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Posts posted by bonanova


You are both on the right track, but I realize now that I misstated the OP. I didn't ask for what I wanted.
What I wanted to get at was the average location, that is the average of all the possible ending location coordinates, more precisely, their centroid, and its distance from the origin.
That's not the same as the expected distance of the ending points  which does take sortof serious math. My bad.
I edited the OP.

2 hours ago, CaptainEd said:Here’s a degenerate answer
put C, D, and E completely inside A without overlapping each other, put B completely outside A. This leaves 16 pi inside (red), and 16 pi outside (green).
Good solution @CaptainEd . Are there other solutions? (Or more general ones?)

Al made sales calls at a number of cabins, which lay in a square field, one mile on a side. He drove his car to the first cabin, then visited the remaining cabins and returned to his car on foot, walking several miles in the process. If Al had had a map and perhaps a computer, he could have picked the shortest route to take, (traveling salesman problem). Lacking these amenities, Al simply chose to visit (after the first one) the (unvisited) cabin closest to his current location.
If there were 6 cabins in all, how might they be placed, so that using Al's nearestneighbor algorithm, and selecting the worst initial cabin, Al would be forced to walk the greatest distance; and what is that distance?
Examples:
2 cabins: diagonal corners, starting at either: 2 x sqrt(2) = 2.828... miles.
3 cabins: any three corners, starting at any of them: 2 + sqrt(2) = 3.414... miles.Check out n=4 and n=5 as a warmup.

Yes and yes.

Al, Bert, and Charlie competed in a track and field event in which points were awarded for 1^{st}, 2^{nd}, and 3^{rd}, place only. At the end of the day, Al had accumulated 22 points, while Bert and Charlie each garnered only 9 points. No other competitor earned points. Bert was 1^{st} in the shot put. Who finished 2^{nd} in the javelin throw?
This is a Gold star puzzle.
 1

Sir, I bear a rhyme excelling
In mystic force and magic spelling
Celestial sprites elucidate
All my own striving can't relate.See, I have a rhyme assisting
My feeble brain, its task ofttimes resisting. 
You are given 5 circles, A, B, C, D, and E, whose radii are, respectively, 5", 4", 2", 2", and 1".
Can you find a way to overlap circle A with portions of some or all of the other four circles so that the unoverlapped portion of A has the same area as the sum of the unoverlapped portions of the other four circles? That is, the red area is equal to the sum of the green areas. Circles B, C, D and E may overlap portions of each other as well as a portion of A.

Find a, b, c.
a^{b} x c^{a}^{}= abca , a 4digit number

Here’s a challenge: I think this puzzle can be solved with almost no math at all. Like, think of a single “what if” question to ask that changes the conditions slightly.

I have a strong feeling, and I'm working on a proof, that
Spoilerboth answers are 1.
SpoilerConsider:
(1) z(x) = x ^ (x/2) ^ (x/4) ^ (x/8) ^ (x/16/) ^ (x/32) ^ .... (evaluated left to right)
(2) Let y(x) = Prod_{(}_{k=1, inf) } { x/(2^{k}) }
(3) Then z(x) = x ^{y(x)}. = e ^{y(x) ln(x)}
Question 1:
What is the behavior of z(x) as x increases without bound?
(4) From (3), ln z = y ln x.
From (2) y increases as x^{k} but decreases as 1/(2^{k}). For any finite x, the product terms in y individually go exponentially with k to zero. Thus y itself is clearly zero for any finite x. From (4), ln z increases more than y by a factor of ln x. But ln x is dominated by 1/(2^{k}), as well. So for infinitely large x, ln z also goes strongly to zero.
lim_{x}_{>inf} { ln z(x) } = 0 and so
(5) lim_{x}_{>inf} { z(x) } = 1.
Question 2:
What is the behavior of z(x) as x decreases to zero?
From (2), y(0) = 0 by inspection.
From (3) z(0) = 0^{0}, which is indeterminate without the knowledge that y(x) goes much more strongly to 0 than x does, deciding in favor of the exponent.
Thus
(6) lim_{x}_{>0} { z(x) } = 1

Which way should we evaluate this?
[ [ [ [ [ x^(x/2) ]^(x/4) ]^(x/8) ]^(x/16) ]^(x/32) ]....
or
x^[ (x/2)^[ (x/4)^[ (x/8)^[ (x/16)^[ (x/32) .... ] ] ] ] ]

One approach, among several, is to
Spoilerconsider the conditions under which a given number of passengers move
and
Spoilerwhether it matters that Al was first to board.

