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Ecoris
Promising
Supreme Hero
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posted October 21, 2010 10:36 PM |
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I will post my solution within the next few days; I will have to make some figures and recall the details of what I did.
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dimis
Responsible
Supreme Hero
Digitally signed by FoG
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posted October 26, 2010 05:35 AM |
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Edited by dimis at 05:41, 26 Oct 2010.
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Unpopular Science
Unpopular Science, NYT.
ihor, I guess everyone's been really busy lately. If Ecoris can't make it until this coming weekend I will try to post a solution. I need a couple of pictures too.
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The empty set
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Ecoris
Promising
Supreme Hero
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posted October 26, 2010 12:23 PM |
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Edited by Ecoris at 12:35, 26 Oct 2010.
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My solution to the bug problem:
The first thing one needs to realize is that the movement of the bugs will be rotational symmetric; the path each bug travels will be the same, so to speak, except for a rotation around the center of the square. If the bugs ever meet, it will therefore be at the center of the square. Each bug will continuously need to adjust its course, and the paths travelled become spirals. The more difficult part is to realize which type of spiral we are dealing with. A few other observations one makes are that it doesn't matter at what speed the bugs travel, as long as they travel at the same speed; our only interest is in how far a bug travels before they meet. In addition, the positions of the four bugs will at any time form a square, since the bugs start out that way.
Consider the figure below
Obviously, the figure incorporates the final solution, but it just serves to illustrate the following point: The four bugs have travelled one quarter round the center at this point, and I have displayed the square that they now define. The point is that the movement that will now take place inside the small square will be the exact same as in the large one; it will just be scaled down by the ratio of the sidelengths of the two squares (Compare the green curve inside the small square to the blue curve). After another quarter of movement, we could insert a third square and so on. So if we take the path of a given bug and scale it by a certain factor it will cover itself. Now one has to know that this is the defining property of a logarithmic spiral. (Other spirals (see links at the bottom of the article on logarithmic spirals) do not have this property).
One can express the (x,y)-position as a function of the parameter t (time) as follows:
x(t) = a*exp(-bt)*cos(t)
y(t) = a*exp(-bt)*sin(t)
where a and b are positive real numbers. The center (0,0) of the spiral corresponds to t = infinity. If we start with the green bug, we could let a = 1/sqrt(2) such that at t = 0 we are at the starting position (1/sqrt(2),0). The equation for the blue bug is then
x(t) = -a*exp(-bt)*sin(t)
y(t) = a*exp(-bt)*cos(t)
The requirement is that at any point in time, the tangent to the green bug's curve should point towards the position of the blue bug at that time. This requirement determines b. It's a simple calculation, so I won't tire you with the details; the result is that b should equal 1. The movement of the green bug is therefore given by:
x(t) = 1/sqrt(2) * exp(-t)*cos(t)
y(t) = 1/sqrt(2) * exp(-t)*sin(t)
where t runs from 0 to infinity. The speed at which the bug travels is then
speed(t) = sqrt( x'(t)^2 + y'(t)^2 )
One could reparametrize this curve such that the bug travels with constant speed, say 1. Then the time interval should be changed accordingly from [0,infinity) to [0,t0] for some positive real number t0. Actually, t0 would then be the length of the curve. However, reparametrizing is basically the same as just finding the length of the curve so we may just as well do that. The length is obtained by integrating speed(t) with respect to t from t=0 to t=infinity. I.e.
length = integral_0 ^infinity { speed(t) dt }
The result of the integration is 1. (Just differenatiate x(t) and y(t) with respect to t to obtain x'(t) and y'(t), then calculate speed(t), then integrate).
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ihor
Supreme Hero
Accidental Hero
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posted October 26, 2010 06:27 PM |
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Edited by ihor at 18:36, 26 Oct 2010.
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^^ The solution looks cool! 
My idea was next:
I also agree, that your quote below is the basis of this problem.
Quote:
The first thing one needs to realize is that the movement of the bugs will be rotational symmetric; the path each bug travels will be the same, so to speak, except for a rotation around the center of the square. If the bugs ever meet, it will therefore be at the center of the square. Each bug will continuously need to adjust its course, and the paths travelled become spirals. The more difficult part is to realize which type of spiral we are dealing with. A few other observations one makes are that it doesn't matter at what speed the bugs travel, as long as they travel at the same speed; our only interest is in how far a bug travels before they meet. In addition, the positions of the four bugs will at any time form a square, since the bugs start out that way.
Now let us denote V as a speed of each bug and calculate the time before meeting.
