If my logic is correct, the answer is 1.5707~ times the speed of the man, or pi/2 times his speed.
Reasoning: We have a square area of length n metres to a side.
|G----|
|R |
| |
| |
|M G
|-----|
The ram is in the top left corner represented by the R symbol, and the man is in the bottom left corner, represented by the M symbol. Assuming the ram does not pursue the man directly, but instead heads straight for the gate, we find that the lower bound for the ram's speed to reach the gate at the same time as the man to be sqrt(2) times the man's speed using pythagoras' theorem.
This is not the case, however, and the ram is constantly charging the man, adjusting his direction as the man moves. This means the ram's path will be curved, and the final path drawn after both have moved to the gate will be a quarter circle assuming they both reach the gate at the same time.
From there, it's a simple case of calculating the length of a quarter circle (2pi*r / 4) = pi*r / 2. Since we can assume the length of each side of the area to be 1, that works out to pi/2. If the ram moves at pi/2 times the man's speed, he will catch the man at the gate. If the ram moves faster, the path will be straighter and he will catch the man before he escapes. If the ram moves slower, he will miss the man entirely.
Joined: 12/28/2013
Posts: 396
Location: Rio de Janeiro, Brasil
ShadowWraith wrote:
If my logic is correct, the answer is 1.5707~ times the speed of the man, or pi/2 times his speed.
Reasoning: We have a square area of length n metres to a side.
|G----|
|R |
| |
| |
|M G
|-----|
The ram is in the top left corner represented by the R symbol, and the man is in the bottom left corner, represented by the M symbol. Assuming the ram does not pursue the man directly, but instead heads straight for the gate, we find that the lower bound for the ram's speed to reach the gate at the same time as the man to be sqrt(2) times the man's speed using pythagoras' theorem.
This is not the case, however, and the ram is constantly charging the man, adjusting his direction as the man moves. This means the ram's path will be curved, and the final path drawn after both have moved to the gate will be a quarter circle assuming they both reach the gate at the same time.
From there, it's a simple case of calculating the length of a quarter circle (2pi*r / 4) = pi*r / 2. Since we can assume the length of each side of the area to be 1, that works out to pi/2. If the ram moves at pi/2 times the man's speed, he will catch the man at the gate. If the ram moves faster, the path will be straighter and he will catch the man before he escapes. If the ram moves slower, he will miss the man entirely.
The trajectory of the ram will be a curve, but I don't think it will be perfectly circular. Let T be the time it takes to the man and the ram to reach the corner, and M the side of the square. If you use analytic geometry to see where the ram would be facing to at the moment T/3, considering a perfectly circular trajectory, it would face to the point sqrt(3)/3 of the side of the square, which is approximately 0.577M (considering that 0 is the initial point and M is the final point). The man would be at point 0.333...M. So the ram isn't facing the man, a contradiction.
I got an answer of phi.
My solution: Let t be the man's position along the east side of the paddock, with the length of the sides being 1. Let x be the ram's position along the same side, let v be the ratio of the ram's speed to the man's speed, and let s be the distance between the man and the ram.
Then, dx/dt = v(t - x)/s, and ds/dt = (t - x)/s - v = (1/v)dx/dt - v.
Since x = 0 and s = 1 at t = 0, we get s = 1 + x/v - vt.
At t = 1, x = 1 and s = 0, which means that 1 + 1/v - v = 0. This has two solutions. One of them is negative and therefore unphysical, and the other one is v = phi.
What is the smallest natural number n such that n has some power whose the last trillion decimal digits are all ones?
i.e., nk ends in 111....111 (a trillion 1's) ( ͡° ͜ʖ ͡°)
I suspect that the answer is some ridiculously huge number with more than 300 billion digits in it? I have no idea how to even begin to compute this.
What is the smallest natural number n such that n has some power whose the last trillion decimal digits are all ones?
i.e., nk ends in 111....111 (a trillion 1's) ( ͡° ͜ʖ ͡°)
I suspect that the answer is some ridiculously huge number with more than 300 billion digits in it? I have no idea how to even begin to compute this.
This was answered by FractalFusion, the value is surprisingly 71. A small variation of this problem (2015 digits) was in a mathematical olympiad here. In fact, powers of 71 will end with arbitrarily many 1's in the decimal system.
