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Title: Ghost ships Post by Paul Hsieh on Aug 5th, 2002, 7:16pm Nice problem. The way to see the solution, I believe, is just visualize the paths of the ship in 3 dimensions (where time is one of the dimensions) and realize that the 4 lines must be coplanar. The ship captain has many alternatives to get off the plane. One way is to veer off in a purpendicular direction for a short time, then redirect to the target location. This will guarantee leaving the plane, thus missing "E". |
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Title: Re: Ghost ships Post by tim on Aug 5th, 2002, 10:18pm Better yet, just slow down or speed up. That way you still reach your destination without having to muck about with fiddly position estimations and compass bearings. |
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Title: Re: Ghost ships Post by Paul Hsieh on Aug 6th, 2002, 8:26pm No, that's not good enough to get off the plane. Consider the possiblity that all the boats are on the same line running at different speeds, with F behind you running towards you and going a lot faster than you. You can still satisfy the original statement of the problem, and your strategy is not sufficient to escape a collision with F. (Alternatively, F may be in your path ahead going really slowly with E and G coming towards you in the other direction ... etc.) You really need to get off the plane. A change in direction while maintaining speed is sufficient, but just changing speed is not. |
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Title: Re: Ghost ships Post by tim on Aug 6th, 2002, 9:30pm No, my solution works. Consider the ships as tracing out coplanar lines in a plane in 3-space (x,y,t). If the motion plane is not orthogonal to the t=0 plane, then varying velocity means that the ship leaves the plane. Yes, this leaves open the case where the plane is orthogonal to t=0. However, we know that the motion plane is not orthogonal to t=0: G hit H amidships, rather than in the stern or bow, which means that G's velocity vector is not parallel with H's. Hence varying speed without direction does mean that H leaves the plane. Remember to read all parts of the problem. Sometimes seemingly insignificant words are more important than they appear ;) |
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Title: Re: Ghost ships Post by Chronos on Aug 15th, 2002, 5:52pm It somewhat bugs me that this riddle requires the captain of H to break the rules. It says right up front that all ships pursue their course steadily, changing neither direction nor speed. If Capt'n H can't change his direction or speed, then there's really nothing meaningful that he can do. If, contrary to the problem statement, he can change his velocity, then why should we assume that the other ships can't? |
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Title: Re: Ghost ships Post by Archon on Aug 16th, 2002, 2:56am I have a feeling that captain H doesn't have to do anything, that it is impossible for H to have 3 collisions if both F and G have had 3 collisions. I doubt I have the ability to prove it, but here's a start for someone with more maths ability to pick up. Certainly, the first step is to turn this problem into one involving lines in 3 dimensions. Each line will be straight. The plot of each line is a set of simultaneous equations in x, y and t (t being time) Let collision between 2 ships (say G and H) occur at xGH, yGH, tGH in this 3d space, and denote the event GH I hypothesize that there exists no solution for the event EH, where exists: FE, FG, FH, GE, GH and where xyt is a different value for each event (because you can clearly draw 4 straight lines in which at least 2 intersect with all other 3 lines if they all pass through the same point) and where no 2 vectors are the same (the ships hit amidships, so no collision is due to one ship overtaking another on the same vector) Come to thik of it, this looks like the cigarettes puzzle. |
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Title: Re: Ghost ships Post by Archon on Aug 16th, 2002, 3:11am It is probably also important that both F and G have had 3 collisions *before* the proposed collision between E and H. |
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Title: Re: Ghost ships Post by Archon on Aug 16th, 2002, 4:05am Hm, some more thought... Given that, as has been said, in the 3d space the lines must be coplanar, we should be able to transform this back into a 2d problem. But clearly it is possible to draw 4 lines in 2 dimensions with 2 lines having 3 intersections. But then time becomes a factor again, because we dont just want 4 intersecting lines, we want them to intersect at specific times. Which takes us back to 3 dimensions... ugh :) |
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Title: Re: Ghost ships Post by Chronos on Aug 17th, 2002, 11:35am Quote:
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Title: Re: Ghost ships Post by pa0pa0 on Aug 21st, 2002, 5:55am on 08/05/02 at 19:16:02, Paul Hsieh wrote:
I think he has to not only get off the plane, but ensure that he stays off it by either diverging from it or proceeding parallel with it (otherwise he could be terribly unlucky and collide with E at the very moment when he re-intersects the plane). But he doesn't know the orientation of the plane! In fact, the only thing he knows about the plane (apart from the fact that his current course is on it) is that the plane is *not* parallel with the time-axis (the operative word being "amidships" as noted by Tim). I think this means that his only way of ensuring that he gets and stays off the plane is by changing his speed while *not* changing his direction. i.e zig-zagging doesn't work! He can then either maintain the new speed (diverging from the plane) or resume his old speed (proceeding parallel with the plane.) Afterthought: if he had measured the speed and direction of G or F when he had the chance, then he could work out the orientation of the plane and get off it by zig-zagging. |
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Title: Re: Ghost ships Post by vijk on Aug 28th, 2002, 5:22pm I guess that if H simply follows F, no collision with E can occur. But reaching the destination is more difficult. |
