bonanova
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Looks like ...
Spoilerdy/dx = 1
Spoilerb = xa = ea ln x = eln ln x = ln x
y = eb = eln x = x -
8 hours ago, CaptainEd said:
That was the first-cut answer in the ant-chessboard problem.
SpoilerThe value is too low. It counts only once the (multiple) cases where 8 heads is reached before the "full complement" of 4 tails occurs.
The solution is easier to find if we (reasonably) assume that
Spoilerthe final toss will be Heads.
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SpoilerSpoiler
But seriously, working on it.
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1 hour ago, plainglazed said:
Agree.
SpoilerI think you have the solution at 6.
I've been doing this in my head for two days. I was only able to eliminate 5. -
Just now, Molly Mae said:
Signature updated!
I'm humbled.
btw I sharpened (with humor) the flavor text and posted this on another favorite site. It will be interesting to see how quickly it's solved there.
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1 hour ago, Molly Mae said:
@Molly Mae gets the coveted bonanova gold star for finding an algorithm and getting the answer in less than one day.
So now that the cat (algorithm) is out of the bag, so to speak, it should be trivial to find whether an integer can triple, (or quadruple, or quintuple) under the same manipulation.
It's also interesting that his solution forms a ring that shows the doubling process also occurs for four other ending digits (not 1, so his is the smallest.) But which ones? Incidentally, this also shows that all such integers have 18 digits.
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Fun one.
SpoilerThe pickle solids weigh, whatever. They're now much easier to carry.
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Things are not going well at the Acme Company.
Executive talent is hard to come by, and it is not cheap. Folks at the water cooler have no ideas, and the coffee-breakers can't imagine how to improve things either. But those who party around the teapot, they came up with something. They suggested to the Board that Acme promote the newest hire in the mail room and make him the CEO! We need to shake things up, but good. Qualifications, job experience, brains, judgment, integrity, these are all things of the past.
Some were not so sure. Doesn't make sense at all, the old timers said. Almost like appointing some guy with orange complexion to be President. That's exactly the idea, said the tea-people. Turn things on end, let the bull loose in the china shop, and see what happens. Hey -- how could it be worse than what we have now? Not surprisingly, the debate was long and heated.
Such a risk merited proof of possible gain, so the old guard posed a challenge: produce a concrete example of where the idea had been tried with incontrovertible benefit. In fact, make it mathematical. You know, something that might make a good BrainDen puzzle.
We'll promote the mail room guy, they said, if you can show us an integer that doubles in value when its least significant digit is promoted to its most-significant position.
That is, give us a number { some digits } q that has half the value of q { same digits }.
Spoiler102 doesn't quite work, since 210 does not quite equal 204
That all happened last week, and now we're looking for the mail room guy. Was he promoted? Did the tea people find such a number? Is there one?
We need a number or a proof that one does not exist.
T.L.D.R.
What number doubles in value by by moving its last digit to the first position (if there is one)?
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Al, Bert and Charlie are on the ballot for secretary of the local yacht club, and they finish in a three-way tie. To break it, they solicit the members' second choices, and again it's a three-way tie. Al notes that since the number of members is odd, a series of two-way votes would be decisive. As a reward for finding a way past the impasse, Al receives the bye and will take on the winner of a Bert vs Charlie vote to decide who gets the position as secretary. Assuming voter preferences do not change, what are the winning probabilities for the three gentlemen?
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Alternate approach to the ant-checkerboard problem asks: What is the probability of flipping 8 heads before 5 tails?
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If we place four matches in the form of a square, they form 4 right angles.
If we place them like a hash-tag (#) they form 16 right angles.
If someone removes one match, can we still form 12 right angles?
(No bending or breaking of the matches is allowed.) -
On 3/3/2018 at 10:52 AM, plasmid said:
Suppose you have one guy that’s responsible for putting the bags on the conveyor belt, and the amount of time it takes him to put each subsequent bag on the belt is randomly drawn from a uniform distribution from 0 to 1 second. Also say the conveyor belt is moving at 1 meter per second to make the math easier. So you can work with a string of distances: x1, x2, x3, … and we’re looking for the ratio of (the sum of all terms xn where xn > xn-1 and xn > xn+1) divided by (the sum of all terms xn regardless of whether it's a non-nearest neighbor segment).
For any given value of xn, the probability that xn > xn-1 is simply the value of xn (since we’re dealing with a uniform distribution of values over a range from 0 to 1), and similarly the probability that xn > xn+1 is xn (since these numbers are randomly generated independently of each other). So the probability that xn is a non-nearest neighbor segment is the product of those two, or xn2.