Joined: 12/28/2013
Posts: 396
Location: Rio de Janeiro, Brasil
HHS wrote:
Oh crap, didn't see that. I gave up after checking 31, thinking I was completely on the wrong track, lol.
To not check every number, you can use congruence modulo 16 and 5, to proof it wouldn't work with any number smaller than 71. The difficult part is to proof it indeed works with 71.
I got an answer of phi.
My solution: Let t be the man's position along the east side of the paddock, with the length of the sides being 1. Let x be the ram's position along the same side, let v be the ratio of the ram's speed to the man's speed, and let s be the distance between the man and the ram.
Then, dx/dt = v(t - x)/s, and ds/dt = (t - x)/s - v = (1/v)dx/dt - v.
Since x = 0 and s = 1 at t = 0, we get s = 1 + x/v - v.
At t = 1, x = 1 and s = 0, which means that 1 + 1/v - v = 0. This has two solutions. One of them is negative and therefore unphysical, and the other one is v = phi.
I got the same answer, but by looking at both the x position and y position simultaneously, which probably complicated things.
At our MtG playgroup we often use funny dice combinations to decide who will play first. Not just your regular old 2d6 or 1d20, but like d20-d12, or even d6*d6*d6, or things like that. Just for the fun.
Yesterday I suggested that we use the median of 3d6. I casually commented that this makes ties more probable than when throwing just 1d6.
For some reason many of the others didn't believe it. After I very informally argued that with the median of three d6's you are much more likely to get a 3 or a 4 than a 1 or a 6, and that's why ties are more probable, some of them still believed that the difference in probability is very small.
So, I was wondering. With n players, each throws a 1d6: what is the probability that two or more players are tied for the highest value? And if instead each throws 3d6 and the median is taken from them, what is the probability then?
The probability that nobody gets a 6 is (5/6)^n, and that one person gets a 6 is n*(1/6)*(5/6)^(n-1); therefore the probability of a tie at 6 is 1-(n/6)*(5/6)^(n-1)-(5/6)^n.
Then the probability that nobody gets a 6 or a 5 is (2/3)^n, and that one person gets a 5 and everybody else gets less is
n*(1/6)*(2/3)^(n-1); therefore the probability of a tie at 5 is
(5/6)^n-(n/6)*(2/3)^(n-1)-(2/3)^n.
The pattern is apparent, ending in the case of a tie at 2:
The probability that nobody gets above 2 is (1/3)^n, that nobody gets above a 1 is (1/6)^n, and that one person gets a 2 and everybody else gets 1 is n*(1/6)*(1/6)^(n-1); therefore, the probability of a tie at 2 is (1/3)^n-(n/6)*(1/6)^(n-1)-(1/6)^n.
The final case of a tie at 1 includes a couple of degenerate cases: (1/6)^n-(n/6)*0^(n-1)-0^n=(1/6)^n.
Adding up these cases yields 1-(n/6)((5/6)^(n-1)+(2/3)^(n-1)+(1/2)^(n-1)+(1/3)^(n-1)+(1/6)^(n-1))
For the case of the median of three rolls of a d6, the following probabilities for one roll hold:
1: (1/6)^3+3*(1/6)^2*(5/6)=2/27
2: (1/6)^3+3*(1/6)^2*(5/6)+6*(1/6)*(1/6)*(2/3)=5/27
3: (1/6)^3+3*(1/6)^2*(5/6)+6*(1/6)*(1/3)*(1/2)=13/54
4: (1/6)^3+3*(1/6)^2*(5/6)+6*(1/6)*(1/3)*(1/2)=13/54
5: (1/6)^3+3*(1/6)^2*(5/6)+6*(1/6)*(1/6)*(2/3)=5/27
6: (1/6)^3+3*(1/6)^2*(5/6)=2/27
Then the probability that nobody gets a 6 is (25/27)^n, and the probability that one person gets a 6 is n*(2/27)*(25/27)^(n-1), so the probability of a tie at 6 is 1-n*(2/27)*(25/27)^(n-1)-(25/27)^n.