On 3/25/2018 at 4:52 AM, harey said:A large colony of squirrels dug holes during the summer and in each hole, they put between 1 and 100 nuts (each quantity has the same probability).
If each squirrel has to eat 100 nuts during the winter, how many holes must he find (in average)?
Each hole contains 50.5 in average, so 2 should be enough, right?
I guess I've always been a little confused about what is being asked.
Sharing is not permitted, yet an "average" result is requested. If average values are used, then two holes is enough. If there is no sharing or averaging, survival is assured only by the (very unlikely) worst case of 100 holes. If the question is what is the expected number of holes that together yields at least 100 nuts, we have an answer from simulation.
Is there a way to say precisely what else might be needed?

SpoilerLet's think about it this way.
Say the bag has 99 white marbles and 1 marble that has a probability p of being white. If I'm giving 5:1 odds that all the marbles are white, I'm betting that p > 5/6. I'm basing the bet on a single event where the first 100 draws from the bag were white. That probability is Q = (.99 + p/100)^{100}.
If we knew the value of Q, we could deduce p and then knowledgeably take the bet, or not. But we don't. And to be on the conservative side, we conclude, from a single success, only that Q > 0.5, which corresponds to p > 0.31. (The 100^{th} root of .5 is .99309.) That means Q is favorable even when p is as low as 1/3, where we should be getting 2:1 odds rather than giving 5:1 odds. To be confident of p being > 5/6 we'd need to be confident that Q > (.99 + 5/600)^{100} = ~0.85. A single favorable outcome doesn't support that level of confidence.
Still not taking the bet.
@harey, some guidance about completing a formula would be appreciated. I'm interested in the "surprise."

23 minutes ago, flamebirde said:Yes, but I can't tell you where (assuming there's only one path). Depending on how the holy man's speed varies between the two trips, there must be a place where the two trips are at the same point at the same time, but you couldn't say where without more information.
The proof is simple: if we imagine that there are two holy men, one going up and one coming down, there is no way that the two men do not meet each other (in order to get to the other end of the mountain, they must both take a continuous path from top to bottom or bottom to top, which is impossible to do without meeting each other). In order to meet each other, they have to be at the same point at the same time. Hence, the two different paths on the two different days must meet at some point.
I'm certain there's another way to do this via calculus theorems (maybe intermediate value or mean value, something like that), but I just can't think of it at the moment.
Yes you cand do that. The elevation of the two men as they ascend and descend are continuous functions of time. If at one time one is greater and at a later time the other is greater there must be a time when they are equal. Since they both stay on the path, and if the the slope of the path never changes sign, then they meet at that time.
If the slope does change sign, then multiple points on the path will have the same elevation and there can be multiple times when the elevation of the two men are the same. They meet at one of those times.

Good start  That's shorter than 3, which comes from any three of the square's edges.

Al and Bert are among 100 passengers assigned to one hundred seats on an airplane. Al was first to board, and Bert was last. Strangely, the first 99 passengers ignored their boarding passes and took random unoccupied seats. Bert liked the seat he was assigned and is not happy with the situation. If he's lucky, his seat is unoccupied and there's no problem. Otherwise, he insists the passenger erroneously occupying it move to his own assigned seat. The displaced passenger must then move, possibly displacing another person. This process continues until all passengers are seated.
What is the probability that Al must move?

Recently we considered the shortest roadway that connects the four corners of a square.
Here we seek the shortest set of line segments, one attached to each of a square's corners, that need not connect with each other. Instead, what we ask of the line segments is that is they will block any ray of light attempting to pass through the square.

Consider a random walk in the plane where each step is taken, beginning at the origin, in either in the positive x or positive y direction, i.e. either east or north, each choice being made by the flip of a fair coin. The length of each step is 1/2 the length of the previous step, and the first step has length √2. After infinitely many steps have been taken, what is your expected distance from the origin?
Edit: Ignore the original text in pink. Instead,
What is the distance to the origin of the centroid of the possible termination points? You find the centroid of a set of points by averaging respectively their x and y coordinates.
First correct answer wins, but style points will be awarded as well.

@plasmid Nice solve. The puzzle itself was more a math exercise than a puzzle  that part was thinking it through and setting up the calculation. I thought is was interesting in that you can sort of envision the setup and know that it had to be e.g. greater than 1/3, but maybe not as great as 2/3, so the question was would it be greater than 1/2?

