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fabpig

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Posts posted by fabpig


  1. Aah. I have been thinking long and hard about this. But no solution comes to hand.
    I'm intrigued as to what "a woman like you" is. But I find it refreshing that a lady (such as you) can be so open about their lives.
    BMAD, I think I love you
    Just FYI, In my avatar, I'm the one one the right. Or left, if you prefer.
    xxxx


  2. Wow. Sorry if I spoilt it for yourself others, pg. But it had a sort of ring to it (that's a pun that will become apparent), and I came over all nostalgic as it reminded me of one of the first riddles of yours that I answered,

    In fact, I have a feeling that that riddle was the reason I signed up for brainden all those three and a half years ago (if not, it was certainly one of yours just before it). So you're responsible for me annoying the hell out of everybody since.

    Thankyou, and more power to your elbow, my friend.


  3. This is a variation on the problem of 2 people trying to split something. [where A cuts and B chooses which half].

    One solution that solves this problem, but can be used for any practical value of N people

    Arrange people in some order [ example: we will use sitting around a table and that things are passed clockwise].

    • Person 1 [arbitrarily chosen] cuts a piece of cake, puts it on a plate and passes it to Person 2
    • Person 2 can either keep the piece [if it looks good] or pass it along to Person 3
    • ...
    • Person 5 can either keep the piece [if it looks good] or pass it along to Person 1 [who must keep it]

    After the 1st slice has been kept by someone. The knife is passed to the next person [that does not have a piece already] and the same process occurs. After someone takes a slice, they will not longer participate.

    This process continues until someone cuts the 4th slice and the remaining person gets to choose that slice or what is left.

    With this process, every "cutter" is encouraged to cut a slice that is fair, but not too big

    Not sure that this is what TSLF is aiming at. I think he means "How do you divide a round cake equally into 5ths using 4 cuts". But I've been known to be wrong before...

    Edit... or more specifically, how do you ensure that each of 5 people gets an equal amount.

    Ahh k-man, where were you when we needed a modicum of common sense? You're THE man.

    Oh and, BTW, I did so inform y'all thusly. :)


  4. This is a variation on the problem of 2 people trying to split something. [where A cuts and B chooses which half].

    One solution that solves this problem, but can be used for any practical value of N people

    Arrange people in some order [ example: we will use sitting around a table and that things are passed clockwise].

    • Person 1 [arbitrarily chosen] cuts a piece of cake, puts it on a plate and passes it to Person 2
    • Person 2 can either keep the piece [if it looks good] or pass it along to Person 3
    • ...
    • Person 5 can either keep the piece [if it looks good] or pass it along to Person 1 [who must keep it]

    After the 1st slice has been kept by someone. The knife is passed to the next person [that does not have a piece already] and the same process occurs. After someone takes a slice, they will not longer participate.

    This process continues until someone cuts the 4th slice and the remaining person gets to choose that slice or what is left.

    With this process, every "cutter" is encouraged to cut a slice that is fair, but not too big

    Not sure that this is what TSLF is aiming at. I think he means "How do you divide a round cake equally into 5ths using 4 cuts". But I've been known to be wrong before...

    Edit... or more specifically, how do you ensure that each of 5 people gets an equal amount.


  5. man look tough.

    it seems to me though bonanova is on the right track.

    take the five lists look at numbers either in the same position, or in the same order.

    that will tell you the original order.

    I'm not entirely sure why 5 is the magic number though.

    lets say we have 9 numbers.

    4 1 7 8 2 3 5 9 6

    2 4 1 9 7 8 3 5 6

    2 4 1 8 3 5 7 9 6

    2 4 1 6 7 8 3 5 9

    3 2 4 1 7 8 5 9 6

    can you deduce my original sequence?

    2,4,1,7,8,3,5,9,6

    woohoo


  6. You are given five lists. For any two numbers in the series, call them A and B, you will know that in one of those lists A has been moved, in one other list B has been moved, and in three lists neither A nor B was the number that was moved -- meaning that in those three lists A will be to the left of B only if it's to the left of B in the original list, and to the right of B only if it's to the right of B in the original list.

    So if you find three (or more) lists where A is to the left of B, then you know that A is to the left of B in the original list. If you find three (or more) where A is to the right of B, then A must be to the right of B in the original list.

    Repeat that for all pairs of numbers (or howevermany you need to figure out the list's order).

    This would be the reason why having (at least) five lists, regardless of how long the list of numbers is, is important.

    That's what I thought I was saying. :unsure:

    Damn my northern accent!


  7. Bad wording... I said that whichever number came first in the lists, when I meant whichever number came to the left of the other one - not necessarily immediately left of - (more than once)



    But even the "more than once" is wrong. EG. take original list 1,2,3,4,5 for simplicity's sake. It is possible for the following 5 lists:

    2,1,3,4,5 /* Here "1" has moved to column 2 */
    2,1,3,4,5 /* And "2" has moved to column 1! */
    1,2,3,5,4
    1,2,4,3,5
    1,3,2,4,5

    In the above, the occupant of column 1 in the original must occupy columns 1 or 2 in the 5 lists on at least 4 occasions. That applies to "1" and "2". 2 comes left of 1 twice, vs 1 left of 2, 3 times. So it's whichever is the leftmost more often that determines that number 1 has column 1 (in this example). This could be nikolas's reference to the number 5 (majority 3 to 2).

    So. Lets move on to column 2. It's possible for the inmate to occupy columns 1,2, or 3 in the 5 lists on >= (n - 1) occasions in the 3 columns. Any occurence of number "1" can be ignored as it has already been allocated to column 1. There are 5 occurences of "2" , and 4 occurences of "3" in the 3 columns. As 2 precedes 3 in the 5 lists on more occasions, it can be allocated to column 2.

    Same applies to each subsequent column, ignoring numbers that have already been allocated.

    Think That about covers it.

    I'm going to try to write a program to see if it works for lists with more elements (you have been warned :) )

    This could get (even) ugli(er)!

  8. for example: 5 lists={[1,2,5,3,4],[1,5,3,4,2],[4,2,1,5,3],[2,3,1,5,4],[2,1,3,4,5]} and the original is Original=[2,1,5,3,4].

    I understand that to go from [1,2,5,3,4] to [1,5,3,4,2] can be easily achieved by moving the 2 to the rear but how does one then go to [4,2,1,5,3]? Can we slide two numbers at once?

    Think I can answer that.....

    Each list has one number moved from the original list...not the previous list.



  9. The number in position 1 must always be in either position 1 or 2 in every list apart from the list in which it's moved. If this applies to only one number, then that must be pos 1.
    Now, it's possible that more than one number meets this criterion (for example if list 3 was [2,1,4,5,3] and/or list 4 was [2,1,3,5,4] then both "2" and "1" fall into this category). In this case we need to check the order in which the two numbers occur more than once. Whichever occurs first (more than once) is pos 1.

    Similarly, the number in position 5 must always be in position 4 or 5 apart from the list in which it's moved.

    I suspect that the same procedure can be applied to positions 2 onwards (+-1 of actual position), but I've yet to try it exhaustively.

  10. @fabpig the numer of elements each list contains is n(variable). In this case n=5, and n>=3 always because of the number of permutations. e.g. for n=2, n!=1*2=2<5, but 3!=1*2*3=6>5 and as a result we can have the 5 lists.

    Not sure if that's what I'm asking :) . If there are eg. 9 elements, will there be 9 lists (that way, we know that each element has been moved once)?

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