The probability that nobody gets 6 or 5 is (20/27)^n, the probability that one person gets 5 and nobody else gets above 5 is n*(5/27)*(20/27)^(n-1), so the probability of a tie at 5 is (25/27)^n-n*(5/27)*(20/27)^(n-1)-(20/27)^n.
A similar telescoping argument yields the following probability of a tie:
1-n*((2/27)*(25/27)^(n-1)+(5/27)*(20/27)^(n-1)+(13/54)*(1/2)^(n-1)+(13/54)*(7/27)^(n-1)+(5/27)*(2/27)^(n-1))
I seem to get the chance of no tie for 3d6 median as usuallyhigher than for 1d6.
___
Now on to the FiveThirtyEight Riddler...
The basic problem for this one is straightforward, but the Extra Credit is tough for those of us who like closed-form solutions: http://fivethirtyeight.com/features/can-you-bake-the-optimal-cake/
Follow-up on arflech's answer:
A surprising result from this dice tie problem is that for 2, 3 and 4 players the median method results in higher tie chances, whereas with ≥5 players the normal method has a higher tie chance.[1]
(Which also fits with my poorly made graph that uses data from a quick test program)
Warning: Might glitch to creditsI will finish this ACE soon as possible
(or will I?)
A surprising result from this dice tie problem is that for 2, 3 and 4 players the median method results in higher tie chances, whereas with ≥5 players the normal method has a higher tie chance.
Well, that's indeed surprising, and rather counter-intuitive.
Is there an intuitive informal explanation of why this is so?
We only care about extreme values because we're determining the winner. We don't care about the number of ties, just whether or not the top result it. Extreme values in 1d6 are more likely, so although ties are rare, the likelihood that the top result is uniform. As you increase the amount of players, the probability you have a tie increases as well.
On the other hand median of 3d6 you're more likely to get the middle numbers, but we care about the high ones. So when someone rolls high, that is an improbable event, and it's more likely that someone will not replicate it. But it's also less likely that you'll roll high to begin with. So with fewer people you're more likely to all roll in the middle, and result in a tie, but as you add more people you're more likely to get just one person who rolls high by chance.
It's all about which effect dominates.
Build a man a fire, warm him for a day,
Set a man on fire, warm him for the rest of his life.
I like this problem a lot. I started working on it, then realized that I'd already posted a more general math challenge.
Let yi be the y coordinate associated with the top of the ith layer (or said another way, where the boundary is between the ith and ith plus one layer). A little work will show that the volume of this layer is pi*R2*(1-yi/h)2*(yi-yi-1). Let's factor out the constants of pi*R2/h2 to leave the "meat" of the expression: (h-yi)2*Δyi. I've re-written yi-yi-1 as Δyi for convenience.
If you haven't already done so, check out my earlier post, linked above, as well as the post with the solution. We wish to maximize the sum over i of (h-yi)2*Δyi. That means our discrete Lagrangian is (h-yi)2*Δyi. The discrete Euler-Lagrange equation is:
Δ(dL/d(Δyi+1)) - dL/dyi
Plugging in our Lagrangian and simplifying leads to the recurrence relation
yi+12 - 2hyi+1 - 3yi2 + 4hyi + 2yiyi-1 - 2hyi-1 = 0
We have y0 = 0, so for i=1, this becomes
y22 - 2hy2 - 3y12 + 4hy1 = 0
On the other end, this Lagrangian does not depend on yn+1, so bookkeeping is a little tricky here. There is a yn dependence as well as a Δyn dependence, but no Δyn+1 dependence. The net effect is that we're left with a modified Euler-Lagrange equation for the end point:
dL/d(Δyn) + dL/dyn = 0
Which eventually gives us:
3yn2 - 4hyn + 2hyn-1 - 2ynyn-1 + h2 = 0
So in summary...
For i = 1:
y22 - 2hy2 - 3y12 + 4hy1 = 0
For 1 < i < n:
yi+12 - 2hyi+1 - 3yi2 + 4hyi + 2yiyi-1 - 2hyi-1 = 0
For i = n:
3yn2 - 4hyn + 2hyn-1 - 2ynyn-1 +h2 = 0
I believe all of those equations are correct, giving us n equations and n variables. As for using them to solve for the yis explicitly, well, that's left as an exercise to the reader...
