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Posts posted by bonanova


On 10/5/2019 at 12:10 PM, plasmid said:I'm pretty sure this isn't optimal, but it's a start.
I think the solution for the previous problem (where the maiden could turn the boat instantaneously) came down to realizing that she could go in a circle of some radius R that’s smaller than the entire lake and be able to move at a radial speed faster than the ogre (she moves a number of degrees around that small circle that’s just slightly more than the number of degrees that the ogre can move around the edge of the lake in the same amount of time). As long as she can move around that inner circle faster than the ogre can move around the perimeter, she can set it up so she’s opposite the ogre when she makes a dash for the shore.
If the maiden tries that same strategy in this problem, first she needs to find a circle of radius R where she’s moving at a radial speed that’s faster than the ogre. Define the lake’s radius as 1 and the ogre’s speed as 1, so it will take the ogre 2 pi time units to run around the lake. For the maiden to paddle around a circle of radius R she would need to make a 360 degree turn, and the amount of time it takes to rotate the boat 360 degrees is 1/g times the amount of time it takes the ogre to circle the lake, so the maiden spends (2 pi)/g time units turning. The amount of remaining time she has to move the boat forward is [the amount of time it takes for the ogre to circle the lake] – [the amount of time spent rotating the boat] = (2 pi) – (2 pi)/g = (2 pi) (g1)/g time units to move forward. Her forward speed is f, so she’ll travel a distance of f(2 pi) (g1)/g. That’s the circumference of the circle she can paddle around, and its radius R is that divided by 2 pi, so R = f(g1)/g.
She can position things however she likes within that circle, but then she’ll have to bolt for the shore. I suspect that just charging forward in a straight line is not optimal, but I’ll go ahead and solve for that strategy.
Her boat won’t be pointing toward the shore when she decides to bolt, it will be pointing parallel to the edge of the circle where she’s paddling. If she’s at a distance R from the center of the lake and just paddles forward in the direction she’s facing, the distance until reaching the shore would be sqrt(1R^{2}), or sqrt(1(f(g1)/g)^{2}). As long as the maiden bolts when her landing spot will be opposite to the ogre’s current position, the ogre will have to travel a distance of pi to reach the landing point. With that strategy, the maiden can be sure of safety if the distance she has to travel is less than f times the distance the ogre has to travel, so the condition for victory is
sqrt(1(f(g1)/g)^{2}) < f piFor the case of g=2,
sqrt(1(f/2)^{2}) < f pi
sqrt((4f^{2})/4) < f pi
sqrt(4f^{2}) < f 2 pi
4f^{2} < f^{2} 4 pi^{2}
4 < f^{2} 4 pi^{2} + f^{2}
4 < f^{2} (4 pi^{2} + 1)
so f > sqrt(4 / (4 pi^{2} + 1)) or about 0.31435Kind of surprisingly, that strategy is only very marginally better than sitting in the center of the lake and rotating until she's facing away from the ogre and then taking off. With that strategy she would travel 1 lake radius and the ogre would have to run pi lake radii, so she wins if f > 1/pi or about 0.31831.
@plasmid My bad. It's not polite to post a puzzle and then go dark for two months. Apologies.
In the first version there was no acceleration cost for either contestant. They both could stop on a dime, turn, and resume at full speed instantly.
So here I've added a cost for angular change of velocity (for the boat only) but none for linear acceleration. Before I finished my solution my hard drive fried. I replaced my computer but I lost my work. I'll share how far I got and maybe we can finish this off collaboratively.
SpoilerWe're in agreement on the general approach. Find the largest circle of safety, pick a landing point and sprint to it.
For the first version this had to be the right approach. It gives the boat its most advantageous starting point, at zero cost. The optimization problem was to select the best landing point head straight for it. EventHorizon's beautiful and succinct solution, here, which I have not verified but which matches my best calculation, gave the landing point with the greatest arrival time differential to be where the tangent intersects the shore.
For this version the first question is what is the radius of the (new and smaller) circle of safety? And, as you found, its f (g1)/g. So to make things simple I set g=2 making the radius f /2.
The second question is what is the optimal landing point  with now a third question, namely, if it's not the intersection of the tangent with the shore, then what is it, and what is its shape, given the cost of changing the boat's heading? (I think this is what you began addressing in your 2^{nd} post.)
My only progress on that point was an intuition that waiting for the boat to turn 90^{o} before sprinting (radially) for shore wouldn't be a good choice. Why wait at all, actually? At least begin along the tangent, and perhaps veer (slightly) away from the ogre.
But ... I found an ugly wrinkle, the last bit of progress I had made. It's this:
SpoilerIn the first version I was certain the best choice was to pick the landing point and head straight for it. The point being that, if the ogre anticipated that strategy, he could get to any landing point (other than the radial one) faster if he reversed direction after the boat began to move. With the result that his angular path would now be less than 180^{o} instead of greater than 180^{o}.
But that does not work for him. Why? Because when the boat sees the ogre reversing direction, the boat immediately draws a new circle at present (larger) radius and heads to shore tangentially in the opposite direction herself. And that is a better starting point, because the starting radius is larger. In fact every time the ogre reverses direction things get worse for him. I think I worked out that if ogre reversed after each of his steps, the boat could actually zigzag to shore with the ogre remaining a full lake diameter away. So in the version where the boat has no turning cost, the ogre cannot reverse direction to his advantage.
But if there's cost for the boat to turn while ogre has no cost for reversing, that argument has to be revisited. I don't think I made any useful headway on it, and maybe it's more work than fun to pursue it?
So I recall considering a simplification: (which maybe we should now adopt)
Give the boat four instantly switchable settings: FSA, CW and CCW, as now, plus Full Speed Astern. That is, boat can reverse direction without penalty. The wrinkle now is that the boat's new tangent won't line up exactly with the old one. A bit of turning is needed. I'm betting it can be proved for a small delta of motion before reversal the cost of a then needed turn will turn out to be smaller than the benefit of a larger radius. If that can be proved, the ogre loses by reversing.
It would then only remain to calculate the new tangential path length (from a circle of radius f/2) to the shore, which gives in turn the optimal value of f .
Does that make sense?
========
I didn't analyze your second post, so maybe it's better than this.