On 3/26/2018 at 10:33 PM, plasmid said:If you do that and the (uv) term is x(F^{3}1), then at x=infinity wouldn't you end up with infinity times zero?
If there's a rule about the limit of multiplication between x and functions with e^{x} as x goes to infinity, then maybe I could use that in the route I was taking.
Yes there is a rule, L'Hopital's rule. Basically you can just replace functions by their derivatives to resolve indeterminate values. Or, you can just evaluate expressions and see how they behave:
SpoilerHere's a plot of uv = x (F^{3} 1) = x {  3e^{}^{x} + 3e^{2}^{x}  e^{3}^{x} } vs x for x in [ 0, 10 (scaled x10) ]
It's intuitive that the exponential dominates for large x if you think of say x / e^{3}^{x}, instead of x e^{3}^{x}.

I guess we can compute expectation value as well:
SpoilerE = Sum_{k} { outcome_{k} } x { probability of outcome_{k }}
One million simulations:
# holes % prob product

1 hole 1.01% 0.0101 0.0101
2 holes 50.40% 0.5040 1.0079
3 holes 32.85% 0.3285 0.9856
4 holes 11.98% 0.1198 0.4790
5 holes 3.04% 0.030427 0.1521
6 holes 0.60% 0.006024 0.0361
7 holes 0.10% 0.000993 0.0070
8 holes 0.01% 0.000125 0.0010
9+ holes 0.00% 0.000021 0.0002
======
2.6792 
9 hours ago, harey said:SpoilerOne million simulations:
# holes % of squirrels that find 100+ nuts in their holes
1 hole 1.01%
2 holes 50.40%
3 holes 32.85%
4 holes 11.98%
5 holes 3.04%
6 holes 0.60%
7 holes 0.10%
8 holes 0.01%
9+ holes 0.00%SpoilerEvery hole has a 1% chance of containing a single nut.
If you want to guarantee all squirrels survive, they must each dig up 100 holes 
The King has decreed that his daughter the Princess shall marry the most wonderful Prince in all the land. One hundred suitors have been selected from their written applications, and on a certain day the King arranges for them, in turn, to interview the Princess. Each suitor must either be chosen or eliminated on the spot. If the Princess does not choose any of them, she will marry the last Prince to speak with her.
You have been chosen as the Royal Advisor to the Princess and tasked with implementing her best strategy to choose the Most Wonderful Prince of the realm. You devise an evaluation scheme by which the princess can assign a unique "wonder number" to each prince as she meets him. The strategy then is to have the Princess reject, but record the highest score of, the first N princes that she meets. The Princess will then choose the first Prince that she subsequently interviews whose score exceeds that recorded score.
That's it. The puzzle is basically solved. Except, of course, to decide on the optimal value of N. It requires some thought. If N to too high, the most wonderful prince is likely to be eliminated at the outset, and she ends up with the last guy. If N is too small, the Princess will likely settle for a fairly undistinguished prince.
What value of N optimally balances these two risks? What is the probability that the Most Wonderful Prince will be chosen?
Disclaimer: I recall this puzzle being posted before, with different flavor text. And it's somewhat of a classic. To give it a fair play here, I'll ask not to post any links and not to just give the answer if you know it, at least not without "showing your work."
Fettered random walk
in New Logic/Math Puzzles
Posted · Report reply
The locus of points equidistant from the origin is a circle, using standard Cartesian distance.
Distance traveled by delta x and delta y (socalled taxicab metric) travel changes the "circle" into a diamond shape of 4 lines, 2 each at 45 degrees and 135 degrees. In the first quadrant it's the line @plainglazed describes, whose centroid (midpoint) is at a (Cartesian) distance of 2 from the origin. That's the answer I originally had in mind  a result that can be obtained just from the knowledge that the series 1 + 1/2 + 1/4 ... converges to 2.
What I ended up asking for was the average (Cartesian) radius of a Taxicab circle. You can get it by integrating the Cartesian distance to the diagonal line as x goes from 0 to 2 sqrt(2) (which one definitely can't do in one's head) or by simulation, as @plainglazed did. It turns out to be a fundamental constant associated with the parabola  the parabola's version of pi, if you will.
Which makes sense, because the distance from the origin to the 45degree line is a parabola if plotted vs angle from 90^{o} to 0^{o}.
The exact value is ln{ 1+sqrt(2) } + sqrt(2) = 2.2955871...
@Molly Mae blazed the trail and @plainglazed nailed it.