bonanova
-
Posts
6975 -
Joined
-
Last visited
-
Days Won
66
Content Type
Profiles
Forums
Events
Gallery
Blogs
Everything posted by bonanova
-
Hats on a death row!! One of my favorites puzzles!
bonanova replied to roolstar's question in New Logic/Math Puzzles
Hi shijumonantony, and welcome to the Den! What's needed is to determine the parity of the black hats and update as we go through the line of prisoners. Whether we count black hats [posted solution] or assign them '1' and red '0' and do the artithmetic [your solution] what we're doing is determining the parity [oddness or evenness] of the black hats. The solutions are logically identical, and it works for any distribution of hat colors and any number of prisoners. -
I wouldn't worry. I think the analysis "has to be" right. I suspect possibly doing 10 million successive exponentiations - even with 32 bits of accuracy - could result in a systematic drift of the order seen here. Altho I scaled back to 10 thousand and the numbers were the same. [No doubt of convergence.] *still scratching head* I'd take the calculations [as well as the exquisite simplicity / symmetry] of your result as verification, not contradiction.
-
Would Prof. Heisenberg care to comment at this point?
-
Nice analysis. It's interesting that when Euler did this analysis in 1778, he determined the convergence range as 1 <= x <= e^(1/e). Why he didn't see convergence for any values less than unity is not clear. Anyway, I started to play with the derivative but switched to brute force calculation. I calculated p(x) for x = unity downward, to see where the calculated values began to diverge. I verified that alternating terms of the sequence converged either to the same or different stable values. Again, 10 million exponents compared to 10 million + 1 exponents. Then I successively subdivided the range where divergence set in. These graphs show the last two ranges of x that I looked at. The result is interesting. I calculate 1/e^e to be 0.06598803585.... The calculated point of divergence is slightly larger : 0.0659928... - a difference of 0.000004 ... The calculated 1/e^e value lies to the left of the second graph entirely. Here are the plots of p[x] vs x - the second plot is a blowup of the indicated region on the first plot.
-
Well, we know that p(x) is bounded above by e, so p will never reach an integer value greater than 2. What are the integer values? Here are a few. I evaluated the exponent stack to 10 million terms, and compared to the 10-million-and-oneth term to be sure of convergence. ... n n^(1/n) ... p[n^(1/n)] ... 1 1.000000000 1.000000000 ... 2 1.414213562 2.000000000 ... 3 1.442249570 2.47805268 ... 4 1.414213562 2.000000000 ... 5 1.379729661 1.764921915 ... 6 1.348006155 1.624243846 ... 7 1.320469248 1.530140119 ... 8 1.296839555 1.462501432 ... 9 1.276518007 1.411381742 .. 10 1.258925412 1.371288574 . 100 1.047128548 1.049519190 .1000 1.006931669 1.006980222 10000 1.000921458 1.000922309
-
But you should demand a reason - from someone - before declaring this solved, no?
-
Yup, and I couldn't tell you with without spoiling... Nice one.
-
Hey Noct - welcome back. btw, I'm not going to even touch this one, but I'll watch for an answer.
-
Your counterexample seems to do a random 1-dim walk along the numbers 1-6. Using a coin toss to move forward or backward, you'll stay put after 2 tosses with 50% probability. What are you saying this is a counter example of? Certainly not of a fair die, where any roll can give any result. I'm clearly missing the point.
-
I accept without proof that you do not accept this as a proof. If you want me accept that it is not a proof, I will. Just point out where it's invalid.
-
Step 1 to some answers. Consider: x to the x to the x to the x to the x to the x .... = e. What value of x, if any, satisfies this equation?
-
Let's try. Knots are of the first type [does not need an end] or the second type [needs an end] [1] You have a knot of the first type. [2] You split the knot into two knots of the second type. [3] You reverse the motions in step [2] to produce a knot of the first type, which is then removed. Step [2] is impossible by definition. Knots of the second type need an end to create.