Since the length of each segment is generated from a uniform distribution from 0 to 1, we can do an easy integral. To calculate the fraction of the conveyor belt that’s covered by non-nearest neighbor segments, I think we would want to take the integral of [length of the segment] · [probability that it’s a non-nearest neighbor segment] (which is x · x2 = x3) and divide that by the integral of [length of the segment].
Int[x=0 to 1] (x3 dx) / Int[x=0 to 1] (x dx)
[x=0 to 1] (x4/4) / [x=0 to 1] (x2/2)
(1/4) / (1/2) = 1/2That answer would very likely change if you used a different method to generate random distances between suitcases on the conveyor belt.
The answer probably does change with the nature of the distances.
SpoilerYour answer seems lower than expected.
SpoilerI wonder what you'd calculate assuming a Poisson distribution, where the probability that any interval is smaller than say a distance x can be taken, with appropriate scaling, to be 1 - e-x. Equivalently, the probability density function for an interval of distance x would just be e-x.
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Regarding the 5 probabilities after 12 ant moves:
SpoilerEnumerating paths and treating them as equals gives { C(12,8) C(12,7) C(12,6) C(12,5) C(12,4) } / (3498) multiplied by { 1 4 6 4 1 } / 16 and summing for a result of .22995. But Al (the faster ant)'s paths involve from 8 to 12 choices, so they are not equally likely. Probabilities derived by dividing each path by the sum of all paths (3498 in this case) will be incorrect. What is the right approach?
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@plasmid augments paths that touch an intermediate border point, where a choice is eliminated, by multiplying by 2 at each point. The effect is to create virtual paths that increase the count to each end point by 299, making 4096 = 212 total paths and leads to the correct probability of .20551.
I have three other approaches. The first is the one I thought would be found here. Another is quick and dirty. The last one occurred to me as I'm writing this.
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This one seems evidently correct but is the most involved of the three.
Call the Final points { F1 F2 F3 F4 F5 } with probabilities of { f1 f2 f3 f4 f5 }.
Start with a result from the first problem, the probabilities { d1 d2 d3 d4 d5 d6 d7 d8 d9 } of reaching the 9 diagonal points. Then (by inspection) determine the probabilities of reaching each Fi from the one-to-five possible starting diagonal points Dj. Call those p(Dj, Fi). There are 25, but really only 15 due to symmetry, and some are trivial. Others take just a little thought. Clearly p(D1 ,F1)=1. Then, since p(D2, F2) (4 E moves) = 1/16, we have p(D2, F1) = 15/16, and so on. Then we have
fi = sum(j) { dj x p(Dj, Fi) }
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This one involves the least work.
Paths to { F2 F3 F4 } don't hit an edge. So { f2 f3 f4 } = { C(12,7) C(12,6) C(12,5) } / 212
Then f1 = f5 = 1/2 ( 1 - sum { f2 f3 f4 } )
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This one recognizes an equivalent calculation.
What is the probability of flipping 8 heads before flipping 5 tails? I don't know whether this is an easy question or not, but the answer gives f1 and f5.
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@plasmid augments paths that touch an intermediate border point, where a choice is eliminated, by multiplying by 2 at each point. The effect is to create virtual paths that increase the count to each end point by 299, making 4096 = 212 total paths and leads to the correct probability of .20551.
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Case by case results, inviting algorithm falsification, while limiting lengths to 10.
SpoilerRows: values of a; Columns: values of b; Elements: triangle counts.
OBTUSE 10
0 0 0 0 0 0 0 0 0 0
0 1 1 1 1 1 1 1 1 0
0 0 1 1 2 2 2 2 1 0
0 0 0 2 2 2 2 2 1 0
0 0 0 0 2 3 2 1 0 0
0 0 0 0 0 2 1 0 0 0
0 0 0 0 0 0 1 0 0 0
0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0
Row sums (obtuse triangles for 10 values of a)
0 8 11 11 8 3 1 0 0 0 <= showing no possible obtuse triangles when 2> a >7Total triangles
42More matrices, for maximum lengths of 2-9:
SpoilerOBTUSE 2
0 0
0 0
OBTUSE 3
0 0 0
0 1 0
0 0 0
OBTUSE 4
0 0 0 0
0 1 1 0 <= meaning a=b=2 permits (only) one obtuse triangle: c=3
0 0 0 0
0 0 0 0
OBTUSE 5
0 0 0 0 0
0 1 1 1 0 <= meaning a=2 b=4 permits (only) one obtuse triangle: c=5
0 0 1 0 0 <= showing that a=3 b=4 c=5 (right-)triangle was not counted
0 0 0 0 0
0 0 0 0 0
OBTUSE 6
0 0 0 0 0 0
0 1 1 1 1 0 <= counting (only) 223, 234, 245, 256 for a=2
0 0 1 1 1 0 <= indicating a=3 b=4 c=6 obtuse triangle (only) was counted; not c=5,7
0 0 0 1 0 0
0 0 0 0 0 0
0 0 0 0 0 0
OBTUSE 7
0 0 0 0 0 0 0
0 1 1 1 1 1 0
0 0 1 1 2 1 0
0 0 0 2 1 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
OBTUSE 8
0 0 0 0 0 0 0 0
0 1 1 1 1 1 1 0
0 0 1 1 2 2 1 0
0 0 0 2 2 1 0 0
0 0 0 0 1 1 0 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
OBTUSE 9
0 0 0 0 0 0 0 0 0
0 1 1 1 1 1 1 1 0
0 0 1 1 2 2 2 1 0
0 0 0 2 2 2 1 1 0
0 0 0 0 2 2 1 0 0
0 0 0 0 0 1 0 0 0
0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 -
I get fewer ...