Joined: 12/28/2013
Posts: 396
Location: Rio de Janeiro, Brasil
Warp wrote:
Masterjun wrote:
A surprising result from this dice tie problem is that for 2, 3 and 4 players the median method results in higher tie chances, whereas with ≥5 players the normal method has a higher tie chance.
Well, that's indeed surprising, and rather counter-intuitive.
Is there an intuitive informal explanation of why this is so?
It's actually very intuitive if you imagine a really big N.
Following OmnipotentEntity's logic, say N = 10^6. It's almost impossible that no one got a 6 in 1d6, and it's also extremely likely someone got a 6 in 3d6.
In which one do you think it's more probable that someone else got a 6? It's still extremely likely on both, but it's easier to see it's more likely on 1d6.
Experimenting with also the median of 5d6, I noticed that it, too, became less likely to produce ties for the winner, as the number of players increases, both compared to 1d6 and to the median of 3d6.
More specifically, with 7 players the median of 5d6 has a probability of about 47.1% for a tie, while for 1d6 it's 48.7%. (For the median of 3d6 it's 43.4%.)
At 16 players the median of 5d6 has a probability of 58.9%, while the median for 3d6 it's 59.4%.
I suppose it's safe to generalize that for any n > 1, the median of nd6 will eventually, as the number of players increases, produce a tie for the winners with less probability than with any smaller values of n.
I'm really disappointed in the answer they gave to last week's problem. I thought they wanted a closed-form solution and I was somehow missing something. I also think my analysis is more thorough than what they cover in the written solution. It basically just says, "Take the derivatives and set them equal to zero." Well, duh!
Anyway, I noticed the comments at the bottom indicate for the camel problem that you can drop bananas off in the desert (something I wouldn't have considered, as it seems like a bad idea...) and pick them up later. That makes the problem non-trivial.
I think the best one can do is 750. First, take 1000 bananas and leave 500 in the middle. Then, take 1000 bananas and leave 250 at the 3/4 way point. Then take the rest to the market, along with the bananas you left.
I think I have a solution that gives 833 bananas.
split the 3000 bananas into 3 piles of 1000:
take each pile 334 miles and drop them
now have 1998 bananas and 666 miles to go
split into 2 piles of 999:
take each 499 miles and drop them
now have 1000 bananas and 167 miles to go
Take them to market. You have 833 bananas left.
The idea is, at each stage to split the bananas into as few piles as possible to minimise consumption. The more piles you transport bananas in, the more the camel eats. So you stop and drop as soon as you're able to reconfigure in to a smaller number of piles.
Isn't it possible to cheat in this problem? Or maybe it's not clear to me the whole situation...
I mean: carry 1000 bananas and walk 0.999 miles. Since the camel will eat a banana after an entire mile (carrying at least one banana), it won't eat this time. Go back and take another 1000 bananas. Repeat the process another time. So, you just took 3000 bananas 0.999 miles away and none was eaten. Repeat this whole process until 1000 miles is traveled.
If it's not considered valid, we have to know if we can travel only an integer number of miles or if the camel memorizes and accumulates the miles. Or maybe, he eats at the beginning of each mile.
I am still the wizard that did it.
"On my business card, I am a corporate president. In my mind, I am a game developer. But in my heart, I am a gamer." -- Satoru Iwata
<scrimpy> at least I now know where every map, energy and save room in this game is
As a Riddler Extension, alter the terrain. The first half is far more treacherous, and demands 2 bananas per mile. This would reduce the overall max haul to 332 then, using the previous best method. If the bad terrain extends any further than 832 miles (168 from Bazaar), then you cannot get any bananas across.
For this week's problem, I got an answer of 5/11 (the ones to your side have 2/11 chance each, and the ones furthest from you have an 1/11 chance). If there are N statisticians, we can calculate your chance of winning as . There is probably a better way of solving this that doesn't involve trigonometry, although I can't think of any.
For an infinite number of statisticians, one would have to compute , which I don't know how to do. Another approach I tried ended me with having to solve the differential equation with f tending to 0 at infinity, and taking f(3) as the answer, but that one too seems hard.