-
Describe a single roll of a fair die. [1] the likelihood of a particular number showing is the same as the likelihood of any other number showing. [2] one of the numbers 1, 2, 3, 4, 5, 6 will show. Roll a fair die 6 times. ei is the expected number of appearances of the value i; i=1, 6. From [1], e1 = e2 = e3 = e4 = e5 = e6 etot is the expected total number of appearances of any number, regardless of value. From [2] and the fact the die is rolled 6 times, etot = 6. Since 1-6 are the only numbers that can show, etot = sum {ei} = 6 ei = 6, for all values of i. ei = 1 for all i. e6 =1. For six rolls of a fair die, the expected number of appearances of a 6 is 1. That appearance is its first appearance. The expected number of rolls of a fair die that will produce the first appearance of a 6 is 6.
-
Start with the ends of the string fastened to opposite walls of a room. Or a loop, but I like the fastened ends idea. Knots in the rope are of two types. Knots of one type can be made starting from a straight string without use of an end - for example pulling a loop of string through your fingers and tying a knot in the resulting double strand. Clearly this type of knot can also be removed from the string, without use of an end, by reversing the motions used to "tie" it. The other type of knot requires the use of an end of the string; knots of this type are impossible to tie in our case because the ends are not available. The first type of knot could be argued to possess an anti knot, as two of them could be brought together and removed. But that's specious - they could be removed individually without bringing them together. So let's turn to the type of knot that needs an end to create. If two such knots - one of them being the anti-knot to the other - could annihilate each other, without involving an end, then by reversing the annihilating motions they could be re-created without using an end. But by definition the only type of knot that can be created without use of an end is the first type, and this contradicts the condition that we are only considering knots of the second type. That's the best I can do off the top of my head.
-
Sorry, I misspoke. The additional question, rather, is ... let x to the x to the x to the x ... = 4. instead of = 2. Then is not x = sqrt(2) a solution to this equation as well? i.e., xy = 4 where y = 4 leads to x = 4th root of 4 = sqrt(2) again. Therefore, what kind of creature is sqrt(2) to the sqrt(2) to the .... - that it can equal both 2 and 4.
-
Or have I misunderstood something? Nope, you're spot on. Convergence of the infinite exponent stack was not addressed. It was only asserted, in violation of bonanova's prime directive. The equation LHS = 2 is meaningless if LHS is undefined. To validate the puzzle, I offer the following proof that LHS converges. If the proof is valid, then we can discuss solutions to the equation LSH = 2. If the proof is flawed, then the puzzle is solved by identifying the flaw. Note: LHS = left hand side of the above equation LHS is the limit [if a limit exists] of the sequence sqrt(2), sqrt(2) to the sqrt(2), .... Write the sequence as a1, a2, a3, ... where i-1 is the height of the exponent stack of term ai. Since ai+2 = sqrt(2) ai+1 > sqrt(2) ai = ai+1 . then 1 < ai < ai+1 for each i >= 1. If the topmost sqrt(2) in any ai is replaced by the larger value 2, ai itself becomes 2. So the sequence is increasing and bounded above; therefore it has a limit and converges.
-
They're not different questions. To see this, ask the same question for 1, 2, 3, 4, and 5. The answer, for a fair die, for each number is the same: one occurrence each for 6 rolls. If there is on average one occurrence, it is the first occurrence.
-
1. I will take without proof your assertion that you do not see this as a proof. 2. It would not be a fair die.
-
Very astute; and inf-1 is in fact inf, for the purposes of this riddle. Before I declare Y-san, octopuppy and woon as solvers, however, and this is where it really becomes tantalizing: Is there yet another value of x which satisfies the equation?
-
-
Yup. Nice job!
-
Fair enough. We have co-solvers, and I'll edit the puzzle tag line to so indicate. Good job all!
-
-
Exactly.
-
That's step 1, now to get there.