Spoiler47308.
To avoid double counting, I restricted a <= b <= c.
I let a be in [1, 100] and b in [a, 100]. I then counted the integers c in [ sqrt {(a2)+(b2)}, a+b -1 ].
I also noticed 52 right triangles (Pythagorean triples) .This is a plot of possible obtuse triangles for each value of a between 1 and 100.
When a = 27 the number reaches a maximum of 1166.
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13 hours ago, plasmid said:
Flipping the same coin twice would be indistinguishable from not flipping it at all to your buddy. So you would just be exposing yourself to more nyaning without accomplishing anything.
Naively.I might want the first n-1 coins to be heads if the valuable one was tails in the nth position. I'd have you keep flipping them until they showed heads. But /// I watched the video, so ... nah.
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I can ask you to flip a coin for a 2nd time?
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17 hours ago, Pickett said:
Approach
Let's start by identifying the 7 different groups of people involved and label them
- Total people that had only Wine = W
- Total people that had only Beer = B
- Total people that had only Water = H
- Total people that had Wine and Beer = WB
- Total people that had Wine and Water = WH
- Total people that had Beer and Water = BH
- Total people that had Wine, Beer, and Water = WBH
Now, given that, we can write our system of equations based on the problem statement:
EQUATION 1 (total cost of all drinks):
5*(WINE) + 2*(BEER) + 1*(WATER) = 293
5*(W+WB+WH+WBH) + 2*(B+WB+BH+WBH) + (H+WH+BH+WBH) = 293
Simplifies to:
5W + 7WB + 6WH + 8WBH + 2B + 3BH + H = 293EQUATION 2 (total glasses used):
W + B + H + 2WB + 2WH + 2BH + 3WBH = 106EQUATION 3 (total non-water drinkers):
W + B + WB = 18EQUATION 4 (total wine drinkers):
W + WB + WH + WBH = 39EQUATION 5 (total non-alcohol drinkers):
H = 9EQUATION 6 (ending assumption that wine-only was half the number of beer/water drinkers):
BH = 2WSo, now that we have our 6 equations, we can start doing our algebra:
Plug equation 5 into equation 1 and 2 to simplify slightly:
1) 5W + 7WB + 6WH + 8WBH + 2B + 3BH = 284
2) W + 2WB + 2WH + 3WBH + B + 2BH = 97Plug equation 6 into equations 1 and 2:
1) 11W + 2B + 7WB + 6WH + 8WBH = 284
2) 5W + B + 2WB + 2WH + 3WBH = 97Re-order equation 1 and 2:
1) 2(W+B) + 9W + 7WB + 6WH + 8WBH = 284
2) (W+B) + 4W + 2WB + 2WH + 3WBH = 97Plug equation 3 (W+B = 18-WB) into equations 1 and 2:
1) 9W + 5WB + 6WH + 8WBH = 248
2) 4W + WB + 2WH + 3WBH = 79Plug equation 4 (W=39-WB-WH-WBH) into equation 1 and 2:
1) 4WB + 3WH + WBH = 103
2) 3WB + 2WH + WBH = 77Solve for one of the remaining variables (WB):
WB = (103 - 3WH - WBH) / 4Plug into equation 2:
WH = 1 + WBHAlright, so we now have 1 equation with 2 unknowns, where one of the unknowns is what we're trying to determine! So, we can list out all of the possible values of WBH and WH...which happen to be:
WBH = 1, WH = 2
WBH = 2, WH = 3
...
