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Request for help - math problem using binary code

  1. Sep 29, 2005 #1

    I have math question that I am struggling with that apparently can be solved using code (or reasoning - I tried that and mine didn't work. GO figure :confused: ). Can anyone help me out?

    Here is the question:

    A coffee club with 8 membres meet every MOnday Wednesday and Friday. At eahc meeting, 4 people sit inside at a table and 4 people sit outside. IS it possible in the course of a week, for each member to be inside at least once, outside at least once, and be at teh same table at least once with wach of the other 7 members?

    Thanks for the help
  2. jcsd
  3. Sep 29, 2005 #2
    Nice question :) But ithink it is impossible because:
    Let's name this eight people from 1 to 8

    IN OUT
    First day: 1234 5678 - A Standard Beginning(Changing places wont make a difference as numbers can be changed too...)

    Second D: Assume that we need to make only 2 people(In my sol. they are 1 and 5) meet each people and go both inside and outside, we would just replace them to meet most people and this would be 5234(I)-1678(O)

    Third D.: Now we must make 1 and 5 meet each other and lets put them like this
    1537 - 2468(There are two contradictions 1st: Its asymmetric, it cant be a solution, there is no sense in putting 1 and 5 to inside not outside(symmetry in solutions for problems like this is VERY Important) and 2nd: 2-7, 4-7, 3-6, 3-8 havent met)

    Another solution trial:(two-to-two meeting system-doesnt works again)
    (By this System, we make two people groups like 12, 34... and have only 4 elements)
    1st Day:1234(I)-5678(O)
    2nd Day:1256(I)-3478(O) (Now we have just 12-78 and 34-56 left)
    3rd Day: 3456(I)-1278(O) (All people met; however 7 and 8 has never gone inside)
    (The weird thing here is there is no symmetry(Why 7 and 8) and this makes me think im doing a mistake somewhere...)
    (Any explanations to this would be great-just mail me if you have any comments)
  4. Oct 3, 2005 #3
    I tried exactly what you tried, and to no avail. (switching two people at a time and moving one person at a time) I feel like it may be impossible, but by the way my prof was talking in class, it sounded like there was a solution! Hmmm....
    He did mention using binary code and pigeon hole principle but I am at a loss as to how to use those (both pretty knew concepts for me, so don't fully understand how to apply to problem!)
    Thank you for you help!
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