try your heads on this

This is a discussion on try your heads on this within the A Brief History of Cprogramming.com forums, part of the Community Boards category; You peepl are not getting the point here im not saying that there are 1 or two people to start ...

  1. #46
    Registered User jasrajva's Avatar
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    You peepl are not getting the point here
    im not saying that there are 1 or two people to start with that is just a hypotheses

    if only one person has the mark then he is eliminated in a day

    if two then in two days etc

    so if all the people with the mark are gone in 7 days
    it follows that seven people originally had the mark

    ask betazap or nvoigt or NICK they all agree with me

    and justin if there are three to start with each sees two people with the mark and believes that there are only two sinners so thy should stop coming in two days so they all come back on tuesday and realise that there have to be three people with the mark

    and so on

    and natase

    you still dont get it at all read my prev posts properly again

  2. #47
    Registered User Natase's Avatar
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    you still dont get it at all read my prev posts properly again
    Oh, believe me, I did... now how about you go back and read my second point properly.

  3. #48
    Registered User jasrajva's Avatar
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    natase
    the point you dont get is that they all (no matter how many sinners they are to start with stop coming on the same day
    ie if 5 sinners were there then they would all come till thursday and stop coming on friday

    (read back for the explanation and you'll see why)

  4. #49
    Registered User Natase's Avatar
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    That makes absolutely no sense... I believe the maths, I just don't think it's being implemented properly.

    If there were 5 sinners, and each day they come to church they see 4 other sinners, why would they assume the others are coming because they see their own mark?

    Surely each of the 5 sinners would assume that there are only 4 sinners and wait for them to figure it out...

  5. #50
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    Exactly.

    and justin if there are three to start with each sees two people with the mark and believes that there are only two sinners so thy should stop coming in two days so they all come back on tuesday and realise that there have to be three people with the mark
    Why does the third person leave then? Why not number 1? Or number 2? Or all three? Put yourself in the shoes of the fourth person. Don't assume you have the mark, wonder about it. Three people with the mark show up the first day. Then they show up the second day. Why do you assume that they are waiting for you to leave instead of one of them?
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  6. #51
    Registered User jasrajva's Avatar
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    why dont you blieve that

    look i'll try to explain it again to you


    heres your first step to thinking out the answer

    imagine there is just one person with the mark
    all others see him/her and hope that he is the only one with the mark so they come back the next day
    the person with the mark on the other hand sees no one with the mark
    he also knows that atleast one person must have a mark so he knows that the mark has to be on him/her so doesnt come back on monday


    now think of two people
    sunday :

    they each see a mark and assume thats its the only one around
    and come back on monday

    monday

    they come back and again see the same person with the mark
    now assume that A and B are the two peepl

    A: why did B come back?
    : he must have seen a mark on someone head
    : now i saw a mark only on B's head and if he sees a mark too then it must be on me as i can see no other mark except B's

    b thinks on the same lines

    so both A and B stop coming on Tuesday

    now for three people with the mark

    each sees two others with the marks and expects them to go thru the reasoning given above for two people and stop coming by Tuesday

    but when they each come back on Tuesday they each know that
    it could only be if the others had also seen two sinners and thought on the same lines as they had

    so if the sinners you can see are each seeing two marks as well again you dont see any one else with a mark so the 2nd mark that the other two sinners see must be on your head

    so the three stop coming on Wednesday


    for four people

    each sees three marked peepl and expects them to go thru the motions and stop coming on Wednesday

    but when they come back on wednesday
    it could only be that they have all seen three marks each and expect the people they saw to stop coming by wednesday

    so each now knows that there must be 4 people wioth the mark and the 4th person must be the person him/herself coz he sees only three marks around him and so the mark must be on him/her

    and so the 4 people stop coming on Thursday


    thats enuf i hope you should get it now or i dont know how too explain it to you





    aND THEY ALL ALWAYS LEAVE TOGETHER

  7. #52
    _B-L-U-E_ Betazep's Avatar
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    >>ask betazap or nvoigt or NICK they all agree with me

    I agree with you 100%... and it is BetazEp not BetazAp dammit

    Plus the final explanation works much better I think. I hope they understand it from there.

    This was an intriguing puzzle.

    I told my father and he got the right answer... (I found out later that he had heard one similar)
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  8. #53
    Registered User Natase's Avatar
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    The explanation falls flat after 2 sinners. The reason this works initially is because they know how many sinners there are (if there are some, that means there has to be at least one).

    If a sinner saw 5 people with a mark he would no more assume that he was a sinner than someone else in the room who sees 6 sinners... UNLESS THEY KNOW HOW MANY THERE ARE.

  9. #54
    _B-L-U-E_ Betazep's Avatar
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    >>If a sinner saw 5 people with a mark he would no more assume that he was a sinner than someone else in the room who sees 6 sinners... UNLESS THEY KNOW HOW MANY THERE ARE.

    You are wrong. You are basing your decision on a single momentary merit where one person would be able to assume he was a sinner simply by looking at a room of people and seeing some sinners.

    The interesting thing about that is the word "assume." This isn't an assumption... it is a deduction.

    The answer is layed out before you in Jasrajva's previous post. It isn't simple. You have to envision it, but once you see it (which will require dropping the path you are travelling on) you will come to understand how true it is.

    You don't need to know how many sinners there are... only that there are sinners and that the number of days that passed after the initial day are seven (one week).

    >>The explanation falls flat after 2 sinners.

    You are wrong.

    On the first day, I see two sinners. I expect that sinner A will see sinner B and vice versa.

    I expect that sinner A and sinner B will come to church the next day because they see each see a sinner.

    They both come. Now I expect them to not come anymore because they are highly logical. Sinner A sees that sinner B came back. Sinner B sees that sinner A came back.

