BrainDen.com - Brain Teasers

# bonanova

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## Posts posted by bonanova

1. ### Make Three 7's equal 10

57 minutes ago, Heureka said:

You can add, subtract, multiply and divide, use parentheses, square root, exponent and factorial.

Using the number 0 or extra 2's does not fit the allowed operations.

Only Sqrt(4) + 4 + 4 = 10 seems allowable.

2. ### Finding a fiver

You are given an exhaustive 5-letter English word list. How would you determine the probability that 7 randomly chosen letters form at least one of those words? Or the expected number of words they form?

3. ### impossible result

Spoiler

His penultimate line evaluates to (-1/2)2 = (1/2)2 which is correct: 1/4 = 1/4.

But this makes the error stand out: we can't conclude that -1/2 = 1/2 from the equality of their squares.

4. ### Gift wrapping the Puzzle blocks

So the problem is to fill a cubical box of side S with
convex blocks having integral-length edges {L W H}
whose values are taken
without replacement from
the set { 1, 2, 3, 4, 5, 6, ..., S-2,  S-1, S } ?

Or must they be taken from the set that excludes S?

That is, can one of the blocks have an edge length equal to S?

5. ### two box paradox

Assuming you meant things of value (perhaps coins?) into the boxes, then ...

Spoiler

Say the box you chose has n coins. There are (at least) four ways to look at the question.

1. Comparing probability of advantage.
The other box has
fewer or more coins, with equal probability.
That is, I chose the box with more coins with probability of one-half.
The decision is binary -- advantageous to switch, or not?
There is no advantage (or penalty) to switch.

2. Comparing probability of advantage.
The other box has
fewer or more coins, with unequal probability.
Fewer coins is four times more likely (one fewer initial heads flip vs. one additional initial heads flip)
There is a substantial advantage not to switch.

3. Comparing expected value.
The other box has
n/3 or 3n coins, with equal probability.
The expectation of switching is (
n/3 + 3n)/2 = (5/3)n > n.
There is an advantage to switch.

4. Comparing expected value.
The other box has either
n/3 or 3n coins, with unequal probability.
If the other box has 3
n coins it took an extra head flip - half as likely as your n-coin box.
If the other box has
n/3 coins it took one fewer initial head flips - twice as likely as your n-coin box.
The expectation of switching is (2
n/3 + 3n/2)/2 = (13/12)n > n.
There is an advantage to switch.

I lean towards the first analysis, because symmetry is simple and beautiful.

Each analysis has its merits, tho, so I'll give the problem the status of a true paradox.

6. ### Gift wrapping the Puzzle blocks

The LxWxH block dimension seems to rule out

Spoiler

The Soma cube, which has block surfaces that are not strictly convex, but do completely fill a 3x3x3 cube. That size is a lower bound in any event since 2x2x2 can't be completely filled with unique convex blocks.

Since OP does not

Spoiler

explicitly require the box to be completely filled

it seems the smallest box would be

Spoiler

2x2x2 which contains three blocks with dimensions 1x1x1, 1x1x2 and 2x2x1

But that requirement probably was intended.

Also, I don't see that unique coloring imposes any limits beyond that of unique dimensions.

7. ### The JayCees

Jane, Janice, Jack, Jasper, and Jim are five high-school chums. Their last names are, in some order, Carter, Carver, Clark, Clayton, and Cramer. What are their full names? Here are some clues.

1. Jasper's mom is deceased.
2. In deference to an influential family member, the Claytons agreed that if they ever had a daughter they would name her Janice.
3. Jane's parents have never met Jack's parents.
4. The Cramer and Carter children have been teammates on several of the school's athletic teams.
5. When he heard that Carver was going to out of town on the night of the school's Father and Son banquet, Cramer called Mrs. Carver and offered to "adopt" her son for the evening. But Jack's father had already asked him to go.
6. The Clarks and Carters, staunch Republicans who are very good friends, were delighted when their children began dating each other.
8. ### Once more with the Maiden

On 10/5/2019 at 12:10 PM, plasmid said:

I'm pretty sure this isn't optimal, but it's a start.

Hide contents

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) (g-1)/g time units to move forward. Her forward speed is f, so she’ll travel a distance of f(2 pi) (g-1)/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(g-1)/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(1-R2), or sqrt(1-(f(g-1)/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(g-1)/g)2) < f pi

For the case of g=2,
sqrt(1-(f/2)2) < f pi
sqrt((4-f2)/4) < f pi
sqrt(4-f2) < f 2 pi
4-f2 < f2 4 pi2
4 < f2 4 pi2 + f2
4 < f2 (4 pi2 + 1)
so f > sqrt(4 / (4 pi2 + 1)) or about 0.31435

Kind 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.

Spoiler

We'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 (g-1)/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 2nd post.)

My only progress on that point was an intuition that waiting for the boat to turn 90o 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:

Spoiler

In 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 180o  instead of greater than 180o.

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.

9. ### Photochromism 326 BC

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.

• 1
10. ### Swapped numbers

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

11. ### Diameters

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  Rmax is the unit diameter circle.

Now imagine a rectangle with unit diameter (diagonal has length 1.)
That can be made to fit into
Rmax .

The question is does every unit-diameter region into Rmax ?

12. ### Diameters

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 Rmax,  encloses the largest area.

Can you prove, or disprove, that Rmax also encloses all other regions of diameter 1? That is, that all other regions of diameter 1 can be made to fit inside Rmax?

13. ### Once more with the Maiden

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 lake-radius per minute, and Maiden chooses the boat's initial heading at the center, what's her shortest time safely to shore?

14. ### Ten Islands

On 8/28/2019 at 8:27 PM, mtngoat said:

Hide contents

Isn't it possible that a bridge could split, connecting 3 islands instead of 2? Just a thought.

Yes it is possible. The Triborough Bridge in New York connects Manhattan, the Bronx and' Queens.

15. ### Kill Bill redux

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?

16. ### The Maiden and the Ogre

On 8/27/2019 at 1:38 AM, EventHorizon said:
Reveal hidden contents

Nicely done.

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.

17. ### The Maiden and the 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?

Spoiler

Say the lake has radius of 1.
The boat's starting point
P0 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
P0 to the landing point P1.
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 | P0 - P1 | change monotonically?

.

18. ### American Indian Artwork

What a deliciously challenging and intriguing puzzle.

19. ### The Maiden and the Ogre

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?

20. ### What is my clock telling me to do?

That's some two-year-old. And yes, it's time to ..

Spoiler

.. change his diapers.

21. ### Probability 101

@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 long-running Teanchy-Beanchy 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.

22. ### Probability 101

11 hours ago, Thalia said:

Reveal hidden contents

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.

23. ### Ten Islands

@CaptainEd Brilliant.

24. ### Kill Bill redux

@flamebirde Yes. The gun’s owner did the deed. Implied but not stated. Good catch.

25. ### Trains that pass in the night

If it takes the 10-car Silver Streak passenger train twice as long* to pass the 93-car 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.

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