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Title: Re: Ghost ships Post by mattian on Jul 13th, 2004, 10:07am Let's inject this discussion with a diagram. |
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Title: Re: Ghost ships Post by Grimbal on Jul 16th, 2004, 10:55am Here are my thoughts about this problem. ::[hide] As mentionned, without a course change, H is guaranteed to collide with E. Any course change in space (move a bit left) or time (wait a while) will prevent the collision. Unfortunately, it will not guarantee that you will not collide elsewhere. In fact, for every course change for H, there are possible courses for E that still intersect the new course of H. You need to take into account the speed of the other ships to know how to get out of the way. If H's course change is to stop a while, the problematic courses for E are all the coursese that follow the same path as H, at a different time. But in that case, F and G must also follow that path. So, if the 2 collisions happened by one ship overtaking the other along the same straight path, you can avoid the collision by moving out of that line. For any other situation, where the ships actually crossed their path at an angle, you know that the last ship is expected to also cross the path at an angle, and at the very time you reach it. In that case, you just need to stop a while or change speed to reach that crossing before or after E reaches it. [/hide]:: |
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Title: Re: Ghost ships Post by mattian on Jul 16th, 2004, 8:27pm In all there is potential for six collisions - I believe that in order for the first five to occur, the shape formed by the paths of the four ships have to correspond to the diagram given below - in which line segments a and b equal, and c and d are equal. It also turns out that d = 2g and c = 2h. The lengths of these segments directly corresponds to the time it takes for each ship to cover the distances shown. In order for three of the ships to collide as the problem statement describes the path diagram has to look like this. Explanation in subsequent post. |
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Title: Re: Ghost ships Post by mattian on Jul 16th, 2004, 8:56pm Explanation. Let H be travelling in the direction P to Q Let G be travelling in the direction R to P Let F be travelling in the direction R to midpoint PQ Let E be travelling in the direction Q to midpoint PR The first collision must occur at R between ships F and G. The next collision must occur at midpoint PR between E and G while F is at the midpoint of d. The next collision must occur at intersection of g, h, c, d between E and F ... AND SIMULTANEOUSLY... at P between G and H. The next collision must occur at midpoint PQ between between F and H while E is at the midpoint of c. The final collision will absolutely occur at Q between E and H unless H slows down or speeds up. The collision will absolutely not occur between E and H if one of them changes their speed and the ships are of close to zero size. It is important that they are very small to ensure that small changes in speed correspond to successfully avoiding the collision. I think the real problem here is that E and H are in the same predicament prior to their collision with each other. And any decision that H makes could be made by E as well which might revive the collision potential. |
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Title: Re: Ghost ships Post by mattian on Jul 16th, 2004, 9:05pm Think of the units of time in my explanation as proportional distances travelled or time units which are not necessarily equal from unit to unit but consistent for all ships at a given time. In other words, the first time unit may be 3 minutes while the next is 5 minutes, but this is true for all ships when it applies. |
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Title: Re: Ghost ships Post by Grimbal on Jul 17th, 2004, 3:03am As I mentionned in the post earlier, simply avoiding the expected collision doesn't guarantee that E and H won't collide elsewhere. Changing speed does not work if all ships follow the same path at different speed. They will just collide earlier or later. |
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Title: Re: Ghost ships Post by mattian on Jul 17th, 2004, 6:22am By my argument which references the diagram, if any two of them are on the same line, then they're all on the same line. Ship H would know if he were on the same line as F and G when they collided, and would therefore know that E would be on that line too. In that case he could simply branch off from the line and reestablish his original heading parallel to his original course. When E passes him by, he can get back onto his original course, if necessary. |
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Title: Re: Ghost ships Post by rmsgrey on Jul 17th, 2004, 9:03am Lets simplify things right down. Pick the inertial frame of reference in which H is stationary - the other ships will still move at a constant velocity in this new frame. Since H is stationary in this new frame, G and F must both be moving directly towards H. Since they collide with each other they must be following the same line at different speeds. Unless E is also stationary, or collided with both F and G simultaneously, the two points of collision must be distinct, so E must also be moving on the same line as F and G. If E is approaching H on that line, then collision is inevitable if H does nothing. If E is receding, then the only way collision can occur is if H accelerates towards E. Back in the original inertial frame of reference, H can avoid collision by accelerating briefly in any direction other than the bearing F and G approached along. If they're not all moving along the same line, then a simple change of speed will suffice, and avoid any navigational complications. |
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Title: Re: Ghost ships Post by mattian on Jul 18th, 2004, 10:31am But surely, in the original reference frame in which the ships are moving with respect to a stationary ocean, H's brief impulse accelaration should be in a direction which differs from the directions of all three of the other ships (E, F and G) and not just those with which H has already collided (F and G). If that is true, then how will H know E's vector in order to avoid this? |