I sort of hesitate to relay a tongue in cheek report I read a while back that photochromic materials are not of recent discovery, but were actually known back in the time of Alexander the Great. A black substance could be ground into powder and dissolved in water. Alexander’s troops would soak strips of cloth torn from their togas in this solution and tie them around their wrists. As the sun rose, traversed the sky and then set, the treated cloth would change color, and by glancing at them his men could tell the approximate time of day.
They called it Alexander’s Rag Time Band.

In each case is there a swap? Or are there longer cycles as well?

10 hours ago, CynPyn said:Here goes...
Rmax must be a circle of diameter one. To prove this, assume it isn’t. Then all other possible diameters of Rmax must be one or shorter than one. If this is true, then Rmax can fit inside a circle of diameter one, which has greater area than Rmax. Therefore Rmax doesn’t enclose the maximum area. By contradiction then, Rmax must be a circle of diameter one.
Not my cleanest proof, but I’m thrilled to be the first response.
Hi @CynPyn and welcome to the Den.
Let's accept from this that R_{max} is the unit diameter circle.
Now imagine a rectangle with unit diameter (diagonal has length 1.)
That can be made to fit into R_{max} .The question is does every unitdiameter region into R_{max} ?

The diameter of a closed, topologically bounded region of the plane is the greatest distance between two points in the region. Example: the diameter of a rectangle is the length of its diagonal. Of all the regions whose diameter equals 1, one of them, call it R_{max}, encloses the largest area.
Can you prove, or disprove, that R_{max} also encloses all other regions of diameter 1? That is, that all other regions of diameter 1 can be made to fit inside R_{max}?

Previously, Maiden’s boat could change its heading instantaneously. Ogre’s heading could change only by virtue of following a circular path along the shore at his current speed. His rotational speed was thus far from infinite, and perhaps that disadvantage was unfair.
So in this final puzzle iteration we’ll limit the boat’s linear speed to be f times that of Ogre, as before, but now we’ll also limit the boat’s angular speed to be never greater than g times Ogre’s top angular speed.
A moment’s thought tells us that unless g is greater than unity the boat’s best strategy is to run at full speed from the center to the shore, keeping its initial bearing, no matter where on the shore Ogre initially stands. That is, never to turn the boat. That sucks for Maiden (e.g., she loses if Ogre initially stands at the boat's initial heading) and it sucks as a puzzle. So we’ll say the boat can change heading faster than Ogre can. For clarity we’ll set g = 2.
We’ll implement that limit by giving the boat’s motor three discrete settings that can be switched instantly an unlimited number of times: clockwise (CW), full speed ahead (FSA), and counterclockwise (CCW.) In the two turning modes the boat turns but maintains its position; in FSA mode it moves forward but does not turn. Boat’s path is thus a succession of arbitrarily short line segments joined at angles of Maiden's choosing, with the time cost of the angle depending on its size.
If the boat starts in the middle of the lake, how large must f now be for Maiden to escape?
Edit: Extra credit (tough):
If Ogre's top speed is 1 lakeradius per minute, and Maiden chooses the boat's initial heading at the center, what's her shortest time safely to shore? 

