m****i
发帖数: 159 | 1 i think ur answer is wrong.
i've got a accurate answer:
p(x) =
when r < 1/2
2*(1-r)^(n-1) + 2(n-2)*(1-2r)^(n-1)
when r >= 1/2
2*(1-r)^(n-1)
answer
model when | m****i
发帖数: 159 | 2 integral (1-x-r)^(n-2) dx, from 0 to r;
integral 2*(1-2r)^(n-2) dx, from r to (1-r);
integral (1-x-r)^(n-2) dx, from (1-r) to 1;
when r >= 1/2, neglect the middle part.
add them together, then normalize it. | S*********g
发帖数: 5298 | 3 For the minimum neighbot distance one, we can think it as the ball
has a radius x and calculate how many ways to arrange it.
This is to say we have 1-(n-1)x spare space to attach on n balls
Suppose we attach xi on ith ball, we have
p(x)=int dx1....dxn delta(x1+x2+...+xn-(1-(n-1)x))
assuming L = 1.
【在 m****i 的大作中提到】
: integral (1-x-r)^(n-2) dx, from 0 to r;
: integral 2*(1-2r)^(n-2) dx, from r to (1-r);
: integral (1-x-r)^(n-2) dx, from (1-r) to 1;
: when r >= 1/2, neglect the middle part.
: add them together, then normalize it.
| m****i
发帖数: 159 | 4 ok, lets imagine there r n hard balls. each radius r/2, with the 2 exceptions
of the leftest and rightest balls with only half of the ball( the curve side
is towards inside ).
now it is easy to c that the number of placing these n balls is the
distribution of placing n points with no neighbor distance longer than r.
lets imagine that we get rid of the part of the line which is covered by these
n balls. now the length of the line is reduced to [1-(n-1)r]. the prob is
equivalent to placing n poi
【在 S*********g 的大作中提到】
: For the minimum neighbot distance one, we can think it as the ball
: has a radius x and calculate how many ways to arrange it.
: This is to say we have 1-(n-1)x spare space to attach on n balls
: Suppose we attach xi on ith ball, we have
: p(x)=int dx1....dxn delta(x1+x2+...+xn-(1-(n-1)x))
: assuming L = 1.
| S*********g
发帖数: 5298 | 5 实际上这个只是把1-(n-1)x切成n段，只需要n-1个切点
所以最后结果是(1-(n-1)x)^(n-1)
【在 m****i 的大作中提到】
: ok, lets imagine there r n hard balls. each radius r/2, with the 2 exceptions
: of the leftest and rightest balls with only half of the ball( the curve side
: is towards inside ).
: now it is easy to c that the number of placing these n balls is the
: distribution of placing n points with no neighbor distance longer than r.
: lets imagine that we get rid of the part of the line which is covered by these
: n balls. now the length of the line is reduced to [1-(n-1)r]. the prob is
: equivalent to placing n poi
| t******y
发帖数: 239 | 6 I went through the problem again and realized that the equation
n(n-1)[1-(n-1)x]^(n-1) was indeed for the distribution of the
smallest distance, instead of the nearest neighbor distance. When n=2,
they are equivalent to each other, but when n>2, they should be
different (or not? I still cannot picture it clearly).
How did you get your equation? It seems very simple. Yesturday I got
an integration equation, and I am still working on it. It's likely that
even I can find the answer, the result will
【在 m****i 的大作中提到】
: ok, lets imagine there r n hard balls. each radius r/2, with the 2 exceptions
: of the leftest and rightest balls with only half of the ball( the curve side
: is towards inside ).
: now it is easy to c that the number of placing these n balls is the
: distribution of placing n points with no neighbor distance longer than r.
: lets imagine that we get rid of the part of the line which is covered by these
: n balls. now the length of the line is reduced to [1-(n-1)r]. the prob is
: equivalent to placing n poi
| H****h
发帖数: 1037 | 7 See 2015 in Math board.
For the case of ring, the problem is equivalent to
put n-1 balls in the interval of length 1, and ask any
two of them have distance greater than r, and the
distances from the end points are also greater than r.
The probability is (1-nr)^{n-1}.
The density is n(n-1)(1-nr)^{n-2}.
的
即
【在 m****i 的大作中提到】
: ok, lets imagine there r n hard balls. each radius r/2, with the 2 exceptions
: of the leftest and rightest balls with only half of the ball( the curve side
: is towards inside ).
: now it is easy to c that the number of placing these n balls is the
: distribution of placing n points with no neighbor distance longer than r.
: lets imagine that we get rid of the part of the line which is covered by these
: n balls. now the length of the line is reduced to [1-(n-1)r]. the prob is
: equivalent to placing n poi
| m****i
发帖数: 159 | 8 你计算的那个是说，给出一个n点分布，然后看一下所有相邻距离里最近的一个，如果是x
，就在x的登记薄上划上一笔。
superstring现在说的那个是，给出一个n点分布，然后看一下所有相邻距离，这样的距离
共有(n-1)个，每出现一次x，就在x的登记薄上划上一笔。
我计算的那个是，跟踪一个质点，然后看一下在n点分布下离它最近的距离，如果是x，就
在x的登记薄上划上一笔。
我现在估计superstring算的那个可能是tigerguy的本意。
圆
的
。
情
【在 H****h 的大作中提到】
: See 2015 in Math board.
: For the case of ring, the problem is equivalent to
: put n-1 balls in the interval of length 1, and ask any
: two of them have distance greater than r, and the
: distances from the end points are also greater than r.
: The probability is (1-nr)^{n-1}.
: The density is n(n-1)(1-nr)^{n-2}.
:
: 的
: 即
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