At any time four bugs will form a square with the same center as initial square. The direction of the bug will be straight to the neighbour vertice, which means that at any time the angle between bug's current direction and line through bug and square center will be 45 degrees.
That means that bug is heading to the center with constant speed V*cos(45°) (basically this is what physicians say - changing of the frame of reference).
The initial distance between bug and center of the square is 1/sqrt(2). So the time he gets to center is t = (1/sqrt(2)) / V*cos(45°).
Finally the needed distance S = V*t = (1/sqrt(2)) / cos(45°) = 1.
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ihor
Supreme Hero
Accidental Hero
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posted October 26, 2010 06:43 PM |
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@dimis
Briliant blog.
That was indeed geeky and funny. Hehe.
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Warmonger
Promising
Legendary Hero
fallen artist
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posted October 26, 2010 09:05 PM |
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I've had bugs problem in school, at physics  It's not about integration (I didn't understand it by then), just shift of coordinate system.
Imagine that you are a bug. There is another bug in front of you, at distance of one unit. You move towards him all the time, he stays still at axis perpendicular to yours. The path does not change from your perspective, so it's just about moving forward exactly one unit.
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The future of Heroes 3 is here!
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Ecoris
Promising
Supreme Hero
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posted October 26, 2010 09:47 PM |
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Your solution is much simpler than mine, ihor. And more elegant too. At least I could use the equation to make the graphs.
Perhaps Warmonger's way of looking at it is even better. I had to think about it for a while to fully understand it; at any point in time the speed at which move towards the other bug is the same (you could also say that the distance between the two bugs differentiated with respect to time is constant; the velocity vectors of the two bugs are at all times perpendicular and the velocity vector of the first bug points towards the second bug. Therefore the velocity of the second bug has no influence on the speed at which the two bugs move toward one another).
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alcibiades
Honorable
Undefeatable Hero
of Gold Dragons
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posted October 26, 2010 10:37 PM |
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Quote:
I've had bugs problem in school, at physics It's not about integration (I didn't understand it by then), just shift of coordinate system.
Imagine that you are a bug. There is another bug in front of you, at distance of one unit. You move towards him all the time, he stays still at axis perpendicular to yours. The path does not change from your perspective, so it's just about moving forward exactly one unit.
Quote:
Perhaps Warmonger's way of looking at it is even better. I had to think about it for a while to fully understand it; at any point in time the speed at which move towards the other bug is the same (you could also say that the distance between the two bugs differentiated with respect to time is constant; the velocity vectors of the two bugs are at all times perpendicular and the velocity vector of the first bug points towards the second bug. Therefore the velocity of the second bug has no influence on the speed at which the two bugs move toward one another).
That is, in fact, an ingeneous solution to the problem. 
I did realize the bugs would move in a spiral motion, but my head imploded when I even considered trying to do the math that Ecoris did. 
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What will happen now?
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dimis
Responsible
Supreme Hero
Digitally signed by FoG
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posted October 26, 2010 11:24 PM |
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Edited by dimis at 23:40, 26 Oct 2010.
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Interesting. I certainly didn't expect the solution by Ecoris. I bow.
My approach was in the end Pythagorean Theorem.
Initially I wanted a rough approximation of the curve to get some intuition. So I drew a 10cmx10cm rectangle (a x a), and then for each bug, in cyclic order, I made steps of say x = 0.5cm (heading towards the last position of the next guy in cyclic order). If we are patient enough for a couple of minutes we can get a rough approximation of the curve until the point where x becomes comparable to a and the shape is funny. However, in the limit, as x -> 0, this construction gives the real curve, and the total length, is the number of steps we made times x. The other observation in the construction was that the bugs are always on the vertices of a rectangle that is rotated and shrinked.
Let's see what happens, after one step x. By the Pythagorean theorem the new edge of the rectangle a_1 is
a_1 = sqrt( (a-x)^2 + x^2 )
Now, initially I expanded everything below the square root, so I get
a^2 - 2ax + x^2 + x^2
and my first idea was to throw away the two x^2 since in the limit their contribution is going to be negligible. But this doesn't give a nice formula now, does it ? So, looking again the original equation (before the expansion), it really makes sense to throw away just one of the x^2, since that one actually goes to 0, and leave (a-x)^2 untouched. Here, if you are not familiar with these sorts of approximations you might think I am cheating, but anyway .. :] All I am trying to say is that x^2 is much-much-much smaller than (a-x)^2, so I ignore it. This further implies that a_1 =about= a - x.
In other words, the steps are so small, that the triangle is almost a straight line. Essentially this is it, and now you can find many ways of justifying a as the total distance.
* By counting steps (in conjunction with the following).
