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plainglazed

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Posts posted by plainglazed

  16. A Lovely Little Riddle

    in New Word Riddles

    Posted

    Had originally thought

    Spoiler

    diamond (engagement ring) - which fit a little better before the last line edit

    And another observation re site traffic:  Think there is less spit balling of answers these days.  Most now seem to reply only when highly confident of being correct.  Tho guess that could be a consequence of lower traffic as previously there was more of a race to the solve.  Still do think that adds to the feel of the lack of involvement.

  18. Turnabout

    in New Logic/Math Puzzles

    Posted

    No fooling you three, y'all were all over this one.  MM recognizing the slip fall right off, the good Captain doing the heavy lifting, and bn wrapping up the actual question which may have been apparent to you all just vaguely asked for in the OP.  Cheers

  19. Turnabout

    in New Logic/Math Puzzles

    Posted

    You and I each have a coin, select its orientation (heads or tails), and reveal to each other simultaneously.  If both are heads, you win a dollar.  If both are tails, you win a quarter.  But if both are different, I win fifty cents.  Fair enough?
     

  20. Coin hunt

    in New Logic/Math Puzzles

    Posted

    an attempt at an explanation by example

    Spoiler

    Let's say there are twenty four coins.  Label 0 for tails and 1 for heads.  Thus a string of twenty four ones and zeros describes

    the coins in play.  Let's also say the tenth coin is the money coin.  So something like this:

       H H T H T T T H T T H H H H T H T H T T H T H T
       1 1 O 1 O O O 1 O O 1 1 1 1 O 1 0 1 0 0 1 0 1 0
     

    Listing each coin position in binary and corresponding coin value:

          1 - 1(heads)
        10 - 1
        11 - 0
       100 - 1
       101 - 0
       110 - 0
       111 - 0
      1000 - 1
      1001 - 0
      1010 - 0
      1011 - 1
      1100 - 1
      1101 - 1
      1110 - 1
      1111 - 0
     10000 - 1
     10001 - 0
     10010 - 1
     10011 - 0
     10100 - 0
     10101 - 1
     10110 - 0
     10111 - 1
     11000 - 0


    We use the bit positions that are powers of two as parity checks to our string of coins.

    Define the parity at bit position 1 as the sum mod 2 of all bits that are in positions where the least significant digit of that

    position in binary is a one (the ones column if you will).  Thus the sum mod 2 of the bits in positions 1,3,5,7,9,etc.  In our

    case that is (1+0+0+0+0+1+1+0+0+0+1+1)mod2=1

    Define the parity at bit position 2 as the sum mod 2 of all bits that are in positions where the second least significant digit

    of that posistion in binary is a one (the twos column).  Thus the sum mod 2 of the bits in positions 2,3,6,7,10,11,etc.  In our

    case that is (1+0+0+0+0+1+1+0+0+1+1+0)mod2=1

    Define the parity at bit position 4 as the sum mod 2 of all bits that are in positions where the third least significant digit of

    that position in binary is a one (the fours column).  Thus the sum mod 2 of the bits in positions 4-7,12-15,20-23,etc.  In our

    case that is (1+0+0+0+1+1+1+0+0+1+0+1)mod2=0

    The parity at position 8 is the sum mod 2 of the bits in positions 8–15,24–31,40–47,etc.  In our case that is

    (1+0+0+1+1+1+1+0+0)mod2=1

    The parity at position 16 is the sum mod 2 of the bits in positions 16–31,48–63,80–95,etc.  In our case that is

    (1+0+1+0+0+1+0+1+0)mod2=0

    etc.


    The values (heads or tails) at bit positions that are powers of two (positions 1,2,4,8,16) are 1,1,1,1,1

    And we've calculated the parity at bit positions that are powers of two to be 1,1,0,1,0

    The bits at positions 4 and 16 of the calculated parity do not match the actual bit value of our coins in those positions.  But

    if we flip the coin at position 10100 and recalculate the parity, it matches the value of our coins.

    Now if we flip the money coin, our partner can calculate the parity at positions 1,2,4,8,16 and see that bit values do not match

    at positions 2 and 8 and know the money coin is at position ten.

    If the total number of coins is a power of two, the steps of making the calculated parity match the actual bit values and then

    flipping the money coin can be combined into one flip.


    For a better explanation wiki Hamming Code

     

     

  21. Coin hunt

    in New Logic/Math Puzzles

    Posted

    Been extremely interested in this one plasmid my friend, and for the longest time thought the above post must be a mistake.  Now think am on to you.  Just need to figure out how to simply explain the solution if indeed my most recent process pans out.  Just wanted to let you know am still on the case.  Know it can be frustrating when a thread goes on without response, especially when it's a fine puzzle such as this one.  I'll be back...

  22. Fettered random walk

    in New Logic/Math Puzzles

    Posted

    more musings

    Spoiler

    If you flip the fair coin and it happens to land all heads or all tails you'd travel along the x- or y-axis a distance of 2sqrt(2).  The line segment connecting these two points, (0,2sqrt(2)) and (2sqrt(2),0), describes the endpoints of all random walks.  The centroid of this segment is (sqrt(2),sqrt(2)) which makes sense because after the first step of sqrt(2), all subsequent steps combine for a distance of sqrt(2) thus making the distance from the origin to the centroid the minimum distance.  All points northwest of the segment's centroid and all points southeast are the inverse of one another so either half of the segment can be considered in search of the expected distance travelled.

    So we can look along the segment from (sqrt(2),sqrt(2)) (which is the minimum distance from (0,0) and is equal to 2) to (2sqrt(2),0) (which is a maximum); to find the expected distance travelled.  The arithmetic average of these two numbers is sqrt(2)+1.  But that does not jive with the midpoint of this segment which is a distance of sqrt(5) from the origin.  So the distribution along the segment is not linear? And my explanation for my third guess is foolishness and is somewhat debunked above.

    I did a simulation which for me means a spreadsheet.  Only sixteen paces (the last was still pretty small at .00004316) and only 100,000 trials and came up with an expected distance of 2.33668.  For me that number leaves no clue as how one might back out an explanation.  Having difficulty understanding how repeated irrational distances could clump at all but I have fractally little maths.  Have found this very interesting and hope someone with better maths can explain.  Thanks bn

     

  23. Fettered random walk

    in New Logic/Math Puzzles

    Posted

    or maybe

    Spoiler

    (2/3)sqrt(10)

    after the first two steps you're sqrt(10)/2 from the origin.  continuing on in the same manner you're then an additional sqrt(10)/8 then sqrt(10)/32 then sqrt(10)/128... from the origin.  for every step you take that is not alternating north and east, the inverse of that path balances out?

    yeah, so one of those three (not the first though)

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