WBH = 18, WH = 19Anything above WBH = 18 will cause equation 4 to not work...so we can stop there with those 18 possibilities. Now it's just a matter of substituting in each of those values into our equations to find the lowest one that works. I'll spare the gory details, but WBH < 11 results in negative values for other variables at some point...so let's just show trying WBH = 11:
WBH = 11, WH = 12:
Substitute those values into our 4 equations and simplify:
- 5W + 2B + 7WB + 3BH = 124
- W + B + 2WB + 2BH = 40
- W + B + WB = 18
- W + WB + = 16
Now substitute 4 into 3 to solve for B:
B = 2...so we know B = 2 and W = 16-WB
Plug those two into equations 1 and 2 and simplify
- 2WB + 3BH = 40
- WB + 2BH = 22
Solve (WB = 22-2BH)
You find BH = 4...so given BH = 4, WBH = 11, WH = 12, B = 2, and H = 9, we can find the other two values of WB = 14 and W = 2...at which point we can verify all of the original equations hold true. so therefore, WBH (total number of people who drank all three drinks) is at least 11 given the assumptions we stated.@Pickett It may not impact your solution, but the OP did not intend to say that all of us drank something.
"Class of drinkers" was not meant to preclude anyone from drinking nothing.
Meaning there are eight "classes" of drinkers.Sorry for that. I have this thing about insisting that zero is a number,
rather than a denial.
Also, I get a different answer. I'll check my analysis against yours to see why.
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2 hours ago, rocdocmac said:
Does it necessarily mean that all 9 of the teetotalers drank water or is that an assumption? What if some didn't drink at all that day and therefore never touched one of the 106 glasses?
- It means that all of the teetotalers drank neither wine nor beer. Let's say also that the others did consume alcohol.
- One class of friends may have drunk nothing, and perforce not used a glass.
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Thinking. Meantime I'll call your 10 minutes of cat flight and raise you 25 minutes of Roomfull of Teeth, as they explore new uses for the human voice, which you might actually enjoy.
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A bunch of friends went to the sports bar and got a group rate on the drinks: $5/glass for wine, $2/glass for beer, and $1/glass for water. When we left, the waiter asked me to sort out the bill. There was enough uncertainty in what people remembered that I could not be precise. So we happily just threw in enough to cover the bill, which came to $293 and we went home.
But it got me thinking. None of us had multiple glasses of the same beverage. The waiter said 106 glasses were used, once each. 18 of us did not drink water. 39 people had wine. I was certain that 9 of us were teetotalers. If I had known the sizes of just three classes of drinkers I could have figured out the bill, as it was, I could not.
But it did occur to me that if those who drank beer and water but not wine were as many as possible, and if those who drank only wine were half as many as that, I could say the smallest number of us who drank all three. Can you?
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Let there be n red cards and n black cards.
SpoilerWhat is the smallest n that permits a winning strategy?
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On 2/22/2018 at 9:18 AM, harey said:
Maybe an idea. There is an infinite number of numbers between 0 and 1. I have the opportunity to discard an infinite number of numbers between 0 and 1.
How much are you willing to bet that nothing remains between 0 and 1?
I just re-read this, and my reply wasn't totally responsive. But let's see whether Charlie even makes us address this question. Remember, Charlie doesn't get any more coins to deal with than Albert did. If we believe Albert's box (can) get emptied, we can't say discarding Charlie's coins is an impossible task. We simply have to show that none of Charlie's coins is able to "hide" indefinitely, in hopes of hearing the clock strike while still being in the box.
Charlie's coins are not inscribed, so we have to treat them collectively. We can't trace the destiny of any individual coin, as we could in Albert's case. Can we still show they are all eventually discarded, and not face the task of depleting an infinite set along the way? We claim that we can do that.
We can refer to each of Charlie's coins as "one of the coins that entered Charlie's box at event n that was also not discarded at event n." The n coins in his box at any such time are indistinguishably equivalent. Their "fate" is countably infinite events with non-zero probability of being discarded. We have shown in a preceding post that each of Charlie's coins has a survival probability of k/n where k is the event when the coin entered his box and n is the current event number. k is fixed. n becomes infinite. The probability of the coin that entered Charlie's box at event k still being in his box after midnight is limit (n->inf) k/n = 0. Since this holds for each k it holds for each coin that entered Charlie's box.
Using other words, every coin that is in Charlie's box at midnight is not a coin that entered Charlie's box. A contradiction.
Also, at no event time did Charlie have an infinite number of coins.
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7 hours ago, Molly Mae said:
It takes 8 moves. All of the second player's moves are forced, but he usually has two ways to respond. In either case, though, the second player can't prolong the game any longer than move 8 (that I can find).
1. f6, f3
2. d6, g3
3. d4, g7
4. e4, g4
5. g6, d3
6. e6, c6
7. e3, resignsOne of 8. e2 and 8. e7 is unstoppable.
Spoiler4 in a row with one end unblocked, or 3 in a row with both ends unblocked, force the next move. A forced win must do this on each move. Nice.
Heads up
in New Logic/Math Puzzles
Posted
Equates with the 3x speed version of ants on a checkerboard:
probability of reaching either of the ends of the five meeting points = 794 / 4096