers-by.

Answer reworded from this open-source (books contributed by the community) book.

Hope that helps.

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  • $\begingroup$ Deliberately left un-spoilered since it's the original answer from the puzzle book which was the origin of this puzzle. OP has declared this answer unsatisfactory without actually telling anyone what it was, so I've provided it. Personally, though, I'd say there's an elegant sort of charm to its rustic simplicity that glosses over all the mechanical details that everybody else has been fussing over (and OP persists in not providing). $\endgroup$ – Trevor Powell May 17 '15 at 8:17
  • $\begingroup$ (Incidentally, if anyone else is interested in making a comparison, this puzzle/answer is #4 in the book; it's on page 17. The puzzle hasn't been reworded in any substantial way, but this answer has been condensed from a couple paragraphs into what I think is a much clearer explanation of the answer given by the book) $\endgroup$ – Trevor Powell May 17 '15 at 8:24
  • $\begingroup$ I can not provide what I do not have. I do not know what answer, if any, this question has. I do feel that the answer from the book is just too simplistic. But I do not have a better (well supported) answer. $\endgroup$ – user12496 May 18 '15 at 1:00
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Assumption: Each passer-by is only counted once.

The moving counter will count more people. He will count the same amount of people who never turn back. People who turn back once are as likely to be counted by only the stationary counter as only by the moving counter. However, people who move in a closed loop are more likely to be counted only by the moving counter.

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