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in New Word Riddles
Posted July 20, 2018
Posted July 19, 2018
I too, have been mulling over this one since you posted. Best I've come up with so far
Don't think I'm there yet. Also, has crossed my mind that this one may be different than Sir Shakee's typical wai.
Posted June 20, 2018
then how about a
Posted June 19, 2018
okay, time to purge - been thinking along the lines of
flypaper or a roach trap
but still thinking am not quite there. much fun plasmid!
Posted June 18, 2018
if not an
escalator, perhaps an excavator?
trying to chime in here before Wilson
too late. and of course his is excellent. rats!
Posted June 15, 2018
in New Logic/Math Puzzles
Posted June 14, 2018
36 minutes ago, plasmid said:
square formed 45 degrees from orthogonal using the black king?
Rats, thought I might have it with
but peaked at Wilson's latest salvo which looks more like a direct hit.
Posted June 5, 2018
ah man - thought for sure she had it. nice entré ash9898!
Posted June 4, 2018
Posted June 3, 2018
Gotta think ^^ that's ^^ it. Nice reentry Wilson.
Nice one as always monsieur 'puddn. plasmid steered me in the right direction with his interpretation of the first two lines. Must say I especially appreciate the double entendres in the second stanza. Very nice!
Posted June 1, 2018
Don't know if this is still relevant but my first two thoughts when looking at this mix of seemingly unrelated characters is their keyboard position or the ascii representation of the characters; and then some kind of manipulation of one of those relative to the 81 or 08/01 clue. Good luck.
Had originally thought
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.
Posted May 31, 2018
well hello good sir - great to see you here
Posted May 16, 2018
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
Posted May 9, 2018
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?
Posted April 19, 2018
an attempt at an explanation by example
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
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
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
Posted April 18, 2018
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...
Posted April 6, 2018
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
Posted April 5, 2018
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)
sqrt(5) - still lacking style but working on it...
zero style points (for this guess, for now...)