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'There is a row of 9 consecutive shops, John will visit a shop for 14 consecutive days. John moves venues daily to a shop directly left or directly right (end of row means forced move). John moves completely randomly. I want to find John in a shop on one of the days, I am not aware of the shops John has visited. Is there a strategy that assures that I will meet John in a shop in the 14 days?'

I think there is no strategy as John could just visit two shops for the duration of the 14 days and I don't think you can find John if he chooses to only remain in two shops.

What do you think?

amWhy
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anon1234
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  • Maybe I misunderstand the question, but if you just start at one end and work your way to the other end one shop at a time, how can John avoid you? – Rob Arthan Feb 16 '18 at 20:34
  • @RobArthan The way I understand it, John can jump over you. That is, you search shop $A$ on the first day, not knowing John is in shop $B$. On the second day, you move to $B$ while John jumps to $A$ and so on. – lulu Feb 16 '18 at 20:35
  • There is no way to succeed. Suppose John has second sight, and knows which shop you will visit. He just visits a different one, unless he happens to have visited a shop on the end, so he'll never do that. – saulspatz Feb 16 '18 at 20:37
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    What's the source of this problem? I have a feeling that the rule set is incomplete. – lulu Feb 16 '18 at 20:41
  • @lulu: I agree that the rule set is incomplete. I dismissed the possibility of John jumping over me because I assumed that I have visibility of who is in each shop when I am in a shop and of who is in the street when I am on the street. You and saulspatz obviously had other (reasonable) alternative interpretations of the question. – Rob Arthan Feb 16 '18 at 20:45
  • @anon1234: I see. You could usefully edit the question to include the extra information. – Rob Arthan Feb 17 '18 at 15:19
  • anon1234 Do not vandalize a question you've posted. Upon submitting the question, it is no longer yours to delete or keep. You accepted an answer, and to try and delete now is an insult. – amWhy Feb 22 '18 at 23:54

1 Answers1

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Number the shops $1-9$ and then visit the shops in the order: $2,3,4,5,6,7,8,8,7,6,5,4,3,2$.

  • During the first half of the walk ($2-8$), if the first day John was in an even shop, it stays in the shop of the same parity as you all along, and because that prevents him from "skipping" over you, you will catch him.

  • Otherwise, he is in the shop with the opposite parity than you on the first day, and up to $7$th day, but on the $8$th day he must be in the shop with the same parity as you (as on the 8th day you switch parity), and so you will catch him on the way back.

A similar problem was recently posted on Math.SE: Find a bug under 6 tiles.

  • Nice strategy (+1) – lulu Feb 16 '18 at 20:50
  • So the assumption here is that it takes as most one day to find John in any of the shops and that he is invisible when he walks from one shop to another (see my comment on the question for why I say this). That's a reasonable assumption, but the OP needs to clarify the question. – Rob Arthan Feb 16 '18 at 20:52
  • @anon1234 I don't really know. I am worried that the diagram may be more confusing. At least if it up to me to draw it...:( –  Feb 17 '18 at 12:54