On 8/28/2019 at 9:05 PM, mtngoat said:But unless the problem has been edited, it is stated: "along with the fatal gun that belonged to one of them ... the murderer." But not absolutely necessary for the solve.
Hmmm. I hadn't considered that. Is there a different solution if the murderer did not own the murder weapon?

On 8/27/2019 at 1:38 AM, EventHorizon said:Nicely done.
As for adding some wrinkle to the problem, how about this?
Suppose we have to add some time for her to get out of the boat before she starts to run. We could say the ogre must have an angular separation of s radians from the boat when it lands, and then minimize f. Hmmm. I'm guessing the same path just minimizes f to a slightly larger value.
On 8/28/2019 at 6:23 AM, The Lonewolf Brand said:The minimum value of f is 1.0.
Hi @The Lonewolf Brand and welcome to the Den.
That would mean the maiden's boat and the ogre have the same speed. She can escape from a faster ogre.

15 hours ago, EventHorizon said:Another interesting addition might be, once the minimum f is found, to find the minimum travel distance (e.g., amount of gas) needed for some f a little higher than the minimum.
Interesting idea. How would we pose that question exactly?
SpoilerSay the lake has radius of 1.
The boat's starting point P_{0} is on a shared diameter, 1 + f from the ogre and 1  f from shore.
The shortest travel distance is the length of a straight line from P_{0} to the landing point P_{1}.
Among the landing points that make sense, one minimizes f , as asked by OP; another minimizes travel distance, namely 1  f .
What if, between those landing points, both f and the distance  P_{0  }P_{1}  change monotonically?
. 
What a deliciously challenging and intriguing puzzle.

If you think you've heard this one before, read it carefully. It's not the standard puzzle.
A beautiful maiden sits in a boat at the center of circular lake. On shore waits an ogre anxious to have his way with her. Being an excellent sprinter she knows she can outrun and therefore escape the lumbering ogre if only she can land her boat safely. But should the ogre reach her landing point first, alas, all will be lost.
The boat is propelled by a motor capable of only a fraction f of the ogre's speed.
What is the minimum value of f that will permit the Maiden to escape?

That's some twoyearold. And yes, it's time to ..
Spoiler.. change his diapers.

@EventHorizon said,
But once Plato says that "one die is a 4," it doesn't merely prune out the ones without 4's.
That’s the crux of it.
EH is correct that this reads on Monty Hall. But more directly on the longrunning TeanchyBeanchy post (one of his two kids is a girl, what’s the probability he has two girls.) First people said has to be 1/2. Then others (including me) said (all that matters is that) it can’t be BB so it’s 1/3. Both wrong.
so I made this one up to show that the informant’s algorithm has to be known.
Nicely explained.

11 hours ago, Thalia said:And so thought poor Aristotle until today's class.
He knew the probability of two dice making seven, until his teacher told him the value of one of his dice. Then he reasoned it to be different. Ah, the magic of conditional probability, he thought. But then he reasoned further that it was not the knowledge of which value one of his dice had, for it did not matter whether that value was 1 2 3 4 5 or 6. It was seemingly only that it had a value. But what kind of conditional probability is that? Was he not already aware of that?
One of his dice has a particular value. Six values engender six cases. In each particular case he reasons the probability of seven changes to a new value. Worse, there are no other cases. Therefore in every case it changes to a new value. How then could it have the value he originally imagined?
So there's the question that lurks within the flavor text. Beneath the surface perhaps, but now fully revealed for all to ponder.



If it takes the 10car Silver Streak passenger train twice as long* to pass the 93car Big Boy steam freight train going in the same direction as it does going the opposite direction, what is the ratio of their speeds?
*Meaning the time during which there is any overlap of the trains.