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Title: Re: Ghost ships Post by rmsgrey on Jul 19th, 2004, 4:07am But in the original ocean-stationary frame, the line along which the ships are moving in the H-stationary frame is moving at a constant speed, and at any point in time, all four ships are on it. If you resolve vectors into horizontal components parallel to that line and perpendicular to it, then any acceleration parallel to the line will keep H on the line and potentially headed for a collision, while acceleration in any other direction will have a non-zero component non-parallel to the line, so will move H off the line. As to the problem of other ships also changing their motion, only H has been in two third collisions, so only H knows that E and H are on a collision course, so H is the only ship with a reason to take evasive maneuvers. |
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Title: Re: Ghost ships Post by mattian on Jul 19th, 2004, 5:26am I'm still working on understanding your first statement - I am resigned now to the fact that I'm a bit slower than you and it takes me a little while to catch up. :-) Re: Your second statement; I follow you. Though E, like H, has in fact had two collisions, E did not hear the other captains say, "that's our third collision". Agreed. |
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Title: Re: Ghost ships Post by mattian on Jul 21st, 2004, 12:50pm RMS is absolutely right. H has to get off the straight line formed by the ships E, F, G and H in the reference frame in which H is stationary. In the standard reference frame - i.e. such that the ocean is stationery - E is either always directly ahead of H or never directly ahead of H. If E is directly ahead of H, H simply needs to move over a little bit. If E is not directly ahead of H, H simply needs to accelerate or decelerate briefly. H will know that E is directly ahead, if its collisions with F and G were directly head on. |
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Title: Re: Ghost ships Post by rmsgrey on Jul 23rd, 2004, 3:48am Actually, E could also be directly behind throughout - in which case the other two ships will have overtaken H, with G moving faster than F moving faster than E, in which case H could just accelerate to match F's speed in which case E can never catch up. If E is ahead, then the only way to avoid collision and still reach the original destination is to move to one side, and do whatever recalculations are necessary (and hope the collision doesn't happen anyway at the destination...) |
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Title: Re: Ghost ships Post by jonderry on Aug 17th, 2004, 10:17am It is true that the ships all move along one line in the inertial frame of H, but I believe it is impossible for H's captain to compute the direction of that straight line; the line could point anywhere in 3D area-time. Unless I am missing something, the only surefire solution appears to be making the earlier assumption that the ships are not moving in the same direction in space at different speeds (from the fact that the collisions are amidships), so that slowing down and speeding back up again in the same direction takes you off of the common plane. |
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Title: Re: Ghost ships Post by rmsgrey on Aug 18th, 2004, 4:50am The captain of H can observe the real-space motion of the other ships as they pass through his. Knowing that, he knows the direction of the line connecting the four ships (which in real-space sweeps at a constant speed in a constant direction) and so can easily avoid accelerating along that line. |
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Title: Re: Ghost ships Post by mike_uk on Nov 10th, 2004, 5:57am I just stumbled across this page, Is it not possible to suggest that as the two ships f and g have already collided 3 times which means that they have collided with every other ship as you can't collide with your self. so f has hit e , g, h g has hit e, f , h as both f and g hit h as their last collision it would suggest to me that e is not on a collision course with h and that if it ever does cross the path of h it will either cross its path before or after it. and that the captain of h needs to do nothing as there is no chance of e hitting him. thats just my interpretation of this |
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Title: Re: Ghost ships Post by rmsgrey on Nov 10th, 2004, 6:21am Lets look at a special case: Assume H is sitting still. Call the point where E and F collide X, where E and G collide Y and where F and G collide Z. Since F collided with both E and H, F must be moving along the line that connects X and H. Similarly, G must be moving along the line connecting Y and H. Since F and G both pass through Z, it must be on the lines XH and YH. Since both lines have 2 points in common, they must be the same line. Finally, since XYZH is a straight line, and E is moving along the line XY (which is the same line), E must be heading either directly towards or directly away from H. Now, if H isn't sitting still, but instead moving at constant velocity, then the same thing applies. If you imagine plotting the ships' positions on a series of transparent charts, then laying the charts on top of each other so that H's positions sit directly above each other, then, by the arguments above, the other ships' courses will form a single dotted line passing through H's "position" |
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Title: Re: Ghost ships Post by Hippo on Dec 17th, 2007, 1:52am Sorry I didn't read the whole thread. The proof with coplanarity is nice, but it seemed to me noone noticed that the ghost ships are sailing on Earth - a ball. Not on a plane. I didn't thing about it so far, but it seemed to me that this little difference can play an important role. Does not the method actualy proof they will not colide when doing nothig? |
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Title: Re: Ghost ships Post by JiNbOtAk on Dec 17th, 2007, 4:08am on 12/17/07 at 01:52:04, Hippo wrote:
If the distance between the ships are relatively small, it wouldn't matter, rite ? After all, we approximate flat surfaces all the time, when it should be a ball. |
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