* By requiring the approximation to be "good" until they form a square of size a/2. This one requires a little bit of extra justification, but it all amounts to reals being a continuum and having the ability to choose arbitrarily small x's.
* You realize, as Warmonger said, that the guys are walking on a straight line in the limit as x -> 0, so the total length of the path is alpha.
That's where I stopped, since I was convinced about the answer.
Indeed, ihor, very-very interesting problem! Thanks!
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The empty set
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Binabik
Responsible
Legendary Hero
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posted October 27, 2010 12:19 AM |
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My approach was similar to Ihor's, but I was never able to resolve some things and didn't have very high confidence in it.
My problem was this. At any given time the direction of travel of two adjacent bugs is perpendicular to each other. One bug moves one increment toward the other, while the second bug moves away from the first at a 90 degree angle. You know they must close a gap of .7 units. Using Pythagorean you could look at the "efficiency" of movement. With the direction of movement being 90 degrees apart, for every increment they move, they will close the gap by .7 * that increment. So you simply multiply 1.4 * .7 and get your answer of one.
However, if you picture their starting points, you do not have a nice isosceles triangle as I was picturing above. You have a very long acute angle. This is the point that I could not resolve.
Having seen Warmonger's reasoning, this makes more sense now. With a long acute angle, for every increment traveled, they close the gap a high percentage of that increment. As the unit of movement approaches zero, the narrowing of the gap approaches that same unit. So to close a gap of one, the distance traveled is also one.
If I'm remembering my definitions correctly, as the increment traveled approaches zero, the acute angle approaches zero, the cos approaches one, and the length of the hypotenuse and long leg approach equality.
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Binabik
Responsible
Legendary Hero
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posted October 31, 2010 08:58 AM |
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Edited by Binabik at 09:06, 31 Oct 2010.
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I play an MMO and there is a location I go to regularly. I go there at random intervals for random lengths of time, but on average I probably spend about 30-90 minutes per day there. Whenever I'm there, there is another person also there.
What conclusion can I make about how much time the other person spends there compared to myself?
I thought this was pretty simple, but there is someone who doesn't agree with me
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dimis
Responsible
Supreme Hero
Digitally signed by FoG
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posted November 01, 2010 06:07 PM |
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So Binabik, you walk into a bar regularly, and *every time* you go there you see the same bartender. And now you are asking how much time the bartender spends in the bar, compared to your time spent in the bar ? Well, at least as much time as you spend in the bar, because *every time* you go there, he is also there and serving you drinks.
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The empty set
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Corribus
Hero of Order
The Abyss Staring Back at You
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posted November 01, 2010 09:51 PM |
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Bin,
If you had some real data to play with, and made the assumption that the times you and this other person are there are random, then you could determine, within a certain degree of confidence, how much time the other person spends there on average. Given the fact that you're data set would probably be fairly small, your confidence interval would probably be large almost to the point of ridiculouslness.
However, because gaming times are unlikely to be random, then all that goes out the window, and pretty much all you can say is either he is stalking you or he's there more than you are.
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Binabik
Responsible
Legendary Hero
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posted November 02, 2010 12:44 AM |
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Grrrrr, this is bugging the crap out of me now and I've gotten myself so trapped that I can't even form a proper question. Or maybe I'm just being stubborn and want to find my own way out of something I think should be easy. But it's frustrating the crap out of me.
Of course the original question is easy, but it leads to a bunch of other questions.
It ain't funny, it's pissing me off. Time to go eat and forget about it for a while.
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Binabik
Responsible
Legendary Hero
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posted November 02, 2010 04:18 AM |
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If someone has the opportunity to do something, it's not the same as the odds of doing it.
If I'm in a location 1 hour per day and another person is always there when I am, I would conclude with near certainty that he is there longer than I am. With time being continuous, the odds of him being there EXACTLY the same amount of time are like 1:infinity. (ok, I probably just violated some rule by saying that but I did it anyway)
If we sample the location once per minute, the person can either be there or not. Having two states is not the same as saying he has a 50% chance of being there at any given time.
If I'm there for 1 hour per day and during that hour he is there for 45 minutes and not there for 15 minutes, I can not conclude that he has a 91:92 chance of being there longer than me (23 hours / 15 minutes). He must be there more than 15 minutes out of the remaining 23 hours. He has the opportunity to be there. But what is the likelihood of it?
If I arrive at 0100 and he is there when I arrive, are the odds of him being there at 0059 any different than the odds of being there at 1300?
Even though he has the opportunity to be there, can I form any conclusion whatsoever on the likelihood of him being there at any given time? If I do not know the likelihood of him being there at any given point in time, how can I draw any conclusions about the total length of time he's there?