Yes.
SpoilerI just tried to start with the * operator, seeing that two of the numbers ended in 0,
and looking at prime factors..... seems like a slew of possibilities.
Thought of writing a program. 
Now, what was it we were talking about? Oh yes ... wherrzzz Beeel? I'll tell you. You know why I tell you? Because Bill would want me to. But this time the old man was not talking to the impossibly gorgeous Beatrix Michelle Kiddo. It was to a guy. A guy who was on a mission to kill Bill. A guy with the name of Al, Jack, Joe or Tom. Which one? Well, it's your job to determine which.
Here's what we know from police interviews that followed the discovery of Bill's cold body on a lonely stretch of country road late last night, along with the fatal gun that belonged to one of them ... the murderer. So read on and see where the evidence leads. But fair warning, exactly half of each of the suspects' four statements are lies.

Al:
I didn't do it.
Tom did it.
Sure I own a gun.
Joe and I were playing poker last night when Bill was shot.

Jack:
I didn't do it.
Al did it.
Joe and I were at the movies last night when Bill was shot.
Bill was shot with Joe's gun.

Joe:
I was asleep when Bill was shot.
Al lied when he said that Tom killed Bill.
Jack is the only one of us who owns a gun.
Tom and Bill were pals.

Tom:
I've never fired a gun in my life.
I don't know who did it.
Joe doesn't own a gun.
I never saw Bill until they showed me the body.
Who was the rat who done poor Bill in?

Al:

Plato: Good morning, class, today's lesson is on probability.
Aristotle: Fantastic.
I'm headed to Vegas this weekend, and I can use some pointers.P: Curb your enthusiasm kid, this is serious stuff.
Here, roll this pair of dice, but don't look at the the result. OK, good.
Now without looking, tell me the probability that you rolled a seven.
If you're going to play craps this is important.
By the way, I can tell you that one of your dice is a four.A: Hmm...
So I could have rolled 41 42 43 44 45 46 14 24 34 54 or 64,
all with equal likelihood, with 34 and 43 making seven.
That's a probability of 2/11.P: You're on a roll kid, now let's do it again. Great.
Again without looking, what's the probability you rolled a seven?
By the way, I can tell you one of your dice is a one.A: So I could have rolled 11 12 13 14 15 16 61 51 41 31 or 21,
with 16 and 61 making seven. Hmm... it's the same as before  2/11.P: And if I had told you one of your dice was a five?
A: Well ... I guess it really doesn't matter what number you tell me.
It will always come out the same. The probability will be 2/11.P: So what can we deduce from that?
A: That the probability of two dice making seven is ... 2/11.
But wait... Hey, you're not really Professor Plato, are you?P: No. I'm an insurance salesman.
So ... what exactly is the probability of making seven?

1 hour ago, bonanova said:After each man, at random, was given one of the boxes, they were given the following test. Each in turn, and out of sight of the others, was to blindly remove two of the balls from his box, read the label on his box, and then endeavor to tell to the others the color of the third ball, which remained in the box.
Yes. The restriction is only that they can't read each others' labels.

SpoilerPrime factors, FWIW
790 = 79 x 5 x 2
1650 = 7 x 5 x 5 x 5 x 2
464 = 29 x 2 x 2 x 2 x 2 
Four identical boxes, each box containing 3 balls, each ball being either black or white, were placed before four men: Al, Bert, Cal and Don. No two boxes contained the same selection of ball colors. Each box bore a label that correctly identified the contents of one and only one of the boxes, but no individual box bore its correct label.
After each man, at random, was given one of the boxes, they were given the following test. Each in turn, and out of sight of the others, was to blindly remove two of the balls from his box, read the label on his box, and then endeavor to tell to the others the color of the third ball, which remained in the box.
It did not seem a difficult task, but the results were a bit surprising:

Al:
I've drawn two black balls.
I can tell you the color of the third ball.

Bert:
I've drawn one white ball and one black ball.
I can tell you the color of the third ball.

Cal:
I've drawn two white balls.
I'm not able to tell you the color of the third ball.

Don: (who was blind, and could not read his box's label)
I don't need to draw.
I can tell you the colors of the balls in my box.
I can tell you the color of the third ball in each of the others' boxes.
And then he proceeded to do so. How did he tell?

Al:
The JayCees
in New Logic/Math Puzzles
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Jane, Janice, Jack, Jasper, and Jim are five highschool chums. Their last names are, in some order, Carter, Carver, Clark, Clayton, and Cramer. What are their full names? Here are some clues.