bonanova
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Posts posted by bonanova
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2 hours ago, tojo928 said:
Think I have the ideal solution here
A (0,0); B (1,0); C(1,1); D (0,1); E (cos(30),sin(30)); F (cos(60),sin(60))
This solution assumes if you are excatly the same distance from points you get to choose. Very slight tweaks to the points can force the longest path
Start at A. B,D,E&F are all 1 mile away so pick E (+1 mile)
At E. B and F are both same distance away (2 sin(15)~.5176) so pick B (+.5176 mile)
At B. A,E,&F are all 1 mile away so pick F (+1 mile)
At F, D is closest (2 sin(15)) (+.5176 mile)
At D, C is closest (+1 mile)
At C, return to A (2^.5~1.414
Total distance walked= 3*(1)+2*(2sin(15))+2^.5~5.449
SpoilerExtremely close to optimal. So close I'm declaring it solved.
SpoilerThe best solution I know of has length of L = (5+√2+√5+√6) / 2 = 5.549+ by moving one of your points (E) to the center of side AB and starting by going from that point to B. Check it out.
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I ask people at random if they have two children and also if one is a boy born on a tuesday. After a long search I finally find someone who answers yes. What is the probability that this person has two boys? Assume an equal chance of giving birth to either sex and an equal chance to giving birth on any day.
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@plasmid Not sure why derivative failed, but you're right about the result. Nice solve.
SpoilerIt's exact for infinitely many princes, otherwise stop at the floor function of N/e.
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Spoiler
Let Tk = kth triangular number = sum(i=1,k) { i } = k(k+1)/2
y(x) = prod(k=1,inf) { x/2k } = lim(k->inf) { xk/2Tk } = lim(k->inf) e[k ln(x)]/e[Tk ln(2)]
(1) y(x) = lim(k->inf) e[k ln(x) - Tk ln(2)]
Then if the continued exponent is evaluated left to right,
z(x) = x ^ (x/2) ^ (x/4) ^ (x/8) ^ (x/16) ^ (x/32) ^ .... = x ^ y(x) = e ^ [ y(x) ln(x) ]
(2) z(x) = e ^ [ ln(x) lim(k->inf) { e[k ln(x) - Tk ln(2)] } ]
For finite x>0, lim(k->inf) { xk/2Tk } exists and is zero, since Tk dominates k, and y(x) = 0, and z(x) = 1.
Otherwise, from equation (2) z(x) is indeterminate: z(x) is 1 for finite x, but goes with x to infinity for finite k. The problematic term is k ln(x) - Tk ln(2). Since there is no satisfactory extension of L'Hopital's rule to multiple variables, we would have to arbitrarily assign a relation between x and k to get a definitive answer. If we let x = k, for example, we still get the above result. But if k increased more slowly, say letting x = ln(k), then z(x) diverges to infinity. Similarly different behaviors occur when x goes to zero.
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On 4/2/2018 at 2:17 AM, Molly Mae said:
Ah.
The odds of Al moving seem to be 50%. If we test it against a smaller number of passengers, we can see how often these closed loops form and how often Al is in his own assigned seat. It seems as the likelihood of the latter decreases, the likelihood of the former increases appropriately.
Here I have the layout for 3 and 4 passengers. 1 is Al and the last number is Bert (3 for 3 passengers, 4 for 4 passengers). Any time 3 can take the third spot, Al stays put. Any time Al is already in his assigned seat, Al can stay put. Any time the displaced person finds their seat, Al can stay put. Al staying put is denoted with a 'y'. Al moving gets an 'n'.
Al stays put 3/6 times with 3 Passengers:
123 - y
132 - y
213 - y
231 - n
312 - n
321 - nAl stays put 12/24 times with 4 passengers:
1234 - y
1243 - y
1324 - y
1342 - y
1423 - y
1432 - y
2134 - y
2143 - y
2314 - y
2341 - n
2413 - n
2431 - n
3124 - y
3142 - n
3214 - y
3241 - n
3412 - y
3421 - n
4123 - n
4132 - n
4213 - n
4231 - n
4312 - n
4321 - nThese cases are the ones I calculated as well, and led me to the right conclusion. (I had to solve it as it's not mine originally and I did not receive the solution.) It took a little insight to get me thinking along productive lines. See the spoilers in my April 1 post - btw not an April Fools post - to get there.