If I arrive and he is there, I think the odds of him being there immediately prior to my arrival are extremely high. But how do I conclude that? Can I conclude that because the odds of him NOT being there are extremely low? How can I conclude THAT without knowing the likelihood of him being there at any given time? Newtonian physics?
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dimis
Responsible
Supreme Hero
Digitally signed by FoG
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posted November 03, 2010 01:59 AM |
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Edited by dimis at 02:19, 03 Nov 2010.
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Balloons
At a zoo people have to enter through the west gate and exit through the east gate. John and Lillian sell balloons outside the zoo near the gates; John near the west gate and Lillian near the east gate. Every day exactly 500 kids visit the zoo with their parents. The parents are willing to buy at most one balloon per kid. The demand at the west gate is
500 - (price in cents of each balloon at the west gate).
The kids who did not buy a balloon at the entrance, as expected, clamor and cry while they are at the zoo since they do not have a balloon. Hence the parents tell their children that they will buy them a balloon at the exit, provided that the price is "fair". The demand at the east gate (exit) is
(The number of kids without a balloon) - (price in cents of each balloon at the east gate).
(a) If x and y are the prices at the two gates, what will be the gross income to John and what to Lillian ?
(b) Find the values for x = 200 and y = 150.
(c) If Lillian finds out that a balloon is priced 200 cents at the entrance, how should Lillian price the balloons at the exit gate ?
(d) If John knows that Lillian gets this information and acts accordingly, and if Lillian knows that John knows that Lillian acts this way, then what could Lillian expect from John ?
(If necessary, assume that both John and Lillian act selfishly; i.e. there is no co-operation between the two.)
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The empty set
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Corribus
Hero of Order
The Abyss Staring Back at You
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posted November 05, 2010 05:13 PM |
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Edited by Corribus at 17:15, 05 Nov 2010.
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My mind was wandering earlier and I thought of an interesting problem.
Suppose you are invited to play a game of darts, and the target looks something like this (image A, to the left):
The target is comprised of squares. The four sides of eaah square are bisected by the four corners of the next smaller square. There are an infinite number of squares.
Assuming your darts have infinitely small tips, what is the probability that you'll hit a red area if you hit the target?
Now. Since you're playing for money, a friend of yours allows you to modify the design of the dart board within certain limitations.
The allowed design is shown in "B", to the right. Here, let each square be rotated by X degrees with respect to the next larger square. [The image shown in "A" is a special case of "B", with X = 45 degrees.] Is there a value of "X" for which your chances of hitting a red area are maximized? [Note: the Figure shown in B is for a value of X = 20 degrees.]
EDIT: Oops - I didn't notice that dimis has already posed a problem which hasn't been solved. Well, I guess now we have two.
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ihor
Supreme Hero
Accidental Hero
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posted November 06, 2010 04:54 PM |
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@dimis
I think I don't understand your explanation of the problem, especially the "fair" thing, how parents decide which balloon to buy?
@Corribus
Sounds funny but seems its the problem on geometric probabilities where we should use geometric progression. The quotient in A will be -1/2 (1 - 1/2 + 1/4 - 1/8 + ...). The quotient in B will be from -1/2 including to -1 excluding.
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dimis
Responsible
Supreme Hero
Digitally signed by FoG
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posted November 06, 2010 05:03 PM |
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I am just playing with words when I write "fair", and the explanation of "fairness" is on the next sentence which describes the demand on the east gate. That's all.
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The empty set
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alcibiades
Honorable
Undefeatable Hero
of Gold Dragons
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posted November 06, 2010 05:39 PM |
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Quote:
My mind was wandering earlier and I thought of an interesting problem.
Suppose you are invited to play a game of darts, and the target looks something like this (image A, to the left):
The target is comprised of squares. The four sides of eaah square are bisected by the four corners of the next smaller square. There are an infinite number of squares.
Assuming your darts have infinitely small tips, what is the probability that you'll hit a red area if you hit the target?
Now. Since you're playing for money, a friend of yours allows you to modify the design of the dart board within certain limitations.
The allowed design is shown in "B", to the right. Here, let each square be rotated by X degrees with respect to the next larger square. [The image shown in "A" is a special case of "B", with X = 45 degrees.] Is there a value of "X" for which your chances of hitting a red area are maximized? [Note: the Figure shown in B is for a value of X = 20 degrees.]
EDIT: Oops - I didn't notice that dimis has already posed a problem which hasn't been solved. Well, I guess now we have two.
Ad 1: Chance to hit red area in A is 2/3 = 66.7 %
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What will happen now?
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