Timeline for Points in a Rectangle
Current License: CC BY-SA 3.0
24 events
| when toggle format | what | by | license | comment | |
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| May 22, 2015 at 12:52 | comment | added | Mark N | @zlobi.wan.kenobi " pick 4 random points inside it" where 'inside it' can be implied as any real number coordinates. As for L and W, they can be mentioned as extra (non-required) information as most questions can have. | |
| May 22, 2015 at 12:49 | comment | added | zlobi.wan.kenobi | @MarkN , oh, this should be specified if the point coordinates are integers or can be any real number, or did I miss part of the specification? I assumed that our space is not continuous. By the way, isn't it strange to specify the parameters L and W when they are not needed for the solution? | |
| May 21, 2015 at 15:44 | comment | added | Ewan | I get the same answer, is the question wrong when it says 9/16? | |
| May 21, 2015 at 12:05 | comment | added | Mark N | @zlobi.wan.kenobi The size of the rectangle is independent of the probability since there will always be an infinite number of possible locations to put the points (i.e [0.0001, 0.0001]) in a LxW rectangle. So the size doesn't matter. | |
| May 20, 2015 at 13:07 | comment | added | zlobi.wan.kenobi | all you say seems logical, but my intuition tells me that the final probability should also depend on the length l and breadth b of the rectangle. For example, the case with 2x2 rectangle and 2 points, where the probabibility that the points are on the same side is actually 1, because the only invalid 2-point combinations in this case is: [(1,1),(2,2)], which happen to be also the only combination with points in 2 different halves. All other combinations contain both points on the same side => 100% probability. Or am I missing something? | |
| May 20, 2015 at 2:52 | comment | added | Dipped Bits | @leoll2: You almost got it. But how is your P(4) = 1/2 ? Check that again. | |
| May 19, 2015 at 13:01 | history | edited | leoll2 | CC BY-SA 3.0 |
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| May 19, 2015 at 12:47 | comment | added | JiK | @leoll2 Technically, one must be careful with something like "bottom/top, left/right one, as you prefer" to make sure that there is always at most one good separator line. For example, "the side where point $(l/4,b/4)$ is" doesn't work. Put something like what you've written in the comments to the answer so it'll be better! | |
| May 19, 2015 at 12:35 | comment | added | leoll2 | @JiK Once you draw a line, you should actually consider only one half (bottom/top, left/right one, as you prefer). Indeed, each separator can be imaginarily be split in two adjacent segments, each one referring to a half. If you multiply by k then divide by k it doesn't matter, the result will always be 2^(n-1) | |
| May 19, 2015 at 12:21 | comment | added | JiK | @leoll2 Hmm, there still seems to be something odd: "Now, what's the probability that each point is in the same half of the previous? Of course it is $1/2$ for each point!" Wouldn't that give $2^{n-2}$, not $2^{n-1}$? This will be countered by the fact that in this way, given $n$ points on the same half, there are two separator lines from different points which split the rectangle in two halves such that all points are on one half. | |
| May 19, 2015 at 12:17 | history | edited | leoll2 | CC BY-SA 3.0 |
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| May 19, 2015 at 12:17 | comment | added | leoll2 | Anyway, I'll remove that note if it makes confusion. | |
| May 19, 2015 at 12:13 | comment | added | leoll2 | @JiK Why they are disjoint? Simply because the half determined by one point is always different from the one determined by another point. Ah, and 0!=1, do you agree? | |
| May 19, 2015 at 12:11 | comment | added | JiK | @leoll2 The idea seems now in principle correct, but I probably wouldn't believe it correct without figuring out the answer by myself. For example, you don't mention why it is possible to sum the $2^{n-1}$ $n$ times; compare this with dmg's answer that mentions that the events are disjoint so their probabilities can be summed. Removed my downvote. | |
| May 19, 2015 at 12:11 | history | edited | leoll2 | CC BY-SA 3.0 |
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| May 19, 2015 at 12:07 | comment | added | leoll2 | @DippedBits That's a rare case! The points are picked randomly! Btw, I've updated my answer | |
| May 19, 2015 at 12:06 | comment | added | leoll2 | @JiK Answer updated, do you like it now? | |
| May 19, 2015 at 12:05 | comment | added | leoll2 | @JulianRosen My answer has been updated, check if it satisfies you | |
| May 19, 2015 at 12:05 | history | edited | leoll2 | CC BY-SA 3.0 |
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| May 19, 2015 at 3:19 | comment | added | Dipped Bits | @JiK: Exactly the situation I was referring to. +1 | |
| May 19, 2015 at 2:55 | comment | added | Dipped Bits | @leoll2 you are assuming that the first two points lie at an angle close to 180 degrees with respect to each other and the center. what if the angle was just over 0 degrees? Your approach is correct just need to consider 'all' the cases. | |
| May 19, 2015 at 2:43 | comment | added | Dipped Bits | What if the first two points chosen are very close to each other? Will the probability of the third still be 1/2? | |
| May 18, 2015 at 14:35 | comment | added | Julian Rosen | If two points are picked at random, there will typically be many ways to divide the table in half so that both points lie on the same half. For three points to lie on the same half, we need to be able to find some half containing the first two that also contains the third. So the probability for three points should be somewhat larger than $1/2$, and similarly for four points the probability should be larger than $1/4$. | |
| May 18, 2015 at 12:27 | history | answered | leoll2 | CC BY-SA 3.0 |