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13 hours ago, DejMar said:
My intuitive placement and path of the cabins would be thus:
Let the vertices of the square be A, B, C, D with the diagonals being AC and BD.Let the first cabin [a] be at point A.
Let the second cabin be a walk of 1 mile from A along the diagonal AC.
Let the third cabin [c] be perpendicular to diagonal AC on side CD.
Let the fourth cabin [d] be at or on toward point D at no less distance than point C from cabin ;
this should be approximately 0.4 miles distant from cabin [c].
Let the fifth cabin [d] be at point C.
Let the sixth cabin [e] be at point B.
Finally, a return to the car at cabin [a], point from cabin [e] A requires another walk of 1 mile.
Accumulative walk is approximately 4.8 miles.Not a bad result.
SpoilerThe locations (four corners, side and circular arc) are all correct. But adjusting the last two a bit, and changing the first cabin, in order to alter the sequence, the distance can exceed 5.
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12 hours ago, plasmid said:
I’m going to make the math easier by
not assuming that there are 100 suitors, but assuming there are a “large” number of suitors. Let’s say it’s the whole kingdom, K, so modifying my previous answer accordingly we have:
P(GCD=1) = N/K
P(GCD=2) = (K-N)/K x N/(K-1) ~= (K-N)/K x N/K = N(K-N) / K2
P(GCD=3) = (K-N)/K x (K-N-1)/(K-1) x N/(K-2) ~= [(K-N)/K]2 x N/K = N(K-N)2 / K3
P(GDC=X) ~= N(K-N)X-1 / KXAnd the odds of picking MWP for any given GCD are still 1/(GCD-1) so the odds of picking MWP are approximately
Sum[GCD=2 to K] N(K-N)GCD-1 / (GCD-1)KGCD
That doesn’t make it easy enough to just calculate by hand, but it does make it easy enough to make a spreadsheet to handle the calculations. If I make K=100, then this estimate says that the best number of chumps is 37 of the hundred, which would leave (ironically?) about a 37% chance of getting the MWP.
Great job! Want to look at the reciprocal of that number and guess the exact result?
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Spoiler
The locus of points equidistant from the origin is a circle, using standard Cartesian distance.
Distance traveled by delta x and delta y (so-called 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 45-degree line is a parabola if plotted vs angle from 90o to 0o.
The exact value is ln{ 1+sqrt(2) } + sqrt(2) = 2.2955871...
@Molly Mae blazed the trail and @plainglazed nailed it.
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You are both on the right track, but I realize now that I mis-stated 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 sort-of serious math. My bad.
I edited the OP.
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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?)
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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 (un-visited) cabin closest to his current location.
If there were 6 cabins in all, how might they be placed, so that using Al's nearest-neighbor 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 warm-up.
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Yes and yes.
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Al, Bert, and Charlie competed in a track and field event in which points were awarded for 1st, 2nd, and 3rd, 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 1st in the shot put. Who finished 2nd in the javelin throw?
This is a Gold star puzzle.
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1
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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 un-overlapped 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.
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Find a, b, c.
ab x ca= abca , a 4-digit number
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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.
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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/(2k) }
(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 xk but decreases as 1/(2k). 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/(2k), as well. So for infinitely large x, ln z also goes strongly to zero.
limx->inf { ln z(x) } = 0 and so
(5) limx->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) = 00, 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) limx->0 { z(x) } = 1
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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) .... ] ] ] ] ]
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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.
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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?
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Spoiler
Let'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 100th 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."
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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.
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Good start -- That's shorter than 3, which comes from any three of the square's edges.
The Most Wonderful Prince
in New Logic/Math Puzzles
Posted · Edited by bonanova
corrected Rs formula
So this guy Thomas Bruss solved the general stopping problem.
(Check out section 2 on page 1388 of the pdf file available in the above link.)
Bruss assigns 1 or 0 to each event in case it's a best case so far. Then the problem reduces to finding the final 1 in the sequence. The probability pk of say the kth event being a 1 is 1/k, making the odds rk = pk/(1-pk) = 1/(k-1). He then summed the odds starting with the last event and working backward, stopping when that sum first reaches (or exceeds) unity: (Edit): Rs = 1/(n-1) + 1/(n-2) + ... + 1/(s-1) >=1. He proved that s is the optimal stopping point. And doing the math he showed that s/n = 1/e and that ((s-1)/n)Rs (the probability of success) approaches 1/e, as well as n -> infinity.
It's really a beautiful result.