    So to my perspective, they had better understand that they are both sinners on day two (which is the first day of the week).

    They won't be back (but I will have to come back tomorrow because I see a sinner still)... unless of course they see another sinner. Who would that sinner be if I see no other sinners? Well... that would be me.

    So they do come back on the third day (second day of the week). I say WTF... why are they back? There must be another sinner. I don't see any, so I must be the sinner.

    The interesting thing is that this perception is only from my POV... so I have to expect that they are thinking the exact same process as me...

    Though in my scenario they do not have to be because I never said they were and I never stated that they figured it out (a very long loop if they don't figure it out)... in jasrajva's they do. Why? Because you were told that they figured it out in a week... therefore they did figure it out... therefore they all had to think from the same POV as me on the same days for the same number of days.

    If the riddle said that the sinners were gone on the third day of the week (the fourth day including the initial meeting) then I would be able to say with great certainty that there were three people that were sinners in any size group over three.

    That is where the riddle faulters in my eyes. If one assumes a group of less than seven, the riddle fails to take effect because how can you get rid of more people than physically exist in the group.
    Last edited by Betazep; 11-06-2001 at 12:33 AM.
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  10. #55
    the hat of redundancy hat nvoigt's Avatar
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    jasrajva is right. I couldn't figure it out or believe it myself at first.

    You have to see it from the sinners perspective. Plus you have to take in account that each day that passes gives MORE
    information about the number of sinners.
    hth
    -nv

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  11. #56
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    Talking

    aND THEY ALL ALWAYS LEAVE TOGETHER
    If I can find one value that does not work, doesn't that make the whole function invalid?

    Case 1 of infinite: There are six people in the congregation. They wait one extra day to see if they can psyche the priest out.

    I see what you're saying jasrajva, but it seems like circular reasoning to assume that of the x amount of people, all follow the same path of reasoning. I think the question is meant to confuse people by its wording. Instead of sinners who are "flawed" people anyway, it should be, "A program that has copied itself x amount of times needs to discover what is x within the server...". And then give a whole lot more detail, if detail is what solves the problem.

    In other words, I can find a flaw with any assumption. 1 sinner: goes to church the next day and sees no mark, assumes that the sinner is gone. Goes home and meets once a week, oblivious to the fact that he has fated the rest of the congregation to returning to church every day (or just twice a week until sinner is gone again?).

    Two sinners: one sees the other one and figures that, as a sinner, the other is prideful and will not admit to being the marked one - or, being actually prideful, the sinners stay an extra day just to be sure. The next day, nobody but the priest comes because they all decided to suddenly follow your logic.

    A million people in the congregation (no limit right?). It takes the one sinner seven days to search through them all (didn't mention how long it takes to search through an infinite amount of people). I don't mean to get technical on ya, but you got technical on me first so.... (yes I know, I am clearly one of the sinners.)

    In the end, seven sinners in a room see each other, (especially if they are in mensa), they aren't going to figure they too must be a sinner no matter how many days go by.... at least not without a lot of extra wording. (Not all functions have an inverse function, mathematically.)
    Last edited by Justin W; 11-06-2001 at 01:17 AM.
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  12. #57
    _B-L-U-E_ Betazep's Avatar
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    ooohhh... this gets better...

    I say that there is the slight possibility of eight people that are sinners depending on how you decode the first line....

    "among those of you who have gathered here are some sinners"

    logically I take that to mean that there are an number of people that are not sinners.... but as written it could mean "some sinners" in respect to the globe or other (possibly... depends how much worth "among" is in the sentence).

    So knowing how it is strongly solved for seven... imagine a group of eight and all eight of them being sinners. They would all not be there on the eight day which would mean that they would have left on the seventh day.

    I also say that the structure of the priest leaving and the people following on the first day doesn't definitively make that day as an exclusion from the "week". This could make it the first day of the week that will end on Saturday.

    Inevitibly this would make for six members that are sinners instead of eight. (or if you combine the two... would be seven again )

    What do those of you that understand this think? Is the first line strong enough to keep it from being eight sinners? Is the day before the week definitive enough to keep it from being six?

    Just figured I would give you something to think about....

    ~Betazep
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  13. #58
    back? dbaryl's Avatar
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    The one thing that has to be in order ofr all this to work is for all the sinners to follow the logic. While it's hard to believe that it could actually happen, I can understand where the solution's coming from...

    Then again, this is all in theory, and that all sinners must be extremely logical/good at math...
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  14. #59
    _B-L-U-E_ Betazep's Avatar
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    >>Case 1 of infinite: There are six people in the congregation. They wait one extra day to see if they can psyche the priest out.


    You have to go off of the data that was given. You can't make stuff up from what you believe 'could' happen. You can't put our religious values and viewpoints of sinners into their definition of sinners.

    >>>1 sinner: goes to church the next day and sees no mark

    He wouldn't go to church the next day. If he was the only sinner, only the congregation would come to church the next day because they are following the data given.

    If he went home and doesn't return the very next day... it is over. So him returning once a week is not even a part of it.

    >>A million people in the congregation (no limit right?). It takes the one sinner seven days to search through them all (didn't mention how long it takes to search through an infinite amount of people). I don't mean to get technical on ya, but you got technical on me first so.... (yes I know, I am clearly one of the sinners.)

    That one is great! You are smart and funny.
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  15. #60
    Registered User Natase's Avatar
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    Well thank you, Betazep... I believe I have been enlightened.

    Actually, I thought the answer made sense at first... it was after thinking about it for while that I decided it must be fuzzy logic... took your explanation to clear the vision I s'pose.

    hehe, I think your answer converted Justin W as well... he seems to be throwing up a number of desperate loopholes (valid real-word hypotheses nonetheless)...

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